HIGHER ALGEBRA - Lucknow Digital Library

COMPETE SOLUTION TO

HIGHER ALGEBRA (Hall & Knight)

[Chapter I-XVIII]

By P. N. MARATHA

M. Sc, L. T.

JSV

ST oP cn9P(

u, Am*0*11"

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^ _ f i t , s i n - ft • »*** • . , , /i

TBI'-|s y

CHAPITER t

RATIO

• EXAMPLES I (on Page 10)

1. We know that if ? » , the compounded latio of these

two ratios is r X ~= j - , and duplicate ratio of a i 0 h & : &*. b a ba r

Hence—'- - ^ (i) ThjSlflicate ratio of 92>a : <# is 81 6* : fi-ea

/ . Thejgffiffcounded ratio will be = ~TX ••,, =* — i.e. 546 : a An.3, 3b arb* a

(ii) Subduplicate ratio of 64 : 9 is y/{U) '• V 9 '•'• § : 3

8 27 .'. required compounded rat io=:rX^ i.e. 9%1 &£.».

3 5o

<iii) Duplicate ratio of ^ : ^ a* is / ^ Y

4b% 3ax Compounded ratto=~g—AXJT '•*• bx '• W ^m*

x4-7 25 2. We have 27x+14)~64T ^or t*10 ^ u P " c a t e ts*'10 °*

5 • * is Is 64 or 324+224=^25x+350 i.e. x ^ I g Aae.

^ 3 * Let 7x and 12* be the two numbers in l&e ratio of 7 ; 12 then as given *e fcave

l5je-7*«=>275 or X=»55 .'. The required numbers are 385. 660 Ans.

^

2 Algebra ' i

4. . Let x be added tp each term of, the ratio :>: 37

TheS -ig*-i % x+37 3 or 3x+l5=x+37 i.e. x = I I Ansr

5. - « 7 .'. 4x=3r or x=-j v JJ 4 ' 4 '

N0W W6^^^^0^ • . . V y - 4 y „ 5^_5 .

ly+y 13 n , .i.e. 5 : 13 Ans. 6. 15 (2x*-y*)<~lxy

or 30x2- 15j>8—7xp=0

or 3 0 ^ - ^ - 1 5 = 0 (Dividing by>>*)

.Putting r—zweget

30z a -7z-15=0 7jr-l/(49+450><:4)„7^43 50'

60 66~~ " 6 0 - 3 6 5 - 3 "60- = 6 ° f " J

x 5 —3 • • -

i.e. The ratio is 5: 6 or — 3 : 5 Am.

7. If ?=4=4~ . prove that o a f

2a*y+3c*e»—5o*,/_fl«, 2fc6+36 s /-5/6 M-

or * 2 a s = ^ i Also . e ' / « r ^ - •

Examples I (on Page 10)

Putting these values in the given Expression we have/'

Ans. 2 6 » + 3 6 I / 1 - 5 / i **(26»+36" /"—5/*) ~¥

.', bc=ad; and b*=ac

.-. &c*=<Pd* ' (i) OT Me'—Pc' x 62=a>da x dc or b*cx=&d* (ii)

Now putting these values of (i) and (ii) in

ra //a3(a34<?M-ac')\

~ 2 A / 1 \ fla+^2+acZ7J

=-5 since 6*=ffc and b2c=ac* Ans. a

^ " 9 , Let - = ^ ~ « - - 2 — ~ — ? — = f c ( s a y ) • 7+r—/> r+p—q p-\-q—r

:. x=k (q+r-p);y**k (r+p-q) ; -?=£ (/>+?-r) Now ( ? - r ) x-Hr-p)j>+fp-9) z

«(*-r).fc ( g + r - p ) + ( r _ p ) . f c ( r + p - g ) •' +(*>-*)& (|»+*~r)

« * \iq-r) (q+r - p ) + ( r - p ) (r+p-q) +(/>-?) </>+*-*•)]

• = ^ [ 0 3 = 0 ^ Ans.

10. Vekftow that if— ==?==-'then _ •

file:///iq-r

Algebra

Hence if ^ _ = £ ± * = * we get *—z z y 6

2(y+x) „ (JV+x)

• 2 _ «£+*=,* = 2 X — Z - Z V I

From 1st two we have.y+x=2z or y+x—2z=0 (i)

And by cross multiplication we have . y _ * z x y z

4 - 2 2+2 -1+2 "*• 4 2 3 Ans.

* pb+qc pc+aq ap+qb j^ff'jn.Q. 10, here also we have _£+z__=_z+x _ x+y 2_(Jt+y+g) pb+qc pc+aq ap+qb p(a+b+c)+q(a+b+c)

_ 2 (x+y+z) ip+q)\a+b+c)

Now multiplying the first fraction by a in the numerator d denom'inator and likewise second by 6, third by c and^.--Iding as above, we have, each of the^aboye equaHo ^ "

=__ °y+or+^g+ *>•*+cx+c> abp+acq+bcp+aby+cap + bcq

' ~ x(b+c)+y(c+a) + z(a+b) , p {ab+be+ac)+q (ab+bc+ac) , ^x(b+c)+y (c+a)+z(a+b)

(P+q) {ab+bc+ac) ,,

. 2 (x+y+z) ^x (b+c)+y(c+g)+z (a-^b) '' (p+q) (a+b+c) (P+q) {ab+bc+ac) ' Hence 2 U + J ' + z ^ ^ (b+c)+y (c+*)+z(a+b)

a+b+c • • ab+bc+ac Ans.

• Examples I on Page 10)

12. Let ~^^-=k (constant). a b c

,\ x=ak ; y—bk ; z=ck. Putting these values of x, y, z in the given expression

*3+Q3 y*+b* z*+c* (x+y+z)3+(a+b+c)>

We get the a3fr3+a'J b*ks + b3 c3k*+c9

L. H. S.=

g f f 8 + 0 6(* a +l ) c(fca+l) " A2+l + fc'+I " + k'-+l

(/<-3+I) .(fl+A+c).

(fca+l) Now R H g ' (ofc+6/c + efe)3 + (a-l-f>4-e)a * ' "* (ak+bk+ckf+tf+b+cp

k2 (a+b-tcy-tja+b+cy ""A:3 (a+b+c)*-t(a+b+c)*

~(fca+l) ta+64-c)8^(fc8 + l) We se^that L. H. S.=R. H. S. Ans. 13. Cross multiplying the first two ratios of the

expression 2y+2z—x _2z+2x—y_ 2x+2y-z

a ~" £ " • c We get

2by+2bz~ bx=2az-\-2ax—ay "or x {b+2a)-y (26+a)+z (2a—26)«0. ...(i>

Also fr<3m last two 2bx^2by—bz^2cz+2cx-cy.

Algebera

or x {2b-2c)+y {2b-c)-z (6+2c)=0. Solving (i) and (ii) by cross multiplication we get

,__ £_ ' U4+2C) («+2&)-2 (2b~c) (a-b)]

y_ _, ^[2 (a-b)\2b-2c)+(2a+b) (6+2c)] ' £ -

a\{2b-c) (2a+*)+(fl+26) {2b,-2c)1' On simplification we get

x y _ z 2c+2ft-fl~2tf+2c-6=*26+2«-c

a a c .*. x=afc ;y=bk;z=ck. Now putting these values we have the L. H. S.

* R.H.S.~*{ax+by+ez)*

.*. L.H. S.=R.H S. Hence proved. 15: We have t(my+nz—lx)=m (nz+lx-niy)

"*n{!x-\-my—nz). It can be written as

my-i-hz—lx nz+lx—my Ix+my—fiz . 1 / / ~ I/m ^ I ^

2ar Vx 2my

l^m m n n I , 2/mnz 2/mnx 2/mrrv


or

i.e.

or

y+z—x z+x—y x+y—z 2/ ~ 2m ~~ 2*r

y+z—x z+-y—y x+y—z l ~ m ~ n I m n

y+z—x ~z+x—y~x-\-y~z 16. ax+cy-\-bz=Q

cx+by+az=Q bx+ay+cz=§*

From (i) and (ii) le1 ax+cy+bz=0

cx-\-by-\-az=0t we get x y z =k (say).

ac—A8 bc—tf^ab—c2' Putting these values of x, y, z in (Ui) we g«t

k\b{ac—fc»)+a (&c-a»)+c (ab—c2)]=0 or 3a6c—^-a3—^=0 or a»+6»+c*—3a6c«0.

17. 4x+by+gz=0 hx+by+fz=0. g*+/y-fcz=0

From ax+Ay+^2=0 Jix+by+fz=G

we get

^S£=gT^f=rt=&=k (say) ' Putting values of JT,^> Z in (iii)

ft[* (¥-bg)+f(gh-af)+c (a6-/i?)]=0 o r g {hf~bg)+f(gh-af)+c (a&-Aa)~0

*or gfh-bg*+fgh—<if*+Qbc—ch*=(} or • # abc+2fgh>-ap—bgz—cA2=0.

18. The given relations can be written as • • *—cj>—&z=0 .

bx+qy—z=iO „.(iii)

•as in Qs. 16 and 17, from (i) and (ii) we get .

Putting the values of x,y, z in (//i),,'we get ft (ft+flc)+<3 (a-f ftc)~-(l—c3)=0

or • a3+6a+c8+2fl6c=i ...(iv) Again '

x8 y\^ z3

or

or

or

(6+flc)8=a i(o+6cja==Cl—caja

6H2a&c+aacJ=V+2a6c-f6Sca~^l-c*>2

l -a^-^+aSc^l-f t3—C2- | -63c2~(l-C2> a

•x3 y* • zl___ ( l -o*)U--c")"( l - f t 2)a-T<: 2) _( l -c*) ; !

or I—a8~"I—68~ 1—c2' A n s ' " < 19.' We have (*

fl 0>+z) = x ; ft (z+*)»j> ; c (x+y)<=*z or x—ay—az=>0 , „.(i)

ftx—^+ftr=0, , ...(ii) •cx+cy-~z*=Q. * ,..(iii)

From (i) and (ii) we have

TV* = -rn.ss' i L—* (say) -ab+a cft-fft 1—aft v "

Putting values as obtained of xty, z from this in (iii) k [c (oft+a)+c(aft+ft)-(l—abfl^Q

or c (aft+a)+c (aft+ft)—(l-aft)=0 /.e. flft+ftc+flc+2rtftc= 1 f • Ans.

20. 3*—4y+7z=0.

from

or

2x~ 3* 3 -

(i) and (ii)

•

Examples I (on

-y-2z=*0 •f+z**=*\? we get

x y 8+7 14+6 x y z

Page 10)

z — .5 + 8

...(ii)

..(iii)

or f ^ ^ ^ f c (Say)

x=3fc, y=4k, z=fc Putling these values in (iii) we get

3(3Ar)3-(4fcj3+&3~I8 or J8&3=18 .'. k=>l

Hencex=3,y=4,2=1 Ans. 21. x + ^ - z = 0 ...(i)

3x-2j>+17z=0 • ,..(ii) *»+3y«+2z»=167 ...(iii)

From (i) and (ii) we get __£ y z 17-2 ~ - 3 - 1 7 = = - 2 - 3 x y z x y z , , .

0 1 T 5 = ^ 2 0 = - r 5 or 3 = - ^ = - , = 4 (say) x=3&:, j»=—4fc, z=—fc

From (iii) we get 27fc3-3(64£3)-2fe3=l67

or — I67fr3 = 167 or J t = - 1 .t=^ — 3, ^=4, r = 1 Ans.

22. 7yz+3zx-4xy=:0 „.(i) 21yz-3zjc— 4*j>«0 ...(ii)

• ^+#2^+3r=iy • ...(iii) From (i) and (it) bv cross-multiplication

Algebra

yz xz xy - 1 2 - 1 2 ~ 2 8 - & 4 = s - 2 1 - 6 3 yz zx xy .. . . 6-=T4=2l= A : 2 ( s a y )

yz=6k2, zx=Uk*, *y«21fca

Multiplying these all together we get x2v8

22=6x 14x21 xA6=1744£8

xyzoy/(\144k* ^42k3=lkx 3kx2k (say) x=7k,y=3k, z=*2k

Putting these values of x, y, z in (iii) 7fc+6fc+6A:=19 / . Ar^l

Hence x~7, .y=3, z=2 23. We have 3x*-2y*+5z2=Q

7*8-3.y*-15z8=2' 5x—4>-+7z=6

Froni (i) and (ii)

30+15 35+45~14-9

x2=9fe2;^2=16fc2;z»=Jt* A=3Jt; y=4k \ z=k

Putting these values of JC, y, z in (Hi) we get 6k*=6 .'. * » 1 '

Hence xs=3,j»=4, z*=l

24 / - , m r " =0

By the process of cross multiplication we*hayc

« Examples I (on Page 10) U

/

[y/a-^bj ( v c + V a ) Wc-V*i \VC+V*>)

m

W ^ V " ) Wfl+v'*/ Wa-y/b' Wc+s/a) n

m \2c~2jiba) \2a—2*/\,bc\ 1 ((&-<:; (c-a) \{c-a) (a~~b)y

• l2b-2Viac) \ \{a-b){b-c)j

Multiplying the denominator of each fraction by

(a—6) (b—c) (c—a) we have / m

{a-b) {2c-2\/(ab,}~(b—c) {2a-2y/{bc)} n

~(c~a){2b-2y/(acj

I m . {a—b) {c— y/Kab)} (b-c) {a—:\£(b

Aris. ' " i t i l_ A (c—a) {b~y{ac\

25. ffX^-bjp+cr=o ,..(i)

bcx+cayj-abz**0 ».(ii)

x>7+a&7 (0°*+*^ 4- c»z)=0 ...(iii) From (i) and (ii) we get

—•k (say) <!&*—ac2 »bc*—ba2 ctP—cb*

file:///2c~2jiba

12 Algebra

x=ak (62-c2) y^blc (ct-a*) i z=cfc(a2-6*)

Putting these values of x, y, z in (iii) we have abck* (fc6-c*) ( c 2 - d 2 ) ( f l»-6*)+f l6c {fl*fc (fc3-ca)

+b*k (c*-a')+c*k (o"^b*)}=0 or ft2 (62-c2) (cs-a*) (b2-62)=~.Z:a* (&'-c2) .; =(£2_c2)(c3_ f l=)(a2_fr»)

Aa=l or fc«±l Hence x=±a(b*—cz)

! v=±6(C2~a*)

26. Here we have a*+6y+cz=0 ', ...(i)

a*x+bzy+c2z=0 „.(ii) x ^ + « + ( f r - c ) (c-a) (fl-fc)«0 ...(iii) *

From (i) and (ii) we get

be*— Pc^T&^ox^aP-bQ^ (Say* x & z i,

£c (c—£) air (a—c) ah (b—a) or x=fc6c(c—b)

y=kac(a—c) z=kab(b—a)

Putting these values in (iii) andsimplifyihg for k we get -k (b~a) {c-a) {a—b) = -{b~c) (c-a) (a-b)

ie. k=l Hence x=bc(c—b)

y=ac(a—c) ' z=ab(b—a\ t • «, Ans.

27. x=a(y+z); v=£(r+x) ; z=c(x+y$ These can be written as • .


x—ay—az=0 y—bz—bx^=Q z—cx—cy=Q.

From (i) and (ii) we have x y z

•=*k (say). a(\+b) b{l+a) (l-ab)

Substituting these proportional values in (iii) ac {l+b)+bc {\+a)-*{l~ab\*=Q

or ac+abc-\-bc+abc+ab~ 1 =0 i.e. cc-f&c+ad-l-2a&c=0.

Also we have

or

or

or

fl3+2fl^+fl2^~^+2^2+a^~(l-a6)2

x8 ' yz z a ia+2ab+abi)~~b (b+2ab+aib) (l—ab)9

x2 y2

[i—ab—bc , ,_1 , fl— ab — ac 1 a — \-ab2) b — r^b)

or £ y*

£ {\-ab~bc+cab*r b-\\^tzaS+a>b a c 1 c

or

( 1 - O 6 J 2

x2 yz

~(t-0bfO-bc) -(\-ab)('\-ac)

0~ab)*

14 , ti * Algebra

• \ \ • '

JC" \\ y* z2

or r c

Ans.

28. (i) ax+hy+gz~Q *' , ' • ...(i) A x + ^ + / r = 0 ...(H) gx+fyj-cz=0. , -»(Iii)

From (i) and (ii) !,

~ i - . _ ' y z -k - ..(iv) hf-bg gh-âf-ab-IP ' v '

Substitiiting these values of x as obtained from (iv) we get g (hf-bg)4-f(gh-af)+c (ab-h*j~0 or fgH-bgi+fgh-ap+abc-cfrÔ or a£c+2fgA-fl/2-&g8-cA2=0. ...(v)

Again from (iv), we get .,• .v» - ' j ' 2 z»

W-bgfi^{gh-af)*-\Qb-Wr or -

hy*+b*g'-2fgh-'g2h*+a*f*-2afSh z8

-(aft-A s)8 ' But from (v) we hav

bgi-~2fgh=abc-ap-cbz . .". ab*e-abf*-bck***b*gt-2l>fgk '

and af1—2fgh~abc~bg%^chs

or aHt—2afghâibc—abgi—achi. Hence putting these respective values we get

# y

fcxampics l (on Page JUJ

or (be—/s)"(flc—g2)~~(ab—/i*)" (ii ax+ky+gz=Q

hx+by+fz=Q gx+fy+cz=*o.

From ({} and (ii) we have

hf-bg-hg~af~~ab-}{~ *' From (ii) and (iii) we have

- * y i k. bc-f2~fg-ch~hf-bg

From (iii) and (i) we have x y z

fg—eh ac-g* gh—af x *>

Now from (iv) r—hf—bg I

£=**-«/ \ r<=ab-h* K

Similarly from (v) and (vi) we have

*p=fg-ch \,.

->=k

"and

k'~"J "* J

p-/«-r* "I

16 ' Algebra

Now from (C), (A) and (B)

(fg^ch) (gh-a/) {hf-bg=)^m. £ 1

xyz_

Similarly from (B), (C) and (A), we have

(bc-p)(ac-g*)(ab-h*)~j2~.

, Hence

{b».-P) (ac-g2) (ab-n<) =(fg-ch) (gh -af) {hf-bg)

CHAPTER II

PROPORTION

EXAMPLES II (on )

1. Let the fourth proportional isx. .'. 3 : 5 : ; 27 : -v

or 3x=27x5 i.e. x=45 Ans. 2. (i) Let the mean proportional be x, then 6 : x : : x: 24 or xs=6x24 i.e. x^X2 Ans.

(ii) 360 A* : x :: x : 250 a2 b* or x 2 =360a*x250a^ i.e. x=3G0a3/j Ans. 3. If we take w as the third proportional Ihen

(H)»?= X - : m y

or (H)--$- l^)-? .'. W=»-T-VT-J: Ans.

y (x*+y2) 4. Since a : b=c : d

Now from u3c-f acE: bH+bd2^-. a-f f)3 : (6-W/j3 we yet

b*d + bd*={b+d,3

~r i u c ocfa.-r-c) blC.dk ib+d).k . , • . bd(b+d) bd\b-\-d)

http://blC.dk

18 Algebra

Hence L. H. S.^=R. H. S. Ans.

3. Let r«--}=A: then b a ; a^bk, c=rffc

_ pat+qb* pc2+qd* . From —^-^-r.^-^1-^ we have * pa%—qb* ptf—qd2

pb*k*-qb*-b* (pk*-qrpk*-q , „ , , „ „ „ pd*k*+qd* d*(pW+q) pk*+q •

Since L. H .S .=R.H. S. P ^ = P ^ when « J • Ans. pa%~qtr pc2—qdz b d

6, Here r = - . or - = - . or fr=5 MI c=V w c = d a—c c a

± . a c a* c3 a* b* Again 3 = 5 . . p - j i o r ^ - g s ,

or

or

c2 ~~ rf*

c - d

...or*.

Ans.

V(o s+c a) c a a—c -

-.. ±ZS_ /J7f l3+cg\l •" ft-rf~A/UA«+dVJ

7. In the above question No. 6 we have proved that

o d e a . t

i Examples'II (on Page 19)

i.e.

Now

a—cky b=dk

JK-) / JKfJ

\Z(a*+c8)

8.

or

y/{b*+d*r^{bd+{d*lb)}

v a b c b c d :. c*=bdtad^=bc

ada=(bc) d=*{bd) c a c~

Now from

R. H. S.=

b+d' /-a

C*d+d* ad2

we have the

3 -oV+rf bd3+d*~b+d

b + d~C*d+(P

9.

or

T ta b c . b e d

a=bkt b=ck, c=dk a=bk^ck*^dks

L H <? ^ 2 £ ± ^ 2 * ! ± i < ' /2ft»+3\ * ' ' 3fl+W3rfjfc»- 4rf* ^ 3 F ^ 4 }

and R H <; 2fl»+36» 2ft»fc'-4-3ff> 2kM-3 and ^R. H ; S . - 3 ? - - i g i l l B 3 i 8 f t l - ^ - j p - j

R. H . S.«*L. H. S. Honce proved.

2? Algebra

10. Let, r = - = T = * b e d

• Then a—bk, b=ckf c=dk L. H. S. of the given expression

=(fl2+£2+c3) (A2+c24-</2) =(W+c 2 / f c 2 +r f 2 / t» ) (A3-}-c2+rf2) = £ 2 (£2+c2-f rf*)2

and R. H.S. = (a&-f&H-V</)2

= (AA:-6+cA-.c+^.rf)3=A:2(^+c2+rf2)2

Since L. H. S.=R. H. S. Hence (a24b2+c*) (a2+A2+rf !)=(^+*c+tt/)2 Ans. 11. If A is the mean proportional between a and e then

we can write as • a: b :: b : c .*. A*=flc '

fl2_/,2 + c2 fiS + ^ + A2 <J2 + C 2 -£ a

Now a-2—/>-2+r-2~l 1 l " c 2 + a 2

(for b2==ac

al+Cl-b2

We have ^ and f ^

a c e £ " b~d'f~h

• 1

~A» and A*=a2c2)

Ans.

12.

Multiplying both ae eg bj^dh

By componendo and dividendo we have ' ae+bf cg+dh ae—bf cg-^dh

Examples II (on Page 19) 21

u - r a c ,v a+b c+d . because if ~ - then — -r = ——r

b a a—b c—d Ans. * 2x»-3x2—x—1 3*3-;c2-5;e-fl3

By the process of componendo and dividendo we have xH2x-3)_xz(3x-l)

x+l (5x—13) • or *a (2x-3) (5x- 13)-x8 (3jt— I) ( .Y+1)=0

or X 2 [7JC 2 -43^H-40]=0

.'. x=0 or lx2—43x+40=0 which gives (x—5) (7x-8)= j.e. x=£, 5

'Hence x+0, f, 5. Ans, 14 3.X*+.Y2—2A-_ 3 _ 5**+, t _ 7x+3

3x4-^a-r-2x+3 5x4~2xa+7-v-3 By the process of componendo and dividendo we have

3x* 5x^ ' x*-2x-3 2x*-7x-{-3

or 3x* (2x a-7x+3)-5x* (JC2~2X—3)=0 or x* [6xa-2I*-f 9-5x«+10x+15]=0 or x*{xz-\\x+24)=0

x=0 from x»- I lx+24=0, (JC-8) (*-3)=0 /. x=0, 8, 3. Ans. , (m-f n) x—(a—b) _ (m+n) x+a+c ' (»i—n) x—(a-\-b) {nt—n) x-j-a—c'

By the process oi componendo and dividendo we get (mx—a) _(mx+a) . (nx+b) (nx+c)

or (mx—a) (nx-\-c)=(mx+a)(nx+b) or * mnxz+x (cm—an)—0c=x*nin+x {an-\-bm)+ab or {cm—an—an—bm) x=ac-\-ab—a (c+5)

" * (c—b)m—2an Ans.

22 Algebra ,

k Let a + ^ f c + c + i i ^ ^ a

_ab+ac+{a—b)(a~c)_ab-{-ac+ai—ac—ab-\-b.* a ' 'jo

or fl,+^^=a*+ac+fls—ac—ab+bc :

or ad=bc i,e. ? = -

f.e. a, b, c, rfare proportionals. Ans.

17. If T ~ - = • % = - then we have b c a e a_ b e d •y/(a84-fta+^+^g)_ ,_ , .

i.e. a«M: .\ ab==b2k '' b^ck .*. fic=c*A: c=dfc ,\ cd=d2k

* //=£& ,\ de—exk Again' (fla+6s4-c8+d*)=fc8 (62+ca+^2+e s) / . L.H.S.=(a&+6c+c</+<fe)a

=(fc&*+fcc3+JW2+*e2)2 •

Hence . (ab+b:+cd+de)*=(a*-{-bi+c*-\-d*) (&2+c2+d*2-f e3) AnS. 18/ Since (x—1) men do (he work in (x4-l) days. .*. 1 man will do the work in '(*-f 1) (x— I) days.

'• (3^2) ( x - I ) 10 !

or x3—9*+8=0 or. (x—8)(x—1)=0 .*. x=8 Ans. • •

19. Let the four-proportionals are a, b, c,d

.*. T ~ J »•*• a: o : : c : a or ad~b<? a <a

Examples II (on page 19) 23

Here a+b=2l (i) £+c=19 (ii)

a2+6 !+c2+rf2=442 (iii) Squaring (i) and (ii) and adding together we get

a3+62+ca+tf2+2 (a</+6c)=441+361 or 442+2 (2arf)=441 + 36I

.'. W = 3 6 0 or arf=90 .\ 6c=90 Now 0+*/=21) m b+c=>19)

ad=90$ ; bc~90\ ;

From these we can easily find out that-a=l5td~6t b=lO, c=9 Ans.

20. Let the cask A contains sherry in the ratio of 2 : 7 and B in the ratio of 1 : 5. Let;> gallons be taken from A and z gallons be taken from 5.

Out of.y gallons the quantity of 1st kind=£ y and „ ,* 2nd kind«£,y Similarly out of z gallons quantity of 1st kiod = * z and „ „ „ 2nd kicd«§ z Now £ y + £ z - 2 and ly+%z=9 Solving these two equations for y and z wc get

j = 3 ; z=8 Ans. 21. Let the cask holds x gallons of wine in the begin

ning. The cask, after taking 9 gallons of wine from it, is left with (x—9) gallons of wine, But since 9 gallons of water have been added to it, the ratio of wine to water will be (x-9) : 9.

Again from this mixture 9 gallons have be&n taken out .*.• quantity of wine drawn in this mixture

x-9 0 9JC—81 = —- x9 ••

. • x x

24 . Algebra

Remaining quantity of wine in the cask

x „ .Ya-18x+81

X

Quantity of water drawn=-x9~-=~-. x x

,\ Remaining water in the cask=9——="£niLL x x

Since again 9 gallons of water have been added. / . Total quantity of w a t e r = — ^ + Q=lMzi5]

x T .v Ratio of wine to water from (i) and. (ii)

„ * * - 1 8 . Y + 8 1 m j&c-81_J6 x x 9

.. **—I8x+8] 16 b r . — i S ^ S T " T or 9 X ? - I 6 2 X + 7 2 9 = 1 2 8 J C - 1 2 9 6 .

It gives . x=45. Ans. 22. Suppose the four quantities in proportions are

• a a „

Now if («r«-r^) > 3 ( rtr-p- \

Then lar*-l)-3 (ar— \ should be +ve

or « [ ( > 3 - p - ) - 3 ( ' ' ~ ) | i s + v e

• o r «(>-^)[rE+~a-2jis-hve

or o^r-4-^r-^j2is+ve. • ,

Examples II (on Page 19) . 25

or a I r -—1 is positive.

1 / i \i Since r is always > - •'• i r ] w ^ always be

positive. Hence the result. Ans. 23. Let the population of the country in 1871 bey and

that of the town be x, The increase in the population-of town is 18% and in that

of the country is 4%. . . , . ,_ 118* 59JC

.'. Increased population of town=7^- — -^r-P * I 0 4 52y

and population of country=-j—• y^-^f-.*. Total population of the town and country

52 , 59 = 50' 4 16*-

But as given the total population has increased by 15%. 1159 .". Total population=(x-f^) y ^ .

. „ 59.x ,50^ 1159 ; , , Hence - ^ + - 3 5 — - g - < x + , )

21 119 0 r 1000 X~ 1000 y

x M or - + v - i.e. x :y;;17 : 3. Ans. 24. Let the consumption of coffee is x, then that of tea

would be 5x. Since tea is consumed a% more and coffee is consumed

b% nwre. , .*. Increased consumption of tea t

• * — 5x (100+<»)„* f 100+a) 100 2U ...(1)

26 Algebra

Increased consumption of coffee _ x (100+6)

i .100 ...f2) / . From (i) and (ii) on adding

• • 6G0x+x(Sa+b):

WO . ' Total consumption is 7c% more of both' . no, ct 6*(100+7c) • .'. 7c% more of 6^;=- - i — — • - '

600x+x(5a-\-b) _6JC( IQ0+7C)

100 100 • ' ...(3) Again consumption of tea is by0 more i.e.

,_*( 100+6) . , . ' 0 /

= = n " — » t n a t °f coffee is a% more _*_(100 + a)

100 • As given we have

x f IQO + fr) , x (W0+a)_6x nOO+3c)• 20 ICO 100 "

600.V+.Y (564a)^tx (100+3c) ° r 100 100" ' ...(4) From (3) 5a+6=42c From (4) _ Sb+a=18c.

Solving these we get ~^z=T. ;

i.e. a=*b Ans. 25. Let the fused mass contains y parts of bronze and

x parts of brass. /. Quantity of copper will be =(x+j>)x "i^r^lS (x-\-y) Zinc in the mass will be*=-2 (x+y). , ',# •

.*. Quantity of copper only in bronze«=^-r^—=-—.

Examples H (on Paee 19) Ti

Ay y and that of zinc o^'y^THn^^*

Quantity of copper only in brass 17

(37 .Y-3J0 . 100%

50 yx^y) 5 50

o r 50* ~~ ...fA) Quantity of zinc only in brass

25 U + W 25 "~~2T~

25.x / 0 .. (B) From (A) and (B) otv.addUion we get

(37x-3j>).100 4 . ' 4 J C + 3 3 ' ) . 1 0 0 _ 50* "*" 25* " " " 1 U U

or 37x—3y+8xH-6y=50x or 3v=5.v /. x/T»3/5 Putting x/y=3/5 in (A) and (B) we get

copper=64«/, zinc=36%. Ans. 26. If the boat moves with velocity V ft./minute and

steam of u ft./minute in still water, then the velocity of boat up the stream will be (V— u) ft,/minute.

.'. Distance moved by the boat in 84 minutes = ( r -H)x84 .

.". Time taken for rowing in still water

= -^ minutes.

Time taken in rowing down the stream

. r+ti ...(0 Thus as given, we have

. ^ ~ ^ x 8 4 _ ( F - » ) x 8 4

28 Algebra

, On its simplification we get 9K2-75rw-J-84u2«0 or (9V-12u)(V-7u)-0 .'. V=*7u, 4u/3.

4w Substituting K=7«, r - in (i) we get the required time

=* _ _ =63 minutes

(f--W"_ o r i—.—_ —12 minutes, Ans.

• r

CHAPTER

VARIATION EXAMPLES HI (on Page 26)

1. If x oc y then x=ky i.e. k=xfy x=*S,y=15

Now -v=i\-x 10-5J Ans. 1 k

2. P o c ^ ,. P» -g -or PQ=zk where k is constant. Now 7 x 3 = £ i.e. £ -21

21 Hence p= — ~9 Ans.

3. x2 ccj>8 or x*^ky* or k~^x2fyz

But x = 3 , r = 4 .-. fc=9/64

V 3 ' 64-"

or 7 = 3 = 13 A n s -4. A KBC or A = AT5C /.«?. K=AjBC But ,4=2. 5=3/5," C=l0/27 .*. * = 2 x 5/3x27/10 = 9 Hence 54=9.3,C or C=2

(-.• ,4 = 54, 5=3) Ans.

5. If v4ccCthen A=KC Also if 5ccC then B=mc :. + A±.B=KC±mc=c (k±m) or A±B=N. C where N~K-km a constant or A±Bc£C*

30 Algebra

Similarly •s/(AB)=V(Kmcz)=V(Km) c of ^/{AB) oc C Hns. ;

«. If A<xBG then A=KBC . •__ 1 A^\IK_ " T'~C CfA

0F BCC Cjl Ans, 7. pocgand />xl/tf .-. P=K{QfR); where AT is any constant.

? & 7 9 7

• 3 - - * - * - o r H^H •'• V(48,='-7§3J or Q = 60 Ans. 8. If xccy then* x=ky :. x*±y*==kzy*+yz=y* (I +fc2) Again x 8 -^=fc 2 3 ' 2 -y 2 =/ (fc>— 1) ' . xS-y8 _ygf A:2 + I) _ fca +1 • »

(where TV is a constant) Hence jfa+j'*«iV(jca-^*) or y*+x* oc x2—y2 Ans. 9. Suppose A and 5 are two quantities, then

y oc {A+B) or >>=/: (i4+B) (fc is a constant) Now Accx .*, A=mx

1 where m and m' are also constant

1 y=m [mxA—I ... * \ x x ; ...(1) #

We know that x'=4, >'=6

% examples III (on Page 26)

and y=* 10/3, x=3 Putting these values in (1) we gd

«-*(<»+£) • or 24=fc 16m+m') ...(2)

and y~k[3m+*)

or J0=A(9w+w') ...(3) From (2) and (3) we get

io_16m+wi' 5 9m+m'

Simplifying it we get m'=--—4m

From 6==/: ( Am—^— J we have

mk=2

Now j>=fc l m x + — U / : U x 1

=mfc ( x

=-2 (x~4/.v)=2x-8je Ans. 10. Let A and 5 be the two qtiantities, then

yccA-\-3 or y=m (A+B) k'

Also A=kx, B=—z x2

J>=m (**+£)• ( ] ) Given x=2, y=19

x=3, y=\9

Putting these values in CO we get

• 1 9 « « ( » + ^ ...(2)

I 32 Algebra

1 9-"l3*+f) ...(3) Dividing (2) by (3) '

2k+k'l4

or \08+4k' = 12k + 9k'

or k ==•?- k

Putting this value of k' in (2)

\9=mUk + ^-4k\

or 19=m{Sk+lk) or km=5 Now from (i) we have

' = " * ( - v + ^ )

1J. We are given that Arc^/B and Aoz];C3

Given A*=3, £-256. C=2

, v/(256) 3 3=m •i—=— or m^

8 2

Now /f —m.-^-j

or 2 4 - ^ . Y ^ '-e- 5 - 4 Ans. 2 I/o

12. If x+y « z-f-l/z then .v+j^Jt (z+.I/z) and x—y x z— \\z then x—y=k' (z— l/z) Now jf—3#)'^1, r=2 • #

^amples 1H (on Page 26)

, . ,. 4-k(D or k=*

Eleminating j) between these we get - 44z^ 4 '22 , 2 * " l T + I £ ° r 2 A ^ I 5 2 + 1 5 ;

13. lt\4*:J?Cthen ^ A ' . f l C

Also 5oc/)s and C oc —

A

A

where m and m' are constants from (i) A--K mD2x~ or A- = K.mm'Dz-N.Dl

A -/>. -42ocZ>2 or /*<x/> 14. Let /I, 5. C be the three quantities, then

^ y<x(A + B+C) or r - ktA+B+C) Also' J?--w.v, C=/»'.\-Given 1—0, A-- 1 ; v --1, x=2\ .x-3, v=4 Hence from (I) we get

1 k lA + 2m+4m') 4=k(A + 3m+<)m')

From (2) ^-f/H+m'=0 i Subtracting (3) from (4) and (2) from (4)

4 \ ^ i2m-t-8ro') 4 x * i>«+5m'J—3xk (2m+tor

3** Algebra '

or 2m=—4m' i\e, m ^ ~ 2 m ' . j

'Putting m«— 2m' fn ^H-m^m^O we. get

Y., (. 3=A fm-f-5m')=jfc (-2m'+5m') or km'=\. ..From (i) j^=/c (m'-2m' x+m'*2)

= l . ( x - l ) 2 ( (sincem'£=I)

* or y=(7~ 1)B=36. Ans. 75; Let. the distance fatten in t sec. is h ft.

[i . j -

s .*. Ji=kt (where k is a constant) S05 , , c , e , . . . • , 161 ' .'. ^-<=£.(5)8=k.2,5 OP £ « j < p

.'. , Distance fallen in 1G sec . ^ - 1 . 100= 1610ft. i ' i

» Qistance fallen in t h e 9 & c . = | ^ xS l .

.'. Distance fallen in the Htth sec=i£10~ j | l .81

. ; ; - { f x 19-305-9«.. Ans_

16. . Suppose V is the volume of the sphere pf radius r.

' K^/rra or 1?9§=*( 3 j Y .

i '*e* A-—-si

', = 22 i i cu . ft. Ans.'* ^-47- Let r stands for radius, t for .thickness audit* for

weiBW* • * •vtfpn * is constant.

Examples III (on Tage 26) ^35

ji>ccr when r is constant wozr't when r and t are variable-

w / • -

9 f-»2

According to the given data ^ = 2. -™

, . 2 - « /.,. ' ' = 4 : 3. i*s 3 r2

18. Let xth, (x+ L)th aud (x+2jth be the three consecutive days and z be the total number of days.

If m be the number of races then 6=xk ( z - x + I ) , 5~k (x+1) ( r - x ) 3=*(x+2)(y- .v+2 + l) 6 _ x ( z - x + l ) 5 _ (x+1) (z-x) 5 ( x + l ) ( z - x / 3 ~ ( x + 2 j U - x - l )

or 6(xz-x 2 +z~x) = 5xz—5x2-Kx ...0} Also 5 ( X + 2 ) ( Z - X - 1 ) - 3 X Z - 3 Z - 3 X 2 - 3 J C

or 2xz 2 -2xE - 12x+7z= 10 -...{ii; Subtracting twice of (i) from (2) r

2+z lOx—5*=*i0 or 2x=2+z /. .r=--,--.

Putting this value of x in (1) 2±:

2,

Solving for r we get z=6 .'. x-—-=—=4. * !

* Hence the three consecutive days are 4th, -5th and 6th-** • * Ans.

19. Let x be the#wt. of equal rings and £ P be the price of one carat gol<i aftd £ P' that of diamond

As eiven in the nnrs f inn

M^-"^)-0

3« ..Algebra r

jErom these pn subtraction we get * -wpv. ^Lr^^-c .; ;Frbm (i) and (ii), equating the coefficients of k ' -;'f ' ].'• \' 7 / ^ r ' 2 p = I 6 a - ^ ' "

and, ,[ I Vpx—2p=a^;b—cl px = \0a—156+6c ;. • 2/>i=9a— 16fr+7c - ,j

[i . . .* - -s - ri * fi • Putting the values piPx, 3/> in (i)'we get

9 t + C 1 0 f l ^ I 5 f c + 6 f ) _ ^ - 2 5 6 + | i L c ) - f l

. .SojviagTpr fc we get • ,' < „ !'

j; ^rrK+h -J |i "and fr«|+|-A , ;i

20. jlet the first person had served,v year. - / . 2nd had served for (y-£Q) yrs.jj

./. p=*k</y, (p+50)=k^/{'y^9)y- , t Again let the first had served for OM-Y') yrs-

fi!e would have get p'. i.e. p'—k\({(y-\-*Z)}

>J("%

Nor I!v-;16

t o r 9 J / /4y+l7 V> k" ' ° r "4 "*.^/\ ,.;4y

U.(ii)

) -

Ans.

or 81_^4yjH7 64*~ Ay

Ans.

\ Examples Hi (on Page 26) 37'

>'„> Service of the second persoo«i6+9—25yis. Pension of lst=p=-*V<i6)=4k;, ^^{)' \, - „ 2nd=(/>+5.0)==VUlV+?HSJfcV %.(\i) .;, From (2> and (i) 5=4 (/>+50)iV.i>«f250 Ans. 21. Supposing /? as the force of attraction we get

or D9==KMTS ; where fc fs a constant

*' d^km2.h*~m2h* Now putting these values we get

tf dfms 2 (35)3xf 27-32)* /?*-rf2

8m, ° r * "~ (31)3X(343> „ /K35)3x(27-32)^ U715SO0x27j2'

° r / , ~ A / 1 ( 3 1 ) 3 X ( 3 4 3 ) J " 7315723 = 1 day 18 hours 28 minutes Ans.

22. Let'Fbe the velocity of train per hour and C tons be the consumption of coal per hour.

Then C x Vs or C= P2 where £ is a constant. As given in question 2=k(\6j2 or &=xk-

" c 28 V

.;, Expenses of coal per hour^lGXTas' Yz=h V2

Expenses of coal per mile -^-rr Vzx y-.=— 64 V 64

. . . , -, 45 I 45 •Other expenses per mile ^4~x j>—Y7>

P i *

• I' 5KJ45 *

It can be least when ( ^ , n y^Ml)^ z\e. y*=\%

.*. Least expanses'per milc=\ft X HY^A ^ Total jcxpenses for tbe purney—^Xjl00

t CHAPTER JV " l ^ L l ^

A RITHMETICAL PROGRESSION 5 ^ " ^

EXAMSLES TV (a) (on Page 31) * V<iV

1. s=%-{2a+(p—\)d}' '

Here fl=8 <*=3£—2--=£, «=20 A Sum of 20 terms~p

sft [ 4+^ E ]= f i f f i =277^ 1 ^ E S e37^

Acs. 2. Here a=49, rf^—5, n« l7

;. l ? i e - ti :

a [ 2 x | - I 8 x - , \ - ] = ¥ x [ | - | ] ^ 0 Ann.

4. Here #^=3, £/=—§» «•=«

.'. Sn=2" [6—(«—l)|]=3n j - ^ = — Y A f l s-

5. Here«=3-75, rf= —25, n^Ifr ;. S u ^ t2(3-75)4-I5(~-25)!=60-30=30 Acs. 6. Here a= — 1 \ , d~\, B*=24

.". 5 2 f t - ^ [ - 2 ( - ^ ) + 234]—1B0+I38 *=—42 Ans.

7. Herea*=I-3, d=~44, n=l0 A & 0 « t f [2(1-3)—9(4-4)]= 13—19S-0»-185 Acs.

8. Here a = 4 ^ ^ = 3 \ / 3 — l » V 3 . r t « 5 0

.'.- & 0 = ¥ ' [ 2 x ~ + 4 9 ; < V 3 ] ~ 1 3 2 5 v r 3 Ass. • v £

11.

If

40' • i Algebra

. . ' ' 2 S \ ' 3 24 V 3 " -u* , , ' A m r •• ;f^2:l2^75+v5Jrv5=75V5 • ' Ans^ ;i0.j Here first term«=(a—36), ,- !, - ...j' Common diff. ' «(a—26) and No. of terms^'40

'.-. k .=^-{2(a-3&)+39(f l-26)}

> rFa820a—16^06 -, I , * Ans. ' Here first term=(2a—&),"d='(2a^-2b), n«»

.i*2{4fl r2*+(»--l)(2«-426)}

t=n (2o-/»)+(«—l)(a-6)-n=n[(n+l) a—«6]

J c=n(n+l) a—n*6 ] Ans-.

. 12. Here niJ^'. c m.^^J^*.»~V "

I ^ [ i l a - 9 6 ] - I "

13. Let i4i, ^ t . i4i...i4i« ^e the nineteen A.m's between

Ant.

A.and-9 | . X a = i , (19+2)«21st term

common difference be d,

= —9j. Let the

_ 9 J = J + ( 2 1 - 1 ) ^ '•«. }\d=—$

_ai! 4

- \ * *Ans. ' • The required ^.m's are.i,— J —*£ rtU*' fl4. As in question above•{ here a»-3J, n+2:=19th

ternt=~4H ( let common diff.be d. . -

http://diff.be

Examples iV (a) (on Page 31).

.*. 1st A.M.=4-j}= — 1

2ndA.M.=i + 2 ( T ! j §

17th A. M.»f+17+ (—8)---39 Hence the required A. M's are —lfif —39 A( 15- Here a=- Sxt 18+2=20th term=3x. '

c diff.=rf. Then 3.v=— 35x+ (20—1) d or 38x= I9</ or d=2. :. 1st A .M.—35x+2x—33x

2nd A.M.=--35x+2 (2x) =—31x *

18th A.M.=-35x+2 (18)x=x , / . Regd. A.M.'s are =—33*, — 3J*f , x AJ 16. Here first term is x2, (x+2)th term«l,

c. difT. be=<*. /. l=x'+(x+2-l)d or d=l— x t .". Al=xt + {\~x)~x2-x+l '

^3=x*+2 ( l - x ) = x a - 2 x + 2

A*=x*+x (\-x)~x N , Hence the reqd. A.M*s are =x»+x—1, xa—2x+2, ,

Ai 17. The 1st n odd numbers are 1, 3, 5, 7, 9, II,

Here #«»J, </»£, **»#

/. 5 . - j { 2 + ( i i - l ) . 2 } = n " Ans.

18. 1st term a=2, last term /=29, 5n—155

42

- » * . ,

Algebra

Now let d be, the c. di£f\' then 10thterm=J2+(I0-l),tf=2+9rf .V* 29-2+9ff i.e. d=3". - Ans. 19. Here H ^ 1 5 , 5,5=600, p. diff.~5, let the 1st term

^e ff.

or 600«~j[2a+(15-]) .5]«15a+525

or 15/i=75 A = 5 Ani 20. 3Jd term=18, 7tMerm=30, Sii=*1 '[ Let a be the 1st term anil d be thee. (Jiff, of the A-P.

ten lS=£-k2<2 30=a+6rf

17/

solving these we get a«j2 , r f=3

Si7=~,[2 . 12 + (17—1).3]=612 Ans. 2jji

21. Leta—rf, a, a+<f be the three numbeis of an A. P; given, we get!

(a~d)-i-a+(a+d) =27 i.e. a=9 Also (a—jj (a.) (a+rf} = 504 'i or ^(a3—^2)=504 or 9 (81-rfa) = 504 i.e. rf=±'j , Hence Ihe th'ree numbers are 4, 9, 14 'i Ans. 22. Let thejthree numbers of an A. P, be 'a—d, a, a-*rfi-then [a—tl)^a-\-{a-)-d)'=\2 or fl=4 I; / ' Also , (a-4)3+fl3 + {o+tf)3=408 ",!' / or a3-3flV^3arf2-rf8

Tffls+a3+3fl8rf-f-3fl(/+jK=408 ;. or 3«2+6^408 )

3 (4)3+6 |(4) c/2=408' or 24/8«216 ,;or <f«±3 /. Reqd. Nos. are 1, 4, 7 ' Ans. / ! , 23. nth term *. 1st terra

JV=4+1 T i=4 . 1+1=5

I E x a m ^ IV (a) (on Page 31)

*2ndtetm ' iTa=4-2+l«9 3rd term r 8 « 4 - 3 + l ~ J 3

Thus the series is 5, 9, i3, Its common diff. is 4.

.'. S i s«^{2-5-KI5- i ) .\}Jy [10 + 56]«49S Ans.

24. /jthterm Tp*=?+2

1st term

2nd term

3rd terra

T 2 ^ 9 l 6

3 17

The series is —-, —, s-, „ Its c difF. is =-l i t i

=,= 160 Ans.

25. /ith terra r „ » - + £ a

/. 1st term 7 W - + * a

2od term r a ^ - + & a I

3rd tmn Tz=~+b a

Hence cosamon diff. is l/o

vY>« '

4% 4 ( '' ( Algebra,

26. Here1st lerm^-? a

Ans.

2nd term=4a--=;**.- ._ ! . ^ • *• , .|, . : ?i

. . c. din. </= —: — —a— = *•-- (a1— I)

j"- EXAMPLES IV (b) '(•n page:35)

' 1. ,Here a=—2, 4«4 , S^l60,'n=?

.-. 160='| [-4rf (n-l)4']«2ii8-4H

-or1 n 2 — 2 H - 8 0 = 0 or («—10)(n+8)=0 i.e.. n = 10

.1.. . ^.; 'i i \ . . or - 8

-,' Interpretation of (~8) : T If We count at the last of 1st HMefrris and count backward 8 terms, the sum will also belfiol '• • \ . , ? v." - . ;

2.[ Let n be the required number of terms ; then • 208=£ [24+<n~Ij 4]=J2«+« ( n - l > 2 *

, or . '« a+5n^ 104=0 which giv;'esn=8.-13, AW., 3. Let the series be a, a+d, a-f-l&ia+Sd,..^ Its3rd'term=W2</' !j Its isCtermna ' and 6th terms=a+5</

<• . ' . • "

As given in question wc have, .1

Examples IV (b> (on Page 35) 45

a+2d=*4a or a=~-, . '

fl+5rf«17 or ^+5rf=17 /. rf=3, c=2

Hence the required series is 2, 5, 8, 11. Ans. 4 Let the first term is a and the number of terms in

the series is ny then ' ' •

7Wa, n - ^ A c d i f T . ^ ^ - a )

Now 3]stterm a+(3l—1) tf-o+3ol~a j - ^ i

or _ 2 9 a + ^ = 4 or ff^=S

31 I v .". c. diff.rf v - - 8 — 1 --f" 4 , 4

Last term — ^ fl-j-(„_|)rf. 8+(/i~l) (—*)

13 29 or ( n - l > ( - i ) — J 8 — -or n— I--58 /.P. rt^-59 Hence 1st term is 8, number of terms is 59. Ans. 5. Let a be the tst term, n be the number of terras and

be the c. diff. then Ti=a, n=a-i-3d^Q :..(i)

r«=a+41r f=-95 .(ii)

From (i) and (ii) we get «-==-, tf~ —-

/. Last term 7V=af(/i—l)rf-^+(/i=-l) f - | )

-But 7;—12#5 /. - 1 2 5 ~ + ( i i - l ) ( - | )

which gives (n— 1) ^53 or n 54

46 ' Ajlgebra'

'• ,' 15 ' Hence 1st tersn^-.'No. of terms=54 .) Ans.

j • . 1 • A (

6. ' Letfl,a'4d,a+2dt ...,....:etc. bethe 40 instalments forming an A P. • • 'j *

2 : f

,/.<?. j3600-20 (2a+39<f) 'i ...(i)

arid, 2400«^[2o4-29rf] ; . ' ;,. >2 •- '. ,! '

[or; 2400=15 (2o+29</) ,J '...{») Solving (i) and (H) for a and d we get 0=^51, tf«2 Hence the,, first instalment is £"51 Ans. 7. Let the first number is x. ,'

Hencertne second/Will be (rrrf-* I i

' • ' " • • M 3 \ '

Now let,means inserted between x and [•=•—x 1 be 2n r< t > ; \2 /

.'. (2/;+2)th term^ / — - x ). .If,'c.(diff.' is d then

~—k^x-\-{2n+2-\) </=.Y+(2«+iD rf .!

2/ithf mean i4i« = x+2«-</=Jf'+2n. ( j ^ ^ )

12ri,Y+6x+26n—24n,v ,,12/^-6. ^

26/i+6x^l2n,Y i '12n4r6

Examples IV (b) (on Page 35) 47

.'. Sum of2n means •

-= }'2nx+13-6x 26«4-6X—Utix] '' n \ Un-hS + I J2n+6 J

J13(2«+l)l 13

Since - > - = 2 H - H

.'. 13/J—lZn+6 or n=6 Hence the No. of means=2n=2x6=l2 Ans. 8. In 2, 5, 8,

a*=2, d=3 and Sa=950

:. 9 5 0 - j {2.2+(n-l).3}*j--[3«+I3

or 1900=3n2+» or 3n2+n—1900=0 or {3«4-76)(n—25)=0

It gives i = 2 5 Ans. /other value of n is inadmissible)

9. In nr—/-»i—:.> j ,- . ••• we have

~2*t i^c ' f

10. Let«' a-H*, a+2^, a+3d, v . be in A. P

7 • 49=^(2a+6<0

Algebra ii

or . t f. a+3di T ,j , ..-(!)

and* ' I 2 8 9 ^ p 2 a + l 6 < / ] '} '

or " f a+8ii=17 jj ...(») 1 Solving (i) and (ii) for a and d we get <i I, d -2

Hence th<e series is I; 3, 5, 7, 9, ... I''

• Now .y/ " ' ,{2. |+(w , r - 2 ] «* ' ;' \ - Ans. i * (i 2 . "I ' -, '!•

I I . "Suppose .\\ \ + </, .v+2<J, .v+ 3«/...be in A. P; /Hl^term a+iit—\)d ji

' •"- Tit\<-.x\\p-\U a lj i • »-<i) '/:,) A-1 l tf'-U'l h .1 ...(») '/•J' A I [r - \)d c , j ! l..(iii)

' MuftlpfyiiiK (it byt</ "r>; (ii^'hy ir-j/»), (iii) by (/>—4) HIUI nHHingfihrm HII we pet ' JJ

f t . i r , . . . / i ) ( i / l - l )+( r - ( / ) . ( r - l ) l

or ( > p u ( ( / - r ) f A(r -/») H*'( /J—J/) : Proved. IZ. Let «, <i hi/, ii f 2./. <i .[.-V, ...ibe in A. I*.

' I) or r ' / ^ + ^ i ) ^ ^ ';' •• „,(.i)

.•iiiull>B«2|12a + ( , / ^ . l ) ^ - e - -'J " -

or j ^ ^ a - H t f M W .J ...(ii)

.Subtracting (ii)from (i) we get ',

" f„!„, 2( ^ ^ = 2 <!z£li?i£L~i!l£^(£±£l"'

' '' />? pq -n />¥ ^

P5 ':

, j

Examples IV.(b) (on Page 35) 49

Dividing by (2)

(f ^ pq pq

On simplification S;J+1?^ — (/>+?) Ans. 13. Let a—ldâ—d a+d, a+id be the four integers

in A. p. then their sum is 24 • .". a~3d+a—d±a+d+a + 2d=24 or s*=6 Also their product is 945 .'. (a-3d) (a*~d2) (a+7d)~94S

V (a2—9d*)(a*~rf*)-*945 pr a<—10a2tf2-}-9rf*-945 or fi4~ 10x36d3+9rf'*=945 or 9rf*-36Prf«+351«0 or (9<tfa~35l> (<*a~ 1)^0

.-. f/2-39 or rf2~l /.e. rf-i%/(39) or ±1

Hence the numbers are 3, 5, 7, 9,,..when rf--^l i , Other values of rfare to be neglected as numbers will .'not ilbc integers. •' .

J'4 The four Nos, in A. P. are a — 3d, a — d, a-{~d, a-f 3tf

.*. Siimâtr-3#+tf—d+a + d+a+3d^20 o*r 4a^20 i.e. a~5

' Produ<<of Utând'4th=aI-9rf2

„ 2nd „3r^=as—d2

_ a»~9rf* 2. * 25-9rf* 2 = ^ or rf8 ~3 25-</2 " 3

50 • ' Algebra . • I , ' ' \ or f. 7S—27d2=50-2tf2 i.e. rf»«l .\ d'=*±[

I i ~

Hence the four required numbers are | ' 2,4,6, 8. < Ans.

15. Let a, a+d, a+2d be in A. F[.' T/ien Tp-a^(p-l) d=q i „;(r)

!, f T^a+Xq-l)d=p ,i ...-(ii) Subtracting (ii) from (i) ;•'

f- lP-<l)'d~rti-p=^(p~qj\ ;'. ,/«—1 Now a-KjP— 0 </=? .'. <*-(>— 1)=? or tf^/>+#~l Now Tm=[p.+q-\)+{m—\){^\)=p+q—m Ans. 16. Let«terms sum to 306 of the series who'se -f ' '' . ' z>=9, J=3

(' ^ ;; - - ' or 3wE-M5/j—612=0 or ./i*+5n—204«0 ,or (n+]7jk (n—12)=6 i.e. 1»«12, —17 Ans |17. J J ,«2B4-3«« . " ,j ' | We know that rth term 7V=&-Sr-i j.V. Putting n=0; r—1 we get,;. | ' Sr=2r+'3r* ;l

& l i«2 ( r - l ) + 3 (r4-l) s=2r—2+3r»-6r+3 • =,3r2^4r+l 'i

.".. . , rr=(2r~3r2)~(3r*-4r+l)»=6r-l Ans. 18. let the A..P. is a+d, a+2d,-6-h3dt ...

. s w | t2fl+(«-"l)'tf]- ' ' . '

Examples IV (b) fon Page 35) 51

Moiv r " ' - g + ( m - 1 ) d^a^t2m-2)d . . « .

Tn fl+(«-l) d 2a+<2fl-2) d "'W

Comparing (i) and (ii)

If we <putm~(2m—1) in (i) and n = 2«— 1 in (ii) we get 2a+(2m—2)<t_2m-\ 2a+(2n~2) d 2n-l

Ans, T^^lm-X Tn 2 n - l

19. Leta, «+(/, fl + 2rf,—be an A. P. of (2n + l) terms

•'. S t t + i * ? ^ [ 2 o ( 2 / i + l - l ) r f ] ...0)

= ~ ^ [2a+2«rf]-=(2rt+l) (fl + m/) .. (i)

Now middle term will be

i.e. (n+l ) th term .*. 7 n + i = a + (n+I + l) rf=tf+nrf Ifis clear that (i) is equal to (M + I ) th term multiplied

by (2n + l) i.*. number of terms. Proved. 20. Given that S„=n (5n—3) .'. S;,=p (5/»—3) = 5ps— 3p; {putting n=p)

and S ^ j « ( P - 0 ( 5 p - 5 - 3 ) * ( p - l ) ( 5 p - 8 )

• • I p^Op— Jp—l 1 ~(Sp*-3p)-[(p-1) (5p-S)]

«{5p 2 -3p) - (5p*~13p + 8) = lOp-8 ..Ans. 21. Let a, a+d, a+2d,...be-the arithmetic • series and

let 2n be the number of terms, then T$n=a+(2n-l)d

* Ttn-i=ra+{2n-2) J

. ; S u m o f o d * terms=" {2a+{2n-2) <*}

Algebra i

=n{a+(n-l) ••rf}«24jgiven) and sum: of even terms=(fl;M)+(a + M) + k..

j. +[a + f2n-lW] *=?[d+rf+fl+.(2«-li}r/]. " *

- " • [2f l+2nr f ] ' ! ' ! '

=n(fl+i«0—30 (given) Thus ' rt[fl+(n—l)rfi=24(' ,]

1 'n[a+flrf]=30;- J Also , T 2 n -Ti=10J(g i v e n ) il

21 'or a\-(2n-\\d—a~.-7r~ ' ;)

• ;-'br (2n-\)d-^- ; !i

i : - •' 2 . • • ' i

Subtracting (il from (ii) 'i •I-nd^6 n

6 . Putting d— - in'(iii) we get

(2n-l)..^±i./e .-nr=4 (ii . n 2

* ' ^ 4 = 2 ' !

' ...(0 ...,0i)

;..(iii)

Putting these values (ii)'we have ,.'

" 4 ( f l -4 .x | . ) = 30 ,V a=\

• Heace the series is f, 3,f, ..and, No.; of terms is .8 Ans. 22, |'Let (a—d). a, (a+d) be the 1st ot th»ee

nunjbtôfan A.P. ' " ' , ] ' ' _ " SIrfte .their sum is 15 ,\ â-dâ^-a^^.15 or a=5

Examples IV (b) (on page 35) 53

Afso 2nd set is a—(rf—1), a, a+(d— I) • Product of 1st set=a (a2—V3) Product of 2nd set=fl {a3—(<i-l)z)

a(az—d2) 7 , . .

or 8 ( a 2 - ^ ) = 7 {a*—(d- if} or a2=8rfV-7 (rf—1)*=*8./*—7(/a + 14rf-7«</2+14rf-7 but fli=5 .'. 25=tf2+14</—7 or rf*+14</—32=0 or (rf+16) (rf-2) = 0 /. rf=2 Hence the 1st set is 3, 5, 7 and 2nd set is 4, 5, 6 Ans. 23, Let /*i, ^3, W3...A be the n A. ATs inserted between

x and 2y. If c. diff. be d then Tn+2^x-\-(n+l)d=2y

• ,/ 2>'~x

n+1 r (2y—x) nx-\-x-\~2ry—rx

Af*=x + n+1 n+1

»*+#«rx-\-2ry .-(0 11+1

Again 2x, A, f2, ,..^n> ^ are in A, P.

«+ i

A *> i/J v~~2x} 2nx+2x+ry—2rx ... ^ ,=~*+ M + J - ^ j >M(iij

• ,\ From (i) mid (ii) we see that the rth mean in both the cajps is the same

• "•*+*—rx+'2ry . 2fix+2x+rp+2rx « + l • ~ «+l

f or ry=itx^x—rx or ry=(n+l—r) x Ans> i

54 • /' . Algebra J • i f

24.'. Let a, a+rf, a+2rf, ... be in A. P. • since Sp--=Sq . j

I * z • ! or 2fl(p+ (/>*-;>) rf=2fl?+(^ff) rf or '2a (p-q)=-d (p2~p-q2+q) -I « - tf{0>s-<?2)^(/>-<?)}

p'r —2a={p+9—l}.rf ,/

Now 5 P + 9 = / ^ { 2 a + ( p + ^ l ) . r f } . '

, ; 2, ' .'

' CHAPTER V

GEOMETRICAL PROGRESSION

EXAM PL£S V (a) (on Page 41)

1 2 1. Here u = - , /•-- =

a (I— rn) Sum of/i terms in G. P. = —-.

1—r

... i J H f U . ^ ^ 3

2059 ^1458

2. Herea=~2, r = - £

. ' c - 2 | t — ( — f ) ° ] 128! . . 5„— ^ i = 5 l T - Ans. 3. Herea=a, r=2, « = 8

= 191i Ans,

4. In 2, - 4 , 8, ... to 10 terms a=2, r=>-2, n= l0

- ri r *m<n 2 ^ . \ 51 0-2 [ J — ^ i - J ^ f [1-1024]—682 Ans.

Ans

56 »- J Algebra

f ^. .'• Here fl=i61.2,.r-J/3, n - 7 ;/

» • f 1093 ,j

1. 'i • 6,j Here «= 1, r=5, /*=/?' ;,

Ans.

7i. Hecefl=3,r=-|> n=2n Jj "

8; Heretf=.I,,r=V3, « = 12 !i

!-• ? -1 J t V 3 ) " - M . ™ : i . 728(V3 + i) * ;j!>»--11 V3--I J~V3f i = (V3i r r ; 773- j r i ] -

| Q 7 2 ^ W U 7 2 ^ 3 ± n l 3 6 4 ( v 3 + v) ^

i> ' • -: i • - i • -,9. Here a = - ^ ; r=--2*/2t h=7 !' •,' ,, y ^ J/' '

" " ^ ^ V ^ T l + 2 V 2 i f 2 ^ 8 5 ^ 2 - 2 9 2 > A n s '

1 3 '' ' 10. Here a^~~jy r = = " 2 ' n^1

'rsJ*fjfto-4{«ffl 463

"192 ^Vns.

\ Examples V (a) (un Page 4 t t . 57

9 4 Tl. Let the thiee G. Means between -&a.nd.=- are

•4 - 9 ' • Gi. G-.', Ga

9 4 Then -,t d, G2, G3, •? will be in G. P.

So 5th term-^f, first term- ^, let c. ratio be r

.'. nth term^ or" -1

5 4-''4 o r H - ¥ i or Ms I A r = 5

9 2 3 3 ' 2

*> 2 G3- l x - -r Ans.

12. We are to insert 5 G.M's between — a n d r -

8*1 ' So 7th term will b e ~ . -Let thee, ratio be r

81 32 - = ' r - . r 6 or /

~ 32 3 16 Hence Gi——-• 5-- ^~

04 \ 2J " 2

Ans. . Cs ^ 4 S,Ca-8x| 12...G. H I ) ' - 2 7

13. Let 6'u Gs,......Ga be the six G. Means between 1.4

•and - 7 ? then 14, Gu G« , Gc»-r i will be in G. P. 64 64

~ 7 If e nuio be ;•, then 8th term —7-.

• 64

58 Algebra

7 / r 1 \'fi 7

. r 2" ^ M ' ' H ) 3 < 14; Here a'- -c. r -— •

i 5 i t>

Sum to infin

"i * "I—r ' l - & 100' ?i-.5"8 I' - _ i - i i|: • ->

16. Here a ^ l ' 665 , r= ,--,,,.-'* -— £

_ I -665" 1-665 3 ^

Iff. Here fi=}~l^t r^yi=f~'=\

- \ I • «- 1 — 3 3

C 1 IS. Here a 3, r - - , '

. l i \ •'

f —

3 _J>/3 3 ( j 4 - y 3 )

VJ

9. He,,, 7. „.><£> j (< ) , . I '' 'l " I'. e = Z . 7 ^ 7 i ' 7 r7 j / , ^ , i

A ns.

A ns.

Ans.

Ans.

Ans.

Ans.

A lis.

1 Examples V (a) (on Page 41) 59

20 Let a, ar, arzt r'\ ar\ ... be in G. P.

V . c fl(I-r°) , _ a ( l - r 3 ) . .*. 5 G = - ^ — ~ - a n d 5 3 = ^ — ^ - '

Since 5 * G = 9 . 5 3

••• ^ = 9 . ^ o r , - , o = ( 1 _ , . , ) 9

or ,-3_9r3+8 = 0 or ( r a - I ) ( r 3 - 8 ) = 0 i.e. r = l , 2 Hence r = 2 ( r = l will be neglected as- the denomi

nator in the sum will be zero) Ans. 2J. Let a be the first term and r be thee, ratio of the

G. P. then nth term=ar.n~1

.*. 8 l=ar* ...(i) v and 24=ar ...(ii)

From (i) and (ii) we get or4 81 . 27 / 3 \ 3 3 Tr=24 °r ' ^ r l l ) -I- r=2

3 From (ii) a-^=24 or a~16

Hence the required G. P. is 16, 24, '36, ..'. ' Ans.

22. Let the G. P. be a, 3a, 9a... let n be the number of terms in it. Here r = 3

.*. Hthe term T„=a(r)n-1

.'. 486-a3"- 1 ...(i)

and 728 = ~ J : ~ . . \ (ii) (sum of« terms)

From (ia .3rt-ix3=3x<tgv or a . 3 " ^ 3 x 4 8 6 Putting this value of a -3" in (ii) we get

fl3"-a 3x486-a T ^ T - ~ — r — r - = 728 or. (i 2

»• .'. 1st term is 2 Ans. 23. First term is 7. Let the c. ratio be r, then the

60 j " Algebra

G. P. will be 7, 7r, 7r2, 7r3,".- • -If this/series contains « terms then the, nth term will be last term /.e.^48 r"^

I' < / . 1 .448=7V"-1" • } • ...(i)

and S^m^-11—r-J- ,-- |; , \~r I ' .-(10 : From (i; 7r"=448r " f; ' ; From (ii) 889 ( l - r ) = 7 - ' 7 f l = 7 - i - 4 4 8 r

•' on 889-889r=7-448 / - or '44lr=882 i.e. r=2 Ans. 2<J, Let the three -numbers iri;G. P. be a/r, a,art their

producUl728 0Oi~} O } ' • " ' " ' .'; - ' X a x / v = ] 7 2 8 or fla-=l728/.e.fl=J2 '

(• r ' " ; ; '

TJheir sum IS~F38

;•. " f c + ^ - 3 8 or^4-12+'12r=--38 \ r r j( or 122-M2r+12 = 3';>f or 12r2-26r-f-12=0

pr ( 6 r - 4 j ( 2 r - 3 j - 0 : a - . ' ' r J | / - = |

(Hence the three num'bers are 8', 12, 18 > Ans.

1 ,li25. Let - > , ar.be in1 G. P . ;

j; • ' a ) l Their /product is 216 .'. -.a .' a r = 2 l 6 i r

[' \ • 'J °r* a , = =2I6/ .e . a = 6 ' tf Q i' '

|- Also-- a.+ a • a r+a r *=-="l56 • /!'" r' ' r i fe-r + ]) = J65or r 36 . l 'C l^±T) = i56 j or a

i

I, or 6rE-f-6r-f6 = 26r or-6rE,-20r-f6=*0 |; or/ ( 6 r - 2 ) ( r - 3 j - 0 / e. r = 3 , & ^

j ' Hence the numbers are 2, 6, 18 - Ans,

f 26. Here a = I, c. ratio r=4rD > ' ..

http://ar.be

(Examples V (a) (on Page 41) 61

1 2

But in 1, — rP, + r2^,...a= I, c. ratio r= — rP *

v. ^ f = 7 , ...(ii)

Adding (i) and (ii)

a ] S . ' + ' - 2

I i

But 5 a p = ] : i r 2 a .*. 5 p + ^ = 2 5 2 i , Ans. 27. • Let x be the first term and y be the common ratio.

. Then as nth te rm^y"- 1

a=xyp-1 ...(i) b^W1 ...(ii)

, c=xyr\ . ...tiii) Raising powers (i) to (q—r), (ii) to (r~pj, (iii) to (p~q)

'ive have

a 9 - ' = (X}'P-l)<7-'

CP-9=( .Y) ' r - 1 )P. - q

Multiplying these together w'e get aq~T _ ty-p CP-q = xQ _ j , 0 = ^yyjO--. |

Hence a?-' . 6*"'' .c'-«=l Ans. 28. Let a, ar, arz, ar3, . be in G. P.

| .". Scf=~ =4 (given) ;..(i)

I Also sum of their cubes i e of the series a3, a3 r3, a3 r9 .. 'to x is 192 1U as '

From (i) and (ii)

62' , Algebra

, " i pr l_3r+3r 2—r 3~3-3r 3 ' •.j or , 2/^+3^—3r-2^0 j

Factorizing by remainder theorem we get r ( r - I )< r+2) (2 r+ l )~0 , .'. r - 4 - ] , r r - ^ 2 , r — 4

a* i; Taking c = — |Trom Sa—7-7-r=4 we have"a=6

i 1+4 'j (We have taken m ft as it is less than 1 numerically) Hence the series is 6, — 3, 3/2...', • Ans.

EXAMPLES V (b) (on Page 45) • 1 , 'Let the sum be represented by S'. Last term will

be va"r%

.',. S=l + 2a4-3a2+4fl3+.-'.'+«^"J •• (0 , Multiplying both sides by the c*? ratio i.e. a we get

S- a^a-h2fl2+3fl34-...+(rt-l);flB-1+«fln •••(»') Subtracting (U) from (i) '.\ S{\-a)^\-\a+a* + a!i+~nd'>i 1 =?d Geometic series of n'.terms

i • l - »

^Multiplying both sides by the c. ratio J we get

Subtr (cting (ii) from (i) we get

f« ?s~,+2+i+S+IS + 4 e

!• • - • ' - . . ' !

00

Examples V (h) (on Page 45) 63

l - J 3 Ar#. 3. Let 5'=l-f-3.v+5%2+7A-:j+9.v4+ . cc .. (I) or Sx=x + 3x* + S\*=*lx*-±... oc ...(2) or S-Sx M+2 .Y+2 .VM-2X 3 +. . . cc

or <l—.r)S= 1 + 2*0+*+** +. . . oc)

I — . V 1 — X

or 5»J±«_ (l- .V) Ans.

4. U.5-I+I44+- + 5 ^ _(i)

or i S - i + J 1 +J ,+ -f^ (.;) Subtracting (ii) from (i)

/. s - i 5=1 + 14-^+4+ l * 22 2 3 ^ 2"-1 2'

or 1 5 - ! - f » y — * r 2 ^ " k 2«

or S - 4 ( 1 - ? ) ~ 2 ^ Ans.

5. Let £ = 1 + 5 + 5 + ^+' * .. (i)

Subtracting (ii) from (i) we get

^ - , + H + i + * ' = 1 + 0 + 2 + * + « )

^ l + l f — i ) 1 3 ^ 2 ^ 3 or,S,=3x2c=6 Ans.

64 , I' , ' ' Algebrat 'J 6. Let 5'=l+3i:+6x24-10.v3-f >...; cc '. .. fi) •

j ' .*, x. .S=*+3xa + 6jcs Jioc '• . ' ...(ii) " ^ Subtracting (ii) from fi) we get ' • , [ ! "

f U^x)S=l+2x+3x2f4*3+j- . ct . . ...{iii) Multiplying again both sides of (iii) by x we get • |' x(lW)S===x+2x2-f3x3-f 4x*|f oc' " ...(iv)

Subtracting (iv) from (iii) 1| T 5'(I-2x+1x3J=H-x4:x2+.v3j+ oc i ^ 1 l',, „

or 5 f l ~ . x ) 2 ^ ~ - / .£ : I—x " • ( I - x j 3 , Ans.

7. Let the common ratio be r. lst,term is a- -Then '

Also'Ietx be thee, ratio of 2nd G.}p.

'A liT^-Ta'^i , Ans.i »

8. /let the 1st G. P. be «, ar, flr2,!

i a ( l — t 2 n \ .', •! Sum 1 2n lerms—5««" —~—--Let [the 2nd G. 'P. be ft,, 6r2, brU ...}..

' & « ] - > *

As *' given we nave —:— -r-r= - . . ... ) '

or [ail +r) = b or b—a—ar ;] Am. ,$. . L e t ^ s l + i l + A j r + d ^ + ^ ' K 2 4 - *• .*. 5 - r=r+(l+£>)r2+, cc ) Subtracting we'get '; - | s-S- r=l + br+bir2+ A , . . x .

Examples V (b) (on Page 45) 6,5

or ( l - D S - j i ; .-. S = [ T - F ^ - 7 r ^ ,

10, L5t-, a, ar be in G. P.

.'. -r+a+ar=10 or a \j + 1 +r) =70 ...(i)

4a Also —, 5a. 4ar are in A. P

r

''• or 5r=2ra-f 2 or 2rz—5r+2 = 0 /.*. r = 2, * ^Putting r = 2 in (i) we get a ( | + l - f2)=70 i.e. a=2$ •^T.ljus the numbers are 10, 20, 40. Ans.

'•I 1. Let a, ar. ar- be in G. P. ' .'. a + ar^5 ...(i)

Since a~3. — - (given) I

l~r—3r or r={ !". from (il a (14- Jl —5 or a = 4 Hsnce the required numbers are 4, 1, i Ans. 12. Let 5-(A-h(3) + (.vH2a) + (A3 + -^) + " terms Let t"his can be written as 5|U(.v+.t2+.-«3+ n terms)4-(«+- 2.-j + 3u-f -••«) terms)

= x( l+.v-r .v l+ n terms)-j-« (l-f-2 + 3-K..?i tenn>)

L = » - i i r = ^ - ! + « - 2 - [ 2 + ( » - i ) . i :

' .v f.v—11 , n , =?,—^^j— +2 (" + 1) * An*.

|^3. l; Let s-.v(.x-J-*v)-j-.v2 -ys.. _,.-,_ x3 lAa . r 3 | +....„ i e r m s

= v- + .vi'-f .v*-J-\;v--r \8 - \?i s r . . J! terms

6 6 I AJgebra

jor S=(.x~-hx*+x*+ « terms)+ (.YV + J ^ 3 + * 3 ; ' 3

. , + n terms)^ I ^.vs (J + . V 2 + A ' 4 - n termsj+.vv {\j-xy+x*y* I ' + . n terms)

; . . g i ^ + ^ , ( f c i ) An, ; 14. I . e t 5 = ( a + i ) + ( 3 a - i ) + C5i+1»a)+. 2/>Jerms J It can be split up into separate series as \ J«(a+3o+5fl+ 2pterms) + fJ-H.i IJ+...2/> terms)

J . * + £ : = 4 ^ + | ( , _ ^ ) Ans,

j 15. Le«5=?44444+ «

' 2 (H4+ «) ^( •44+ «)+

16. Letfe* l + i _ I + i _ L + « 7 7»773 7* 7* 7* +

or J- , f t+i+t+ \_ - '

•Examples V (b; (on Paye 45) 67

4 K - 4 + • • • • « ) -

M-44+ *) ' iH)(<44+ «)

17. If (J, /), r, rfare in G. P. then b e d , , , . , , — = _ = _ _ o r i->- = ac, C' — nd, bc = tjci a b c

Now (b-c)* + (c--a)2 + icl—by2

-~bt + c2-2bc-l~ci + (r'-2ac-\-l!1 + b2—2bd

**a*-\-dz+2b1-!r2cn—2bc~2ac-2bd ^a2 + d2 + 2ac+2bii-~2bc-2ac—:bii ^.a* + d2-2bc ( 7 />5-=tJc, v'-^bd) —/Jl-rd*—2ad (v hr -oil) =«(a-rf)a Ans.

18. A . M . « ^ f G . M.«x'(<;M

As given we have <~- -2\'(ab)

or (a + b)~=\6ab or a=—l$ifr + />7^0

OI U~) ~~'6 i5") + 1~0'II 'S 1 u a d r a t i c i n r ncnce~h ^ ^ — .

a 7 - 4 , '3 or •— = ~-—^~ {can be written)

o 4 — J

- _7~-<V3 _ T24V3-.' - 2 + v / ?

a + ^ 3 i ( 2 - V 3 ) i2 + v ' 3 M 2 - % ' 3 j 2 - V 3

I Algebra

Hence a : ^ = (2+^/3) : (2-A '3) , Ans. 19. i rr==i2r+l).2r

Put , T = \,2, 3, 4',...we get '7-1 = 3.21 72 = 5.2; ' r3=7-23

• r 4 -9 .2*

7W2«+n.2« A S=3.21+ 5.28+7.2 s + 9.2* + ..: + (2n + l).2-6r 2.S"=3.2*+5.23 + 7.24 + ...

+ ( 2 M - 1 ) 2 " + ( 2 « + I ) . 2 B + 1

Subtracting we get S—25=3.2»+2 . 22+2 , 2S+.. . '

i " +2 .2"—(2n + \)\ 2"-1

or ; -5"=«='3 . 2l + 2 (2* + 23+....+ 2")—(2w+I) . 2"+»

| =3 . 2 + 2 [2« ^ £ Z ± . ) ] - ( 2 B + 1) . 2"-1

; -=6+2*+2—S—i2«+l). 2**1

or , —S=2B^ (1— 2n) —2

S^n . 2"+8-2"-1 + 2. Ahs. 20. Let the series be ] t a, ac, ff'c. dV*,- a3c2 .. 2?i terms Sa n-(l ,+a+a scB + ...n terms)+(a+a!o+as<rI+...n terms) J 1 . (I—aV) a fl— aV*) _ ( l - f l V l .'l+.a)

1—ar 1—ac 1—ac -

or $ • - » ~ ^ — J ; — An&

21 a; a r , ^ ; or3,...'in G. P.

.Examples V (b) (on Page 45)

. c o ( I - r i )

a{\ — r1)

Now 5 ! + ^ 3 + 5 s + ...

=<* + —z—i—+—:—r-+ •••• r - l r-l r - l

^ " " f ^ T T + ~ i + - t 0 ("~" terms 1

r - l _ o (» —h i q r = jr"n~2—\

(rf + /-5-f-...to ( H - 1 ) terms]

= a -\n—i} ar- IT-"— —

r—1 + f - l [ r*—l ^ar—a—an-{-a, ar* !r2"~2—l\

I "TTTi — J

^ L ^ - 1 r - l

a "r^l _

22. Sl = l + i^+i+...x = r U i -2 2 1 '

S.-2 + 5 + g+...x-jLj-J 5 J=34+ix. . . ; c = ' .4

A

0 ; Algebra

.'. •' Ji + S- + i W . . . 4 - V - 2 + 3 + 4-t-... + (/>+!)

j «f[2 + i/M-l)]-[v S ^ («+/,] ,

J * Ans. 231 Let ' 1<> r, r2, rz

t...r""lt rm, rm+\ rn +2,...r*m be in

- • p . j We know that (1— r"')" > 0 .'.J \ + rSm + 2rm > 0 or i + r3m > .2/-'" Thus in the above scries making pairs (1st and last term)

.2nd and last but one term* and so on we get I !-}-/•••»< > ^x")

\ r+i->"-i > 2r'" | r*.].}••><>-% > 2 r ' "

Adding both sides l + r+r8 + r«" > 2 (r"»+r'» + i-*+...ftr times)+rf

i i — - r •

,'or (2in+\){r"-)(\-r) < 1—r3'" [Multiplying.both sides by rm~l we have j(2m-H)/' l in+1 ( 1 - n < rm+,(I—r1"**1) j Putting n=2mJt~\, we have .

j «f> (1—r) < r - X(l—r") j Making « indefinitely great

j Eil ,' r 2 is indefinitely small anil therefore tit" indefinitely

small when n is indefinitely great. • ,

CHAPTER VI

HARMONICAL PROGRESSION

EXAMSLES VI (a) (on Page 52)

1. (ij Corresponding A. P. of the given series will be 1 2 L 2' 5 ' iO""

Here ««= p ^ 5 — — " f t

.-. 4lh term of A.P.«} + 3 /—L) = I-

Hence 4th term of the given series—5 Ans, (ii) GiveQ series is 2, 2\, 3,,.. . It is in A. P. I t sa«2, rf«i .-. 4th term=:n+3^=2 + i = 3i Ans.

O'iiJ 2>2 » g"»---are in G P *

Its (j*=2, r«£ .*. 4th term«=2r«->«2 (£;*

.-2 Ans. 2. Let between 5 and 11 the two H. Means are H,

and m, ;. 5,Hu ^s» U are in H. P.

T' W li«' iT m v' Its fl=I/5, rf=? and 4th term is 1/11

^

• ' Hx 5 55 5 5 a D Q ffa 5 ~ Z * 5 5 Js

9 «7 Ans. 3. Let the four H. Means be tf:, Hz, Ha, Ht

.*. =-, Hi, Hz, Hz, Hi, -Q.are in H. P.

Or »y, £ - , ^ r , j p ^ - , j - will be in. A.P.

If d be the c. diff. then J V ^

.-. L3 = j -+5rf or rf«i

1 3 M I

5 rr 2

•'• HrT+l=2 o r i / l = = r • 1 3 7 2

2 4 - H 3 . . l 4 or' ft-l- '.

. 1 4 + 4 . 1 4 or ft4 ' A n s .

4. Let the numbers are a, b, then

'.G. M.= \'{ab) ; H. M . = 2* 'a+b - •

As civen wc have \/(fl0)*=12f-7—7=37-* v ' a+b 5

or 00= 144 .', r *=-=- or a+b = 3Q a+b 3-

Now a+Z;=30 and <JA= 144

Solving we get a*='4, 6 = 6 ' Ans

Examples VI fa) (on Page 52)

5. Let x and y be the two numbers.then

H . M . = ^ ^ , G . M . - v ' { ( - v n }

According to the question v-t have

0r ' ^ = = 12V(.w) o r l M ^ = 6

or .I3-V'(jrj,}=6 (x+y) Squaring both sides \69xy = 36x2+16y-=l2xy or 36.ta—97.vy+36r2=0 or i9. \ -4y) (4.v-9y>«=0

•'ie. - - - or .v : v=4 : 9 Ans. y 9

6. V a : (a—b) = (a-{-c) : (a—c)

— = or a2 — ac=Q2 — ab4-ac—bc — b a—c

2ar or b (a + c) = 2ac or b— •

a -t- c Hence a, £, r are in H. P. But this is gi\en .'. -a : (a—b) = {a+b) : (a—c) 7. Let the corresponding A. P. is a. a + rf, o + 2c/

Then T„s = a~-{m— 1) i / = -

i

. . — — = (m —m (/ or-/J m v »1/J

= (»i—H) (/or d— — inn

74 Algebra

r- l ,-, , .'"—1 1 1 /»!-] 1 J-rom (ij a + •=- or a = - — !=_i_

• i "'« n 4l ' « m« MMI

.-. ;Ttn+n=a+(m+^l) rf« J-+»L+B-1«»!±« • Hence the corresponding term H. P.== J ^ _

j ?«+« Ans. '8. Let the corresponding A. P. is.v, x+d,x + 2'd, then TP = .v+(/>—l)d=l/a (given) (i)

JTfl=.\-+•.<£—l)rf=l/6 fii) ]V = x+(r-\)d=:\lc (iii)

Multiplying each equation by abc we get j xabc+abc(p— 1) il=bc (iv) I ,xabc+abc(q— I) d=ac (v) ;• xabc-\-abc(r— 1) rf—a6 (vi)

Now multiply (iv) by iq—r), (v) by (r—q) and (vj) by p—q) arid adding together we get

' be (q-r)-irac{r—p)+ab{p-q)

j C-ff)(?-l)+(/»-?)(r-I)]' 1 =0 + 0

Hence &c(£—r) + ac(r—p) + ab(p~?)=0 Ans. j 2. If c,&, c are in H. P. then i 2 1 , 1 . , lac

Z> a c a-tc ' \ .' 1 ». * ^fe-c+fi-fl _ 26-£q+£)

i/j-a 6 —c lij^aj/J-c) (i~s) /) —c)

b 26=—?gf

/,-_-6 (o + cj-J-dc 7> 0- —2(iC + fl;'

2 i£*-ac) 2 1 , 1 . = rrn ; — 7 = - T - from (n

b (« ' — ac) b a c '

Kxamples VI (a) ton Page 25)

1 Ans.

75

Hence - ^ r ^ + M ^ <J «

10. T„-3nft-w

But we know that ^ - l i ' i l H - f ! ^ ^ .„d 1 ^ 4 («+D

Hence $,--3.*- — 6 2 l" ' U

, ! ( „ + ! ) [2«+l - J ] -«Mn+J) Ans.

11- T „ - n 3 + r "

But 2rn»-[^l«+l)]«»^«^(«+l)

... s f l - ' ^ « + i ) a + 2 - ' j - t » f n

= 4 ^ + 1 ) [(«+ D + 3] 4

4

12. T„=/n» + 2)----«, + 2« • ,$„«=£ i ^ + ^ U - W + ^ r t

Putting the values of .Tn8 and Zn we get

^ n " " 6 -

fi yi +j j i2« + 7)

13. r«=sniC« + 3)-2n" + 3/i* .-. 5-J-,s:i2«3 + 3ns)!=--fl3 + 3- '1

Ans.

Ans,

Algebra •

or S . - 2 . £ ( « + ! ) fa[*!±ng»±h

', -"r(a+l)["("+»f"+»] ! _ n (n4- lUn24-3/txl)

: 2 Ans.' 14. TB--=3"-2" Putting n=l,2.3,4 we get 1 Ti = 3 l - 2 1

T a «3*-2 3

] T8 = 3 3 -2 3

i

T f f=3n-2" Adding all the terms columnwise we get ;\ Sfl=B(31 + 3s+33+344-.».,..+3'*)-*

I [21'+2n-+2:i+2* + +2") J -=3(l + 3 + 32+ n terms)-

' 2 ( l+2+2 2 4- n terms) „ . , | 3^ ;1 ) , 1 2 « - l \ " " " 3 -1 " 1 2 - . /

• ; = ^ V : " T l + 2=" ^ 2"*i + i Ans.

Note ': — When n is in the powers we can not make use 1 of the sigma notation as we have done in i Qs U, 12, 13.

is T„ = 3 (4" 4- 2«?» - An* = 3. 4"+6n= -- 4na

:. Tx -3 . 4] + 6 • I 2 - 4 . I3

Tg--=3 . 4* + G . 2 2 -4 . 23

i

Examples VI {&) (on Page 52) 77

r 3 = 3.4D+6 32-4 .31

r„ = 3 4"-H5H- —4«3

.'. 5n - 3 (4-M=-M3-r -f 4") + 6r«*-4:S/i* ?=\2 ( I -M + 4S + .....// terms) + 6i:n2-4£ri3

" \ J - l ' ' 6 4 - 4 " - 1 —44-/J (//-Hi {2 / i4 -n-«- [n+lY-

s s =4"-i-4 + n (n-f I) ( ] + « - « = ) A n s -16 Let the A.P. be fl. a+d. a±2J. ff+3<£ , then

rn_i = a + * " 4 - l - l ) d=a-nd r , - i = fl4-fr-i-l —1> i/=n + rtJ

Now if fl + m<f: a + «J : a + r J are in G, P. then (tf4-M(/)2—(fl-fTe/) (l-rind)

or fl- + 2f;/;«/-f-;i-(f- = fl::-raJ (r+m) + tnrd'1

or 2and± n-d- = ad (r-f- /H) -r wni :

or 2afl + «V-rfl rr-f m)-r"ird 2an—a (r-J-m)=lmr— »-* tf

c nir—«a

</ 2/i — r — ?«

7 m, H, r are in H- P. .". n = z

ni-rr or n (m + r) = 2mr

or - , - , „ , **—= Ans.

Hence g . , " " - " 8 — ^ ^ ^ ' " "

or a-

78 Algebra

! 17. Since /, m, n, are in G. P. .*. m'^/n U Let ff.o+rf, tf+2t/, bean A. P. | .'. 7 }=a+( / - l ) rf

! S nee r,, r„ , r„ are in H. P. /. Tm = ^ &

; 2[g+(/- l t r f ]Cf l+(/ i -Mrf I 2<z+rf(/+«—2)

lor 2a=+2a(m-l)rf+rf(/+«—2)a-f^(/4-«-2)(m-l) 1 = 2 [a2+<7</ ( /+ n _2) + rf4 (/--U (H—1)

Dividing by </ ; 2o(m-n+( j ( /+n-2)+</( /+n—2) (m-1)

= 2a(/-fH-2)+2rf(/-l)(«—1) or a [2m—2—f-n+2] =*2rf [*/-1) <«-1)—(/+n-2) (m-1J

., a 2fa—2/— 2A 4-2 — w/— »w+2m + /4-H^2 ';' rf ~ 2 w - / - / i f~

! 2m%—m}—mnA-7m—l—n \ - 2m-n-l

' . ~ (2m — I-h)

or i -, =

2m — i — n a ffi + 1 rf= 1

i

Hence a : rf=fm + 0 : 1 18.( S ^ a + aw + fn"

' '=a+6«—A + cn2 -2r«+c v=^4-- i / r+ f« 2 ) -^4-2rw-r )

But kh term Tn-Sm-Sn-i

' =b+2cn-c , ,

Ana.,

Examples VI (a) (on Page 52) 79

Putting n=\, 2, 3 T1 = b + c = b + (2A-\jc 7a =* * J-.V=/•-*-( 2 .2-1) c T3 = b+5c = b + {2.3-l)c

T»=b + c (2n-\)c Ans

19. Tn = 4n (H ! +l ) - (6 / i 2 + l) = 4/j3+4/i-6flSi-l Using sigma notation we have •S'n=4I7i3-f-4Z,/>-6.r«2 —TI

s = 4 - ( n + l ) s + 4 . ; r (,,-fl) i _ _ n

^ « s ( n + l ) = + 2/i(rt + 1)- / i ( n - 1 ) (2n + l ) - f l = n ( n + I U / i ( / i+l) + 2 - ( 2 n + M ] - / i = " (n + l ) [ / i + « 2 - f - 2 - 2 n - l ) ] - « =*« fn+1) [ V - n - f I ] - / i ={n3+n> (« '—«+n-H=i i 4 - / i 3 +*i*+n s —n 2 + « -H

= «• Ans 20. Let the two quantities be a and b

Then a, / i j , ^2 , b are in A. P.

.'. Ax-a=b~At or / l i+ / f 2 =a+f t .. (ij

If a and A are in G. P. then a, Cu G-, rf are in G. P.

... <* * or G,Gs=fl*» ...(») a Gz

Also if ox; ft ; Ht ; ^ afe in H. P.

then - , 77, TT • t' w i n b e i n A- p-

I 1 I J_ J_ _^_] 2 ^+* •'• ifn^b'Ht or #i ^ " o + r a* From (1) and (2) we have

) i Algebra

° r '( HiH2 ~ G1G2 ° r H)H*'~Hi + Ht

Hence GiGa : HiHz = (Ai+A2) : (H1+H2) Ans. 21. Let A and/> be two numbers and Ai, A->, As...An be

1 N

arithmertc me.ins between them. 'If dbe the common diff, then a, b is («-r-2) th term

6'«a + t«+2—I) rf=a-f (H-J-1) d ' b -a „ , b—a an-i-b iwi - i - l " y « + l / i - r l

. . ^ 1 1 1 1 1 • , D

Again • - . •—, ~-r 77. r are in A. P. • a • Hi H) H„ b

Ifrf'be/c. diff. thin

I+OM-l^-l or * - ( £ _ I ) . ^ - ^ J ^ <7=lst harmonic mean

,, _ I ah (H+_1 ) " I > a~A ~~ °+* f l

b" 1 "^ ( / i+l)

If o does not lie between p and • -rrr, p, then'§ should

? less than p\ or greater than I - ^ T ) P*

_' _an + b ab (»4-1) " ^ * ? ~ f i 4 - T _ -a + bn Simplifying it we get

• ' t f ,^-* ' . \2 aw _ « f £ - f c . ' e , ^ y (n-M) <a+6n) m- rU ^ - r M .*. p is less than 9

I , J ; '»+ l \ 2

Hence q can not lie between p and I ^ 7 7 1 V- Ans. 22. Let a\ a + d, a+2d be in A. P.

Examples VI (b) (on Page 56) •

Sum of n terms Sw=*-£ [2« + ( « - l ) d]

•.. .(1

2 ha+n{n— 1) rf

2

Also sum of the cubes of n t e rms=Z[a+(n~ l ) d]3

=-£{a3+3ff2 ( n - 1 ) rf+3« <n- l ) 2 rf3+(n-l)3 rf3} =Sa3 + Za-dS [n-\)+2ad*2 m - I ^ + r f 8 ^ - ! ) 3

I 4

By " actual division you will find that (2) is exactl; divisible by (1) ^ Proved

E X A M P L E S VI (b) (on Page 56)

1. No. of shots in a square p i l e ^ " C w + 1 j ^ i + l l , 6

H e r e n = l

.-. No. of shots r e q u i r e d = 1 5 X * 6 x 3 1 ^ 1 2 4 0

2. No. of shots required as above l i ^ l ? ^ ° « H 4 0 o

• 3. No. of shots in a rectangular pile «(n + l) (3ro-«+ ' l j „ „

-2 . Here «=28 , m=50 o

XT f u • • JX 2 8 x 2 9 x ( 1 5 0 - 2 8 + l) .'. No. of shots required= --. —^ o

28x29x123 ,CCA, — , = 16646

o 4. We see that if we put on the given pile triangula

pile having 13 shots in each of its base side, we gel a completi triangular pile of 25 shots in the base.

= 2925

, , , : . . 13x1 / l H o H m i d : ,

6!

82 ; Algebra ! ' ]

;.*. No. of shots in the complete pile i 2 5 x 2 6 x 2 7 I 6 i , . . , , . , , . . : ., I 3 x l 4 x 15 . . -Also No. of shots in the added pile= —r.—-—=455

/ . No. of shots in the complete pile=2925-455=2470. ' i . Ans.

5ii Here again if we put a square pile having 13 shots in each'iside of the base we get complete square of 40 shots

i' VT r- i. • L » . -. 4 0 x 4 1 x 4 2 ,\ No. of shots in the complete pile= T — I 6

• - • =22140 ji j 3 X 14x27

Also No. of shots in the added pile=- 7 = 819 'r

/. ' Ko. of shots in the complete pile , =22140-819 = 21321 Ans.

6.', Let then be m shots in the length of rectangular pile. .'. i No. of shots in the complete pile

• ' _ n ( * i + l ) ( 3 m - « + l )

! 6 But K = 3 4

. I" XT r 1. * 34x35 (3m—34+1) „ . , n , , . . .'. , No. of shots=- —^ —^-=24395 (given)

' or 134x35 f 3 m ~ : 3 3 ) =6x24395 1 " D

or |?rt— 11 = ——=r=—„- =41 '•• -• .24395 X 6-34x35x3~

.'. >m=41 + l l = 52. Ans. .1

7. The lowest layer contains 1089 shots.

.'. No. of shots, in one side of the square— y/(lO%9)^'Z3

Also] shots in one side'of the top layer=^(469) = 1 3

'/ Now if we place on the given pile a square pile having — 2.shots in each side of the pile we obtain a complete square pilehaving 33 shots in each side of the base

F' .'. No. of shots in the complete square pile .33X34X67 %

6 12 x 13 x 25 Also No. of shots in the added pile= —'—, =650

.*. No. of shots in the complete pile ./ =12529-650=11879 Ans.

8. Being a complete rectangular pile of 15 courses it 'will have 15 shots in the shorter side of the lowest layer.

7 .'. No, of shots in the complete rectangular pile

/ „ ( „ + ! ) O m - H ^ ; H e r C „ = 1 5 , H i = 2 0 ' 6

15x16 (60-15+1)_ 15x 16x46 = 1840 Ans.

11 and 18 are the shots in the sides of the upper layer and there are 30 shots in the shorter side of the base. So there will be 37 shots in the longer side of the base. Thus we have got an incomplete pile of 20 courses. If we place on the given pile a rectangular pile having 17 and 10 shots in its sides of the base, we get a complete pile. Now number of shots in the complete rectangular pile

30X31 (3x37-30 + 1) , „ = • — - — - - g ! — (Here «=30, m=31)

= 155x82x5=12710 -,

. Also No. of shots in the added pile = ^ ^ ~ 2 ^ - = 770

Hence No. of shots in the incomplete pile ^ =12710—770=11940 Ans.

.84' Algebra

10, For having a complete pile we shall require a rectangular pile having 14 and 5 shots in its longer and shorter*^ sides respectively of the base.

'' " . T> - A * r i, * - 5..6 (14 - 3—5+1) \ . . Required No. of shots= = =——r-—

\ • J 5 X $ X 3 8 _ 1 9 Q A n s >

V 11. Suppose there be n shots in the side of base of the rectangular pile.

\ V Layers in both the cases are the same, there will be the same number of shots in each side of the base in the square pile

\ • " ' ("+!) f"f2)_« (n+l)'(2»+l) , \" . 6 12" • . +

or In («4-l) (n+2)-n (n+1) (2n+l)=1800

Solving this cubic equation we get one value of n=24. Hence No. of shots in the lowest-layer of the pile

j = i («2+n)=l (243-f-24)=300 Ans. 12. Let there be n shots in each side of the lowest layer

of the'given incomplete square pile. It contains only 16 courses., Hence its upper layer will contain [n—(16—1)] = (n—15) shots in all

.*. ^=( r t -15 ) a -1005 ",, or 'n2=n3-30/i+225-1005 or n=4l .*. n -15=41-15=26 * Nowiif we place on the given pile ; a square pile having

25 shots in each side of the base, we obtain a complete square pile. . 1

No. of shots in complete square pile =4i*2ii?21??=23821

• i '. .' 6 * No. of'shots in the added yilc^**€£Q*=552S *

6 " i

Examples VI (b) (on Page 56) 85

.*. No. of shots in the incomplete square pile .=23821-5525=18296 Ans.

13. Let there "be/* shots in eactnside of the square pile. , Hence the No. of shots in the pile

_n (n + l ) ( 2 n + l ) ~ 6 .«.(1)

rJ As given the rectangular pile has got double courses so /there will be 2n shots in each side of the lowest layer.

• T i I NT f U • - .U 'I 2 f l ( 2 r t + l } ( 2 » + 2 ) 1 . . Total No. of shots in the pile=- ~ •

. _ 4ff ( j i+l ) (2n+l) " 6 ...(2)

Thus we see that (I) is Jth of (2). Ans. 14. Let there be n shots in each side of the base of the

rectangular pile.

^ . . No. of shots in this pile= g

But since the square pile has ' got double courses, so there will be 2n shots in each side of the lowest layer of its base. Then

xr r u t • i 2n (2n + \) (An+l) No. of shots :in square pilc= : g—•—-

Hence, as given we have \ J i (n+1)(«+2J

6 13

or

' 6 . (ji + l ) (n+2) ___13

2(2n + I)(4«-t-l) 175

or 208na + 156n-f-26=175n*+525n+35O or 33n s--369n-324=0

or l l n 4 - 1 2 3 / ^ - 1 0 8 ^ 0 or ( i r n ^ - 9 ) ( n - 1 2 ; « 0 ie. n=12

86 .' Algebra

.\ No. c'f shots in the rectangular pile 4 12x13.x 14 . . .

:i = . = 3 6 4

And.No. of shots in the square pile= — ~ ~ ~4900"

Hence square 4900, Triangular 364 Ans.

t15. 10s. 6d.«£J£, lcwt.= 112ibs. '•* £U >s the value of 112 lbs. . • r «i -MU *i i , r112x40x51 .. £ 51 will be the- value of— ~-.

= 16x40x17 lbs. 'i .'., wt. of one shot—16 lbs".

\ Total No. of shots in the triangular pile 16x40x17 . . .

• — — ^ = 6 8 0 •

i

Let there be n shots in each side of the base of the triangular pile.

.:. No. of shots in j t=-^~— '^—• o v

1T\ * t- • • rt (rt+1) («+2) • . Hence as given — > -—'=680

Solving'it for n by remainder theorem we get n=l5. :. \ No. of shots in the lowest layer

' = l ( » 2 t ^ J ( 2 2 H 1 5 ) = I'20 Ans. 16.- \ As there are n courses in the square pile, so theie

will be n shots in each side of the lowest.course and the same will be the, number of shots in each side of the rectangular pile in the lowest course. . - • • _ . ,

' .\ No.'of shots in the square p i l e ^ ^ ^ - l ^ + A ^ 6

And No. of shots in the triangular pile' =n-kL±l}iH±Il' \ 6

E x a m p l e s VI [D) \OU r a g C J « ;

*.' Triangular pile is formed from the square pile

.*. No. of remaining shots in the square pile

«(« + !) (2/T+l) „ n («f l")f«+2) 6 " ~ " ~* 6

="("6ti){(2H+ij-(»+2)}='L("6xL). („_i)

= J ^ J . n > J « + 1>BS_N (N + U <N+2)_ where N s ( B 6 6,

But it also gives the No. of shots in a triangular having N=;«—l) shots in each of its lowest course. i

CHAPTER Vll

SCALES OF* NOTATIONS

EXAMPLES Vil (a) (on p a g e 5 9 ) /

i! We see foat the sum of the figures of the first columnl is 4 which is less than 5. Hence it remains as it is Sum of the figures of the 2nd column is 9 which is greater than 5 and equals to'(Ix5+4).; so we put 4 and carry on 1 similarlyiothers. Thus following this method we get

';' 300421

1 2 3 2 4 1 •. -1 _ J 9 ? 2

S^ 1' 333244- Ans. {2? ^s in question No. 1, following the same method

ve get the sum in scale of nine - ,', I 303478 i 150732 1 . 264305 1 ' 721626 ' '• •

3. We. are to subtract 1732765 from 3673124 in the :ale of eight .: . H .

*> '3673124 i J732765 ! - 1740137.

Since we can not subtract 5 from 4, We thus add 8 to 4 id subtract 5^om the total getting 7. Now in the second ilumn we can not subtract 6 from I hence we add 8 to it id subtract 6 from 9 getting 3. Following this method we t the above result, ' .

. Examples VII (a) (on Page 59) 89

4. We are to take 2e 46/ 2 in the scale of twelve from ^ne 756

so Ve 756 » :; 2e46/2 * 1 F7074" ' A n s -

5. Difference of 1131315 and 23 5143 in scale of six. is 1131315 235143 452132

4)452132(11022 4

4

r 8 9 8 8" 8 X

After dividing 1st figure when we divide 5 by 4 we get ] as remainder which when combined with 2 becomes 8 (in scale of six) and so on.

6. * We have to'multiply 6431 by 35 in the scale of seven

so 6431 35^

45115 ?5623_ AnSi

334345

f- 7. Mfultiply 'nonary' numbers .means the multiplication in the scale of nine.

^ ' Algebra

4685-3483

15276 42154 i. <

21072 . ,r 15276

• 17832126 '!

; 8. Here the deviser is 36'in- tfie scale of seven equals to 27=9x3 V (3x7+6=27) ' •

\ So dividing 102432 by 36 we should divide its first term by n and the quotient by 3, thv.*

\ 5511

\ and then 3)5511-i " 1 ^ 1 <., V 10^=1x7+0=7 \ to express 102=lx7 3 +0x7+2=51

1 And 51=5x9 + 6 so putting 5 and carrying 6 and so on.

' *' 9. Ternary" scale means in the scale of three

v ' - 11022201 , i 121012 •, ; 10201112

1201 in the scale of 3 = l x 3 3 + 2 x 3 a + 0 x 3 + l * 4 6

Hence 10201112 is to be divided by 46

.'.i 46)10201112 i "2012^ , (V 10201112=:0ni2)

Its' explanation may also be given'as given above

10.1 Quinary scale means tlie seal* of five.

Examples VII (a) (on Page 59)

300114(342 14

114)1101 1021

1232)3014 3014 Ans. 342

X '•

11. For square / / / / (in the scale of eleve tttt'

9ttt\ 9ttt 1

9 tt t 1 9t t tl tit 90001 Ans.

12, G. C M . of 2541 and 3102 in the scale of 7 2541 ) 3102 (1

2541 231)2541(11

23J 231 231 •

The reqd. G. G. M; is 231 _ ' Ans. 13. In septenary scale, we have the division as

6541 ) 14332216 (1456 6531 44612 36124 54551 45665 55536" 55536

, Algebra' •! \

14. Subtraction . 1 I 103050301 t 2040402p 1 . 6244426f

Square root of 62444261 will be-1 ' '• 7071

V' 62444261

61

15.

nwn 114442 1607 | l4261

16161 l ^ " 1

x '

The reqd. square root is 7071, Ans.

eee

\ Je'et 0* 0 1 - > ;

lte U901

l e " 19*01 x

The reqd. square root is eee Ans.

16. (i) .• 3102)31141(10 3102 • 121)3102(22

242 242 242

V * Hence the reqd. G. C. M. is 121

Examples VII (b) (on Page 65) 9

(ii) In the scale of six we have #

23=3X5 [V .23=2x6+3=15] 3C>=3x3x2 24=4x4 32=5x4 4 0 = 2 x 3 x 4 41=5X5 4 3 = 3 x 3 x 3 5 0 = 3 x 5 x 2

;. Reqd. L. C. M. is^3 3x5 2x4 a=m00ft An:

EXAMPLES VH (b) (on P-ge 65)

1. 4954, in the scale of 7 will be expressed as follows,

7 V4954 \-7)_707 5

_101 0 7)ii_ 3

2 0 Thus4954=2x7 4+0x7 3+3x7*+0x7+5 Hence the reqd. numbers

=2xl0H-0+103+3.Xl02 + 0xlO+5 = 20000+0 + 300-^0+5^=20305 Ans.

2. 624 in the scale of five 5)624 5)_l?i 4

5)2_4_ 4 . 4 4

Thus 624=4x5 3 +4x5 a +4x5+4 Hence the required number is

= 4 x l 0 3 + 4 x l 0 2 + 4 x l 0 + 4 . =4444 Ans. i

3. For expressing 206 in the libnary scale we have

Algebra

2)206 2)103. 0 2)51_. . . l

„ I :...... 1 _ 0 ^...... 0

...... 1

2 )25_ 2)12

2)± 2 ) 3

1 Thus 206=lx2 7 +lX2 G +0x2 6 +0 'x2 4 +l + l x 2 3

• \ . + lx2 a +lX"2+0 And'the required No. « I X 10?+I08+0x 105+0X 10*

I -plXlO'-HxlO'+JxlQ-MT • \ • '• =11001110' , Ans.

4. 1458 in scale of three " '

1 3) 1458 3)48_£ 0

\ 3) 162 ......0 1 3)_5 4_ ° ; . 3) 18. 0' \ 3 ) 6 0 j 2 ] \ 0'

Thus i458=2x3 c+Ox35+OX34+Ox33+0x32+0x3+0 and the No.' =2x l0?+0+0 =^2000000 Ans . .

5. :. 9)5381 i 9 ) 597 % 8 • ', • 9 ) 66 ' 3

7 . ^ ' thus 5381 =7x9 3+3X 92+3X 9+8 ,

and the required No.==7x l0 3+3xlO a+3x 10+8 ! -7000+300+30+8=7338 Ans.

6. la order to transform 212231 from scale four to scale Jfivft we have

i


5 ) 212231 5 ) 13233 2 5) ^203 ! 0 5) 103_ 4

li • " 3 4

/ . Reqd. No. is 314402 ' 'Explanation—In the-first line of work we have got 21 which ^ .;' equals to 2 x 4 + 1 = 9 = 1 x 5 + 4 / - Hence on dividing it by 5 we puf down one and carry out / 4 and so on. ' 7. 398e in powers of 10

' /)398e t) 4G/_ 7 0 55 8

"6 5 Hence the reqd. No. =6587 Ans,

8. e)j*}2L e)75b_1.....8

e)81 7 _ 8 9

Hence the reqd. No.=8978 Ans. 9. 9 ) 213014

9) 13001 1 9) 100 i

9 ) 2 £ 0 "2 6

Here 1000= 1X 63+0 X 62+0 x 6+0=215 .'. Read. No.=260U • Ans.

• 10. « ) 23861 8) 2663_ 4

8)307 1 -S)34 2

• # 3 7 •

Algebra \

Henee.the reqd. No is 37214 Ans. 11. 5 ) 400803

3) 71872 2 5)13885 4

' .4 5)2534 .......3 5)_460 * 5 J 83 3 5 ) 1 6 0

3 0,

Hence the reqd. No. ^30034342 Ans. 12. f ) 20665152

T)J514I4 3 T ) 50500 e

TT2655 / \ T)J51 0 I T j i o ; i \ "0 7 Hence the reqd. No.»710te3 Ans. 13. ' ) {«ee<?__ \ - ' r)iiH_24 7 \ t) 13862 ......8 \ ' t)U>tl 6 \ t) UTj 4 \ t)23 1 \ 2 ......7

Hence the reqd. No.=2714687 Ans.

14>l, • I 6 X 7 io 2 + i o •'

10X 10 U + 10

\ i o x / io~4 + io

Examples VII (h) <PQ Page 65) 97

After it the digits rocur, .

,\ Reqd. No.=>'2046 AnS-15. Integral part Fractional part

T )£7 '15625 1 S T

1-87500 T_

f.50000 T

b-xsmfo Hence the reqd. No. =15*1(6 Atis. 16. Treating the integral and fractional part of 200,

211 separately we have 9 J 200 '211

T ...0 7-100 9

"3-000 Hence =20'73 Ans. 17. Treating the iategral and fractional part separately

we have .8 ) 71 8 )j_. 5

1 2 •03

8

2-80 8

5-40 • 8

2*80

3& Algebra

\ I .*. • Reqd. n u m b e r = 1 2 5 '0125 ( y Digits after 25 recure) 1 18. Since 1552=lx73+5x72+5;<7+2«625 fand 262$=2x7H6x78+2?<7+6=-1000 • 625 5 1 Hence tae reqd. denary vulgar f racHpn= * I Q O O ' = T v

(19. (i) V -4=-444444

= • ' ( since it is in G. P . and-sum of. I— r \

a \ G. P. to infinity is j — J

. \ =4/6=2/3. • Ans.

(iij -42=-42424242 oc

7 """7a ""73 T" 71 '"76 "*"7* "J"

(f4+£+ «)

20. Supposing r as the radix of scale .\ 222 in the scale r=2r a +2r+2 . Now as given 2ra+2r+2=182 or r2±r*=90i.e. r = 9 Ans.

. 2 1 . Let r be the radix of the scale, then 0302, in the ' scale of r can be written as

. 0 , 3 , 0 , 2 3 , 2 *•• r» ra r* r a r*

But as given we have sJ_,_2_=25 r* " V 128"

or 25r*-384/-fl-256=0 or (25ra+16)0-3-16)=0 >. r=4 Ans. 22. Let r be the radix of the scale then

, 554 in the scale of r=5r a +5r+4 and 24 in the scale of r=2r+4 0 .*. As given, we have

5rB+5r+4=(2r+4)*=4rHlfr+16 or r a - l l r - 1 2 = 0 V. r=12 Ans.

22. Let r be the radix of the scale then 1746335=lxr0+7xr5+4xr1+6xr3-i-3xr2+3Xf-f5

According to question we have c

re-t-7r5+4r*+6r3H-3ra+3r+5=511197. Solving it by remainder theorem (Trial) we get

r=8 Ans.

24. If r be the radix of the scale then 479=4.r*+7.r+9 698;=6.r3+9.r+8 907=9.r8+0.r-J-7=9/'84-7

As given we have (4r»+7r+9)+(9ra+7)=2 ( 6 r 2 + 9 r + 8 )

or r2-Ur=0 or r ( r - I l ) = 2 / . r=U Ans.

25. Let r be the radix'of scale then .

•16 on the scale ofr =—+^-"

*20 „ >» n = —

. . . _ 2 8 *28 „ „ s, — - - r r 3

• •

According to the given condition we have

i.e. 2—??—f|feO or r*-10;—24=0

or r«12 Ans. . 26. In the scale of six, 212542=2X6G-J-lx6M-2x63+

0 5 x 6 a + 4 x 6 + 2 «17486

„AIso in the scale of six, 212542 is denoted by 17486 U*. Reqd. radix is 10 APS. 27. Let the radix be r, then ' •

' | 1 4 8 * 8 4 = r » + 4 r + 8 + T V ~ V + 2 + T y Hence 148*84 is the square of (r+2+2|r)

Hence the given number is the square of 12*2 Ans. •

28; Let the radix of the scale be r, then in scale of sfo we have \

1234321=r«+2r5+3r*+4ra+3ra-f2r+l i «(r»-i-ra+r-fl)»

Hence the given number is the square of 1111, which is expressed by the same four digits namely 1, 1,1,1.

29. Let r be the radix of the scale then 1*331 in the scale of r •

Hence trie given number is the cube of 1*1 An*. 30. 1 ton=2240 lbs. For expressing it in the binary

scale (scale of 2) we have


2)2240^ 2)U20 0 2)560_ 0 2)280 0 2)140 0 2)70 0 2)35 0 2)17 1 2)8_ t

2)4 0 S

s2)2_ 0 .1 0

Thus 2240=2"+2'+2° . Am. • x \

31. As we have proceeded in the above question, We have

\ 3)10000 3)3333 L-l 3)1111 0 3)370 1 3)123 1 3)4£ 0 3)41 1 3 £ I 3)2

f 1 -Thus weights of 3», 38, 3", 1 must be placed in one scale

and that of 38, V, 38, 3* in the other scale at the remainders 9|e of opposite sign.

32. te t the radix be r, then in the scale of r we have 1367631 «r»+3r*+c>*+7'*+«'1+V+l

«=(r»+r+ll*

102 Algebra

\ ^ Hence the given number is a perfect cube of \ * ir*+r+1* i.e. cube of 111 \ 33. Let the number be denoted by \ / . lO^fmlO^+alO^-f. . . . . .+i» . W+q 102+r 10+S V .".' 10s, I0*f105 are divisible by 8 therefore the number

is'divisible by 8 if the rest portion i.e. $10*-r-r . 10+S is divisible by 8. Ans.

• 34. v (/•«$—r) .the number rrrr in the scale of j will be equal to (10000—I) and square of this is ioOOOpOOO+1—20000 ; Hence wc Have the result, since

q==s—2 and r=s— 1 35. Let S denote the sum of the digits ; then by Art.

'i$—S N'—S 82, -^y- and — p are both integers., ' Hence

/ N - S N ' - S ' \ N - N ' ... . . . , A

f—-|7—.———) or -—y—will also be an integer. Ans. 36. \Let 2n be the' number of digits> then the number

may be denoted by Ar^+Ars" - fl+Cr!n-3+... + Cr2+Br+A It can'be written as

,,,- A ( r ^ + D + B ( r ^ ' + D + C (r2 n-3+l)+.. . v which is divfsible by (r+1).

37. As in Q. 35, so here wc have ~~K— and.—j—- as

integers. ' 1 o- -. , ^N-SSa N - S 2 .

Similarly :—«-— or —„ —is an integer

, Hence —W—is an integer. Proved..

38. Let the number be=ABC ABC «A10S+B10HC1Q3+A10af B,. J0+C = A (W+IO^+B (104+10)+C (10*+i)


=A (103+I) 102+B (10H 1) 10+C (103+I) «(10*+1) (A IO*+B . 10+C)

Hence the number may be'divided by iHP+l)** 1001 ie. by 7 x 11 X13. Hence proved. 39. Let N is the number, S the sum of its digits and r

j ^ g the radix of the scale, then _ —1 where 2 is an integer. But (r—1) is even. Hence N—S' is even and therefore N and S' are either both even or both odd. Hence proved,

40. Let f=10, and the number be

By repeating the n digits, themew number will be

=('"4-1) (ffi«"-1+<iA"-*+...+ff*-1*+tfll) It shows that the number must be divisible by the ori

ginal number and also by W"4-l). Again since* n is odd (("+1) will also be divisible by

(/+I) i.e. by(10+l )=I l . Thus 100001 = 11x909!

and 1000000=11 x 909091 and so on giving the quotient 9090 .9091 Proved.

CHAPTER VIH

SURDS AND IMAGINARY QUANTITIES

EXAMPLES VIII (a) (on Page 7a)

} . We can write the given expression as

1 \ td + V2J-y/3]

Multiplying above and below by [ l+V2)+\/3I we get J 1 + V 2 4 V 3 l + V ^ + y ^

\ [(1 + V2J+V31 [U+V2)—v/5] (l + ^ ^ - C V S ) 8

_ \ l + y / 2 + V 3 _ l + \ / 2 + ^ 3 , ( l + V 2 + V 3 ) . V 2 I + 2 + 2 V 2 - 3 2V2. • 2 V 2 X \ / 2

•_ ^ 2 + 2 + V 6 \ 4 Ans.

2 . % V2

V 2 + V 3 - V 5

Multiplying above and below by (V^-f V%i~V5)

^ \ \Z2(V2+V3 + VS), 2+V6+-v/10 (V2 + V 3 - V 5 ) ( V 2 + V 3 + V5) . 5+576^5"

__(2^V6 + y/10) y V6_6-f2V6+. l /60 2V6 V6 2X6 Ans.

3 - I — ! . ... • _ Multiplying it above and below by [Va-fV*—V(«+&)]

Va+Vb-Via—b) a-{-b+21/{ab)^(a+b)

Examples VIII (a) (on Page 72) 105

_ y/a+Vb-V(g+b) V(ab) 2Viab) V(ab)

_aVb+by/a-y/{a*b+b*a) . _ lab

^ 2\/(a+ll V(a-l)_A/(2a)+A/(a-l) 2V(«+D Wia-D+VW + Via-l)]

[Vt«~l)-V(2a)+V(^-l)]X[V(«-l)^V(2a) 4V(*-1)]

„ 2 y V - l ) + 2 (a+l) + 2V(2a'+2d) ^ a - l + a + l + 2 \ / ^ a - I ) - 2 a _V(^-l)+(*?+D+V(2fl'+2fl) . Vftfl'-D}

V i a ' - D . V ^ - l ) __(„*-l)+(a+l) V(oa-l)+Vf{(a ,-l)} . V(2ii'+2a)

<?"+! L(g-1)+V(^-1)+V[2a (o-DJ . .

• : ^ AnS.

V10+V5-V3 {(VIO-V5)-V3) °- y '3 fv lO-V5 {(V10-V5)—V3} _(%/10-V3)'-(-\/5)» 13-2V30-25

12-2x5 . V2 ~ 12-10v*2 V30-12^-y/30-6 6+5^2

^12-10-^2 6-5V2 6+5V2 (V30-6) (6+5V2? 3^/(30)^5^(15)^12-10^2

~ 36-50 ~ 7

* (V3+V5)(v '5+ V2)_V15+y/6+5+V10 ' KV2+V3J+V51 KV2+V3)+V51 {-v/15+V6+V10+5}v[(V2+V3)-V5] ~[(Vr2+V3) + V5] 1(V2+V3)~V5J

Simplifying above and below we get V30'+2>/3+3V2_(^30+-?V3+3V2)XV6

e=3~~ 2^/6 " 2y/6Xy/6

Ans,

106 Algebra \

• 1 x ^ 6 V 5 + 6 V 2 + 6 V 3 _ V 2 + ^ . 3 + V 5 A n S

i 12 1 \ 7. Put 31'8***, 21,2=j>] then * > , / are both rational \ and ^ -y^C* 3 - ; ? 3 ) (x8+j>3) -\ « ( - y ) (je'+^+ay) (* ?+^)

\ The rationalizing factor is 1 *R+*4j'+*y+*3j'3-M;'a-t-;'5

lor 3B,8+34 ,a . 21'*+31'3,y+32,s . 23,2-f3v'sj8+>5'2 Ans. 8. - ^ 5 + ^ 2 . Putx=5 l '8 ,^=21 '3

.*. »", * both are rational and \ x*-y*={X+y) {xi-xyi'y*) {x>-y*) \ ={x+y) (x*—xty+xtyi-x^+xyt-y*)

The required factor is, therefore

—5B/6_52i3 t 21 '3+51 '8 . 2"8—5 l ,B . 2 + 5 1 ' 8 . 24 ,3 ;-2B '3

Ans. 9. Put x=al,a, y~bxn .*. a=x\ b=?y*? *13, y " are both rational and x"-^»=(x«->«) (x*+y6)

1 =(JJ+J>) (*"—^-f-xV—x*y*+xTy4-— x?y6

+x*y6.i.—y1?) Rationalizing factor is =(j( a i-*1 0

1F+*V-. . . - j^1)

" Ans. 10. Put x=*3*I*, y= I then *•, y* are both rational V- *-y*~{x-y) {st+xy+y*) ..". Rcqd. rationalizing factor is (x2-\-xy+y2) or | 3 f ' 8+3 1 ' 8 . 1+1. ' Ans.

Examples VIII (a) (on Page 72) 10:

11. Put x=% y = 7 2 " then a4, y4 are both rational

=(x-y) (x+y) (x*+y*)=(x+y) (x*-x*y+xy*-y*)

Required rationalizing factor is (x3—x*y-\-xy2—y*) or 2 s - 2 3 . 7 l "+2 . 71"—73,i .An;

12. Put *=51'3;*y=3l'« then xia and ^ " both ar rational

• =(x-y) (x*+%y+y8) (jf+^+j'a*B-hy»)

+> Rationalizing factor is (*u+jeM.y+xV+...+j; lx) or 5" ' 3+5" '« . 3i»4.58/3, 31/2^ ..._j_3n 4 ^ns

3 1 / 3 ^ 1

13. In order to rationalise-^-^- , we first obtai

the rational factor of O ^ + l ) Put x=3ll\y=l then x8+J'8«(*+->0 {x*-*y+y*) The rationalizing factor is Xs—xy~\-y* *or 3a,B—3l/*+l

'• Now multiplying by (3a'8—31'3-!-!) above and'belo* we have

(31 , 3-3)f3^3-31 /3+I) _ 3 - 2 x 3 ' / ' + 2 x 3 1 ' » - l (3 l '»+l) O ^ - S ^ + l ) (3 I (V-f(l)a

1—3»/"4-31's

~ - 2 ^ A n s"

14. ^ j q ^ 7 B . As above here put x=91/6, y=8l>8

, TJien tf-yM*8-!-?8) (*»-?»)

08 \ - Algebra',1,

The rational factor is (*B—x*y-\-x*y*.«—y8) or* 95)*«9«tt , 8:'M-93'6,. 81"—...—S** « 1 9i,'fl_8i/8

Multiplying oijaToiTa by this ratiooalizing factor

bove and below and simplifying we get l7\-2x3il3x2}'z+2.. 34'3 . 2*-2 . 3 l . 2*i* __ \ + 2 , 3 i / 3 . 2*-2 . 3 1 ' 8 . 26'2

~ j " 9^8 ;=17—3W1 . 2*' l+3*n .yB—3*., 2fi£-f2 . 31 '3 . ya

v - 3 l « . 2 " a Ans. 15. Pujt 3V3=*» V2=v

- -. _\ =(*+J0 (^-xy+y1) (**->•) >,, - (x+y) (x 6 - r fy+xy-~*y+xy 4 ~y s ) -

The rational factor is (*8—**y+*3y3...'—y5) or • 36 / s-34 ; a . 2V»+3X . 2 1 -3 a ' 3 .•2a 'a+3 l V - > ' / a

Multiply above and below the given fraction by this tionalizing factor and simplifying we get '* 2x,a—36/a . 21+3*/* 21'1—31 2a+3" ,s . 28/1—2* . 31'3

- ; '-~?=r • *=3a. 2 » a , - 3 , ' l . 2+3 4 ' 1 . 2m—3 . 2 a +3 8 ' 8 .

\ 2* '*-3U 3 . 2» ' Ans.

16. Put **=>&*, y~9™ then > . *»-y*=(*8+y*) {**-y*)

=(*4y) (x*-xy+y*) (x*-y*) =(*+>) {xi-xiy+xY-xY+rf-y1)

Thus the rational factor is (x6—x*y+x8ys...~yB) or 3Ha-3a . 91,6+3*'a. 9w-*& , 9 , 'M-31 '".

\ 3«« ' Multiplying ~rrr~£/a above and below by this rational

tor and simplifying wc get •',

Examples VIII (a) (on Page 72) 10*

31S/8 m 3« | t -3 , / i . 3 1 / 1 . 3ll*+Plf . 3V,-3*/» . 31

+3V.3*/a-3»/*.31/1

9 (3«/'-3y+3v»-3'fr+3»*-n

« i (3*/B-3E'8+3y-3Va-f3V8-U Ans, 17. For rationalizing the denominator PutK=81/3»J'=4V9

Then * • - / « ( « • + ; * ) (xa->>a) =(*->-) (Jt»+*y+^) (x*+y*) =(x~y) (xt+xiy+x'yt+xiy'+xyt+y5)

,\ Rational factor is (x5+x*y+ ... -f.y5) or g^+S4 /* • 4V"+89 /2 . 4*/*+8»/1. 41

+81/*.4 ii»+4B/* Multiplying by it the numerator and denominator

of "T^^T wegct

(V<8+^4) (8*/«+8«/». 4V»-f... +4g/a) (87^-41/- ) (86/»+84/8 r ^ / ^ + Z + ^ T 7 )

_2flj-2&6/'+2BO/*+2i5/,l+2w/ft+2a5/g+2»/» (8)9^(4)»

16(26+2 , ,Vfl+2^a+281/fl+21<l/8+2u/6+0 = - 496"^ ~

_2 a+2 eV<+2 t t /<+2»/*+2M / a+2"/ '+l "" 31 ~ Ans.

18 -mr.-jmL^ f . . 9.7.-3./., 3-91/* 3 - 3 1 /

Put jf=3,j '«3V i then xM-y*=(x-y) (x*-bxy+y*) The rational factor is x*+xy+y*

or 3 « + 3 . 3 l / ' + 3 V Multiplying (1) above and below by it we get

<3-31,a)(3s + 3 ,»3v'a-f34'3) ~ 3810 n a 4 - V 3 - L V ' a \

Ts ^ ^ - j - ) (V27«B=*3»»)

__ 3 (3wg4-3«w-|-3ijB) _ '•33ia+3Bio+3»6 ~ 24 , 8 " Ans.

\ 19. Let V(16-2V/(20]-2v'(28)+2v'(35)

\ * =*y/x-«Jy-y/z i. Squaring both sides we get

-2V(«)+2V^) Equating rationafand Irrational parts of both sides \ x+j+*-16 5 •- , ... (i) \ xy=20 ... (ii) \ « = 2 8 ....(m)

'.yz«35 'I., (iv From (ii),(iii),(iv) we get xay iz s=20x28x35

\ or x>-zs=±140 •Then with the help of (if), (iii), (iv) we have

' z = * i 7 , ^ ± 5 , x « ± 4 Butx+y+z=16 .'. Taking -}-ye sign we get * = 4 , 7 = 5 , z=7 which

satisfy this relation (i) Hence the required square root is ±(^4— V5~\ /7)

or ±(2~V5—v/7) Ans. ' 20. Let v /(24)+4v /(I5)-4v'(21 - V(3 5)

=Vx-Vy~Vz or 2 4 + 4 V ( 1 5 ) - 4 V ( 2 i ) - 2 V ( 3 S ) - ( V ^ - V J ' - V ^

»*+y+i-2V{(xy)}-2-v/{(»)}+2V ,{(J'*)}

Equating the rational and irrational parts of both Sides x+y+z'=2A ... . (i)

s>-=21x4=84 ... ' (ii) 7Z=15X4=60 ... (iii) zx=35 ... (iv)


From (ii), (iii) & (iv) we get *yz B =84x60x35 or xyz»420

Thus from (ii), (iii), (iv) we jget ^=7 ,^=12 , 2=5 wtich satisfy the relation (i)

Hence the required square root is ±Vr[(5)-V(7)+2v'(3)l Ans.

21. Let V{6+Va2)-V(24)-V8}=' \ /A--Vj '+ 'v /z Squaring both sides we get 6+V12-V24-V8~*+j '+z-2 'v / <xj ' )

Equating rational and irrational parts we get *+.y+z=6 ... (i)

;cz=3 ... (ii) xy=6 ... (iii) z_y=z2 ... (iv)

From (i), (ii), (iii) we get *3>8z a^3x6x2 .'. xyz=6

Hence *=3 , y= 2, z = I .*. Reqd. square root=±v /(V3— \ /2+ l ) Ans.

22. Let V ' { 5 - V ( 1 0 ) - V ( l 5 ) + v ' « H V * - V J ' + V z or 5 - 1 / ( 1 0 ) — V / ( 1 5 ) + V 6 - ( V / ^ - V J ' + V ^

=x+^+z—2\/jy—2\/J''Z+2V*2

Equating rational and irrational parts we get

or

f z = 5 5

3

15

xYz* = 5 2

15 ' 4 '

3 2 "

15x15 " "4x4

...

•

(i)

(ii)

(iii)

(iv)

112 I Algebra '. ' '

or xyz=^

.'. Rcqd.< square root is ± ( l + y ^ - ^ y ) A n J

23 Let V[«+3*+4+4v/a-4v/(3&)-2^(3fl6)]

!' =^/x-Vy-\-Vz or a+3ft+4+4-v/ff-4^(36)-2v'(3a6)

j ~x+y+z-2y/(xy)~2y/{yz)f2^{xz) Equating rational and irrational parts of both sides we

get , " x+^+z-flH-36+4 ...# (i)

xz=*4a ... (ii) yz=3ab ' : ... (iii) r^ *y=12fr , ... (iv) '

.'. x2y2z*=4a . 3ab , 4 . Sft^lSarf)1

or xyz—Ylab / . x=4, /=3A, z=a which satisfy (i)

Hence the reqd. square root is

j ±EV4-V(36)+Va] Ans, 24. liet v / (2I+3 v

/ 2-6- v /3-6v '7-V24-V56+2 A /21 ! ) = Vfl—V&— V^+y/d

Squaring both sides we have

21+3^2-6^3-6^ /7 -V24-V564-2y ' (21 ) J, =a+b+c+d-2\/(<*b)-2^(ac)+2^(ad)

+2y/{bc)-2^{bd)-2^icd)

Equating rational and irrational parts of both the tides wehaTe !

I; a+&+c+(/=2l ! • •


ad=\S; ab=27 ; ac=63 ; bd=6 ; cd=\4; bc=2l ;

Solving these we get fc=3, c*=7, a=9, d=2 ,\ Reqd. square root is V9—\/3—A/7+ V2 or i ( 3 - V 3 - V 7 + V 2 ) Ans.

25. Let(10+6V3)V3=*+V> - 0 ) Then (10-6V3)V3=*-VJ! - (2) Multiplying (1) and (2)

(100-108)1/3=x2-.v

or x'i—y=~2 y=x*+2 ...(3)

Now cubing (i) on both sides we have 10+6^3=x*+yyy+3xWy+3xy

Equating rational parts on both sides *3+3x(x s+2)-10=0 (V y=xz-\-2}

or 4s 9 +6*-10=0 By remainder theorem we have Jc=l / . J K = 1 * + 2 = 3

.'. Reqd. cube root is =l-f-V3 Ans. 26. Let(38-M7V5)1-'a=^+V'>' »-(l)

Then(38-17V5)1 '9=^;--Vj' ».(2) Multiply (1) and (2) togethei

[(38)2-17»x5]1 'a=xa- iy or (1444— I445)1/3=x*—y or — \=x*—y / . y=x*+l Now cubing (1) we get

3%+\7^/5=x*+3xy+3xWy+y'\/y Equating rational parts

xa+3xj>=38

114 Algebra

or [ jc»+3jc(jeN-l)-38=0 ,' ( V y=xa+l) 'or ' 4x3—3x-38 = 0 ,

.* By remamder theorem *=»2, y—2a+l=5 .". Reqd. cube root is ==2+V5 Ans. 27., Let (99—7tV2)VB=x~\/y .„(i)

..'. ' (99+7(V2)1'9=:<H-Vj' •»("> or - (99-70v '2)V»(99+70v /2)1/3^3-y or (9801-9800)1/3=*a-.)' or i 1=*'—y .". j-=xa—1 Now on cubing (1) we get

99--70V2=xm+3*;'- 3xWy—yVy Equating rational part we get

x8+3xy=99 or * 3 - f3x(* a - l )~99=0 By remamder theorem x = 3

. / . >»=3*-l=8 .". Reqd. cube root is 3—^8=3—2^2 Ang. 28. 38VI4~100V2=-2V2 (50-19VVI

Now (38V14-100v2)1/a«K-2V2J(50-19V7)]1/a

; "(-V2X50-19S/7)1 /8

Now let (SO-WViy-l^x-VI - 0 ) and ($b+19V7j I/'=x+v;>' •••(») Multiplying (i) and (ii) we get

(2500-2527)V3=«3-y or — 3=xz—y

y=x*+3 Cubing <i)

50-19\/7=xa-3x2v'y+3xy-y\/j>

Equating rational parts we have ; x*+3xy*=50

or ;c3+3x(*M-3)-56=0


or 4*3+9*-50=0 i.e. # = 2 (by remainder theorem) .'. y=7 Hence the reqd. cube root is •

-V2(2~v '7 ) = v / l 4 - 2 ^ 2 Ans. 29. . 54^3+41 V5=3V3(18+41/3xv 'S/3)

(54V3+41V5)V8 = [3v/3(18+41/3 . V5/3]*/8

= V3(18+41/3V5/3)^3

Now let (18+41/3v/5/3)1'3=*-tV.V (i)

and (18-41/3V5/3)1 '3=:x-v> (ii) Multiplying (i) and (ii) we have

[(18)*-(41/3 X T/SI3}*}lis=x2-y fi-iA 1681x5\l'» ,

or 7/3=r>-y or ._ y=x2~7l3

Cubing (i) 1 8 + 4 1 / 3 V / 5 / 3 = X 3 + 3 ^ + 3 J C V ^ + J ' \ / J '

Equating rational part we get ' ^ + 3 ^ = 1 8 or x*+3x(x

2-7\$)=n Solving it for x we get x=2 / . y=4—7/3==5/3 Hence the reqd. cube root is

V3 (2-t-V5/3)=2V3 + V5 A M . 30. 135V3-87V6=3-/3 (45-29^2) Taking cube root of both the sides we get

(135V3-87V6)1 '3=V'3 (45~-29V2)1'3

Now let (45-29V2) 1 ' 3=*-V3' (i) and ( 4 5 + 2 9 - V

/ 2 ) 1 ' I = J : + V ' J ' (ii)

Multiplying (i) and (ii) we get $Q25-l6Z2)1l*=x*-y

or 7 = x a ~ y

]. Algebra

/. y=x*-7 Cubing (i) we have •

\ 45~-29V2=xs-3x2Vy^xy-y^/y

Equating rational parts j x3 4-3x^=45 or x 3 +3x(x a -7)-45=0

or . 4x*—21*—45=0 i.e. ' ;c=a3 (by remainder theorem) / . ' > .=9-7=2 Hence (135V3-87V6)1,3=V3(3—V

/2)=3V2—s/6 Ans. . 33. Xet [a+x+i/pax+xpyy-t^^m+s/n Squaring both sides we have

; a+x+T/(2ax+x*)=m-hn+2i/(mn) Equating rational and irrational parts we have

i m+n=a-\-x i 2^/(mn)=A/{2ax+xz)

.*. ' (m~M)s=(OT+n)s^4mM=(a+x)2-(20X-f**) =aa+2<w+*2—2ax-xz=az

m—n=a and m-\-n=a+x Solving these we get m=*fl+*/2 ; n=x}2 .". Required square root

32. 2a-<</(3u*-2ab-bz)

=4 [(2x2fl-2V(3fl2-2a6-£2)]

' - 4 C(3a+A)+(fl-*)-2V{(3«+*) (*-*»]

it

Examples VllI (a) (on Page 72) 117

Required square root is •

l+«2+V(l+a8-H)

Required square root is

=vrrb)-H2+2V(i-»)}

°2V(l-a'> «'+«)+(l-«)+2V(l-«'»

=27ff±35<V(t+<0+v'(i-«»'

Required square root is

A/1£H(^) 1

Jt^^pi ADS*

We are given that a=> 2 - V 3 h 1

118 ' Algebra

Putting these values ia 7a*+nab-7b2 we get

• '* 7 (2=V3) + U * g=V3)" (2+V3)

" > - 7 M - Y ' = __7_ ,11 1_ , 7—4vr3"1' 1 7+4V3

V7-4\/3 7+4V3/^

= 7 V 49=48 r l i

=7 (8V3)+11=11 + 56V3 Ans.

36- ^ V 5 + V 2 V34V2

* \ ' 3 -V2 Multiplying T y ^ ^ ; above and below by (\/3—v'2)

1 V J T V * we get'

* 3^2 J - 5 - 2 V 6

and multiplying f ^ ^ — ^ ) above and below by(V3+V"2)

we. get

' = ~ 3 = 2 1 ^5+2V6 Now 3*2-5^+3j>a=3(*- r)'+*y

=3 [5-2V6-5-2v/6] s+(5-2v '6) (5+2^6)

=3 (~4V6)"+(25-24)=48 x 6+1 =289 Ans, '37, Let y(26-15V3)=V*-\/y 'or 26-15V3«(V*-VJ#«+.y-2V<J90 ..-. x+y=26 and 4^=675


Solving these for x and y we get x=27/2, y=25\2 :. V(26-15V3)=V(27/2)-V(25/2)=3v '3/2-5/V2

Again 38-f5V3=i [76+2 (V?5)] « H " + i + V ( 7 S ) x i ]

/ . Required value of V(38+5V3)=(5V3/2+l/V2)

V(26-15-y/3) = 3V3/2-S/y'2 " 5v '2-V(38+5-v/3) 5V2-(5V3/2+l/V2)

= (3V3-5) _ (3V3-5 ) (9+5V3) 5V2X V 2 - 5 V 3 - 1 (9-5V3) (9+5V3)

_ 2 7 V 3 - 2 5 V 3 - 4 5 + 4 5 _ 2 V 3 _ V 3 81=75 6 T AnS*

38. Multiplying _ ^ above and below by

(33+19V3) we get = 6+2V3 33 + 19V3

33-19V3 33+19V3 _198 + 180V3 + 14„312+180y /3

1U89-1083 6 = 52+30V3

Now (52+30V3)=52+2(225x3)*'2

=254-2y'(25x27)+27 .=(V25+V27)2

/ . (524-30V3)"s=V25+-v/27-5+3v /3 Ans. , 39. We can write 28 -10^3 as 28-2^/(25x3)

or 2 8 - l ( V 3 = 2 5 - 2 (V25x V3)+3=(-v/25—v/3)2

,\ V ( 2 8 - 1 0 v / 3 ) = V 2 5 - V 3 = 5 - \ / 3 Similarly we can write

7-HV3' as 7+2Vl4x3)

120 Algebra "

or> 7 4 -4^3=4+2 . V(4x3)-f-3 [ • i ^ ( V 4 - t V 3 ) 2 = ( 2 + V 3 r ^

.\ V(7+4V3)=2+V3

Hence (28-10v/3)1 '2-(7+4v/3)-1/3

i = 5 - V 3 - ( 2 + \ / 3 ) - 1

1 = 5 - V 3 -

= 5 - V 3 -

2+V3

(2-V3) 4 - 1

! = 5 - V 3 - 2 + V 3 1 = 3 Ans. i

4(). Let (26+15V3)1 '3=*+V^ --(0 j then (26-15V3) 1 ' 3=x-V^ ».(")

Multiplying these together we get

j [(26)2-(15V3)2]1''3=(*+V.>0(*-Vy) or (sie-eisy^x^-y or{ I =x%—y

y=x2-^l Cubing (i) 26+l5T/3=x*+3xy+3xWy+y*/y Equating rational parts of both sides we get

| *3+3*)>=26 or x*±3x (x*-l)~26=0 or 4x3-3ar-26=0 Solving it by remainder theorem we have x=2 .-: J > = 4 - 1 = 3

Hence &6-l5VW*=2+</3

Examples VIII (a) (on Page 72) m

" l ' + * V 3 J . (2+V3) 2 (2-V3j 3

= ( 7 + 4 ^ 3 ) - ^ ^ 3 ) = 8 V 3 Ans. •

41. V 3 + V 5 = i <°>2V5)=J (5+2V5-H)

By symmetry we will have

0~V5)U^^-X .-(2)

Now-- 1 ( V 2 - VIO+3y/2 V18~V{3+V5) V8+VC3-V5)

_ 10V2 VI0+3V2 / I 8 V5 + ! ~ / s . V 5 - 1

_ 20 _2_i/5+6 5~V5 3+V5

« __ 20(5+V5) _(2-/5+6) (3-VS) (5-V5J (5 + V5) (3 f-V5) ( 3 - ^ 5 )

_2Q (5-4-V5>_6^5-6V54-l8-10 25-5 ' " 9 - 5

+2 23607 V (V5=2-236C

=5-23607 Ans.

42. By actual division of *s-H-|-3*^2 by #--1—->;

we have. _

[22 Algebra

x - l + ^ J x s+3*^2+l (^ 2 +x-*^2+V2+^4+l

. , ~ + x a -x 8 #2+3x^ /2 x2 +x$2 —x

• - + ~-*^/2.4-2^/2 H-x -x 2 ^ /2+x^ /2 -x$>4 ' + , ~ +

x^2+» +x^/4 x^2+ x-^4-^2

j ^ - + *^4-=#4+^2+* +1 X^/4-^4 +2

- -f --_ x—14-^2 -xTV±V2

X i

Hence the result. , 43. !Let{9flfc2+(i2+24fta) V i ^ - S f l V ' ^ ^ + V ^ ».(1)

Then {9ab2-(b*+24a2) V(&*—3a»)}"8=.x-V7 .-(2) Multiplying (1) and (2) together we have on simpli-

cation \ {\2a'i~bz)=x'i-y .-. ' ^=x2+(*a-12d2) Cubing (1) on both sides we get

i 9(7fi2+(A2+24a2)V(63-3a2)=x3+3xV7

Equating rational parts on both sides x*+3xy=9aP

or x3+3x (*H/>2-12</2)-9ffda=0. By remainder theorem we get x=3a ,\ ' y=b2-3a2

123 Examples VIII (b) (on Page 81)

Hence the reqd. cube root is =3a+V(fca-3a2) Ans.

44. Since 2 x = V a + - V = ' a —

or JCS=V ' ' Aa

Subtracting 1 from both sides we get *»_i <<*+iy i =_(q+l)3-4a_(a~l)»

1 Aa Aa Aa

Also W ( * - 1 ) - $ T ^ _ a + l - f l + l _ _ _2_«J i

2-^3 " 2ya \/a

Hence " a-\

x - V F T f T 2 Ans*

EXAMPLES VHI (b) (on Page 8x)

1. 2 V ( - 3 ) + 3 V ( - 2 ) = 2 V 3 ' + 3V2' and 4 V ( - 3 ) - 5 v / l - 2 ) « 4 V 3 ( - 5 x / 2 i

[v * W ( - l )

2V3i+3V2t 4V3t-5y ,2i 24ia +12V6'J

24ia +2V6ia -30i a

= - ^ - 2 - ^ 6 + 3 0 = 6 - 2 ^ / 6 Ans. (v i3*=-;

ADS.

124 \ Algebra

*Z We have 3 V ( - 7 ) - 5 V ( -2)=3V7i -5V2i

and 3-v / (-7)+5V(-2)=3v'7/+5V2i

• 63/a-15VHP+15V14i2-7 50i2

i = _ 6 3 + 5 0 = - 1 3 Ans.

3.'; Wehave ^ ^ 1 ) + ^ V ( ~ 1 ) = e, + e " '

and e v v '—e v '=e—e

= 2V(~D„ - 2 V ( - I ) \ e

l + V ( - 3 ) _ ^ , V3/ __2x-HV3/+l 2 " * + ^ ~ 2

: 1 -V(-3)J„ V3i_2x+V3i-l * T*~ ~ 2 2

4 4 =*a—x-fl Ans

s w ••. - 1 _ 1 3 + V 2 t WCDaVe 3-V(-2) 3= 7"2f (3-V2iK3+V2^|

' ^3+V2^3+V2/„3+V( -2 ) ', " "9+2 U H I " " A n S '

6 3y/(-2)+2V(-5)y 3V(-2)+2V(-5) D* 3V(-2j-2V(-5)A3>/(-2)+2V(-5j

_ [3Vf-2) + 2V(-5)P [3V/(^2J1^[2V'(-5)J8

„[3V(^2)]'+[2V(-5)]g-12y/I0 -18 + 20

Examples VIII (b) (on Page 81) ' 125

^ - 1 8 - 2 0 + 1 2 V 1 0 . = _ 1 9 ^ 6 v / l 0 A n s >

2 m

7. We have by putting •/£—])=/' 3 + 2 V ( ^ l ) 3 - 2 V ( - l ) 3 + 2/ , 3 - 2 / 2-5y(^- l ) i "2- l -5V(- l ) = * 2~5i 2+5/

tt_6+19/+IQ/'-f 6~-19/-M0/2 _ 12+20ia

4-25/ 3 4+25 ^ 1 2 - 2 0 ^ . _8

29 29

8. We have - *+*rt~}} - ^ i d |

a—*/ a+xi' _a2+2qx/-x2-(Q ; i-2fly/-jc2)_

42+*a

_ 4gxf _ t o V ( - l )

9 [-*+V(-l)] s . f j r -Vf-DI ' JC-V( - l ) x + V t - l i

_(x+i)»_(x-q'p_ (*+/)3-(*-t)5

* — / jc"+i JCB—*z

_ 6x8/+2/3 / V f 3 « ~ i \ .

6*af-2/ _2/pJc a-l) _ 2(3^-l)-y/(^l) =s~^+i xwr~ ^n •" 10. Simplifying we get

(a+0*-(a- i )« g f l 3+3o 2 /+3f l i a+i 3 - ( f l 3 -3 f l 2 /+3f l i a - / 3 )

(7=*+ 2<rf-r f8—(**- 2«'+/*) ._ 6aai+2i3 ^ ( a a ' + i g ^ S g H j ; = 3 q g - j . ,~ 4<7/ ~ 4<tf 2tf 2ff

Ans. ' ( V * ? « - ! )

! Algebra1'

J 1 . We have [—v ,(- iO]4 ' l f : ,«(-04f l4a=(-0* f ,X(— i)3

«={(-/)4}°X/=lx/=i Ans. - A !l2. We have [ v ^ - f 400+V(9-40/)]8

^ ( 9 + 4 0 0 + ( 9 - 4 0 0 + 2 V ( 9 + 4 0 0 X V ( 9 - 4 0 I )

^18+2V[l9+400(9-40i)]«18+2V(8l+1600) I =18+2V(1681)=18+82=100 Ans. 13. Let V(-5+l2 i )=x+ ' iv ..\ —5+12i=(Jc+iy),«*a—<y

J+2x>'f •Equating real and imaginary parts we get i x2-y*=-5, xy=6 Now(x2+v2) s=(*2-/)2+4Ky=254-144=169 •/. *8-fv3=±13 Now from xa-fva=±13 and JC3~V8*=—5 ,We have x=±2,y=±3 Since product of x, y is +ve=6 •,". Taking -fve values x=2, y = 3 . ,.'. V ( - 5 + I 2 i ) = ± ( 2 + 3 0 Ans. |14. Let V ( ~ n - 6 0 t ) = x - i > '.*. — l l - 6 0 j = ( x - ^ ) 2 = s x a - v s - 2 x ^ Hence xa-v*=—11

2*y=60 Now (x2+va)a«(x3-v2)2+4*^a=12l+3600=3721 '.'. * H v a « ± 6 1 Now from x2—,y2=—11 and Aft4-ja=±61 we have, on solving*

* = ± 5 , v = ± 6 Thus Vt~l l—60/)=±(5-6i) Ans. ,15. Let t -47+8 v ' 30 1 ' 2 =*+O' Then squaring both sides

-47+8V3i=x3--y24-2;o»f

Equating leal and imaginary parts of both sides x a - v 8 = - 4 7 a n d 2X7«8-V'3 . . .

Examples Mil (b) (on Page 81)

But (jca+}>a)a=(;c1-j>2)2+4xy=(-4'7)s4-64x 3=2401 .\ x*+y*=y/{240\)=±49 Solving x2+y*=>±49 and *"—j>t

a=—47 we get * = ± l , J ' - ± 4 v 3

.'. VC-47+8V30=±( l+4v '30 Ans. .'. x*=a2 or x=±aty=±l Thus A/(f l 2 - l+2a0=±(«+0 16. Let V(-8i ')=VlO-80=x—'> Then 0— 8i=(x—i»*=x8~;>"—2JCJ>/ Equating real and imaginary parts of both sides we get

x2—j;s.-=o and 2xy=~8 ' or xy—4 i.e. x=y and xy=4 Solving these for x and y we get x = ± 2 , ^ = ± 2 Hence ^{-%i)=±(2-li)=±2 (1- /) .Ans. 17. Let (a2—l+2aOl'2=a>H> / . u a - \+2ai=(x+iy)2=:Xt-y2+2xyi or x3—y2=aa—1 and 2xy=2a But (x2+^2)2=^2-y; a+4^2>'2=(fl a-l)2+4a a

*=(a2~l)3

/ . x2-f3/2=±(aa~l) Now from xs+>'a=±(<ia—1), x2— y*=(a2—l) we gef

.'. { a a - l + 2 a V ( - l ) } 1 / s , « ± " ± ' - ± { a + V(-D> Ans.

18. Let T/[4ab-2 [a*-b*) i]=x-iy

Then squaring both sides we get [4*6-2 (fl»-6*) !>***-.>'*-2xy/

or jca—>*=4tf& and 2xy=>2 (aa—£a)

But'(xa-f>a)"=(^3-^2)a+4jc2>2

*=16fl2fc2+4(aa-6a)

Algebra

\ =4,(a*+bz)z ) :. U B +J» 2 «±2 (a2+63) '.: ' (2)

*Hence from (1) and (2) we get

.*. V[4ab-2 (az~b2) i H ± [ ( a + 6 ) - ( a ^ 6 ) /] Ans.

19. Multiplying above and below by (2-1-30 we

= 3+5/ 2+3/ _ 6 + 19/-15_—9 + 19/ 2 - 3 / 2 + 3 / = 4+9 13

= ( - 1 + | | ' \ l t is in the form (A+/B)

20. We have as above

V 3 - / V 2 V 3 - / V 2 2-y/3+/-v/2 2V3- /V2 2V3- iV2 2V3+/V2

^ - V6/+2_^ 8- v ' 6 /_ /_4 _v/6^ \ 12+2 14 V 7 14 /

It is in the form A+/B

21. We have

1 + / ^ ( 1 + Q ( ! + / ) _ 1+2/+/2 _ l + 2 / - l , . , . 3_ 1 - ; - ( l - j ) ( l + / ) 1 - / - 1-fl ^ - ' - ^

0+2i 2i . ft A , -

^-2 y^=o.A+l , It is injthe form mA+/B Ans. 22. We have

(1+Q3_(1+/)8 ( 3 + / ) _ 6 / ~ 2 ^ 3 j - j ^ 3 - i ( 3 - 0 ( 3 + / ) 9 + 1 -5

1 — 1 + 1 i 5 5 Ans.

23 ( « + # ) ' _ ( g - f t ) ' - ( g + a y - ( g - / f t 3

* a—ib a+ib {ar~ib)(a+ib)

n»+3a*W+3o (6i)1+(6fi»-{a1—3A2«+3a (W)» - =W>

_6g^f+2Wsl6a»fc/-'263i_2& Qa'-fe') f fft+6* fl»+^ a*+&* .Ans-

24. (l+>y*)« = (-H')*=,li'4*=,IV V l+w+w*=0 .'. ( 1 -HV-)^ -H> • Also M>3=1 .\ K,4=JV3 . H"=l . w==w Ans. 25. (jl.-w+w») (i+iV~)v8)=:4 L.H.&=(—w—w) ( - H > 2 - H > 2 ) = ( - 2 H 0 (-2M>8)

= 4 ^ = 4 . 1=4 Ans. 26. (L-w) (I-u>2) (1—w4).(l—w5) =a-iv) a-ws) (i-w3. iv) a-H*. iv2) = ( l - w ) ( l - i v f ) ( I - w ) ( I - v r a ) ( v w*«l) =( l~w) a ( l-w a)2=(l-2iv+w s) (l-2w2-}-V)

. « ( - 2 w - w ) ( 1 ~ 2 W H H > ) = ( - 3 I V ) (-wa-2>va) • =9 iv 3 =9.1=9 Ans.

27. (24-5H>-f2H>s)5=[2+2H>+2wJ4-3u>]s

=[2(l-f-H>-f>v2)+3»>] =(0+3w)8=36xii'6=38x(w3)*=36xO)8=729 and- (2+2H>+5w2)°«(2-f2>e+2H>3+3H'*)*

-={2(l+H'+H'2)+3>i'2}« " - ( O + S M ^ S " » " = 3 8 . (1)'=729

Hence (2+5w+2n'2j8=(2+2w'-t-5w2)8 An 28. We have L.H.S.=(1-H-+H>2 . ) ( I -wHw 1 ) ( l -^ '+ iv 8 )

...2n factors =(l+w2~w) (1-f w-w2) (1+w*—w)...2n factors =(-2w) (—2w2) (-2w)...2/i factors = ( - 2 r - . ( w 3 r = 2 w . 1=(2)** An*

29. We have R.H.S,«(*+.y+*) {x+yw+zw2) {x+yvP+zW)

130 ' Algebra i ' . ' ._

i = (x+y+z) {x*+.xy .(w+w2)+;vzJ(w+ w2)'

j (V M»+>va=-l,W3=l)

=X*+yz+zP—3xyz Proved.

30. (i) Given that x=a+b, y=aw+bw\ z*=aw2+bw

.*.. xyz=(a+K) (aw+bw*) (aw*-\-b\V)

'..,•, ; =(<*+*>) {aa-fl&+2>a}=a3-t-fc3 A n s -(ii) 22=(aw2+6H')2=^S!w4-f-2fl6w3+5aw2

—fl2w+6fwa+2a6 . - ( 0

Also ,\ x=a-f-&=.*a~a2+2aA+ia - ( 2 1 , 2/"a((7vi>+fcw'2)a=(38iv2+i2iv4+2a6w3 ' ^ }>=&tf+b%w+2jb ...(3)

Adding these three we have ;r-x2+J>a+z3=aa (l + w+u>a)+£2 (l+w+w")+6fl& .

' =atxO-tb2xQ+6ab=6ab Ans. (iii) x*=*(a+b)*=<P-pB*-l-3ab (a+b) \ ...(I)

^=(aw+6w*)3=aV+&sw6+3<i8w8 . fiji^+Sflft2^5 " ' 'i =a3-f63+.3a2iiv+3ff62H'a . ...(2)

. ' . zi=(aw2^bw)s=a3\^-\-b3w:i+3aibwt>{-3ab2wi

' =a3+b*+3a2bwa+3ab?w ...(3)

Adding the three we get , A-3+rJ+23=3Ga+3^+3a2fc'(i+w+w2)+3fl62 ( l + wH-w)

' . . =3fl3+363+0H-p=3 (<23-f 6s) • Ans.

31. X*=cx-^by+~bz\ Z^bx^ay^cz, Y=cx'+by+ai .*. X a +YHZ 2 -YZ-ZX-XY~can be written as ' =(X+Yn>+ZW2) ( X + Y W H f f w ) . .

Examples VIII (b) (on Page 81) 131

={(flx-J-cy-f 6z)-f-w (cx+by+az)-{-w2 (bx+ay~{-cz)} X{(ax-\-cy-\-cz)+wz {cx+by+az)+w (bx+ay+cz)}

= (ax-\-bz-\-cyy+{cx+by+azy+[bx+ay+cz)* H-(M'+H>2) (ax+cy+bz)+(cx+by+az) -f (w+w2) {cx+by-\-az) (bx+ay+cz) + (n'+iv2) (bx+ay+cb) (ax+cy+bz)

= (ax-\-by+cz)2-r(cx-i-by+tiz)2+(bx+ay-{-czy — (ax+cy+bz) [cx-\-by+az)—(cx+by+az) {bx-\-ay+cz)~(bxJt-ay+cz) (ax+cy+bz)

= (a2+b*+c2-bc-ac-ab)(x2+y2+z*-yz-zx--xy) Atis.

! . '. CHAPTER IX

THE THEORY OF QUADRATIC EQJUATIONS

' EXAMPLES IX (a) .(on Page 88) F ' . '

I.;, We know that if a,' p be the", two roots then the quadratic equation is. x*—(cs+p) *+ap=0

t r ~ 4 ft 3

Here * = - £ - » P = y Hence the reqd. quadratic equation is

j *-(4^H(!X-4)=° , or| 35x»-H3x—12=0 Ans.

Hence the reqd. equation is

, »a—x I =— 1,—1=0 • I \ mn J or, mnx*-\-x (n*—ma)—mn=Q 3.'Here **£=* , p ~ * ± *

; JH-? />-?

Sum of roots a + f , t 2 - £ ± ?

} p+q p-q T

I pS-qz p2^qi

Product . p ^ ^ x f - ^ t i V - l

.*;)• Reqd. equation is^ a - .vf ^ - ^ ^ - 1 = 0

or, *3 0> 2 -g ! )+^*+(g»- ,p ' )=0 Ans.

Examples IX (a) (on Page 88) 133

4. Given roots are.tt=5=7+2^5, p=7—2^5 .*. Sum of roots a + p = 7 + V 5 + 7 - 2 y ' 5 = 1 4 Product a£=(7+2\ /5) (7-2v '5H49-20=29 .*. Reqd. equation is xa— I4x-f 29=0 Ans. 5. Here one root a=-f 2-^3—5, another root p=—2V3—5 Sum of roots a-f(3=:+2V3—5—2\/3-5=—10 Product of roofs a(3=(2V3—5)(—2V3—5)

= »(12-25}=J3 Hence the reqd. equation is

J C " - H 0 * - H 3 = 0 Ans.

6. Leta=-/)+2Vi2?) , p = ^/»—2V(2?) .". Sum of roots B + 3 = - ^ + 2 A / ( 2 9 ) - P - 2 V ( 2 ? ) = — 2 p

and Product«[-/>+2V<2tf)] [-p-2V(29)]=^ a~8g .". Required equation is *M-2px+(/>*—fi?)=0 Ans. 7. Let one root a be =—3+5f, and p be. *=>—3—5(

. / . a+p=—3+5f—3—5i«—6 a p « ( - 3 + 5 i ) ( - 3 - 5 0 - 9 + 2 5 = 3 4

.'. Required equation is x*-f 6*+34=0 • Ans. 8. Let a.= — a-\-ib, and p=—a—ft ' / . a+(J=—2a, ap«u«+6* . .*. Equation reqd. is x*~x (—2a)+(as+68)=Q

or xa+2<w+aa-i-62=0 Ans. '9. a « + ( a - 6 ) » , p « - ( a - A ) i Sum of-roots a+p=0, Product a?=(a—b)2

.*. Reqd. equation is x*~x(0)+(a—A)a=0 or *a'-ftf2—2(7&+£»=0 Ans.

10. Here the root given are more than two. The equation whose roots are more than two is written as follows.

(x-a) («-£) (*-Y) ( X - 5 ) =0 where a, p, T, B ... are the roots.

Hence the equation whose roots are —3, £, J is (J f+3) ( * - i ) ( j t - l )«0 or (x+3) (3x-2) (2* - l )=0

I , ,

or 6*3+ n**-19*+6=0 . Ans.

11. jHere the given roots are a/2, 0; — l\a

.'. Reqd. equation is (x 2)^X~Q)(X^ y ^ 0

or (2x-a){x) ( < M + 2 ) = 0

or iax3+(4^a*) x*-2ax = 0 Ans.

12. 'Given roots are (2 +V3);(2-~V3) ;4 .". Requited equacion is'

. [*-(24V3'][*-(2—v/3)][*-4] = 0

or [* 2 -* (2 -v '3 ) -* (2+V3)+(4 -3 ) ] [*-4] = 0 or [x*-4x+\] [*-4]=0

or X 3 - 8 S 2 + 1 7 J C ' - 4 = 0 . Ans. ;

13. Roots of the equation x3—2a*+a2-6a—2=0 will be real if fc*-4oc>0 '

Here b=~^2a, a = l , c=a 2 -£ a ~c* A 4a 2 -4 (o a -6 a -c 2 J>0 or 4'(p%+cl)>0 which is +ye. Hence the roots of the given equation are real (ii) In (a-6+c) * H 4 (a-6) ,x+(a--6-c)=0 We hive 6=4(a-i»), a=(a-b+c)t c=(a-b-c) The roots will be real if b2~4ac>0 i.e. -fve or 16 (a-i-fi)*-4 (a-6+£) (a-r6~c)S*0 or I2(a-b)2+4ct>0_ = a positive quantity. Hence the roots are real. Ans. 14. Equation ax2+bx+c=0 has its both roots equal

if b*-4oc=0. Here in the equation we have

b=—2m, a~\, C=8J»—15 .". If roots are equal then ( - 2 ^ - 4 . 1 (8m-15)=!0 or 4m»-4 (8m-15)=0 or j»3H8m+15=0 or *n=3,5 Ans.

* • i

' I

Examples IX (a) (on Page 88) 135

15. Since the roots of x s - 2 x (l+3m)-J-7 (3+2m)=0 .are equal then as above we have m

[ -2 ( H 3 m ) ] a - 4 . 1 . £ l + 14m)=0 or 4+36ma+24m—56m—84^=0 or 36m2-32m—80=0 or. 9m1—8m—20=0

Solving for m we get m=2, —g— »n s

, , x2—bx m—] l o . = — — — ax—c m+I or (x*-bx)(m±l)=(ax-c) (m-1) or x2 (m+lj— x (bm'-\-b)=>x {am—a)—cm+r or *2 (m+l)-{bm+b+am—a) x+(cm—c)=0 If the roots are equal in magnitude but opposite in sign,

then their sum must be zero.

/. *"»+*+"»-« = 0 ( v sum of roots „+^bla)

or bm+b+am—a=0 or m (fc-f a)=(a—6)

o r m = f l T * , Ans. 17. (i) If roots of (a+c-/») x a +2«-f ; 6 + c - a ) = 0 are

rational then from the condition, that for rational roots of ax2+bx-\-c=Q, (b*—4ac) must be a perfect square we have

4c2—4 (a-f c—b) (b+c—a) be a perfect square or 4ca—4 {c+(a—b)} {c—[a—b)} be a perfect square or 4c2-4 [ci~-(a-b)2\ or 4c8-4ca-t-4(a-6)» or 4 {a~b)z It is perfect square of 2 (a—b) Hence the roots of the given equation are rational. Ans. (ii) Similarly if the roots of", abc2x*+ 3a2cx + b*cx - 6a*-ab + 2Z>2=0 are rational then (3a2c+b2c)3-4 (abc*) (—fa*—ab+2&) must be a perfect

square. • .

136 \ Algebra1 t

]

I • • ' , ' , ' ' . •

• oij 9a4ca+i1c'+6aa6ac2+24iiadc8+4fl36«ca-8fl63c* be -aS, perfect square. '• ' , ' '

or! ca(9fla+64+10flfl6,+24(i3/»—8a63)be'a perfect square or c2 (3aa-f4ij6—ia)a be a perfect square It is a perfect square of c (3a3-\-4ab—b*). Hence the roots of the given equation are rational.

18. If a, p be the roots of the equation ax2+bx+c=0 . I ' —b c >

then a + P = — - and ap= -1-,

! , r- -V-2 i ' ' \ a ) *' a b*—2ac

& > Ans. .,

19; 4'(a, p) be the roots of ax*+bx+c=:Q then

Now7 aJP7-*-a7p*=a«p* (P3+a3) • •' ' ' «^p* ' [ ia+P)3-3aP(a+P)]

! _ c j p—&3+3afrc " | _ c* (Zabc-b*)

, _(<x+P)s Ffct+P)3-4aP]_, &s l>a a J

• {. ' , ' : *

| _(fea-4fle) y ,: a*c* • Ans . "

Examples IX (a) (on Page 88) ' 137

21. Let a, p be two roots then a= l+2 i , p=l—2* , Sum of roots «+p=H-2i-fl—2i=2 Product of root ap={l+2i) ( 1 - 2 0 = 1 + 4 = 5 .*. Equation is x2—2x+5=0 .*. x2—2x+5 is a quadratic expression'which vanishes

for jra=l±2/ Again x3 4-A:2—JC+22

—JC3—2*2-}-5.s-f-3;ta~6*-H5-f-7 -=x ( * 2 - 2x+5 )+3 fj,a_2jc+5)+7 « * x 0 - r - 3 x 0 + 7 = 7

Hence value of x3+x«—.\-+22 is 7 when x = l ± 2 i Ans.

22. As in the previous question let us first form an quation whose roots are 3±/

Equatipn will be x 3 - * l3 + f+3- / ) + (3+/) (3- i )=0

or x a -6x + 10=0 ...(i)

Now we can write given expression i 3 — 3x2—8x+I5 as (x3-6xfi+10*)+(3;c2-18;c+30)-15 * or *(x9-r6x-H0)+3.(;ca^6.x+I0).-I5 or xx0-r-3x0—15 from (i)

= - 1 5

Hence the value of a3—3x2—8X+ 15 whcn*==3±/is - 1 5 Ans.

23. Equation, whose two roots are say (a±a\/3i) is x2—x {a+a^li-{-a—aiJ3i}+(a+a^2i) {a~a\/3i)^0 or ;ca-2flx-f4fl2=0 _ ( i

Now the given expression x*-ux*-h2a2x+4a3' can b< written as

(x2-2flA-2+4a2x)+(as2-2a2^+4a3) or x (x-2a*-H3a) + a (x2-2ax+4a2) or" xxO-hixO=Ofrom (i)

138 Algebra

Hence the value of the given expression when *x=*a±4i/3i is zero. .Ans. 24. Since a, p are the reots of x*-\-px+q*=0 then

Now in order to find but the equation whose roots art (a—P)a, (oc+p)a we' should, find their sum and product. Thus

sum=(a-3)H(«+P)3=[(«+P)1i-4a(31+(*+P)2

! . =2p2-4q . Product= ( a - (3)a (a+p)a

! «t*+mc*+P),-**P}«(-/o»a-i')I-4d Hence the reqd. equation' is

or, x*-2x(p*-2g)+p2(p*-4g)*=b Ans 251 (x-a) (x-b)=h* • or' * 2 ^ x (<j+fi)+(a&—A2)=0, Its roots will be real when (<*+£)2—4 . 1 (ab—h1) be a -f ve quantity f.e >0

ori [(fl+6)2-4fl6]-|-4A2>0 or {a^-bY+Ah2 > 0 This is a +ve quantity i.e. > 0 Hence the roots of the, given eguation are always

real. Ans.

26; Since *„ *2 are the roots of axs-\-bx-{-c=Q then , *i+sra=—&/<J and X\X^=cla

Now («*x+6)-*+(a*ii+6)-a

1 „ * + 1 . " (fl*a+6)2 T (fl*a+fi)2 '

W (nxri-6) (ax2+b)

"" [tf**i*2+«fc(*i+*a) +*>*]» _ flg{f*i+^a-2x1*2}+2g&u14.ag-4-2fca

" [tPxxXi+ab (*i+*2)+62]2 "•

examples ix {a) ton f age 88) 13*

_ «2 ( ^ ) + 2 o 6 ( - W + 2 6 ' , 3 . 2 g c

(ii) {axi+by-t+iaxi+by*. 1 . 1

«3JCB3+3aV*+3^a*2+ft3+o!l*i3+3a"foi"+

= ; 3gA2^ + &3

_fl3f.vl3+A:2

3)+3a^(A:1a4-yg

2)+3gA2(y1+^)+2fc3

Butx13+V=(^+xB)3-3.-e1a;2(^1 + ac2)

A3 . 36c labc-b*

.(0

and *i8+^22=fJli:i+^2)z—2JC1;V2

=51—?£ = fc'—2tfg_ a- a a2

Putting these values in (I) we get

3^c-63+3A3-6flftc-3Z>3+263

or (ac-b*-\-b2)3

!?-Jlb<L.=sJ> (b2-3ac) (acf a3c3 Ans.

27. Let QIW root of »ix8-t-&x+c=0 is a, the otbtt is not .*. sum of the roots i.e. a.-\-na.=—bja ov oe(l-f«)=— b\a ' (i) Product of roots i.e, a . nx=c{a or ntx.i=cja A..'..(ii) JFrom (i) and (ii) we get

Algebra;' ! . • . . ! • ' , . •

jj. fteqd. condition is bH=ac (1+w)3 Ans. 28. If a,p be the roots of ax84f-£x-f£=0 then i «+P«—i/a and «p=c/a , Roots of the required equation are

Their sum • 1 ' 1 1 ' '" o-2-l-G2

i. («.+p., +• $.+£. =* (OCHP3) + ^ f -

=/£L-?£- V l+c2 /o a \ (fe»-2ac) (aa+ca) ! \ a 8 ' rt A cs!a2 J • a2^

Product

,\ Reqd. equation is ' , ; *2-(*+P) x+ap=0 . f ( ^ t e ) ( n ' + a I L , (b*-2oc?

o r *~l &fi~. J,x + S5« or A 2 * 3 — (A^ffc) (AH^ B ) A : + ( * 3 - 2 ^ ; 2 = 0 Ans. 29. Let a, (3 be the roots-of the equation ' 2*M-2 (m+rt) *+(m2-f-n2)=0. Then •

Roots of the reqd. equation are (a+(*)2 ,* («—fi)2

Their.sum=(a+p)z-f(«-li)2=2:(aa+^)

I, =4mn

Product ^ ( a + P ) 1 (*-(3)2

= (*+P)M(<x+P)3-4«P}

= (m+^{(m+^-4(^)} . •=(w+n)2 (2mrt-m 2-n 2) = —(m+n)2(m2—2m/i+n?)=-(m+n) s(m—n)a

.". Reqd. equation is *2—x (sum of roots)+product of roots=0

i.e. xz-4m/ix-~(mi-rj2)t=0 Ans. 30. The given equation is px2+qx+r=Q Sum of its ioots=~gjp Product of its roots=r//> Here we see that sum of the roots is negative. It

shows that both roots are either negative. But since the product of the roots is-f-ve, both roots will be of the same sign that is —ve. Hence both the roots of the given equation are negative.

EXAMPLES IX (b) (on Page 92) J. 2ax {ax+nc)+{n2-2) c2=0 or 2a2x2+2tfnc;c+ca ( n a_2)=0

Tts roots are real, therr.from bs—4ac^0, we have {2anc)*-A (2a2) (n a -2) c 2 ^ 0

or 4aVc 3 -8a S ! c 2 ' (n 2 -2) > 0 or - 2 («*-2f>0

o r _„2_|_4^o or 4 - n 2 > 0 or (2—n) ( 2+n)>0 . It is ^-ve when both factors are

of the same sign. But zero when n = 2 or —2 Hence limits between which n must lie are —1 and 2

Ans.

2. Let - i — r ~ - n = ; ; then

142! Algebra

! x2y-5xy+9y—*-0 or x*y—x(Sy-{-\)+9y=to • I Since all values of a; are real, therefore from b%—4ac^0

we have • i • (5y+\)2~4yXay^Q lor . 24y ,+ lQy+l —36yl>Q or - H ^ - f l Q y + t ^ O ••or llja—10y+l<0 (.". while dividing by — ve sign, J the sign of inequality is

changed) or O l y - r l H J ' - l K O i.e. j = £ - , 1

Hence the limits are —^T and 1 Ans.

f" L e t*q^+T^ then

or ' x*(j—l>+x t r + l ) + O ' - l ) = 0

If * be real for all values then ! (v+0 2 -4 (>— l)[y~l)>0 i.e. can not b e - v e or y2+2y-\-l-4y*j.$y-A>0-or . — 3^+lQy—3>0' or 3j2—lb/+3<0 i.e. should be —ve or (3.y—l)(.>'—3)<0 or should be negative •It is —ve when y = | , 3 Hence the result. A T • x a+34x-71 ' . 4 j L e t -,K«.4.2x-7 " * t h e n

yxz+2xy—ly=xz+24x—71 or . x 2(^- l )+^(2^-34)4-(71~7^)=0 If.* be real for all values; then •

4(>-_t7)s-4 0 ' - l ) (71—7»>0 or! ;y3-34;>+2894-7.y2^78>'+7lS?0 ori 8j>*-112,H-360>0 or J > 2 - 1 4 J ; + 4 5 ^ 0 . or; <y-5)(y-9te0 i.e. y=5,9

Examples IX (b) (on Page 92) 14

It is -f ve only when y takes all the values less than and greater than 9. Hence the given expression .can nc have values between 5 and 9. . Ans.

5. The two roots are

^ and ^a

v ' f l +Vt f l _ *) '. v'a—\/(a~-b) Sum of roots ___ y/a , -y/a ___ 2a _2a

y/a+\/{a^b) V«--V(a-~t>) a~(a—b) b

Product =—r~-~r^= a /*

.Y Reqd, equation is xz~x (2a}b)-\-{a}b)=§ or bx*—2ax-+-'i=Q Ans.

6. If a, p are the roots of x2—px+q=0 ; then a+p=/>, ap=?

(i) Nowa^a^ -PHP 3 ^ 3 *" 1 -* )

-*£-,)+, (C-) _«2(K3-P2) ,p3(E3-«2)

p a

fea—P*) («3-p3) «P

_fa+P)(«-P)a(«8+Pa+ttP) «P

_(«+P) [(a f P)3-4aP] [(a + P)2- aP] ap

„/>0>a-4g)(ps-<?) A n s

9

(«-/>F(P-/>;4 *!•

W+tff^W-ip {(«+P)3-3ap (« + p)} ^ , +fr '{ l> ,-2g}-4Pa( lO+2p*

- ^ gr =

^ - V ? + 4 ? a ^ 2 g 2 - V + 1 2 r t 1 ; +6>4-12pBg-2p*

v Ans,

7.1 Let a, p be the roots of /*2fnx-ffl=0 ; then as given a/p=pganda+p=s—n//, o0=n//' .

Now V(pW+V(cilp) + V(nll)

J « - V(«//i+V(«/0=o Hence VO^)+\/te//0+.\/[«/<)=0 Ans.

„ ' T . (x+m)2—4mn

( x*-{-2mx+m2~4mn=2yx—2yn, or i ^a+,(2m-2^)*+(7w2-4mn-f2^«)=0

I " 'V x is real for all values

T "

Examples IX (b) (on Page 92) 145

4 (m—y)2—4 {ma—4mn+2)w) should be +ve or y*—2y(m+u)-\-4mn > 0 i\e. (should bs -f ve) or y 0>-2m)-2n <>-2m)>0 i.e. ( „ „" ) or (j»—2n) 0>-2ro)>0

It is +ve for all values of y except those-lying between 2n and 2m. Hence the result.

9. Let a, p be the roots of ax* + 2bx+c=*6

then a + P = - —

^ 4/?2—4trc

fl* ' ...(1) Also if a+8, p+S are the roots of

A*2+2B*+C=0 then'

(«+S) (P+S)=-^

But [(*+S)-(p+8)]3={(a+S)+(P+5J}2-4 (K+8) (P -f-

' or ( X-P)^^4g-=4 B 2-4 C A

A2 A As ,..(2) V Difference of roots of both the equations is equal.

.'. Expression (1) and (2) are equal . 4 (b*-ac) 4 (B2-AC)

a* A« 6 2 -ac B*-AC

or A2

Proved.

\ ,146 ' Algebra

! i n T ' px*+3x—4 . . * . 10. Let £—r~ T-»~=y. then

- p+3x—4x3 J* /wa+3x-4=/y+3;cy^4;cV

or xHp+4y)+3x{\~y)-(4+py)=0 -.Since xis real / . '9(I-jOa+4(/H-4j)(4+/yr) .

should be a +ve quantity. .or 90,a+l-2y)+4(p2

iv+4/>+16j)'+4/7;;i!)>0 / (should be + ve)

• or ^9+l<jp)+v(46+4,p8)+(9+16Jp)>0 Hence (46+4/>2)3-4 (9+16»2 must be negative or

zero and (9+16/0 must be +'ve. Now (46+4/?V—4 (9+16/0* is negative or zero as

• [46+4p«+2 (9+16/1)] X 4 6 + V - 2 (9+16/01 , is negative or zero. ^

i.e. ' (4p8+32p+64) (4/>»—32/J+28) is negative or zero. or ' 4 (p+4)2.x4(p—7) (p—l)'is"-negative or'zero. It is negative so long as p lies between vl and 7 and for

such values 9+16p is +ve. It is zero. t > When p— — 4, 7,1 but 9+16p is—ve when />=—4 .". Linking values of pare 1,7 and may take any

other value lying between these limits. Ans. ii T ' *+2.

" • L e t ' 2 ^ + 3 U + 6 = m

or 2m**+3mx+6m—x—2=0 or 2mx»+x (3m-l)+(6m-2)=0 If x be real for'all values, then

• , (3m-l ) a ~4 (2m) (6m-2) > 0 or 9M*—6m+l -4Rm2+16m>0 or —39m2+10m+l>0

^or. 39m8—10m—J <0 • I.e. should be zero or negative.

Examples IX (bj (on Page 92) 147

or (13m-f-I)(3m-l) <Q This expression is —ve so long as both factors are of the

opposite sign. It "is zero when m = § or —rV* Also it is negative for the values lying between —TV and $.

Thus the greatest value of m is \.

12. Let if-b-t ^ > t h e n

x2—bc—2xy+yb+cy=0

or ar—2xy-\-(by-\-cy—&c)=0 If x be real then

4 J , 2 _ 4 (by+cy~bc)^Q

• or >»21— 6.y—cy+ftc>0 I or ,y ( y - 6 ) — c ( . y - 6 » 0

r or (y-b) (y-c)^Q It is -j- ve so long as both factors are of the same sign.

It is zero when y~b or c and negative for all values lying between-6 and c. Hence y cannoi have any real values. between b andc. Ans.

13. If roots of ax34-26;c-f-c=0 are possible and different then 46a—4ac>0 or positive

or 62—ac>Q

Again the equation (a+c) (ax"+26v+c)=2 (ac-6 2 ) (xM-1)

or x2{a(a+c)—2{ac-b2)} + 2b(a+c)x -f-c(<7-K)-2(a<:-2i)2=0

For imaginary roots, discriminant should be negative i e. 4b°(a+c)z-4{a(a-}-c)-2{ac-bn-)}

{c{a+c)~2{ac—bi)} be —ve Simplifying it we get or 4[(ac±-b3) {(a—c)2 + 46*}] should be negative

148 • Algebra^

- ' V b2> ac and (a—e)2~\-4b2 is positive ;.". Whole expression (1) will be negative .*. roots will be imaginary Ana.

, (bx—a) (cx—d) ' K

cross multiplying and simplifying we get xa (ad—mbc)+x (mbd-t-mac—ac—bdH-(bc—mad)=0 Since x is real

(mbd+mac--ac—bd)'i—4 (ad—mbc) (be—mad)^® ie. (kd+ac)2 ( m - l ) 2 - 4 (ad-mbc) {be—mad)

must be +ve or m* (W-ac) a-m {2 (ac-bd)2-4 {ad-bcf)

Hence {2(ac~bd)2-4(ad-bc)2}2-4(bd--t7c)i must be negative or zero and (bd- ac)2 must be positive

We see that (bd—ac)2 is clearly positive (being a perfecl square)

Now ., {2 (ac-bd)2-4 (ad~bcf}2-4 (bd-ac)* is negative

or, zero a${4(ac~bdf—\6(ac—bd)2(ad—bc) v+16 {ad-bcY-4 (bd-ac)*

is negative or zero or 16 (ad—bc)s {(ad—bc) — (ac—bd)2} is —ve or zero

1 V ( 16 (ad-be)2 is -fve (ad—bc)2—(ac —bd)2 will be negative or zero

or (ad— bc+ac—bd) (ad—bc—ac+bd) willbe —ve or zero or {(a+c) (a—b)} {m+6) (d- c)} will be —ve or zero or (as—b2) (dz~-c2) will be — ve or zero or ' — {(a3—b2) (c*—dz)} is negative or zero Jt is possible only when (a2—b2) and (c2—dz) should

have the same signs. . Proved i

Examples IX (c) (on Page 96) 149

EXAMPLES IX (c) (on Page 96) 1. Equating the-'given expression to zero we get • y2+2xy+2x+my—3=0 •

or y+2, («+")+(*+!) '

2x + 5

=**+*(m-2) + (^+3)

The given expression can be split up into two linear factors if the R.H.S is also a perfect square. For it we must have

or 4 * ( ^ ) - * < « - V

or ma-fl2=m8—4m+4 or m=— 2 Ans. 2. Equating the given expression to zero we have

2x2+mxy+3ya—5y—2 =--Q Dividing by.2 •

or ^ ( ^ ) ^ ^ / - i ^

/ , m V a / M 2 ~ 2 4 \ . 5 , ,

if the given expression can be split up into two linear

.û/ta men K..n.a. must be a perfect square. For it we

o r ? ^=^ !~ i5 o r m»-24»=2$ -V . m « ± 7 Ans. « 4 • . 4 .

1 3. Equating the expression to zertf we get ' A(x2~^)-*>-CB-C)=0 or Ax*~xy (B-c)-Ay*=0 It .is a 2nd degree equation and * 2nd degree equation

represents two linear factors if «6c+2/feft—<$/"*—6g*-cA««Q Thus A (-A)XO'-f 2 (O)xOxO-AxOH-AxO-OM) Hence the given expression can be split up into two

linear factors Proved. 4. Let the common root be a,, then, as a is the root of

these equations it must satisfy them .*. <r2-f/>cc+#=0

and > a2-}-/j'a-f- '=s0 From\these two equations we have '• a8 .„_.«. . = ^ L

pq'-p'q q-q' p'-p "

• o r ' « . ^ or *£=i'± Ans. P-P P -P

5. Equating to êro each of tUe given equations we have

fc'+mxy+nyÔ ; Fx*+m'xy+ny*±Q or i ^ j + n = 0 ; r ^ + m ' ^ n ' = 0

Let the common root be a /.e. -—â then we have y

From these we get ;

Examples IX (c) (on Page 96) 151.

a 1 mn'—m'n nl—In' Im'—mV •

nV—ln* . . •mn'—m'n .'. a=j—;—=,— and az=.--—r;—

Im. —I'm. lm— I'm

H vnn'—m'n/nl'—ln' V Im'—l'm \lm'—l'm)

or («/ '-fe'J ,=(mB'-m'fl).(k'-/ 'm) Ans. 6. If 3x*+2pxy+2y*+2ax—4y+l can be split-up into

two linear factors then the condition is abc+2fgh—af1-bg2-chi=0

Here a = 3 , / = —2, 6*=2, g=tf, c = l , /i=p Putting these values we get

6-4ap-l2—2at-p*=0 or /72+4d/>+(2a"+6;=0 It is a quadratic mp

Hence p itself is a root of an equation. Proved, 7. Equating the given expressions to zero we have ax2+2hxy+by*=Q ...(i). a'x*+2h'xy+b'y*=0 ...(ii) Put « y=mx in (i) and x=—.ym in (ii) flx2-f2Arox2+£mV=0

or a+2/mi-f &ro*=0 ' ...(iii) and a'y2m*—2h'myt+b'y*=0 or a'm*—2h'm-\jb'=0 »>(iv) Thus fcm2-J-2ftm4-o=0 and o'ms—2ftm-J-fc'=0 Solving these we have

m' _ m ___ I 2 {hb'+h'b) aa—EF~~—2(bh'+a'h)

ai'-bb' t A6'+A'o ' - 2 (bh' +a'h) ' ~-(6A'+a'A) T aa'-bb' T hb'+h'a

J 5 ^ i Algebra • -V

', or (*a'-M*)»+4 (M'+o'A) (A6'+A'o)=0 Ans. ,8. Considering^ as constant we can write the given

equation as follows:— • < • — w * 2 - * (3j>+2)+(2j>3~3j»»35)==0 which is a quadratic in *.

Thus two values of x from ^ = ^ ( i t f e ) a r e , i 2 , a

" T - ( 3 y + 2 jbV J(3^+2)"-4 , l (2yg-3y-25tt < ., V 2 •"!

', _(3,v+2)±V(9,ra+12,H-4-8j'a+12r+140) — j :

1 &(3r+2):fcy0>«+24>+14_ (3;y+2l±0>+12)

2 " ' 2 ."•"• (3;>+2)+0>+12)_:4.y+l4 -. , x= . ^

*=2.y+7 i.*. x—2.y-7=Q „.(i)

• and ^Qy+^ty+w^y.-riti 2i 2

- Vv i **=y-5 i.e. x—^+5=0 .»•(«). (i> and (ii) are ths two factors. From these factors it

is clear that for every real values of * there is a real value of y and vice versa. Hence the result,

9." As in the previous example, here also we consider y as a constant and write the give'n equation as

?x2+s<2y-92)+(y»-20j-+244)=0 which is quadratic in x. If AT be real then b 3-4oc^0

or (2^-92)a-4'-X9O>a-~204+244)>0 or >2~92>'+2116-9>'2+l80;'-2196^0 or — 8j>2+88y~80>0 or y1—lly+80^0 (dividing by —8) or tf—10) (y—1)<0 . i.e. should be negative It is possible only if y takes values lying between 1 and

10,because these values make the expression (y~ 10) (y—l) negative.

Examples IX (c) (on Page 96) 153

Similarly now considering x as constant we can write the given expression as

y%-y (20-2x)-H9x2-92x4-224)=0 For all real values of y we'must have •

(2*-20)»-4 (9r»r92x-i-224)>0. or 4*2-8O;c-MO0-36xa4-368*-896>O or -8x a - r72*-144>0 or * a -9*+18<0 (dividing by - 8 ) -or (»—3) (x—6)<0 i.e. should be negative.

It is possible- only if both the factors have opposite signs. Again this expression can be negative only if x takes values lying between 3 and 6.

Hence X- lies between 3 and 6 whereas y lies between 1 and 10. Ans.

10. Considering y as constant we have x*{ay+a')+x {by+b') + (cy+c')=0

If x be a rational function of y we must have ftom bz—4ac as a perfect square

or (6v-f 6')*—4 [ay+a') (cy+c') must be 0 perfect square or bty*+2ybb'+b*—4 [acy*+y (ac'+a'c)+a'c'}

as perfect square or va (b*-4ac)+2y (bb'-2ac'~2a'c)+{b'2~4a'c')

as perfect square It can be a perfect square only if

v ' ( (6i_4ac !}xV{^'8-4a 'c ' )}=(W-2flc ' -2a 'c) or (bb'-2ac'-2a'c)2=[b*-4ac) {b'*-4a'c') Simplifying it further-we get or a^c't-\-a'tci-2aa' cc'=abb'c+a'b'bc-b'i.i'c'—acb'i

or {ac'-a'cy={ab'-a'b) (bc'-b'c) This is the required condition. Ans.

CHAPTER X

MISCELLANEOUS EQUATIONS

EXAMPLES X (a) (on P«g« roi) i . •

1.' ^ » - ' 2 x - 1 « 8 o r i - — - 8 = 0 o r 8 x a + 2 x + l » 0

Factorizing we get

(4x- l ) (2*.+l)=0 . .V x » i or - $ .. Ans.

V !9 + x - 4 «10ir a or 9 + ^ = ^ or 9**-iaic»+lM)

Factorizing we get ; ( 9 * * - 1 ) ( A ; « - 1 ) = 0 / . * ~ ± i , ± l Ans.

3. tyx+2jr*l*-S~>ti or V * + " - ^ 5=0

or 2 * - 5 y * + 2 = 0 Factorizing we get

*. ( 2 V » - ! ) ( V * - 2 ) = 0 .\ * ~ i or 4 Ans.

4. 6xs"=7;c»4*-2xr»'* or 6, (x 1 ' 1 ) , -7x u «- -£ r

Putting***'1^ then 6>a=7>»—2/y or <b^—7y,+2-t'0 o r 6>'—4>-*-3>a+2=0 or 2^(3> 3 -2) - l (3^"-2)=0 . or•".- (2j'1-l)(3j''^2)=0 i.*. >»*=tor^orxV«=ior-| , 1 4

* *"T o r T . , Ans. 5. xt,n-\-'6^5x1'H put xlint=y .-. j»*+6«5y or j'^-S.y+e^O oi* 0»-3) (y—2)~0 /.e. ^=3 or 2 /. *""=3 or 2 or x=3° or 2" Ans.

6. 3x1,sn~x^n~2=^0 Let y«x"M then we have 3y—/-2=0

or >2-3y+2=*0 or <>"-2)<y-l)=G \t. y .\ * l , » « 2 , l or x=(2)** or 1 ADS.

Let A / ( - ^ J**y then we have

5^+7/^=68/3 or 15^-68^4-21=0 Factorizing we get (3y—1) (5^—2I)=0

1 21 ' * r - T

1 441 f o r T —

25 °r r i ' 25

147 Ans.

* = 6"

Hence A/(T)"4 or T i.e. *=27 ;

Let AR^H**1

y + J - = ^ o r 6 > » - I 3 y + 6 = 0

Factorizing it we get (3^—2) (2j»—3)=0 2 3

••• * = T ' T

4 or 9x=4—4x or 13x=4 .". * = —

Again ^(n^)^ = T or

or 4 = 9 - 9 * ;. x - 1 3

Thus * - n . J 5 Ans.

1 - * 4 9

156(1 Algebra',

',9. 6 ^ = 5 ^ 1 / 2 - 1 3 t ' Let ^/x^=y then we have 6y=5jy—13

or 6y*+tty~5=0 Factorizing it we get 6^+15^—2^—5—0 or (3^- l ) (2^+5)=0 » ^=4,-5/2 i.e. V*=J, -512

x=l/9, 25/4 Ans. 10. l + 8 ^ ^ + 9 ^ / e = 0 'Let x3 '5=j' .'. xG'5*=v2

Then 1+8^+9^=0 or , (8y+i)0 '+J)=0 i-e. y=-l>B,-\

xW^-VIZ { 1 V'3. l' •

° r . ^ ( - 8 ) = ~ 3 2 -and x 3 / ^ - ! or x = ( - l ) 6 ' 3 = - l Ans. 11. 3B*+9=--10.3* Let "$x=p then we havep8— 10p+9j=0 or (P-9)(J7-1)<=0 Letp=99V 'or , 3a=9=(3)2 / . x=2 ' and 3*=1=(3)0 .*. x=0 Ans.

12. 5(5*+5-*)='26

•or ' 5 ( 5 ' + 1 ) = 2 6 .

Let 5x~y then we have '

or , 5yl~26y+5=0 Factorizing it we get (5^—1) 0>—5)=0 ie. j>=l/5,5 Now 5^1/5=(5)-1 .'. x=—\ and 5*=5=(5Ja .\ x=\ Ans.

Examples X (a) (on Page 101)

13. 22 s + 8+l=32 . 2* or 2 2 e . 2 8 + l = 3 2 . 2* or 256s2*-32 . 2»+l«=0 Let 2*«y then 256^-32^+1=0 or (I6j>-1)2=0 or I6y—1==0 i.e. y=lJ16

2*=1/16=(2)-4 .'. . v = - 4 Ans 14. 2I<r+3-57=*65(2*-l) or 2» . 23-57=65 . 2*-65 or 23 . 22*-65 . 2*+8=0 or 8X2S"-65X2*+8=Q Let 2*=y then 8y2--65.y+8 = 0 or Sy*-64y-y+%=0 or (&y-l) (y~8)=0 ie. y=h* Now 2*=y=$=(2)-* :. x = - 3 and 2*=y=8=(2)3 .*. *=3 Hence x = ± 3 Ans. 15- ^ + v k = 2

Let V(2*)=.V then we have

y+— - 2 = 0

or j'8—2>-+l=0 or (y—l)2*=0 or ^—1 = 0 ;.*?. y=l Now V(2S)=^=1 or 2*=1 = (2)° .'. x=G , Ans. . , _ 3 V(2x)=5g ltK V(2x) 5 10

3 y *9 Let V(2J0=J' ^en -.—£-=*— or . 2j>1+59j>—30=0

I

I

Algebra V

•Factorizing we get (2y—1) (y-i-30}«*0 ikl :]y=± or. —30 • .*. V(23t)=*,v*=J or . 2 x = i ,\ x=>l/8 anil y ^ ) ^ 1 ^ — 3 0 ' or 2x=*900 / . *=450 Ans. 17.' (x-7) (*-3) (x-f 5) (x+1)=1680 : or ;,[(^-7> (*+5)JXl(«+l> (*-3)]«1680 or ; tx a -2*-35) (x 3 -2^-3)«r680 Now let x*~2x=y then we have

(y-35) (y -3 )= 1680 or ya-38y-f 105= 1680. or y»-38y-1575«0 " * • , ' ' or (y-63) (>+25)=0 i.e.' y^=63, - 2 5 Now xs—2*=63 or ',x*—2x-63=0 or ,(*-9) (*+7)=0 ,\ x = 9 ^ ~ 7 ' Again' x2~2x=—25 or' x*^2x+25=0

or xJ^vi£w>^±Vi_24) . Ans 38. (x+9)(x-3) (x-7) (x+5)=385 or [(x+9)(ar-7)]x[(x-3)(*+5)]=385 or (x»7T-2x-63) (x*+22x-15)=385 Let x»-f2x=y ' ...(i) then (y-63) (y-15)«385 or y3^78y+945;=38S or .y»-78y+560=0 Factorizing we get (y— 8) (y—70j=0 l e , y=8 , 70 Nowfrom(i)**+2#=8 or x s+2x-8*=0 or ix+4) (x -2)=0 / . x « . - 4 , 2 Again from (i) x ,+2at=70 or *a-f-2x-70=O '. ' -2j:V(4-V280). . . . * « — ^ - ^ ^ i - f c y ^ j ) Thus * « 4 - 4 , 2 , - l i V ( - 7 i ) Ans. '

Examples X (a) (on Page 101) ' 159

19. x(2*-H) (*-2) (2*-3)«63 or [x <2x -3 ) ]xUx-$ (2x4-1)3-63 or (2xa-3x) (2**-3*-2)=63 Let 2a:1—3x~y ... (i) then y(y-2)«63 or ^—2^—63=0

Factorizing we get (y—9) (^+7)=0 r.e. ,y=9, —7 From(i) 2x*-3x«9 or 2x*-3;e~9=0 ie. (2*+3) (*-3)=0

*=>-3/2 or 3

Also 2x*-3x=-l or 2*»-3*+7=0 - v ^ 3±V(9-Sf l m 3 ± V ( - 4 7 )

20. (2x-7) (*a-9) <2*+5)=9I or (2x-7) (x-f 3) (*—3) (2*+5)=9l cr v< 1(2*—7) jjr+3)] X [ ( J : - 3 ) ( 2 X + 5 ) ] = ? I or (2*«-*-21)(2jca-A:-I5)=91 .

Let 2x*-x~y N(i) ' Then we have (y-*2\)(y-\5)=9l' or j»2-36>-f 224=0 Factorizing we get

Cc~28) (y-8)=0 i.et j<=28, 8 From (i) 2x*~-x»28 or 2 X 3 - A : ^ - 2 8 = 0 •>. (2x+7) ( x - 4 ) - 0 or *«4 , - 7 /2 Also 2*a—x*=8 or 2*3-x—8=0 or x~ ^ V ^ + 6 4 ) „1±V(65)

4 4

Thus , = 4 , - 7 /2 , i i ^ S « . Ans_

l Algebra ,

21. X*+2T/{X2+6X)=24-6X

or (*a+6K)+2V(*a+6-y)=24 Put V(^2+6*)=y;then we have -y*+2y=24 or y2+2y—24=6 • Factorizing we get (y+6) (y—4)—0

/.e- (*+8) (x-2)=0

i.e. Now or or Also or

nr

y=4, - 6 V(x'+6x)=4

X 2 + 6 A : - 1 6 = 0

, *=2, - 8 V(* a +6x)=^6 K3+6JC—36=0 .'.

-6±V(36+144) 2

or > x=3±3V5 Thus x=2, —8, 3 ± V 5 A™5.. -22. 3xa—4x+v/(3x3—4x—6)^18 we can write it as

(3x2~4;t-~6)+V(3x2--4;c--6)=I2 Now let V(3*a—4x—6)=y theri we have y2+y=12 or y»+y-12=0 or (y+4) (y-3)=0 i.e. y = 3 Now V(3« a -4x-6)=3 or 3x2—4x—6=9 or 3x8—4x—15=0 or 3* a -9x+5x- i5=0 or 3 x ( * - 3 ) + 5 ( s - 3 ) = 0 or (x^-3)(3x+5)=0

*=3 , -5/3 Also V(3JCa-4Jc-6)=-4 of 3x»-4x-6=16 or 3x3- 4*-22=0

_4±V(16+264)_2±y /(70) •• *~ 6 7 T

Th„s *=3, -5/3, ^ L 7 ° >

Examples X (a) (on Page 101) 161

23. 3x a -7+3 V/(3**-16A:+21) = 16X

It can be written as • (3*2-16x+21)+3\/(3*a-16x+2l)=28

Let V(3x2-16x+21)=;> then webave yM-3>-28=0 or (y+7)(y-4)=0 ie. ^ = 4 , - 7 Now V(3* a -16*4-2I)=-7 or 3xa-I6*4-21=49 or 3x I-16x+28=0

. , _ 16±-y/(256+336)_8i:-v/(l48) .» * 6 r Again V(3*a-16x4-21)M or 3x*-I6;c+2! = 16 or 3xa-16*4-5=0 or (3x--l) (x-5)=0 i.e. *=£, 5

Hence *=J, 5, 5 = = ^ ! ® Ans.

24. 8-f-9-v/{(3x-l) (x-2)}=3*2-7x or 8+9V(3Jta-7je+2)=3.ia-7*

It can be written as 104-9V{(3*a-7;c-f2>} = (3**~7x-r-2)

Now let V (3JC*-7JC+2)«^ then l b + 9 ^ = / or y*—9y—10=0 or (y-10) (y+l)=Q ie. ^ = 1 0 , - 1 Now VO*s—7x+2s«10 or . 3x 2 -7; t -98=0 Factorizing we get (3.x 4-14) (*— 7)=0

/. * = 7 ; -14/3 Also V(3* 8-7*+2)=-; i or 3x2-7*4-l=0

7±V(49-12) 7±V37 or * = g « — g —

Hence x=7, -14/3, 7 ± ^ 3 ? - Ans.

125. ? ^ i + V ( 2 ^ - 5 J C + 3 ) - ( - ^ y ^

'or 3 (3A:-2)f6v '(2x2-5*4*3)=2 (x+1)51

or 9x-6+6y/(2x*-5xj-3)=2xi+4x+2

It can be written as • \, 1 - 6 + 6 V / ( 2 * 2 - 5 ; C + 3 ) = ( 2 ; C 2 - 5 J C + 3 )

Put V(2*2-5x+3)=y then •±5+6y*=y* or j>2-6>+5=0 or (y-5)(y~l)=0 •\ y = i , 5 Now V(2**—5*+3)=l or 2x a -5x+2=0 or (2x- l ) ( J C - 2 ) = 0 ,\ x=h 2 Also V(2^ a-5x+3)=5 or 2x 2 -5*-22=0 •1 «_.5±vl25+176)_5±V(201 " *~* 4 4

, _ v ( 3 ^ ± i ) = ( ^ + v , ; 7*a—v/(3s2-8x+l)„(8+x)a

Thus

26

or x x

or i 7^ -v ' (3* 8 -8*+ l ) = 64+16;t+*2

or ; 6 ic*-16x-V(3^-8^+l)=64-or | 2 (3x 2 -8x+l ) -V(3* a -8x+l )=66

(adding 2 to both sides) Let | V(3x2-8x+l)=j> then 2ya-j»=66 or i (2^ + 11) O--6)=0 i.e. > = - l l / 2 or 6 Nowj V(3* a -8x+I)=6 or 3*2-8; t-35= or '. (3x+7) (x-5}=0 ,\ x=5, ^

' —11 151 Also V ( 3 ^ a - 8 x + D = ™ or 3 J C ? - 8 X + 1 = - £

*pr 1(2*2-32*-117)=0


„., , 32±V(1024+I404x4)_,8±V(4151) o r i X~ 4X6 6

Thus rt-7/3,Mp Ans. 27. •y/(4x*-7x-l5)-y/(x2-3x) = V(xt-9) or v'I{(4*+5) (*-3)}]-V[W*-3)}] = V'{(Jf-3) (*+3)} Cancelling V(*— 3) from both sides we get

V{(4*+5)}-V*=V(*+3) Squaring both sides we get

( 4 *+5 ) -2 - / {X (4JC+5) }+ ;< ;= *+3

(4x-\-2)=2y/{x(4x+5)} > Squaring both sides again we have

l&t2+I6x+4=16xa+20x or 4X=4 .'. *=1 Also equating to zero the cancelled factor y/{[x—3)} we

y(3—3)=0 or x—3=0 or ' x=3 /. , t = t , 3 Ans. 28. T/(W-9x+4)+3V{2x-l)=y/i2Ji*+2\x-l\)

V{(2*—I) (*-4)}+3V«2*-l)}-V{<2*-D (*+U)}=0 Taking V{(2*—I)} common we get

Vfl2*-!)} [ \ / ( * - 4 ) + 3 - V ( * + « ) ] « 0 If ,/(2x—l)=0 then 2x—1 = 0 or *=J If ry(*_4) + 3—•(.v+ll)]«0 then vX*-4)+3

or x—4+6V(x-4)+9=*4-l l (squaring both sides) or 6-v/(x-4)=6 or V(*— 4) —1 or .v—4=1 or x=5 Hence #=£, 5 Ans. 29. (2xM-5x-7) + -/[3 (jca-7je+6)W(7**-fiX-l)

Taking V(x~^) a s common we get

Algebra i ' '

;If V(^- l )=Othenx^- l=0 or x = l flf W<2x4-7)-f V{3 (Ar-6)}-V(7A:+l)}-3 j.fhen V(2^+7)+V(3(*-6)}=V(7*+i) Squaring both sides we get i 2*+7+3 (*-6)+2-v/{3 (*-6) (2x+7)}=7x+l or,' 2V{3 (X-6) ( 2 * + 7 ) } = 2 X + 1 2

or V{3 (x-6) (2x+7)=x-r-6 or 3 (x-6) (2x+7)=(*+6)3

or • 6x2-I5x-l26=*»4- X2x+36 or 5*1-27x-162=0 or 5x s-45*+18x-162=0 Factorizing it we get

1 (5x+18)(x—9j=0 or x=9 , -1R/5 . Hence ^ = 1 , 9, -18/5 Ans. 30. V(al-\-2ax-3x2)~^{az-\-ax~-6x2)

< =i/(2a2+3ax-9x*) y or <s/[(a+3x)(a-x)]-Vl(a+3x)(a-2x)]

-V[(fl+3x)(2a-3v)]=0 or' V(fl+3x) [V(f l+x)~v ,(a-2x)-V(2c-3x)=0 If V(«+3Jf)=0 then a+3*=0 or x=-af3 If' V( f l -») -V(f l -2x)- ; v / (2a-3x)=0 Then \/{«-x)-V( f l—2*)=V(2a-3x) Squaring both sides we get N

, a—x+a-2x—2v'{(a—*) (a— 2x)}=2a—3x or y{(a -x) (a -2x)}=0 or (a~x) (a-2x)=0 or ! x=a, a/2 Hence *=a, a/2, —a/3 Ans.. 31. v '(2x2+5x-2)-V(2x a+5x-9) = l Let V(2*8+5x-2)=> and ~V(2x2+5x-9)=z

yVz2=(2»3+'5x-2)-(2x2+5x-9)=7 r or iy-z)iy+z)=l or y+z=7 (ii) Kow solving (i) and (ii) we get y=4;' z=3 "

i

Examples X (a) (on Page 101) '* ' x 165

or 2x2+5;c-18=0 and 2x*+5x—l&~0 or (2x+9) (*-2)=0 or *=>2, -9 /2 Ans. _ 32. V(3^2~2^+9) + -v /(3^~2*-4)=I3 Put V(3^2-2x+9)=^ and ' V(3*a-2x-4)=z f r

Then ^+z==13 ...(1) Also^ 2 -z 2 - (3^ a -2x+9)- (3Jc a—2A:-4) = 13 or (y-z)(y+z)= 13 or 13 (>>~z)=l3 i.e. j»-z=l....(2) Now from (i) and (ii) we have .y=7, z=6 ,\ v'(3xa-2*+9)=7 or 3x2-~2*+9=49 or 3x a -2x-40=0 or (3x+10) (x -4)=0 ie. x=4,-10/4 From other values we will get the same value of x. Ans.

y/{2x*~lx+\)--v/(2xa--9*+4)=l Let V(2*2—lx+\)=y and y ' (2^+9x+4)=z Then y-z=\ y ...(i) Also j>a-z2=(2xB~7x+l)-(2x2-9;c+4)=(2x-3) or (y—z) (y+z)=(2x-3) or ,y+z=2*-3 ...(ii)

(.*. y-z=l) Now y—z—l and y+z=2x—3 .*. y=x— 1, 2=x—2

Hence V(2*2-7*-f-l)=x-l or 2x 2 -7x+l =*8-&c-H or x2—5x=0or x Qx—5)=0 i.e. x=0, 5 Ans. .34. - v / (3x a-7*-30)-V(2*"-7x-5)=;c-5 Let V(3^a-7«-30)==>- and V(2* 2 -7*-5)=z ,\ y-z=x-5 ...(i)

=A"-25=(x-5)(x+5) or (y-.z)(y+z)=(x~5)(x+5) or ,y.i-2=x+.S ...(H) (v j - z = s - 5 ) From (i) and (ii) we have y=x^ z=5 / . V ( 3 ^ 2 - 7 J C - 3 0 ) = X or 2 x 2 - 7 x - 3 0 « 0 or (2x+5) (*-6)=0 .-. *=6 , - 5 /2 Ans.

...(0

f, Algebra

''35. x'+x'-4xi+x+i=0 ', ^Dividing by xs we get

*»+*-4+i+i=o;

Put x+ — then ^ + 2 + , - r = z 3

',. ^ ' *

or x 3 + 4 =z a -2

.% Putting in (i) we get j z a+z-6=0 or (z+3)(z-2)=0

or z= —3 ; 2 or, jc+J- = - 3 or x2+3x+l=0 .'. x==*^l

} ' X *

, Also a + ^ ^ o r * 2 — 2 * - H = 6 o r (*-l) ,a*0 .*- JC= I -Y Ans.

x

36: *•+•§-x»+l=3*H-3* ' 1'

Multiplying both sides by 9 we get ', 9x4+8*8+9=27x1+27*

or ^ 9*«-27;t3+8;t2-27;c-f 9=0 Dividing it by x2 we get

1 27 9 9 * 2 - 2 7 x + & ~ + 4 = °

- o r ,1 9 ( ^ + . y - 2 7 ( * + i . ) + 8 . - 0 ^

Let; » + I - J >

'.'. • #+j£-+2=y* or x 2 + ^ = > - 2

From (i) we have 9 6^|-2)-27y+8=0 or 9.ya-27>'-10=O

1 i '.


I 0 or (3^+1) (3?-10) =0 or > '=-£ , V

or 3*»+*+3=0 .-. , - = 1 * ^

Also x + - - = ^ or 3x2~l(M-3=0

or (3x- l ) (*-3)=0 .'. x = J ; 3 Ans.

37. *4+l-3(;rM-*)=2;c3

or r*-3* a—2* a-3x+I=0

or x * - 3 * - 2 - — + ~ = 0

or

1 .. , . 1 . . _ „ * Let * H — = y then jca+ —+2=.?

or * 2+i-=/-2 .'. From (1) we have

(j»*-2)-3 0>)-2«=0 or y*-3y—4=Q or (>-4)(^ + D=0 *e. ;>—4, —I

.'. x+ —=4 or JCS-4;C.*-1=0 . \ A - = 2 ± V 3

Also x-f ~ = - l or *2-f-X+l=0

s 2 Ans. 38. 10 (s*+l)-63x (x2-l)-f 52x2=0 or 10x4+10-63^1+63x+52^2=0

or 10JC1-63X3+52A;2 + 6 3 X - M 0 = 0

Dividing by x2 we have

or l o ( * - + ^ ) - 6 3 ( ? - i ) + » i o ^

Let x—r=^thenx»+4-«7*+2

Putting these values in (i) we get

10 0<2+2)-63j>+52=0 or 10)P-63y+72^Q'. J . . .» ' 63±V(63x 63-720x4) Solving it we get y~ :±-*-i—s« -

' " = ^ i 3 2 _ . 2 0

3 24

1 3 ' I Hence from x - ~—=-^- we get x=— -=-, 2 A ^ r 1 24 c 1 And from * =-r- we get *=J5 , —-£-„ , x + V ( 1 2 q - s ) _ y f l + l . *--v/(12a-*) V a - 1

By applying the process of. componendo and dividendo on both sides we have on simplification -

2x _ 2yg * __ Vo 2V(I2«-x) 2, o r V(12a-*) ~

Squaring both sides

l2ll-x = T or x ^ l ^ - a * or **—ffjc-I2aa=0

Factorizing it we get '(x-Ma)(x-3fl)=0 .'. *=3<7, - 4 a Ans. 40. Applying componendo and dividendo; to the both

sides of the given expression, we get

e x a m p l e s s\ \af \vu i agv I<JI/ IV?

_J2x±a)_ ^x-ha V ( A 2 - 4 X 3 ) 5x -a

Squaring both sides (2x+a)*_(5x-{-ar~ ai-4x* ^ ^ - ^ a

Cross multiplying and simplifying it we get or 50x3-40a2.x-f32*42=0 or 50^ -8^a a =0

or 2x (25*2-4fl*)=0 ... *=0 ; ± j - A n g <

41. x W ( * 3 - l ) _ JC—x/(*B—1) x~V(x2~l) x+y/(x*-\)"

= 8*. \ / (x2-3x+2) or 4*v ' (*2- l ) = 8xV{(x^2) (x-1)} or 4xv ' l(*+l) (*-l)}=8*v{(*-2) (JC-1)} ;

{Dividingboth sides \>y.*y/[x—\)} or V U + H = 2 . V ( K - 2 ) or x + l = 4 (x-2)

i.e. 3x=9 or *=3 Also from X\/(x~ 1)=0 we get JC=0, * = l

.". x=0, 1, 3 Ans.

42. V(* 2 +*)+-£4=^ r =|-Vt* —x) 2 o r v , . V ( , + 1 ) + _ ^ ^ ^ :

or Vx.V(x- f - lH V , { V J ( X + 1 ) } = y

x(x+n+t s

Put V V ( ^ + D ^

170 \ Algebra h >

, lor i j ' ^ j - or 2 ^ - 5 ^ 2 = 0

br {2y-\)'{y-2)=0 .i.e. y=2, ft

!\ V*(*+l)«2 or x(*+,l)«4 /.e. *2+*-4=*0

- . - " ' ± ^ 0 7 ) ' • V I " • .

Also V W « + 1 ) } 4 or x(x-H)=£ ( or ' 4 J C 2 + 4 * - 1 = 0

', " 2 ~ Ans.

or i 6 {x-\)*=x or 6*2 — I 3 J C + 6 ^ 0

Factorizing it we get (2x— 3) (3*-2)=i0 i 3 2 .

' 0 r * = T ' T ; Ans. i

44. I We have -?—=-?- or 2*3~2-v=8=(2)*

Hence x2—2JT=3 (Equating powers of both' sides) or . r s - 2 * - 3 = 0 or ( * - 3 ) ( x + l ) = 0 ..-. * « - l , 3 Ans. 45. i2* (a2 + l)==(fl3B+^)a-or a2a,(a2-f-l)-fl*((J2*+l).<7=0 »..(i)

Taking ax common and equating to zero we have I • a*=§' i.e. x—infinity which does not satisfy the

quation. | Now from (i) ax (<j*+l)_a (<23«-f 1)=0

Examples X (a) (on Page 101)

Put a*=y then aZx=yz

.". y (a*+\)-a (y*+l)=Q or ay2—y (a2 + l)-f a=<3 or (y—a) {ya—\)=Q

i.e. y-ttf±

And ax=—= (tf)-1 .'. x~ — 1

H e n c e x = ± I ADS,

4* 8 V U - 5 ) ^ V ( 3 x - 7 ) 3 * - 7 x-5

Cross multiplying we get 8 (* T 5) 3 ' 2 =(3; t -7) 3 ' 2

Raising both sides to the power 2/3 we get 4 ( x - 5 ) = (3x-7 ) or x = 1 3 Ans. ._ 1 8 ( 7 * - 3 ) _ 2 5 0 V[2x+1]

2 J C + 1 3 v/ ( 7 ^ - 3 )

Cross multiplying we get 54 (7*-3)3<3=250 (2x+l)3 '3

or 27 (7x-3) 3 f 2 =I25 (2jc+l)afa

or (3)3 ( 7 A - - 3 ) 3 ' 2 = ( 5 ) 3 (2x-j-l)3<2

Raising both sides to the power 2/3 we get 9 (7x -3 )=25 ( x + l ) or 13x^52 A x = 4

48. (tf+s)2>3+4 ( a - x ) 2 ' 3 = 5 (a*--*2)1 '3

(f7+*)2/3 4 (a—r)»» _ 5 (a 2 -**) 1 ' 3

o r ~ ~ ( a 2 ^ W 3 (a 2-* 2 ; 1 ' 3 "(a*-*')1 '3

Then we have

, Algebra

I 4 . '' ->+~-=5 or p*—5p+4=0

\ W (j»-4) ( p - l ) = : 0 • \ i.e.. p*=4f 1

then -i?w=l or a-f-x=a—* ;*. *«()

1 then ^±2=64 or 65x=63a .'. * = § *

a—x 65

/ . * = 0 , | ? Ans.

49. V f ^ + ^ - D - v / ^ + A x - l ^ V o - V ^ - ( 0 We have (xz+ax-r-V)-{x*+bx-\)=x {a-b) Factorizing it we get

or from (i) Wa—^b) W{x*+ax—\) • + <s/(x\+bx-l)]*tx Wa-Vb) Wa+Vb)

or V ( x H w - D ' - f V ^ + ^ - D =xWa+Vb) .-.(2)

Adding (i) and (ii) we get 2V{x*+ax-\)=^a (*+D+V* C*~l) Squaring both sides we have 4 (*»+«: - l )=o (x+l)*+* (Jc-l)9+2V(fl*) (xB - l ) Taking terms to one side we have x2 (a+6+2V(fl6)-4)-l-2.x [a-ft~-2a£] ,

+f l+£«-2V(^)+4=0 or x* \Wa+^b?~A)—2x (a+b)

+ (V«+V6)a+4=0


Solving this for x we have 2 (<i+6)± V[4 (a+&)2-4 {(vW V^)a-4}]

. : . x { ( y f l - y^4 -4 } 2{(Va+Vfc)a-4>

On further simplification we get

or 4x2-2=98 or 4*2 = 100 or x2=25 .*. x=±5 Ans.

^ 51. The given equation is **-2:c3-f-x=380. It can be written as

x*—5*8+3*2-15ji:*+15x9-75jr-r-76x—380*0 x=5 ' satisfies this equation. Hence (x—5) is its factor

so by remainder theorem we have *a (x-5)+3x2 ( X - 5 ) + 15JC ( X - 5 ) ' + 7 6 (*-5)'*=0

or ( X - 5 ) ( X 3 + 3 J : 8 + 1 5 X + 7 6 ) = 0

Again A==—4 satisfies this equation. Hence (JC-5) {*a (x+4)--x (*+4)+19 (x+4)}=0

or (.t-5)(x-f-4)(xa-x-J-I9)=0 Equating each factor to zero, we have

x—5=0 .'. x—5 x + 4 = 0 .'. x«—4

l ± V O - 7 6 ) c3-x+19=0 .*. x=

2 _I±V<-75)

i . " (' A l g e b r a v,

j , „ j , _4, l±S^2»- An, ' 52. The given equation is 27*3+21x+8=0 or' 27*9+9*2-9:t3—3x+'24*+8=0 or 'a9jc* (3*+l)-3x (3*+!)+8 ( 3 K + 1 ) = 0

or '(3x+I) (9*?-3x+8)«0> Equating each factor to zero we have

\ • 3x4-1=0 .\ * « - J

W-3*+8=o ,-. ^?rf=288 i=,i±4^52 ' ::. ,=_j , i±^u Ans.

1 EXAMPLES X (b) (on Page io6) 1'. The given equations are

s 3*-2.y=7 ...(i) and ', xy=20 „..(ii) Squaring (i) and adding 24xy to it, we get

9,*a+4^-12x}»+24x.y«49+480 or ' (3x+2;0s=529=(23)a

• >-. 3*+2.y=±23 ...(iii) Now solving (i) and (iii) we get i

x=5> - * . y « 4 , - V Ans. 2. The given equations are

5x~y~Z .„({) ,ai_6xB«25 \ ' „.(ii)

From (i).y=5.v—3 Substituting this value of y in (ii) we get '

(5x-3)3-6* a=25 or 19A-2—30JC- 16=^0

or '(19x'+8) (x~2)=0 .-. x=—rV, 2 and .y=-2£, 7 Ans. , 3. The given equations' are

Examples X (b) (on Page 106)

12^+13^=25 ...(ii)

From{0 y^^~

Putting in (ii) we get

or Ax ( 4 * - l ) - f V (16*a~8*-H)=25 or 1 4 4 X 2 - 3 6 X + 2 0 8 J C 8 - 1 0 4 * + 1 3 = 2 2

or 352*2~140x-212=0 or 88«2_3 5 x_ 5 3 = 0

or (S8*-f53) ( x - l ) = 0

Hence .y=],—Jf Ans.

4. The given equations are x*+x*y*+y*=m (j) * a -*H-j< a «19 (ii)

Factorizing (i) we have {x*+xy+y*) (x 2-^-f^ a)=93k

With the help of (ii) we get (x*+xy+y*) , 19=931

or *a-f*>>4^*=49 (iii) Now adding (H) and (in) we have *2-f^>a=34 And subtracting (ii) from (iii) we have xy= 15 .". (*8-j s)a*=(jc3-F>2)a-4xB>'2

=(34)2-4x(I5)3 = 1156-900=256

Now solving x2+y°=34 and a1—>*s=±16 we have » = ± 5 o r ± 3 , .v='±3 or £ 5 Ans.

5. The two equations are jc*+jry+.y«=84 ...(i) ar—-/A3»+y=6 ...(ii)

Algebra t (i) Can be written as , "•" {*+V (xy)+y) (*- V ^ + ^ ) = 8 4

A 6 { J C + V M + J ^ H B *

or xf^{xy)+y=\A (iii) Now adding (ii) and (iii) we get x-\-y=\§ Subtracting (ii) from (iii) we get xy=l6 Solving x +y—10 and xy^\6 we get

i x=Z, 2 ;y=2, 8 Ans. 6. ' The two equations are

' x+V{xy)+y=65. ' (i) ^*a+xy+y2=2275 ;. (ii)

(ji)' Can be written as *' "• ;• , i*+>/{ty)+y}[x-,/{xy)+y}=m5 or , 65 {*-V(^)+^}=2275 .*. _ x-v'{xy)+y=35 (iii) Now adding (ii) and (iii) we get :x-\-y=50 And subtracting (iii) from (ii).we get .xy=a225 Solving #+y=5J and xy=225 we get

x=45, 5 ;.y=5,45 Ans. 7. The two equations are

*-fj;=V(xy)+7 or * - - V ( ^ ) + ^ = 7 (0 ^-1-^=133—xj» or x3+x.>'+>a=133 (ii)

(ii) Can be written as {i+V(xy)+y}{x-Vixy)+y}=l33

or {x+V(*J')+3'}x7=133 ' , or x+V(xy)+y=\9 < (iii) Now adding (i) and (Hi) we get . ^+^=13 And subtracting (i) from (iii) we get xy=36 Solving x+^=13 and xy==36 we get

x=9, 4 ; y=^4, 9 Ans. 8. We have 3x3-5y*=7 '' (i)

1 3*y-4j*«2 : (ii)

Examples k (b) (on Page 106) 177 N

Let .y=m* then 3jc2-5m«jc2=7 or x* (S—5m*)«7 (iii)

3mx*-4ma;cs=2 or x* (3m—4m*)=2 (iv) Dividing (iii) by (iv) we have

3-5ff l '___7. 3m-4ma 2

or 18m2—2lm+6=0 Solving for m we have m=£, £ /. From (iv)x=±3 and ,\ y~±2 Ans. 9. Putting .y=wi;c in both the given equations we have

JC« (5ma—7) = 17 |...(1) x*(5m-6)=6 ...(2)

Dividing (1) by (2) we get 5m«-7 _17 5m—6 6

or 30m*-85m+60=0 or 6ms—17m-fl2=0 Factorizing it we get (2m—3) (3m—4}™"1,

m=3/2, 4/3 Putting these values of m in 1st equation we have, when

m=.3/2, x*(5.9/4-7) = I7 or » '«4 v ^ ± 2 Also from (i) xz (5X16/9-7)=17 or x = ± 3 when m=4/3 Hence substituting th'ese values of x and miny=mxv/c

get the corresponding values of y iey^^zh ± 4 Hence x = ± 3 , ±2 , y=±*> ± 3

x=±l, ? = ± 3 , ± 4 Ans. . 10. Putting y=mx in both the piven equations we

have x* (3—16m)=-165- «.(*)' * And *t(7m+3m*)=Mk2 ».(2)

Dividing (1) by (2) wc get '3-16m 165 5

178 Algebra

or Cross multiplying we get 15m2—29m+I2=0 -Factorizing it we get (3m—4) (5m—3)=0

i.e. m«4/3,3/5 Now from (i) when m=4/3 we have

x 2 ( 3 - 1 6 x 4 / 3 ) = - l 6 5 %

— 55*2 s

or = ^ — = - 1 6 5 -

or x*=9 .\ x~±3 Again when m=3/5 we have

* 2 (3 -16x3 /5 )= - l65

, or - Y * * = - 1 6 5 or y**=25 -.'. x=±5 i' Now substituting these values"*, of m and x iny=mx we

get the corresponding values of, j ' . , i.e. . J J « ± 4 , ±3 :•

; Hence x = ± 3 , v = ± 4 . * = ± 4 , j - = ± 3

11. * Putting J/«BW in both the given equations we have j" . x* {3+m+m*)*=\5 . ...(i) j. And x* (31m—3-5m*)=45 ...(ii)

I Dividing (2) by (1) we get I. ... • '• ' s im-S-Sro 3 ^ 45,

. 3+m+m a i 15 ', or 4m1— 14m-jr6«0

'.'/. or (4tn—2) (m— 3) =0 i.e. m=J or 3 J Now putting n m } and 3 in (1) we get

., '.. *• <3+i+i)=15 or * = ± 2 'And w3 (3+3+9)^15 or x = ± l

' Now putting these values of m and' x in y=mx we get. the corresponding i-alues of j> i.e. >—±1, ± 3

] Hcncex=±2, v = ± l * = ± I » . y = i 3 ADS.

Examples X (b) (on Page 106) 17!

12. Potting y=m>: in the given equations we have x*(l+f» s-3m)=3 ...(1) *» ( 2 + ^ = 6 „.<2)

. l+m'—3m_ ^ _3_= _1_ 2+m2 " 6 , 2

or /«*—6m=0 or m (m—6)=0 .". m=0, 6 Now putting these values of m in (I) we get

*?<t + 0—0) = 3 .'. J C = ± V 3

And x 2 i l + 36-18)=3 .'. *=V(3/19) Now corresponding values of y from y^mx are <

*=±M)-y~±6M) An./ 13. The given equations are . *

• * + j > = 8 ....(21

Let s=u+v and y=u—v ...(3) Then x±y=u-\-v+u—v=2u ' or 8*=2u /. u=4 .*. x=4-|-v, j> = 4—v Frora(i) ( 4+F) 4 + ( 4 - I » ) 4 « 7 0 6

or 256>256v-t-96v2+l6v3+vt+256-2S6v+96i's

— I6v»+v*=70 or 2 (256 -r-26vB+V*)« 706

o r y*+96v2-97=0 or (v»—1) (va+97)=0 /. v - ± l , ± V { ( - 9 7 ) } From (3) x = 4 ± l , 4±V{(-97)}

y«4=Fl,<HV{(-97)}

180 I . Algebra l •

on * = 5 , 3 , 4 ± V ( - 9 7 ) ) • *==3, 5, 4 ± v (_97) Ans.

' 4-; **+7*-27a and j - ^ 2 Let * H « + V , > ( = w _ p •

or 2=2v ... V£=I

Thenjc=«+i( y=u-\ Putting these values of * and y we get

.(«+l)*+(y-l)««272

or «J+6« ,-135=0 °r <«*-9)(«'+15)=0 .'; »=±3,±V((-I5)} ;'• ' = ± 3 + 1 - ± V ( - I 5 ) + 1 " • . 4 » . -2 , l±V( - i5 ) And ^ ± 3 ^ i ; . ± v ( ~ 1 5 j - l

15. a*— "=992 *—y==2

le t * » ^ v ; .v=u~V Then x—j>=ufV— «+V=2V

2=2V or V » l ' Hence ar^l+B.j^B-l

(«+D*'-(«-l)«==992 w 2(5««+10M>+1)^99 2

or w*+2«8—99=0

/. . - ± 3 - 1 , ±V(^I I )+I - ± V ± V ( "> • ^ ± 3 - 1 , ±V(-11)-1

V. * = 4 , - 2 ± V ( - l l ) + l > « 2 , » 4 ± V ( - l i ) - l A n s ,


16. , + ! _ , _ ( | )

or xy+4=.y #„#(3) xy+4**2Sz ...(4)

Subtracting we get y=25x .'. From (3) we get 25*2+4=25x or 25*2-25*+4=0 or (5» -4 ) (5x - l )«0

•'• x=h$ • :. >-=5, 20 Ans. £,y*_ 9 y x ~ 2 •••(•)

X ^ 1 -..(II)

or x 3 + / = i - * y a n d x+y=$

or {x+y)*-ixy {s+y)V xy

or 27-9 xy=±.xy

2 i» f i

2 27

or y a y « 2 7 .-. xy^2

Now solving * + ^ = 3 and xy~2 we get *=2, l , y « i , 2 Ans,

or 5*+2;>=50 Sx+2y=$xy. Hence {xy=50

' ' *« 60 or ?:y=60 or v=— Putting value of y in, (i)

' x ,60 1 _ x , 12." T + _ . _ = 5 o r ^ y + _ = 5

or x*4-24—10x=0 or ,, (x— 6 ) (x -4 )=0 .". *=6>4,>'=10> 15 Ans. 19. x+v=1072 ...(i)

»i/3+y/3=16 ....(ii) From (i) we have

, . 1(x1'3)3-Hv1/3)8=I072

or {xin+ywy*—3xll' . v1'3 (xV3+yV3)==i072 oi (16)a-3 (xy)1'3 . 16=1072 or .16x16-3 (.vy,V3=67 or 3 (A>0 1 »=189-or •(xv)1/s=63 Now (x»3-y1-?)=V[(*1 '8+*1 '3)s-4 (xv)1'3]

= V(256-252j = ± 2 i

Now JcV3+>a'*=16 and Xw-yiia=±2 From ,these two equation? we have by addition and

Direction . * m = 9 , 7 ; j - V 3 = 7 ( 9

. ie. x=729, 343 ;,y=343, 729 Ans. 20. Wc have * . y*'?+y. x , , 8=20 . ...(i). and **+y*-i*=65 ...(ii)

. From (i) we have , VW) { V ^ + V W ^ O ...(iii)

.(iilcanbe written as <x»1)*+(y ia) , ,=(*1 'M-> , 'V-3V(^)(s1™+> l 'a)=65 U. (xV»+y1 '5) '-3x20=65 ; frora (,-,-{) or Wx+y/y)*=65+4Q=l25-(5)* -,

.'. x'x+Vy=s 20 .'. From (iii) V(xy)=-^=4

Now (Vx-Vy>2={V*+Vy)z-W(xy) =25-16=9

Vx-^y=±3 ' ' »..(iv) Solving y/x+\/y=5

We have \/x=4t 1 i.e. x=\6 ; I \'y~\t 4 ie. y^l ; 16 Ans.

21. We have t/x+y/y^S ...(i)

From (2) we have Vy+Vx=5l6^(xy)

or 5=5/6v'(Jfv) '•«• V t ^ H ^ Now ( V * - Vv) ,=.(\ /* + V ^ ) a - 4 V ( ^ )

=25—24=1 or •s/x— v ^ i l Now solving V ^ + V . ^ 5

We have y/x—Z, 2 i.e. x=9, 4 y/-y=2, 3 ' /.«. j>=4, 9 Ans.

22. The two equations are V<*+v)+v/(*-v)=4 ...<i) * 2 - v ' = 9 ,..(ii)

From (ii) we have > U-v) ' (x+7)=9

or V{(*-J0(*+v)}=±3 Again we have

[V(*+v)-V(*-J>)la

= [>/(*+>•)+v'<x~v)]*-4V((x+v) <*-v)} = 16-12=4

le. V(x+y)-V(x-y)±±2 ..,(i«) Now from (i) and (iii) on addition we have

ty(x+y)=6, 2 or x+y=9,1 and on subtraction 2v%*—y)—t, _ or *—>«!, 9 Now froip these we Have

x=5t — 5 (But —5 is absurd as it does not satisfy the equation)

y=±4 Ans. \ 23. The two equations are

I Squaring (ii). we have (x+l)+(x- l ) -2V(x a - l )=y ,

or 2x—2V(x1—l)«j> , ...(UO. Putting this value of y in.(i) we have

2*-2V(*"-i)4V(*"--:l)«2 Qf . 2 x - v ' ( J f a r 1 ) t * 2 ' or V(*fi—i)=2x—2 Squaring both sides we have

•;;••'. \ * B - 1 « 4 ^ 1 ^ 8 J C + 4 , - or, 3*"—«i+5=0 . • * . . "or ' ;<3x-5)(jc-l)=0 . : . , ' , .

'" -". Y**** > 2 [frpm'(iii)]" Ans.' !< 24. ,The two given equations are

' V(7)+Af(fH: •••<» . *+y=iG _ „,(ii),, ;.:

. -From (>) we have ( .

V(xy) 3 V(*J0=3 or xy=9

Now (JC—>>)*=(.x-t-y)2 —Axy • = l00-3£=64

.\ x—>»=8 (iM)1

Now from (ii) and (iii) we have

x—y—Z 2x=I8 . .\ s=-9 .*. y=\

V*+V.V "*" A / K - V J 4 (I) » 8 _ ( _ > , 3 = 7 0 6 ( 2 )

From (i) we get

(vf*-vO08+(V*+v'>)»..n

or WyT =14" or 8^^=17x-17->' 9

or 9x=*25y or ^ = 2 5 x

Hence from (2) we have

or • **=625 U. »=,±25 .\ x = ± 9

26. * 3 +4y-15x=10(3^-8) x)>=6

From (2) y= — ^ ;

Putting this value of y in (1) we have '

* a + 4 . ^ -45*=* 10 ( 3 . | ~ 8 ^

]r ,.+J^_15x=10(I!_8) or x*-tjl44-15x'—180x+80^2«0 ! • Factorizing it by remainder theorem we have (x-2) (x-3) <*-4H*-6)«tf *A • *~2,"3 f4, 6

A From (2)>=3, "2, 3/2^1 ' Ana.

27. *V-M00—41JO"«0 ' J (0 >*^5xy+4x*=0 / .(2)

From (2) we have ' , (y—4x) (y—*H0 - or y~4*, *

Putting these values o'f y in (1) we get ac«+400^41«,=^0 when y=x

or (**-25)(**-16)=0 .\ * » ± 5 i - i 4 1 . . V y = ± 5 , ± 4 j ...(A)

Again putting y=4* we have 16xa-164x«+400=0

o r 4x*-4U1+100=0 or {x*f-4)_ (4x*-25)~Q *

.V ,*« ±2, ±5/21 .".! v = ± 8 , ± 1 0 / _ ...(B)

, .'. x and'7 are given by A, B, Ans.;* 28. 4x*+Sy=6+20xy-25y*+2x It can be written as

(2x-5.y)»-(2x-5y)-6«0 Let 2x— 5y=z then.z*—z—6=0

1 or ' (z—3) (z+2>=*0 {e. z=3 , -2" Hence 2*—5y=3 or —2 Taking 2x—5>=3 we have 7*—Uy=17 Solving these two equations we have

107 48 ,13 'y=13 Ans.

29. Wehave9a:*+33*-12=12j:y— 4yM-22y

or 9 X 2 - 1 2 K J ' + 4 ^ , + 3 3 X - 2 2 > ' - 1 2 = 0

or (3x-2y)*+ll(2x-2y)-l2=0

Factorizing it we get (3x—2y+12) (3x—2y—I)=0

.'. 3*~2.M-12=0 and 3 x - 2 y - l = 0

3x+12 , 3*- l i.e. y——2 ^n d y~^2— Now substituting these values ofy in the given equa-

lions x2—xy=18 we have

or x* + \2x+3.6=±0 or (x+6)a= 0 • v - « •" » - 3 x 6 + 1 2 • . . x = —o , . v=> s = - 3

3JC—1 , Also y = - j —

•'• ^ - " ( ^ T " 1 : ) s = s l 8 or * ' - * + 3 6 : = 0

. „ 1±V(-143) " x 2 .

... y ^ r 3 ± 3 ^ - - ' 4 3 ) 1 / 2 J 1±3^(-143)

Hence x = - 6 , ^ V t - 1 4 3 ) ; y = s _ 3 > l ± 3 v j - l 4 3 )

30. -(*»->-*) (x-y) = 16*y ...(I) ( x 1 - / ) (*•->*) ~ 6 4 t a V .„;/ .-(2)

Squaring <i) we get s/^"''

Dividing (2) by it p ' ^ (**- / ) (x'-J*) 640x,y>

. (x«- / )* (x-jr)»™2S6xVfc . , > -

(x»-?'Kx-/) 7

Algebra '

or 2x*+2y*=j5x*+5y*-lQxy or 3x*-10xy+3y*=0

i.e. (3x—y) (x-3y)=0 ^ x *

Putting y^3x in (i) we get (x*-9**) (x--3x)=16 x . 3x -or • 16x3-48;e*«0

' or x* (x—3)=0 i.e. x=0 ,3

.'. From>=3x we have y=0, 9

Hence *=0; >»=T) x=3,.y=9 Ans,

31, The two equations are W-xy+y^Zy ...(1)

• 2x*+4xy=5y ...(2)-,

From (2) 7 » 2 * ± ^

,\ 2 x 2 - ^ + ^ ' = 2 ( 2xZ+4xL. \ from (i)

or \0xi^5xy+5yi=Axi-\-%xy or 6x a -13^+5 j ' 2 -0 or (2*-.y) {3x-5y)=0

„ 3JC ; .\ J>=2x, -j

Putting y~2x in (i) we get

, 2xa—2xa+4;c,=4;i: , or x2—s=0 or JC (x—1)=0

.'. x=0, 1 .'. ytr-0, 2 Again putting y=?3}5'x in (i) we get

&» ? £ = f i x . l

. 5 + 2 5 5 ' ^ or 44*2-30*=O

. EMtoP1es X (b). (on Page 106)

*(44*-30)«0 .'. , = 0, I 5

• 22 •Thus v=o - -

'22 Hence »=,o,^=o

22 ' y~~22 ^as'

32. f - ! ± Z ? + ^ 3 - ^ _ 4 3

5x-7j<=4 -Amplifying (I) we get

' - ^ ^ ^ 1 + ^ (*-J»« • 8"

or 0 . + ^ j ^ + ^ ) + i g 1 . « , 2x+3Xy(-l J_W3* •

\x-y x+y) ~T 2 * + - ^ ? « 0_ 6xy^ 27x A

'•«. * = 0 and J ^ L - i ^ n

or 4Bya=27x*-27y* or 7 5 ^ = 2 7 ^ or 25y2=9*»

^ /. 5j>=±3* or y=±3{5x . . irom (2) we have * = l x 5 = 5

10 ,. X:=23 ^30 x=iQ

From ; > = | ^ = o when x=0

^ = 3 when x=5

or

or

or

190 Algebra

33.. y{y*-3^-^)^-24=0 ' .x (va-4x,y-t-2jie9)+8=0

; ^(y^—Zxy-x*) ^—24 . * i

"•' , x (ys~4xy+Zx3) - 8 Cross multiplying we get

By remainder theorem we have : (x-v) (3x-v) (2x-v)=0 g

I .'. x=y,y=3x, y=2x Putting these values in (i) we get

wbenv=x, x (x a -3x*~x ,+24)=a or 3x3=24 or'" x = 2

when y.=3*, 3x (9xa-9x«—X*)=^—24 or 3x3=24 or x = 2

, wheny=2x, 2* (4jca-6xar**)-f24*=0.

or x—^4

, Thus x=2 ,^4 ,2 ; . y=2 , 2..-$'4, 6

3 4. The given equations are 3x*—*xy*+y*+ll=*b

; **(>>-;0=l * 1 Making (i) homogeneous with the help of (2) weNget

3x5-8Jcy.a+va-r-21 (x2v-x3)=0 or 3xs-r-8*y+/»+2I*>>—21x»=0 *

1 or 18xs+8xva -v3—21x ,v-0 . Factorizing it we get

(2x-y)[3x~yy*<=0 .*. v=2x, y=3*

(Putting. y=2y in (2) we get * xa i2x~x)=l . ,". x=l,y=2

! Also when v=3x from (2) we get ix*(3x-x)=l A x*=M» ,\ y=3.y\


Hence x~\ty=^2, x = ^ ^ = 3 . ^ J Ans.

35. WeIjave^!(4*8~-108)=Ar(*3-9^J And 2x*-t-9;r>>-f;>a=108 From (i) with the help of (ii) we have y2 (4X»-2JCS—9xy-ya)^x (x3— 9>3)

or 4xV—t^y^Sxy3—3'*=xj—9xy9

or * « — - I X ^ ' + J ^ 0 or (xi-y2)i=0

/ . at*—j*=0 i.e. x=±y

Putting x=+y in (ii) we have 2*H9*2+x*=108

or - 12*a=108 ie. x?=9 ' x—±3 and ^ = ± 3

Also when y * - * then 2x2—9x2-f*a-I08 or 6JC»^=-108 or jc?= —18

j r = ± V ( - 1 8 ) and >-«=Fvt-l*) Hence x=±3, ± V(—18?

7 = ± 3 , TV(-18) Ans. 36. The two equations are

6**+jeV-H6=24x3+2xp2 .,.(1) And &+xy~j>*=«4 ...(2)

Making (1) homogenous with we help of (2)

e Simplifying it further we get x*-4x»>'+6xV~4xV

3-f / = o Factorizing if h^ r*i».«:--*— -"-

• Algebra ' i

, (x-y)(x~y)(x~y)(x-y)=0 i.e. t , (x—y)*=0 .\ *x*=y Putting x=y in (2).we get

: *»+**-*•=4 .*. ;.*=±2 .\ >=±2 Thus x=±2\ y=£2 Ans.. "' ' 37. If we simplify the given two equations we get them'

ax-by=£y*—x* ...(1) ,ax+by=*y*+x*+2xy . ...(2)

Adding (1) and'(2) we get ax=y%+'xy or , x{a-~y]**y* •;

Now from (1) we get

y*+Xy-by=y*~(J^2

. ' a->> (a—y)1

or 6- * !_= -

(Dividing both sides by y) ,\ y**Q or b (a—y)l—yv (a—y)=y* or bax~ 2aby+by2—ay*+y?=y\ vr ) * {i>~-£ff—2sb^-fhsi%=^0, liis quadratic in y

_2ofe±V{4flt68-4 (b-_g) bg*j ab±g^(ab) " y 2{b-a) • " b-a

aVb{Vb±v'a) .

a-Jb . a\/b , or y=, /. —?-* and v

or

Now when v=0 from #=*-/—'#=a0

Examples X (b) (en Page 106) 193

. aVb by/a (e y2 \ when y = — 7 . - , — - . — ;x=—r-*— ,u I from x— ' — ]

ana wheny=—JT~— ,— ; * = • ^Jb—^/a * ^/h—\'a AnS.

38. The two equations are

xy+ab=2#x ...(1) x!y2-r-a2&2=262y2 ».(2)

From (1) we have xy—2ax=—ab ab

or x = x 2a— y

Putting this value of * in (2) we have

Simplifying it we get a2y2+a3 (2a -y)*-2y 2 ( 2 a - y ) 2 = 0

or 2y*-8o>'3+6a2y2+4a3y-4a4=0

It can be written as

2y*—2y3a-6ay3+6a2y2+4a Jy+4a1=0 or 2y3 (y—a) —6ay* (y—a)+4a3 (y—a)=0 or ( y - a ) (2y3-6a>>2+4a3)=0 or ( y - a ) (2y2 (y-a)—4ay (y—a)—4a2 (y-a)} = 0 or ( y - a ) ( y - a ) (2> 2 -4ay-4a*)=0 or 2 ( y - a ) 3 ( y a - 2 u y - 2 a a ) = 0

Thus (y—a)2= '< .*. y—a y%—2ay-2a2=0

2a±V(4a3H-8a I) , . , ,. , ,„v •• y = • " v

2 — - , = a ± a v ' 3 = f l ( l ± V 3 )

.*. From the relation x~-la—y

We have x = £ when y=a

K = 6 ( - l i V 3 ) when y = a ( l ± ^ 3 ) Aas.

Algebra

39. We have t ^ + Z ^ o ^ ^

I . 1 I • at—b' y—a a—b —(2)

From (Q <*- . )= ^ ^ L

x—a+ A2 (ft-jrj

*3

Subtracting A from both sides we have

; ; * - * - * + — 4 1 *=* & ...(3)

.'. From (2) after substituting this value of {x~b) we

b' _ 1 _ 1 ab+a*b—&y—b* ' .y-a a—6

1 A2 1 or

or

_y--<j ab~\-asb—a'-y- b2 a—b 1 _ B flfra-63-fl6a-a86+a8,H-&3

y - a = (a—iXa^+a'd—?>—A3) ; ' J = a*y-a*b ! >>-a ia-b) {ai?'+a*b^a*y—&)~ 1.

or 0--a)(aV-a ,A) = (fl-*)(afc*+fl26-a2>'^6') or a*y*-a*by~Azy+a*b=Qlb'i+a*b+cPy-ab'&-Hbi

-aW+a*by+b* or a V + ? (-r2aa6)+2fl63~6*=0 or aa.)>2-2a%+2a63-6*=0 . __ 2a86±V,(4fl2fra-8fl363+4a36j) ... * - . 2 ^

a ^ i a V C a ^ - 2 ^ + 6 ^ _fcB

a

Examples X (b) (on Page 106)' 1,95

get

Putting these values of y in (3) we get

b (2a- b)

x=-r ^vhen y = — b ' a

Ans. when >>=

40. The two given equations are bx*=lQ&bx+'3a*y ...(1) ayS^lO&ay+lPx ...(2)

From (2) we have x - ^ ^ V ^

Putting this value of x in (1) we get

/ a3y9-30aV&g+300feVj's-H]00&('a3j'5 \ o r V 27^8 )

"C TP ~ )-3^=0 or aV' - 30fl8A">7+300.i364jB -1OOOWy

-90a3&Y+900a3&8y-819a368>>=0

Dividing throughout by a3 and taking y common we

y[>-8-306Y+3006y-10906y+81968]=0 It gives v=0 And /-3062 /+300&Y-10906y+81968==0 Factorizing it by remainder theorem we get {f-b*) (y»-96») {y%~\Zbl) (y*~7b2)=0 :. y~0,y=±b;±3b;±bVM->±i>V7 Substituting these 9 values of y in (3) we get the corres

ponding values of x ie. x=0 when -y=0 x=±a „ y=±b x=±3a ' t> y=±3b

1 Algebra

x=±«v '13 when y=>±by/\l x=£a^7 „ y=±b\/7 Ans. f

41. 2 o ( i - i ) + 4 f l » - l ...(1)

I From (1) 2a (x^~^yi)-\-4a2xy=xy |or 2a <±s-iy*)=xj' (l-4fl8)- ... (3) and from (2) 'i %a*x*-2y*-2a*+axy^0 • - ». (4) Putting y=mx in (3) and (4) we get ]2fl"<x»-m^2)=*'m ( 1 - 4 A 2 )

•r • 2a ( l -m E )=m( l -4a 2 ) .«• (5) and %atx*~2mzx*+axim=2a'1 •»• (6)

From (5) we have i. 2a—2ama=m—4aam or 2am*+m (1—4aa)~2a^2 1 .". m=±2a I

Putting this value of m in (6) we get 8aV—8dV+2oaje»«2a*' - , -or x=z ± 1 .*. from j>«m*, >•= ±2a Ans.

EXAMPLES X (c) (on Page 109)

1J The given three^equations are \ 9x+j ' -8z=0 ' ... (1) \ 4x-8;>+7z=0 ... (2)

yz~\-zx+xy^47 ' ... (3) Solving (i) and (ii) by cross multiplication we have * i, _£_ _ y ,_ 2

1 7-64 - 3 2 - 6 3 —72-4

or! * - y - z ' • - 5 7 - 9 5 - 7 6 i

Examples X (c) (on Page 109)

x=3&, y=*5k, z^Ak *

Substituting these values in (3) we get 20fc3+12£2-f 15^=47 '

or 47/c2=47 .'. A = ± I Thus *=±3,>>=±5, zs=±4 Ans. 2. The three equations are

$x-\-y—2z=0 4jf—>--3z=0

x*+y3+z*=467 From (I) and (2) we have

0 - 3 - 2 - 8 + 9 - 3 - 4 x y z

.*. x=Sk, y= —k,-z=7k

Putting these values in (3) we have I25*3-A3+343A-3=467 or 467A:*=4&7 i.e. k = l

Hence x^5, y= - 1 , z=7 3. JC—_y—z«2

*3:r->>2-z2=22 *v=5

(2) can be written as xz+y*-2xy-z*=22-2xy or ( jc- j ) J t-z '=a,2-2x 5=22-10

(x—v—z) (:c—v+z)=l2 or 2 (x-y+z)=12, from (i) or x—y+z=6

Now from (1) and (4j wc have on addition

Aigebra i

2.T—2y=8 ior x—y=4 Now x~y—4 and xy=5

;, (x+yy=(x-y)*+4xy=36 i.e. *-*-.}'= ±S

Now from x+y=±6 and *—j»=4we have —J x=5, —1;>-=1, — 5;z=2 Ans.

4. The three equations are x+2 .y -z« l l

i x*—4y*-\rz*=3i xz=24

i Subtracting 2xz from (2) **-2xz+z»—4.y*=37-2xz (x-z)« -(2y)3=37-48 = - 1 1

or (x-z-{-2y) (X-z-2y)=-U or 11 {x—is-2y)=—ll

x-z-2y=—A Adding (4) and (1) we get

2 {x—z)=10" or *~z=5 . Now x—z=5 and xz=24 .*. *+z=V{(*-z) a+4*z}=V(25+96)=±ll Nowfrom, ^ c + r = ± n and •*—z=5 we have * =fc, —3 ; z=3, —8 and y=3 [from (1)] 5. x*+y*^#~l\

3xz+3yz—2xy=\% X+y~z*=S • "-

From (3) *+*=*+5 Squaring both sides

*»+j ,+2x.y=z»+ I0r+25 pr (x*+y*-z*)+2xy=310z+25

or, 21+2xy^l0z+25 or xy=2+5z From (2) we have

Examples X (c) (on Page 109)

or 3z(z+5)-(2+5z)=18 or 3z '+5z-22=0

or (3z+11)(z-2)-=0 ie. ==2,-11/3 Hence x+y=lt xy=\2 *hen r = 2

.'. x=4, 3 ,^=3, 4, ==2 4 49 11

Also x+y=-~-1 Xy=*~~ when z = - - ^

3 » ' 3 • z

6. * a4xj;+xz=I8 Z + y z + j w ^ —17 z'+zx+*y=30

Adding all the three equations we have *3+>!z-^3+2(xj>+xz+zy) = I8-12+30=36 or (x+,v+z)a=36 or x+y+z=±6

or y+z={±6-x)

Putting this value in (1) i.e. *M-*CH-*)«18 wehavex2+^(±6—^)=18 ' or x-±6x — x2=18 i or x=±3

This .y+z=* ± 6 - x from (4) -\ JM-z«±6=F3=±3

Also z+x=±6—y (from 4) But from (21 y*+y (z+x)=-l2

or v3+.v (±6->) = ~ I 2 or ± 6 j ' = - 1 2 t .*. y=±2

Similarly z=±5 ,\ x=±3 Thus * « ± 3 , ^ = T 2 , z = ± 5 Ans. 7. The three equations are

*2+2XJ>-|-3.YZ=50

2y*+3yz+yxi?\0 ..-(2) f 3z*+^4-2z7=fl0 ...(3) (Subtracting (3) from (2) we get i 2j>a-3za+J'z-h)>#—zx=0 i or (y~z) (2y+3z-x)~0' Equating each factor to zero we get, i y—z=0'* ,\ y=z i ' 2^+3z=* ' Now substituting these values in (1) i.e.. '|. x*+x (2^+3z)=50 we get ', JC*+**=50 or x*«25 ,\ x = ± 5

Since >•=£ and 2y+3z=x I" . or 2y+3y~~±5 j .".' y~±\,z=*±Ux=±S Ans.

8j The three equations are j ! (v-z) <z+x)=22 ...(1) i (z-i-*) (x~y)=33 ...(2) | (x^y) (y-z)=6 ...(3)

Multiplying all the three equations we get ] [y~zf (z-i-x)« (x-j)*==22x 33x6=66x66 ,' or (y-z) (z+*) (x-.y)=±66 ...(4)

Dividing (4) by (1), {2)t (3) respectively we get ; x - j p - i - S . "l p-z=?±2

z+x-ill Adding these together we get

9.

2*=±I6 /. * = ± 8 .'- ^ = ± 5 , z = ± 3 Aus.

We have , &y*z*u=*\2

x*yxzu2^% i x ^ z V ^ l | 3xy*z2iiB=4

.»(J) ...12) ...0) ...(4)

Multiplying.all the four we have 3 * y z V = 1 2 x 8 x ! x 4

or {xyzuy = i ? 2 i | 2 5 ± = 128=:(2)' •

*yzM=2 • or x2y*z*u*=4 ...(5) Dividing (5) by (i}, (ii), (iii), (iv) respectively we have

u=l, z=£, y=4, x=3 Ans. 10. x?y2z=\2 ...(i)

*3^z3=54 ...(ii) *Vz a =72 ...(iii)

Dividing (ii) by (i) we have

y 12 2 Multiplying (i) and (ii) we get xR ,y* z4=12x54 Dividing (3) by it we have

x _ 72 __ 1 7? 12x54 9

y = T or v=2x

From (i) 4x5z=12 or s5z=3 From (ii)2x4za=54 *4z8=27 '

* s z _ 3 *' %Hl 27

o r z" 9 .'. z=9x

Hence putting the value of y and z we can get the'value of x. and hence the value of y and z.

We get on simplification * = ± l i y = ± 2 , z=r±3 Ans.

202 I' Algebra

| I I . xy+x+y=23 ; ...(1) « + * + z = 4 1 ...(2)

I V 2 - K K + 2 = 2 7 ...(3)

(I) can be written as i xv+x+j>+I=23+l or *0>+l )'+!(>»+1)^24 or ' <*-H) 0'-H)=24 ...(4)

Similarly from (2) we have (x+1) (z-t-l)=42 ...(5)' and from (3) we have 0>-fl) (z+l)=28 [ ...(6)

Multiplying (4), (5), (6) together we get (*+l)3(>-+D2(z+l)a-=24x42x28

-V .(x+\)(y+\)(z+\)*=±m ....(7) Dividing (7) by (4), (5), (6) Respectively we get

z+ l = ±7 .*. z=6, - 8 I y + l = ± 4 /. v = 3 , - 5

x + l = ± 6 .'. x=5t — 7 Ans. 12. 2xF-4x+y=17 -»(1)

! 3yz-j-y-6z«:52 ...(2) 6xz+3z+2x=29 ...(3)

From (1) 2x t y - 2 ) + l (y-2)=15 or (2x+l)(y-2)=15 ...(4) Similarly from (2) and (3) we have

3z(.y-2)-U (y-2}=50 or (3z-M)(y-2)=50 ...(5) And 3z(2*-fl)+I(2*+l)=30 or (3z+l)(Zr+l)=30 ...(6) Multiplying (4), (5), (6) we have

, (3z+l)a (2x+l)a (y-2)3=15x50x30 V. ( ( 3z+ l ) (2x+I ) (y -2 )= I5x lO=±I50

Dividing it by (4), (5), (6) respectively we get ' * = l , - 2 ; > ' = - 3 f + 7 ; z = 3 , - V Ans.

Examples X (c) (on Page 109) 203

13. *z-fj>~7z=0 ...(1) 7z+*-8z=0 ...(2)

x+y+z=l2 ..-(3) Adding (1) and (2) we gel xz+y-lz+yz+x-%z=0 or z(*+yi+iJc+>-)-15z=0 ...(4) From (3) x+^=l2—z

z(12-z) + ( l2 -z ) -15z=0 or z2-f4z-12=0 or (s-2)(z+6)=:0 .'. z==2, —6

X+.F=12-2=10 or x+j»=12+6«18 Subtracting (2) from (1> we get

Z ( X - J O - ( X - J > ) + Z = 0

or ( x - j > ) ( z - l ) = ^ z When z=2 we have (x—y) (2 - l ) = - 2 or x—y= —2 When z=—6 we have (x—>')—£ Now solving x-j-y=\0 or 18

and x—y= — 2 or 6/7 we get .JC=J4, y=6, z=2

60 66 or x = T , ^ T , z = - 6 A n s >

14. The three equations are x+y+z=a ...(1) x2+ya-fza=fls _ ( 2 )

We know that

+3*/z ...(4) and (x+v+z)»=x»+/+z '+2 (xj'+.yz+zx) ...(5 From (5) we have

(«)i=-aa+2 {xy+yz+zx)

204 ' Algebra- -i

or xy+yz+zx=0 ? From. (4) we have '" j ' a3=tf{<ia-0}+3;cyz . ' . ' as==a8-f 3xyz • * '.\ xyz^o

' Thus we may have x=0, jf==bt z=0 Also from (1) x+y=a—z From xy-\-yz+zx=Q „ . We have z (x-\-y)-\-xy~0 or z (a—z)+0=0 /.«. z—0, <* '

I (whenx=0 o r y~0 ,\ xy=0) Similarly we can find but'that ~ ( *=a, 0, 0 ; y=0, at, 0 ; 2=0,0, a Ans.

15. The three equations are j ^ - j - ^ - f z 2 - ^ ##/(1J-i xy+vz+zx=a8 ... (2)

" I, 3 X - J - + Z = A V 3 ». (3) •

We'know that '. (x+v+z)B=x»+va+z1+2 (jty-fcyz-r-zx)

, .*. j (»+^+z)8*=a*+2a3=3aa

or I x+x+z=±a-)/3 or , 7+z==±av '3—x ,„(4)

From (3) we have y—z=3x—a^/3 ,... (5) From (4) and (5) we have

1 y=x, (x-oV3) and! z=(«V3-2x), — 2x Putting these values in (2) and solving for x •

• we have x=—TX-. V3

', ' a

i i

Examples X (c) (on Page 109) 20$

.. 2a a

and x=fl_V3±|V(r£>

o

j . - S V 3 ± V ( - 9 ) „

z = _ V 3 i V t 9 ) f l ^

. 16. We have xa+j'a+z i !=2Ia2 ...(1) -3>Z+2JC—xy=6a2 ....(2) 3x+7-^2z=3a ...(3)

Multiplying (1) by (2) and then subtracting from (1) ive have

*2+?a-f 2wj+za-2z (.v+^)=9a» or (x+^) a+z3-2z (x+^)=9aa

or (*+ji-z)s=9tf2

or x + ^ - z = ± 3 t f ...(4)

Taking +ve sign we have from 4) and (3) on subtraction x+y— z=3a

3x+y-2z=3o z=2x

Putting this value of zrin (1) and (2) we have

or 5*8-f.>>3=2Ia2

and 2xy+2xs—xy=6a8

or 2xz-\-xy=6a*

Mow put y=mx in both of these equations we get 5*2+maxa=2Ifla

or x2 (5+ma) = 21aa

and 2*3+mx2:=6a2

or x2 (2-f m)=6fl3

206 |, Algebra,

\ • * 2 (5+m*)_21q a

5+m* _^7_ -• f"" 2+m 2

or l0-t-2m s«14+7m, or 2ma—7m—4,=0 or (2m+1) ( m - 4 ) = 0 f.e. " m ^ 4 . ^ i Putting. m = 4 we get

jc=tf,y=4fl, z=2a ' Putting m=—\ we get

.' x=-T2d,y=atz=—4a Aus.

( Examples X (d).-(on Page 113) -

In te rmedia te Equat ions

i. We have 3x+8y=l03 or 3x+6y+2y=102+1

Dividing throughout by 3, we get

1 '! * + 2 y + ^ = 3 4 + l

Since x> y are to be integers, we must

2.7—1 • • —$- = i n t e S e r

4 ^ - 2 . A or r-—- =integer

3y+y-2 . " or —L-\—=integer

y—2 i.e. >-H-^y-=iriteger

y—2 i.\ —=- =integer=/7 (say)

Examples X (d) (on Page 113) 207

Now y - 2 = 3 p ; / . y=2p+2 ...(2)

Substituting this value of y in (1), we have *+6>+4+2p-M=34

or *=29-fy> • ....(3)

In these results if we give any integral vaiue to\p'we get corresponding integral Values of x and.y ; but if,p>3, we see from (3) that A: is negative and if/? is a negative integral, y is negative. Thus the only positive integral values of x, • are obtained by putting p=0, I, 2, 3. (values <3) ' Thus if p=0 , 1, 2, 3 ; x=29, 21, 13, 5 ;

' y=*2t 5,8,11 Ans.

• 2.' We have 5.t+2.y=53 or 4*+,x-f-2.y=52-fl

Dividing by 2, we get 2x+(Jf/2)+^=26 + i

or 2x+.y+*^i—26 ..r(I)

Since x, y are integers, we must have X~^26=p{say)- or -x=2p + l ...(2)

Substituting this value of x in (1), 4p'+2+y+P=26

or ;-=24-5/> ...(3) From (I) and (3) fay giving integral" values top, we get

corresponding values of x and.?, but if/>>4, we see from (3) that y is negative ; and if p is negative, integer x is negative. Thus the only positive integral values of x and y are obtained by putting p—0, 1, 2, 3, 4.

Hence we have if J J = 0 , 1, 2,3,4 ; x = l , 3, 5,7,9 ;

r - 2 4 , 19, 14,9,4 Ans.

208 l Algebra i

' 3 . We have 7x+12y=152 I or 7x+7.y-j-5y=147+5 -Dividing throughout by 1, we obtain

; or x + y + 5 ^ ^ = 2 l ...(1)

Since x and y are to be integers,

. . . -™~=mteger

1 ~±—=integer [on multiplying by 3]

I or {2y~~2)+^-1="integer ;

i" , v - i • 1 ,\ •—=— =mteger=p (say) 1 / • .

' te. y="7p+\ ...(2) Substituting this value of y in (1), we have

1 x+7p + l + 5p=>21 or'. x=20-\2p ...(3) From (3) by giving integral values to p, we obtain

corresponding values of x, y ; but if p>\, we see from (3) that x is negative and if p is negative, y is negative. Thus the onlyi positive integral values of xty are obtained by putting p==0, I only

Hence we have if p=0, 1 ;x=2Q,Z;y=\,S An's. 4. '\3x+Uy=4\4

; or n*+2Atf-11y=407 + 7;-Dividing by 11, we get

'+*+n-37+n


Since xt y are integers, . 2 x - 7

• • —«-T— «=integer

rt I2*~42 . . —Ti— = , n t cge r

x—9 or (x—3)+-TJ—=» integer

*—9 I-p=mteger«/J (say)

or x**\\p+9 ...(2)

Substituting this value of x in (I), we obtain

n j > + 9 4 , + 3f t»_ 3 7 or 7=27-13/7 ...(3)

On giving integral values to p, we obtain corresponding . vaiuts of xf y; but ifp>2 we see from (3) thaiy is negative and ifp is negative integer,* is negative. Thus the only positive integral values of x, y are obtained by putting/?—0, 1,2 only "J

Hence if />=0, 1, 2 ; x=9, 20, 31; 7=27, 14, 1 Ans. 5. 23x+25>»=915

Dividing by 23, we get ,

or ,+,+^-» Since x and y are to be integers, wc must have

2v-18' . 2T" M m t c s e r

24v 216 ' l€' -^2y-—=integer (multiplying by 12)

or

or

y—9

. y-9. .

l >»23/»+9 ...(2) .'Substituting this value of y in <1), we have | *+23p-f9+2/>==39 o r *=30-25p.., ..,(3) In results (2), (3) by giving- p any integral values; we

obtain-corresponding values pf*tj>;'but i f>>I , we see from (3) that, xis negative, and if p is a negative integer,^ is negative. Thus the only positive integral values of* andy are obtained by putting /?=0, I only. .

Hence if ;>=0, I; *==30, 5; >>=9, 32. Ans. 6- c 4IA:+47V=2191 Dividing by 41, we have

. ; *+'+?-*+# ..(.> Since*,y are to be integers, we have

J- • ~~AY~ = integer" ie • 42^-126 .

. *~~4j =integer-(multiplying by 7)

° r - ^ ^ ' ^ r V i m e g e r ;

• ••• -4J--mteger=/7 (say) I; ^ .or- . » ^ ;>=3+41/> V Substituting this value of 7 in (I), we get

• ,, , >-+3 + 4IjH-<ip«53t \ °f ^ 'i • *=50-47;> • , , #< {3)

...(2)


By giving any integral values to p, we obtain corresponding values of x, y; but if/) > 1, we see from (3) that * is negative, and Up is a negative integer, y is negative. Thus the only positive integral values of*y y are obtained by putt ingp=0, 1.

Hence if p=0t I ; x = 5 0 , 3 ; y=3, 44. Ans.

7. 5*-7j>=3 or 5x-5y-2y=3 Dividing by 5, we get

1y 3 2.V+3 n ... * _ v _ _ Z = _ _ or x-y—^g-=0 ...(1)

V A: and y are to be integers, we must have 2;-+3 . t — - = integer

6v+9 v + 4 or ~ ~ = integer or 7 + 1 + -F - =integer

V4-4 . .'. ' 5 ' - = in t ege r=0>) say

or y=5p—A ...(2) Substituting this value of y in (1),

j c - 5 p + 4 - 2 / j + I = 0 ; / . x = 7 p - 5 ...(3)

Relations (2), (3) give the general solution in positive integers and least values of x and y are obtained by putting

.'. Least value of x.^^rre respectively * = 7 p - < 2 ; j y = 5 p - 4 , l

8. 6x~I3 .y=l . -Dividing by 6, we obtain

x-2y-^=\ or x~2y-y+- = 0 . . .(I)

Since x and y are to be integers, we must have

^ ~ = i n t e g e r = / > ( s a y ) ;

*.** Aigecra

.-. y=6p-l „.(2) Substituting this value of y in (1), we get

x— 12p+2-p=0 y / . x~l&-2 ...(3)

\ Relations \2) and (3) give general solutions of x, y and values of x andy are obtained by puttingp=\, from which

\ x=il, y=5-\ ,\ x=l3p—2, 11 ; y=6p-l, 5 Ans.

!. 9. 8*-21;y=33 or 8*-16^-5y=32+l ', Dividing by 8, we get ' 5v '"

or *_2, - -^tJ=4 ...(1)

L *i' x and .y are to be integers, then

5 £ t * integer

or - j y ~ ^integer (multiplied by 5) . I s

or 'Y — = 3^-+ f" = integer

y+5 or =~--integer=/r(*ay);

..; ^ = 8 ^ - 5 ...(2)

Substituting this value of y in (1), we obtain *—16/> + 10-5/>+3=0;

or x=2\p-9 ....(3)

.. (2) and (3) give general solutions for ytx and hence least values are x=12, y—3.

.'. Solutions are x=2I/>-9, 12 and v=8p—5, 3 Ans.

10. We have I7,y-13x=0 or 13*+13>>-J 4.y=0

\


Dividing by 13, we get

V x and y are integers, we must have 4v/l3=integer o r fj=integer=g (say)';

/ . y=\3g ..-(2)

Substituting this value of y h (1), we have - * + I 3 ? + 4 ? = 0 or S = 1 7 ? ...(3)

Relations (3) and (2) give general solutions for *,.>»;„ their least values are obtained by putting <7=1,

ie JC=17, y=f=\3 Complete solution is *=17</=I7; ^=13^=13 Ans.

II. 19;>-24x=7 Dividing b^ 19, we get

4x^7_ y X \9 \$\

} * 19 ... (1> .". x and y are to be integers, \.e must have

4x+7 . , ' - y ^ - = integer

Multiplying by 5, we have -20s-1-35

19 •integer

or J C + 1 + * * 1 ^integer •

y 1 1 f.

i e. ___' .=integer=^ (say) ;

or JC*=19?-16 . . . (2)

Substituting this value of x in (I), we obtain V—199 + 16-47 + 3=0 or y=2Zq-\9 ..,(3)

Ai**!, Algebra

i Hence least values of1*, y are'3, 4 (by putting q=\)

\.\ Solutionis^=19^-16=31 ;^=23^-19=4 Ans. j \}1. 77;>-30x=295. Dividing by 30, we get '

' ' 17v 2* v

if 2,-*+i^=9 _ Since x and y are to be integers, we must have

17^—25 30 , =integer

i 13y+25 . , or, y ~p .=mteger

;.\ ^ 3 ^ - = integer;

1 90M-175 . , or i - — Zfr^— «integer . 30

V-f-25 or > 3 ^ + 5 + 3o""" ~ ' n t e S e r '•>

'. y-*-25 .'. ', - 3Q~ = in tege r=g (say) ;

' ,\ l y=30>-25 ...(2) Substituting this value of y in (1), we have

i 6 0 ? - 5 0 - * + 1 7 ? - l 5 = 9 or \ • jc«77$-74 ...(3) From relations (3) and (2), we have leait values of x and

y *8 3, 5 (putting <? = I) ."." Complete solution is x=77</—74=3 ;,y—30£—25=5. 13v Let x and y be the number of horses and cows

respectively then • • ; 37x+23y^752

DividingWougbout by 23, we get


or , + , + ^ - - 8 2 . . . ( ] )

V s and y are to be integers, we must have 14.Y-16 . .

— 23—"integer

70*-80 . ( or — - — = integer

T—11 or 3^—34.^ = integer;

.Y-1I ' , , .*. —2j-=mtcger=(7 (say)

*=23?+l l .-(2)

Substituting this value of x in (1), we get y f 2 3 ? + l l + 14?+6«32

or y=l5-27q ^ ...(3)

By giving any integral values to q we get corresponding values of x and y ; but if? >0, we see from (3) that y is negative and if q is a negative integer, x is negative. Thus the only positive integral values of x and v are obtained by putting q=Q

:, x~\\,y=15

Thus there are II horses and 15 cows. Ans.

14. Let x be the number of shillings and y the number X six-pence ; then

12^+6^=1200 [ v l s. =12 6.1 £ 1 = 240 d. ] or 2x+y=200 ' ...(1) *.* * and y are positive integers and we s*e that the

quation is satisfied by *=9, i)'=2, i.«. 2x99+2=200 ...<2)

r

216', , Algebra i-

!' Subtracting (1) from (2),

2(x-99)+o>-2)=0

tor (y^2)~2 (99-$ or ^ « | i y-2=*2q /. y=2?+2

•'jAnd 99~x=q and x=99-q

j Solutions are obtained by ascribing to q the values 0, 1, 2, 3,4,„99 ; besides, there is one more possibility of being paid tne sum wholly in shillings.

',Thus the,total number of ways is 100+1 = 101 Ans. i

15. Let * be any multiple of 8 ; then 81—* must be a gi x

multiple of 5, le. — - — is integer. ii

Here we shall grve to x such values as (81 — x) always \ would remain a multiple of 5.

*=16,56 other parts'ate 65, 25 Ans.

26. " Let x be the number of guineas and ;• the number of naif-crowns.

Debt is 10* 6d.=126d

.% 252*--30>'=126.v.a) Dividing by 30, we get

: _,+ t o f !£.4+ |

Since x, y are to be positive integers, so 12x-6 . _ - ^ — = i n t e g e r .

Thus clearly *=3 A! from equation (1), >>=21

1 Guinea=2Is.^=252d. \ Crown=5s.=60d. 1 Florin—2s.:=24d*'


Hence the person should pay 3 guineas and receive ( 21 half-crowns. Ans.

17. Let x be the required number ; then J C - 1 6 . . ' , , —^-«mleger=fl (say)

or s=39<z+16 ...(1)

Also —^—=integer=/> (suppose)

or x=56>+27 ...(2) Equating (1) and (2),

56/7+27=399 + 16 or 5 6 p - 3 9 ? = - l l

Dividing by 39, - ? + J + ^ = " - ^

V q and p are positive integers, 17p+H . t

• 39—=integers If we give to p the value 20, we find that 17x20+11

is perfectly divisible by 39. *=56/7+27 x=56x 20+27=1147

Thus an infinite number.of the. form 1147+39 X 56« will be the other number where n is another integer. Ans.

13. Let x be the number to be paid and y the number to be received.

Now debt is £ 1, 6 s. 6d=(240+72+6)=318 d.. Then 24^-30^=318 ...(1)

Dividing it by 24, x - v - ^ = I 3 + ^

Since x and y are +ve integers, 6v + 6 . . - ^ — *=integer

Putting y=3 f (6^+6) is perfectly divisible by 24

218 ; Algebra1;

Substituting ^ = 1 in equation (1), we get x==l7 Hence to discharge the debt he. roust pay 17 florins and

shouldjtake back 3 half-crowns Ans .

19. Let the two parts be * and y ; ihen ji x+y=U6 (1) I x—2

Also —— =integer=*7 (say) ; .*[' *=5<?+2 . . (a) j v_3

and J~~ ^integer-/* (say) ; , ."J ' y=Sp+3 " -(b) > Substituting these values of * and y in (1),

/J ' • 5<?+2+8/>+3=136 , or 5?+8p=131 Dividing throughout by 5, we get

I 9+^+^=26+1

4 ,+,+^=26 ' :>(A) V q, p are positive integers, so

! --^—=mteger

I. 6>-2 . t

or -£.-—=mteger »—2 . .

P+^-y ^integer

or j ^ - = i n t e g e r = 7 (say);.

./. | p=5 /+2 ...(2) Substituting this value in (A),

9+5/+2+3/+l = 26 ' or ' g=*23-8/ ', ,„(3)

i 'i

Examples IX (d) (on Page 113) .219

From (2) and (3) giving values to / = 0 , 1 , 2, we get ^ = 2 , 7 , 1 2 ^ = 2 3 , 1 5 , 7

From (a) and (b), we have x=117,77, 37;>-.= 19, 59,99

A Required parts are (37, 99); (77, 59) ; (117, 19) Ans. 20. Let x, y, z denote the number of rams, pigs and

oxen respectively ; then we have x+y+z=4Q ...(I)

and 4x4-2^+17z=301 ...(2) Eliminating z from these two equations we have

13jc-f-t5^=379 Dividing by 13, we get-

*+'+£-» + £

v—l •'-jy-=integer=<2 (say); y=l3q+l ...(B)

Putting this value of y in (A), "*+13?+l + 2?=29

or *=28—I5q ...(C) " i Substituting these values of x, y in (1), we have • 28-15<?-f-13<z+l+2~40

or 2=11+2? Here q cannot be > 1. Hence giving to q the values less than one i e. 0 , 1 , we

get x=28, 13 ;y=\, 14 ; 2=11, 13 Thus we may buy in two ways : • . (28 rams, 1 pig, 11 oxen) ; (13 rams, 14 pigs, 13 oXeh) 21. Let x, y, z denote the number of sovereigns, haff-

crowns and shillings; then we have ~*x • *-KK+2=27 ...(1)

' Algebra

and 240x+30y+12z=1206 ..-(2) ^ Eliminating z from (1) and (2), we get

228K-H8J>=882

Dividing by 18, we get y+\2x+{\2xl\$)=49 ...(1)

Since x, y are definite integers,- therefore we must have \2x . , -j£-=integer

2x j or —^integer i

! of -r-=integer 1 D •

I I X

I or . x+^-=mteger or y=integer=^ (say) ;

.". x = 3q ...(2) , Substituting this value of x in (1),

y+36q+2q=49 or ^=49-38^ ....(3)

I Putting these values of x and y in (1), 3<?+49-38?fz=27

,or z=35?-22 * ... (4) Here the only -J-ve value of? is 1. < -V *=3 , /—11, z*=I3 /. 3 sovereigns, 11 half-crowns, 13 shillings. Ans.

CHAPTER XI

PERMUTATIONS AND COMBINATIONS

EXAMPLES XI (a) (on Page 122)

1. There are 3 consonants and 4 vowels in the word COURAGE.

.With one consonant we may choose any one of the four vowels. It can be done in 4 ways.

.'. There are three consonants, the total number of ways will be 3 X 4= 12 Ans.

2. Scholarship to 8 candidates can be awarded in 8CI

{ ways=8 ways, to 7 candidates in T d ways=7 ways and to four candidates in iC1 ways=4 ways.

.". Required number of ways=8x7x4=224 ways.

8 x 7 x 6 x 5 x 4 x 3 x 2 3. 8P7= (8-7)1

i ^ S _ ^ ^ l x 2 = 4 0 3 2 0

a«r _ . 2 5 *24x23x22x21 6 ~ "~ (25-5) I

= 52 X 24 X 23 x 22 X 21 =537500 24x23x22x21

^4 1 x 2 x 3 x 4

L H ~ <~5- l x 2 x 3 x 4 x 5 . n . , fin 8 x 7 x 6 x 5 x 4 x 3 x 2 4. Required no.=8P5= (8=5)1

= 8x7x6x5x4=6720 Ans.

222 • Algebra

5. V 4xnP3=5xn-1P ;

or

or

• X (n~3) ! , { ^ 4 f !

(n-3) (n->-4)!~; (n—4)!

4 5* n - 3 1

or 5H-15==4« n —15 Ans.

6. (i) Required number of permutations will be 8P8

ie. 8 ! =40320 Ans. (ii) t is the first letter of each word and e is the last.

Therefore remaining 6 letters can be permuted in 8P6 wa ys.

or in 6 ! ways=720 • Ans.

7. Selection can be done in 6C„ ways

i.e. in jjyjT ways==l5 Ans.

* Different nunibers can be founded in 6P4

(6 -4 ) ! 2 ! 8, aflCj : nC2=44 : 3

=—-=360. Ans,

or »"Cs_44

(2n)_l 2 ! x ( ; i - 2 ) ! 44 o r 3!(2«^3)! n! 3

(2«)(2w-n(2ff-2) (2/1-3)! 2xlx(«~2)! _ 44 w ' 3X2XlX(2n-3)! rtC^iy^T^f'T or 2(2n~I)Xi=44 or ' 2 » - l * U i.e. , »=6 Ans.

Examples XI (a) (on Page J 22) 223

9, Required number=5P5*= - i - l - « - i = 5 ]«i20 Ans

10. Required number=(7-l) !=6 ! = 6 * 5 x 4 x 3 x 2 x 1 = 7 2 0 Ans.

U. (i) Required number of nights='"C/ = 2 4 ! = 24x^3x22x21

41x20 1 4 x 3 x 2 x 1 ^=10626 Ans.

(ii) As one particular watchman is always to be included the remaining, out of 23, can be taken in 23C3 ways.

23 T 21 x22x21 0 r - in ! C T ! w a y s = — 2 - x T - = 1771ways Ans.

12. There are 7 letters in the word "draught", the two vowels are V and'«' Since the vowels are not to be separated so 'aw'may be considered as one letter. Therefore the number of letters will be 6 instead of 7.

The arrangement can be in CP0 ways.

But as the arrangement of the two vowels can be either *ua' or 'au* so the required number of arrangements will ire

2x°P6=2x.720=1440 Ans

13. 5 councillors, oiit of 25, can be chosen in 2SCB ways and 3 oldermen out of 10 can be chosen in 10C3 ways. Since each of the 1st committee caia be nssodated with each of the second committee, the number of combined committees, each consisting of 5 councillors and'three oldermens can be

"Qx^V^jr^V^y^Tl^1062^12^6375600 Ans' 14. (i) It is given that every word begins with a

capital letter so the remaining five letters can be arranged in 5 ! ways. In the beginning any capital

~4ji* , Algebra

letter out of three can be chosen in *CX i.e. in (3) .ways.,

>. Total number of arfangements=3x5 ! = 3 x l 2 0 ' =360 Ans.

(ii) Since each word begins as well a s . ends wi capita] letter.' So the, number, of arrangements

, will be 3 ! or 6; . -

• The remaining four letters can be arranged in 4 1 ways or in 240 ways. '

1 .". Total combined arrangements=6x24=I44 Ans.

(50)!' \ 15. Reqd. number=50CW==-; (46HX(4>!

.50x49x48x47 =230300 4 ' X 3 x 2 * t

16. V "C1 2=BC a •

. n_! n _ . n\ /__

, or !2+8=n '• ie. n=20 Ans.'

20x19x18 1 x 2 x 3

nC1 7=3 0C„-"C8o-i7= a oC3=^^^i-0===n40 Ans.

• and a s C ( 1 = " C 2 0 = « 8 C a = ^ i I = 2 3 1 Ans.

17. , Two vowels are to be placed in 3 odd places, So the remaining 4 consonants in remaining places.

The reqd. number will be = * P a X 4 ! = 6 x 4 x 3 x 2 x l

1 i= 144 Ans. "18. (i) One officer is to be selected by alL means and

the remaining 5 are to be selected out of 8. .*. Reqd. numbeic=8C5x4C,

Examples XI (a) (on Page 122) 225

_8x7x6x4x3x2xl 1X2X3 1x3x2x1

—56x4=224 Ans. (ii) Hence the number of officers may be two, three

or four, but at least one is always to be taken. Hence for it in selecting 6 persons out of 12, the number of arrangements will be "CB=924. It includes those very cases when no officer is chosen i e. all the six may be chosen out of 8. Thus it is to be subtracted from 924.

6 persons but of 8 can be chosen in 8C8 ways - *X7 ~« or in .n-r =28 ways

.*. Reqd. number=924-~28=896 Ans.

19. At least four are to be chosen in a party but they Ajnay be up to 10.

(2) „ „ (3) „ „ (4) „ ., <5> „ „ (6) „ „ (7) „ „

„ five „ it SIX i)

„ seven „ „ eight „ „nine „ >. ten „

.*. (I) a party of four can be ehosen in =10C4 ways —lop — lor- *

« *~ *-* »» —lor* _ lor1

»t — *-8 »>

lor* » •— * - » rt

Adding these all we get the total number of reqd. ways. .'. Reqd. number of ways

="C t+, 0C5+1 0C6+1 0C,+1 ( 'C8+1 0C1+"Cw

On simplification we get=420+252+I20-f-45+H =848 Ans.

20. It is given that "C,="(%+, .'. r-f(r+2) = 18 or r « 8

226 i- / ,; Algebra 'I .', / -, 'I 21. /Two consonants (can be chosen in 25C2 ways and

two vowels can be chosen in SC3 ways. Number of combined groups each containing 2 vowels and -3 consonants will be

-Further since'each of these groups contains, 5 letters which may be arranged in 5 ! ways among themselves. Hence required number-ofwortfs=£5C2X

5C3x5 1=360000 Ans.

/22. 3 latin' books out of 20 can be chosen in 20C3 ways. 2 from 6 Greek books can be chosen in 6C2 ways, group of

. latin books can be combined with groups of Greek books. .'•Thus number of arrangements each containing 3 latin and 2 Greek books are

/ 2 0C3X8C3

J '.' Each of these groups contains 5 books in total, and these five books can be arranged in .' ways among themselves. So the required number of ways=20C3 X

6C2 X 5 ! / '=1140x15x120=205200 Ans.

23. 12 things can be divided among four persons in _ .' 12! ^ ^ 1 2 x l l x l 0 x 9 x 8 x 7 x 6 _ x 4 x 3 1

3 ! x 3 ' x , 3 ! x 3 ! . 3 x 2 x 3 x 2 x 3 x 2 x 3 ! =369600 Ans. 24. Each word is of 6 letters. Capital letter in the

beginning and 3 consonants and two vowels in each word can be done in SCX X 5C3 x *C2 combined ways Further besides the capital, letter there'are i each word 5 more letters and these five letters can be arranged among themselves in 5 ! ways. Hence' the required number of words

=3C1x*C3Xf lC2x5!=2I600 Ans.

'25. .45 men can be alloted to different districts in -45.'

„ 1 0 ! x l S ! x 2 0 ! W a y s

Examples XI (a) (en Page 122) 22?

Y all the groups will be different, the expression not be divided by 3 !

Hence the required number is 45 ! '

101x15 1x20! ' 26. One English out of 3 and four latin books out of

7 can be selected in 3CX and 7C4 ways respectively. Since each of the first group is to be associated w.th each of the second group the number of combined groups will be 3C1XTC4=105. Further the English book is always, to be placed in the middle of 4 latin books, 2 on each side of it. These four books can be arranged in *C3=6 way*. Asjhere are two books on each side which can be arranged in 2 !x2 1 = 4 ways among themselves.

.*. Required number of ways = 105x6x4 = 2520 Ans. 27. There will be 4 men in the boat. Out of the rest

ie, five men, two will work on one side and three on the other. It can be done in 6C8 or ECg ways respectively.

Further four men on each side can be arranged in 4 ! ways. m

:. Reqd. number of ways«5Csx4 !'x4 !=5760 Ans. 28. 4 works can be arranged in 4 ! ways. It is given

that volumes of the same work are not to be separated. .'. Each work consisting of 3 volumes can be arranged

in 3 ! ways and that of 2 volumes in 2 ! ways. .*. Reqd. number of arrangements = 4 !X3 !x3 !x2 !x2 .' = 3456 Ans. 29. 10 examination papers 1( can be arranged in 10.'

ways=3628800 ways. These arrangements include those cases also when the

worst and the best paper come together. It can be so in 9 1 ways=362880 ways.

228; j Algebra i

Position of two papers among ' themselves can be arranged in 2 ! ways.

s '• .".[ Combined number*such that best and worst papers come together=2x362880=725760 i

' A| Reqd. number of ways=3628800—725760 I =2903040 Ans. i .

,1 30. Two persons can only bow on low side and three ican steer only. From the rest i.e. 6 persons 2 will go to bow •side and 4 on the stroke side which can be done in sCa or *C4 =15 ways. Further pn either side 4 persons can be .arranged in 4 I ways and 3 persons on steer can be arranged in 3 ! ways.

/ . reqd number= 15X4 !X4 1x3 I ' | - » * * X 2 4 X 6 = 2 S 9 2 0 A n ,

31. There can be (.P+l) places for V negative signs which', can be placed in p*\Cn ways. (Because there />+ve signs in a raw and Negative signs may be placed in between Xwo +ve signs together with one —ve .to the left and one to ;the right of -(-ve extremes). Proved,

• "Prn - 3°800

" !"' "/w " I - ' -. i° r (50-r ) !x54. ' - 3 0 8 0 ° '

1 56x55x54 !x(50-r)!x(Sl-r)^ fl | 'o r (50 - r ) ' x54 l " 3 0 8 0 °

j.or Simplifying we get (51-r)=10 .". r=41 Ana.

, 23. There are six coloured flags. The signals may be given by I flag at a time in "Cjiways by 2 flags at a time in lCa ways and the two flags may be permuted in 2 1 ways.

Examples XI (b) (on Page 131) 229

.*. Combined of giving signals by two flags = 8 C 3 X 2 !

Similarly by 3, 4, 5, 6 flags at a time separately t h e w a y s a r e 6 C 3 x 3 ! ; flC4x4 ! f «C5X5 ! ; e C 6 x 6 !

.*. Reqd. Number by adding these all

= 6 C 1 + e C i . 2 ! + 8 C 3 x 3 ! + c C 4 x 4 ! + 8 C 6 x 5 f+«c ,x6! = 6+30+120+360 + 720+720=1956 Ans. 34. 88C2r : a lCa ,_4=225 : 11

1SC2, _225 • ' 2*ca r_4 ( i i

(28) !x ' (28-2r) ! (2 r -4 ) !_225 or

or

(2 r ) !x (28-2 r ) ! x 2 4 ! II

2 8 x 2 7 x 2 6 x 2 5 _225 2 r ( 2 r - l ) ( 2 r - 2 ) ( 2 r - 3 i 11

or 2 r ( 2 r - l ) < 2 r - 2 ) (2r -3) = l l X28x 3x26 or r = 7 Ans.

EXAMPLES XI (b) (on P«ge 13O

1. (i) In the word, INDEPENDENCE, there are 12 letters.

Three n, two d, four e.

. . reqd. a r r a n g e m e n t ^ - , — - ^ " . H

^ 1 2 x 1 1 x 1 0 x 9 x 8 x 7 x 6 x 5 (4)! 3xfxix2xiX(4i •!

= 1663200 Ans.

(ii) There are 13 letters in the word SUPERSTITIOUS Three s, two u, two /, two i .', Reqd. arrangement

1 3 ! 3 ! . 2! . 2! . 2!

^. = 129729600 Ans.

•30 i -

''(Hi)

Algebra

We have 12 letters in the \yord INSTITUTIONS •tf which there are— "

Thiee /, two n, two s and (three t \ !,.*. .Required arrangement

I 12!

2. 3 ! 2 ! 2 ! 3]=3326m A n s-

'Out of 17 balls we have 7' black, 6 red, 4 white i Required arrangement i (17) ' ' "

- :L~r^.= 4084080 Ans. ; I (7)!x(6)!x(4)! ,3. Required No of arrangements -will be .

;:4'2Jy73-rr2!Li-TT2-!T2!=151351200 Ans-•4. 2* 3,0, 3, 4, 2, 3 are 7 figures of which two are 2,

three are 3, one 4 and one 0. llTotal No. of arrangements^—,-s-j =420

This number 420 includes those cases also in which zero -puld be at the ten lac's place.

THence these number must be subtracted from the total number of arrangements. These arrangements are

j (2n l ( ! 3n = 6 0 i n n u m b e r ' ' '.'. i Reqd No. of arrangements=*=420—60=360 Ans. 5. !In the word 'ALGEBRA' we have three vowels and

tour consonants. '. The three vowels (two fl's and one e) can be arranged

<r three rways and the four consonants in (4) ! ways as the relative ipositions of vowels and consonants are not to be chaDged.j .

.*. iReqd. arrangements=3x(4) 1=3x3x4x2=72 Ans. ',6. First day journey can be done in 5 ways ',Similarly 2nd day in 5 ways

file:///yord

Examples %I ,b) (on Page 131) 231

Thus the journeys can be done in 5 x 5=25 ways .'. Three journeys in 5 x 5 x 5 = 1 2 5 ways Ans.

7. There are *r counters of different colours and any v>ne of the arrangements can be done in n ways.

.'. Reqd. No. of ways=(n)" Ans. 8. There are 12 animal stalls in total. Ship load of

cow, calf and horse can be made in 3 ways. .\ Reqd. No.=(3)w=53144 Ans. 9. Any one thing can be given in p ways. Also any

one of the remaining things can be given in p ways. Thus the two things can be given in pxp=p2 ways, three things in py.pxp=pz ways and so on

.'. n thing can be given in pn ways Ans. 10. Any one of the things can be given in 2 ways. Also

any one of the remaining in two ways. Thus two things can be arranged in 2 x 2 = 2 2 ways. Similarly three things in 3a

and so on. Hence 5 things will be arranged in (2)5 ways. It includes those very two classes in which either may receive none. ^

/ . Reqd. number = 2 5 - 2 = 3 2 - 2 = 3 0 Ans.

11. a3 gives three a's, b* gives b's and c1 gives four c's when written at full length. Total No. of letters is

3 f 2 + 4 = 9 9 !

.'. Required No. of arcangements= -r~-,-r-r-t = 1260 Ans.

12 1st ring can be attempted in 15 ways 2 rings can be attempted in 15x 15 = (15)a ways 3 r ings, , „ „ „ 1 5 x l 5 x l 5 = U 5 ) 3 ways

Only one attempt will be successful .'. Reqd. No. of one successful attempt

= ( I 5 ) 3 - l = 3 3 7 5 - f r = 3 3 7 4 Ans.

lit- I Algebra

16.' Reqd. No. of triangles=15C3=455 (V| a A'niay be obtained by joining any three points

out of 15). 14. Total No. of books==«+2$+3c-rf - ' » A xr r * (a+2Z>+3c — rf)!

15.. Any 7 number can fill one thousands place in 7 Ci=7 ways. In hundreds place for every number it can be filled in 8 ways.

.*. Total No. of fillings 100's p l a c e = 7 x 8 = 5 6 ways „ „ „ ten's „ = 7 x 8 x 8 = 4 4 8 ways „ „ „ units „ = 7 x 8 x 8 x 8

=3584 ways Total No. of ways required

= 7 + 5 6 + 4 4 8 + 3584=4095 Ans. 16. 1st classical prize can be given in 20 ways

2nd „ „ „ „ „ 19 ways 1st science „ ,, „ ,, 20 ways 2nd „ „ , , „ „ 19 ways 1 st French „ „ „ „ 20 ways

• '. 2nd „ „ „ „ „ 19 ways All the prizes can be given in

( 2 0 x l 9 ) x ( 2 0 x l 9 ) X ( 2 0 x l 9 1 ways or in 57760000 ways Ans. 17. Signals by 1st arm can be made in 4 Q ways=4 ways

since any one of the four positions may be taken by the arm. 2nd arm can also make signals in-4 ways.

.'. Both, 1st and 2nd arm, can give signals in 4 x 4 = 1 6 ways and so on.

. Hence 5 arms can give signals in 4 x 4 x 4 x 4 X 4 = 1 0 2 4 ways

There will be one position of rest .*. Reqd. No. of ways=1024- l = 1023 Ans.

Examples XL (b) (on Page 131) 233

18. One person can sit any where. Rest 6 persons can sit in (6) ! ways

Hence ring can be formed in (6)! ways

= 6 x 5 x 4 x 3 x 2 x 1 = 7 2 0 ways.' 7 English men can sit in such a way that one seat may

be left vacant between each two and this can be done in (6) I ways

Now at vacant seats 7 Americans can sit in (7) ! ways .*. Reqd. No. of wnys=(6) !x(7) ! = 3628800 Ans.

19. The required number of ways, in which the coins of 7 kinds, by taking all or some of them, can be arranged

= ( 2 / - I = 128-1 = 127 Ans. 20. 3 coconuts can be selected in

3C1+3C2x3C3 = 3 + 3 + l = 7 ways 4 apples can be selected in

dC1-^C2-t-'C3-MC4=4+6+4-r-l=15 ways 2 oranges can be selected in '

zC1 + : !C a=2+l=3 ways

Hence total No. of reqd. ways=7x 15x3=315 Ans. 21 Directly by § 147 of the book, the reqd. numb^

__ (mn) 1 (7Hl,n(n) ! Ans.

22. (i) There are four flags, by taking one of them we can make signals in four ways. By taking any of the four flags we can make signals in 4pt=l2 ways

Similarly with three flags in 4C3=24 ways and with four flags in *P4=»24 ways .*. Total number of signals=4+ 12+24+24=64 (ii) As above here total number of signals will be

5+20+60 + 120+120=325 Ans.. 23. Letters s, s ; e, e ; rti are of four different kinds.

134 Algebra

Selection can be as— (i) AJ1 the three different (ii) Two alike, one different (i) It can be done in 4CS=4 ways !- (!V 3 different letters are to be chosen out of four) (ii) Selection can be done in 6 way's. Now; for finding the arrangements of three one must

permute iri all possible ways the above mentioned groups (i) and iii)

from (i) we get 4x3 1=24 arrangements

•from (ii) we get 6 Xs-'=18 „ •

A jTotal No. of arrangements== 18 + 24-42 Ans. '24. (i) As q points alone are in one st. line so °C2 lines

can be formed by joining pairs of points. Similarly if there bep points alone in one st. line so pCa lines can be formed.

-But'there is only one line. ' •'.*, I Reqd. No. of st. lines formed=I'C2>;0C3

* i -P<P-U y ? ( g - l ) j . , I 2 2 T 1

(ii) j If there had been no three points in one st. line,

we would' have obtained t r i ang les= r C 3 =—^y^xV" But as q points are in one sU line.

,T ,\ I No. of cases when there will be no triangle ii i __ q(q-i)(.q-2)

3 , 1 x 2 x 3 Hence reqd. No. of &s " ,

_P(P-})(p-2)_q (g-\) (q-2) 1 x 2 x 3 " 1 x 2 x 3

rap(p-J) (p-q) __q(q~\) i(£-2) 16 ,6 AnS,

Examples XI (b) (on Page 131) 235

25. No. of planes each containing three points out of

e will be *C,= P("-l)2

(P;2)

» Since q points are in one plane. The planes aC3 can be

obtained by joining any three points. It' gives only one plane.

.*. Reqd. No. of pIane's=*C3—*C3 + 1 ^P(P~l)ip-2) „<?(<?-!? Q?-2)

1 . 2 . 3 1 . 2 . 3 ' + 1

=PiP^P-2)_q(q-inq-2^ { A n S i

26. N books are of different kinds.-' Each book has got its p copies Hence one book can be disposed in (p-}-l) ways

Similarly 2nd book can be disposed in (p+1) ways and so on. It means ali the n books may be disposed of in (p + l) (p + \) {p+l) to n factors-={p + lJn wa>s.

" It includes that case where none of the books are taken. This case should be rejected.

.*. Total No. of ways reqd.=(p + I)B —I A n s * 27. (i) e, e ; s, s ; x ; p, r, i, ot n are eight letters in all.

For finding group of ten we may have them as (1) Two alike, two like of other kind (2) Two alike, two unlike (3) All four of different kind.

Now (i) selection can be in 2Ca=>l way (ii) Selection can be in 2x 7 C 2 =2x2I=42 ways (Since we select one pair out of two pairs and two frorr

the remaining 1) (iiij The relation can be in 8C4=70 ways

(V we select 4 out of 8 letters) Total No. of selections=l+42+70=113 Ans

236 Algebra ,! i i

(ii), In finding arrangements of four letters we have to permute in all possible ways each of the above mentioned groups, J

' I . 4 ' 1 (i)l Gives rise to 1 x 2 iX2 ' =6

! • 4 i ' ' (ii) Gives.rise' to 4 2 x 2 T = 5 0 4 (iii) Gives rise,to 70x4 ! = 1680

' ,'. ( Total number of arrangements

', 6+504 + 1680=2190 Ans. i

" 28.; There are IHetters in the word EXAMINATION For doing selection of four, these may be classified as

' (i)' J Two alike, two others alike (ii) * Two alike, two of other different kinds (iii) ; All the four'of different kinds. ' Fori (i) selection can be done in 3C2 ways=3 "Fori(ii) selection can be done in 3x7C3 ways=63 ;;For''(iii) selection can be .done in 8C4 ways=70

^ For doing permutation of 4 letters we have to permute in all possible ways each of the above mentioned groups.

I " ' ' 4 l (i) gives rise tc 3x 2 i X 2 ! ~ = 1?

| • 41 (ii) gives rise to 63 X -*1,- = 756

• i

(iii) , gives rise to 70x4 !=1680 .'. Total No. of permutations ;• ' =18+756+1680=2454 Ans, 29. At the units place any number will occur as many

times as 4 !! 1.

/ . Sum of the digits occurring at units place , = ( l+3+5+7+9)X4 M600

Examples XI (,b) (on Page 131) 237

Some will be the sum of different digits at ten's place. Similarly hundred's place, thousand's place.

Total sum = 600 + 6000 + 00000 +600000+6000000 = 600 (1+10+100 + 1000+10000) = 600 (11111) = 6666600 Ans.

- 30. Since all the numbers formed are greater than 10000.

.*. Zero can not occur at ten thousand's place.

S urn of all the number occur ring at units place is 2 4 x 0 + ( 2 4 ~ 6 i (2+4 + 6 + 8)=360

Sum of numbers occurring at ten thousands place is 2 4 ( 2 + 4 + 6 + 8)=480

'/. Reqd. sum = 360+3600+36000 + 360000 + 4800000 = 5199960 Ans.

3 . . Out of P things we can select at a time 0, i, 2, 3,...P things. Hence P things can be grouped in (p + I) ways.

Similarly q ihings iv (f/+l) ways and remaining r may be grouped in (2)r ways because one thing may be rejected or selected. Neglecting the case when'all the things may be rejected we have the total no. of ways

= ( p + 0 (</+!)- ( 2 / - 1 Ans. 32. Total number oftf's and fi's is In. Let the a's are

r, then Z>'s will be {2n—r) in number. .*. Reqd. number of permutations

= _ < * ' > J — = * - c r\X{2n~r)l r

2"C, can only be greatest

when r = ( y ) = r t '

i.e. o's and r and fe's will be (2n—r) = n :. a 's=6's Ans.

238 I! " ;• Algebra Jj

i! 33. 0,1,2, 3, ...m can be selected out of m in (m-\-{) ways, each of which will give a different factor of am, Dis-. similar letters maybe chosen in two ways either taken or V rejected. It can be in 2n» ways. There will be one case when all may be rejected.

• .*.' Reqd. No. of ways 1 «(m+I) . 2B-1 Ans.

CHAPTER XII *

MATHEMATICAL INDUCTION

EXAMPLES XII (on Page 135)

1. Let 1+3+5+ + (2n-l)=n 2 when the numbei of terms is n.

Adding («+l)th term which is (2«+l)to bothsideswi get

1+3 + 5+ + (2n-l )+(2«+l)=« 2 + (2n+l) or 1+3 + 5+ + (2fl+l)=(«+l}2

It is also of the same form as the result we have suppose< to be true. Thus the result is univeisally true.

Proved 2. Supposing the result to be true for n then

6 Adding (n + lj2 to both sides we get

la + 2 a+3' + +rt3+(rt+i>»

.JL(«±U(25.+ l) + (M + 1)...

=2_M-3/ia + « + 6H3+ 12«+' 6

^_2^3+9«'+13rt+j 6

_(n + l)(n + 2)(2n + 3) • 6

' It is of the form as we have assumed the result. Henc the assumed result is true for («+I) also. It will be tru for (n+2) also. Hence the result is universally true.

Proved

240 \' 1 ' Algebra | 1

'3. Xet the given result be true for'/i then 2+22 + 23+ M-2"=2 (2"—1) ...(1)

Adding to both sides (n + flth term i e.t 2n+I we, get

'•• 2+22-f-23 + :+2n + 2n+1==2.(2n-l)+2n+1. ; =2n+1 —2+2fl+1

= 2 . 2"*1—2 :; i = 2 ( 2 r + l - i )

It is of the form 2 (2n—1) i.e. 'as the result we have assumed to be true for.«. Thus it is truej for (n + l) also. It will be true for (w+2) also and soon. Hence the result is true for all', +ve values.

Proved. 4.' Let the result is true for n terms. Then

I 1 r 1 , I , , n + i—*+ 5—^ + n terms=—-. I 1 . 2 2 . 3 T 3 . 4 ' t i " ' ° n + l

Adding (ti + ljth term ic. —,, ,-, , , _. to both sides (n+Ij (n+2)

* lTT'2"f2~~3+3~4 + ' " t 0 ^ n + 1 ^ t e r m ?

n , 1 n + l (n +1) '(/i4-2)

__ »2+2«+l (n+l) (n + 2)

(n + l)"(ii + 2) n + 2

N It is of the form as ^T—-. i.e. as the result we have i N + l

assumed true.for n terms. Thus this result is true fqr (n + l) terms also. It will be true for (n+2) terms also.

Hence the result is universally true. Proved. ,

Examples' XII (on Page 135J* 241

5. By the actual division of (xn—yn) by (x+y) , we have

x+y x+y

Now if — - ~ - is divisible by (x+y) ilien xn—yn is x-\-y also divisible by (x+y) But y*+xy is divisible by (x+y) Therefore x*y+'y* is also divisible by (x+y) and so on.

Hence the result is true. Proved,

CHAPTER XIII

BINOMIAL THEOREM

(Positive Integral • Index)

EXAMPLES XIII (a) (on Page 14a)

i. We know that (x+a)n=xn+nC1x

n-1 . a + nC2xn~* - a2

-j-nC3xn-3.a3 + ... + a" -In (x—3)5 we have a=—3, « = 5 .*. Putting a=—3, n=5 in the above result we have

(x--3)B=^6-6C1x4. 3 + 5C2x3 . (3)a-5C3x

9(3)3

+ 5C4x(3)^(3)5

=xB-15x44-90x3-270;t2+405x,-243 Ans.^ 2. As above here putting x=3x,

a=2y and n=4 in (x+a)" we get (3x+2y)^(ix)^iCl [3xy U.v)+4Ca(3x)2 {2y)*

+4C3 (3x) (2J>) 8+(2J0\

^8l*4+2]fo3j+2]6;r2/+96xj3+l<y'4 Ans. 3. ( 2 X . - J O ^ ( 2 * ) H 8 Q (2*)4 (-v)+5C3 (2«)> (-y)»

+ «C8 (2x)a(-^)3+ec4(2«) ( - ^ - f (~j>)a

or {2x-y)s=32x*-%0xiy+Z0x*ys-40xzy5+l0xyi-y5

Ans. 4.. ( l*3aV=(l j 6 + e C I (-3aa)-feC2(-3fl2)a+6C3(-3^}3

+eC4 (_3a*)«+*Qi (-3OB)« + (-3II»)« ' ' «l-18fl2+135fl4-540(3

6+12'15fl8--1455aloH-729aia

Ans. 5. (xH*)6=(*a)5+5Ci («2)V)+BC2(*

a)3(*)2

f6C3(x2)2(x)3-hEC4(*

2) (*)4+5CB(x)W-«x10-f-52&-H0x8-H0s7-i-5x3-j-x5 Ans.

Examples XIII (a) (on Page 142) 243

6. a-JTjO'-cnN-'Ct (-*jo+*ci(-,jtjo*

+7C(-^)s+7C^-.v.)')7

= l -7x j>+21*y- 35^V+35xV ~21jr\yB-f7.*y i+*T Ans.

7. (2'3U=(2)H1CJ . (2j ' (-^)

+*c3 (2)« (-~y+4c3(2) (-?^y+*ct(( - ^ y

= I 6 - 4 8 ^ + 5 4 . x J - 2 7 x 8 + ^ - Ans. ^

8. (3a-|)«=(3a)S-9ClOti)»(^+«Ca(3fl)M)2

-6C3(3fl)3(#)3+6C4(3«7)2{^ -6C6(3alU)B + 8C6(|)*

=729fla-972aB+540fl*-160d3

, 80 , 64a , 64 + T ° ^ 2 T + 7 2 9 AnS*

9. (l+x/2)7«l+7C1(x/2)+-Ca(x/2)8-t-7C3(X/2)3 -+7C4(W2)i+7Cs^/2)"+7C8(x/2)«+7C^v^)'

, , lx, 21xa , 35x3j 35x* 2U« 'TJC6, * T " A = 1 + T + ^ - + -T + " i r + " 3 T + 64 +F28 Ans*

+MH5(-fj^(HKy

„64xB_32x* 20x» n , 135_243 729 ~"729~ ~27~+ 3 z u + 4 x » Sx*"*" 64*". Ans*

I |l Algebra ", ! I !

111.1 (j +</)e=a)8+8c1 $y (fl)+«Ca a)8 (a?

' + 8 C, (J) («)7+8C8 (a)8

'•• 1 , a , 7a2 7a3 35^ 256 16^ 16 + 4 T 8

12.

+10c

= 1 j

+las+JeP+4a7+aB Ans.

+ioc<-Ty+ioc=(-T)3+i°c«(-F)4

H)^(-y+^(-T)'

^ - i - ^ - I ^ P , 210_252 210_220

t * 8 xa^x" Ans*

13. For general term we have, the general term

TV+J^CJC"- ' ' - . ar in the expansion of (*+a)° .*. In the expansion of (JC—5)13 4th term by putting r=3 , ri=13,fl= —5 in the above formula we have T4=13C3 (x)10 (—5)3= —357 50*10 Ans. 14. In'(1-2*)" wehavex^l , a=(-2x), n=I2 For 10th term put r=9t we get T10= l2C8(l)3(-2x)9=-112640*9 , Ans. 15. Here x*=2x, a=- 1, H = 1 3 For itsjl2th term putting r = l l in T r + 1 = n C> n - r . ar


we have T,g= u C u (-1)11 (2x)2=-13C, (4x*) =-312.*2 Ans.

16. 28th term of (5*-i-8;>)30 following the above method we have

=30C27 (5*)3 (8>>)"

^ T T ^ I T T <5*)3 * <*»" Ans. 17. Following the same method, the 4th term of

(u/3-f9fc)10is T4= i aC3 (c/3)7 {9b)3=40a7b3 Ans. 18. 5th term in the expansion of (2a—A/3)8 is "

/4x 5 \B

19. 7th term in the expansion of f -=-—•— J is

^ ( 4 £ > ! H i y , or J!_-Ji! 5° _^ 5a_IO5O0

. *" 6 53 " 2e . xG ~ x3 xs '

20. T 5 = ° C ^ - l 7 ^ {- b 3 i ^ = ^ - Ans.

21. ( x - f V ^ H ^ - ^ ) 4

. - J ' + V , *3 - ( ^ 2 ) + ^ JC" . W2>Z+4C3 x .

+ x'-*Clx*. (v '2)+4CaJC2(V2)E-4C3* .(V2)3+(V'2)1

=2**+2 4Ca.\-a (V2)3-f2 (v'2) l=2 (**+I2*1+4) Ans.

22. {•(**-«*)+*}•-{••(*•-vfl8)-*}5

put V(-ts—a2)=^ we get 0-+Jf)5-(^-x)B«(>-8+fiC1j?*JC+eCj;V+6Ca/.x

a

+3C4)'J?4f^5)-(>'5-6C1v4JC+5C2^

3Jta-5C3;'2A-3

246 : Algebra

= 2 [tCyx+sCzyW+x*] 1 = 2 [5x^+10/x34-*5l

=2 [5* (x2-<?2)a+10:c3 (x^-a*)-)-**] =2* [5 (x2-aa)24-10xa tf—a^+x1-]

• =2* [5*a-10x2a2+5a*4-10jc4-10xV+.v*] = 2* (16s*-2(h;2fl2+5rt4j Ans.

23. ( v /2+l) G - (V2-6) 3

==E(-/2)H6C1 (VV-MQ (v/2)4+?C3 (-/2)HCC4 (A/2)2

+cC5(v,2)+sG6(V2)(']

- - K ^ ) 6 - 6 ^ (V2J5+CC2 (v/2)a-°C3 (V2)3

+&C4 (V2)2-6C5 (V2)+<>C6 (V2?l = 2 PQ ( v ^ ) 5 ^ ^ ) 3 * ^ (A/2)] = [6(V2)f i+20(V2)3+6(V2)] =2 [24^/2+40 V24-6y'2]=2x70V2=140V2 Ans.

24. p - y c i - ^ p + ^ - fVf l -^ ) ] 8

put VU —*)=? We get (2-^B+(2-f^°=[28-DC l (2)= (^°)+aC3 (2)*. yz

- e C 3 (2f . + 3 Q (2)2 OO*-0CE (2) M +6Q., j,.] J - P ^ ' Q (2J5 O0+BC* (2)V+6C3(2)3(>')3+BC4 (2)s (>-)4

•+6C6(2) W + 6C f l/J . = 2 po+'C, (2)V+GC4 (2)2 (y*)+*C^*] = 2 [64+15. I6 .7M-I5 . 4 . y 4 + j 6 ] = 2 [64x240 ( i --x)+60(l-x) s+(l-x)»J Simplifying it further we get = 2 [365—363X+63X3-JC3] Ans.

25. In the expansion of (tf/.v-f x/o')10, the middle term

will be j ~ =6th term z i

/. Ta = 10C6 (*/*)• (*/<7)s = 10C8=252 Ans.

144-2 26: Here —r- =8th terra will be the middle term


•• T,.=»c,(iy.(-f/—™*» ^ 27. Let A;18 comes in the (r-f 1) th term of the expansion

J {*2+f} ,S' then

T,+1 = "Cr (*')»-' . [~j

We want the term involving x18. Here power of x is (30-3r)

.-. 30—3r=18 or r=»4 Hence A18 occurs in the 4 + l = 5th term

.*. T5= l sC4(3i)4 = llQ565a4 ^ Ans. 28. Let xls occurs in the (#+ l)th term of (ax4, —bx)s

Then T r+1=DC r. {ax*)*-'. (bx)r

~"C,{ay-r .(x)*«-*'.'. b" .x" = 9Cr\d)9--'. (^J36-3'. b"

For x18 we must have 36—3r=l$ or ~=6 Hence x1B occurs-.in 7th term T7=9Cfl ( - l ) f l 6°a3=84fl368 Ans. 29. Let x-17 and *32 occurs in the (r+\)th term

* {- -ft" Then T^=^C r f>)'5"' . (~ ^J

=»crx*a-*>{-iy. v3r Xs

= ( _ i ) - - . i»c„ . j ' 0 - " Now for x32 we must have 60—7r=32 or r—4 and for sr17 we must have 60—7r=—17 or r = II

2481 I'I Algebra !

I Hence coefficients of x38 and x^17 are obtained by putting r=4 land i l respectively

icoeff. ofx38=(-l)4(16C4)=1365 ,coeff. of x - " « ( - l ) u (1 8CU)=-1365 ' Ans.

9+1 ;30. As here «=9 the middle terms will be -T— th

and ( ~ - + 1 V h i.e. 5th and 6th terms.

f. T^WKW- Ans. i . - *^ ( - £ ) ' - ^ . Ans 31. Let x does not occur iff the (r+ljth term

(WJ Then V l =»C,(2 , .y- ' . ( - 3 l ) '

i='^'(•§•)" •w** - : ( - f ) '

must have For the fffm independent of x (ie. involving x°) we

Ans.

__ 18—3r=0 or r=*6 .'. Hence the 7th term is independent of x

and > -C (|)° ( - l ) S ^ / i v8

32. In the expansion of 19* —«—,— 1 we haye

the 13th term==18Cl2 (?*)« ( ~ 3 ^ )*' - » C M < 9 * ) ° X - ^ « C 1 S = 1 8 5 6 4 A n S t

Examples XIII (a) («n Page 142) 24

33. Let xr occurs in the O + lJth terra of (x-\—J

then

For JC" we must have n—2p=r

or P ^

.*. Reqd. coefficient of * r="C„-, 2~

n ! Ans.

34. Let term independent of x (having x°) in i\

( 1 V x—- 1 is (r-f l)th term.

/. T r f l =t o C r ( x ) ' 1 ^ ( - J r

13 y = ^ C ; ( - n V ' , - ^ ' for *

equating 3«—3rto zero *e have 3n—3r=0 " r=rt

A Reqd. coefficient=8"C,i(-l)n

<• i»« ' 3 " ) ' ==(-,) VfixW! An

Then

35, Let x* occurs in the (r-f-l)th term of fxa-f—Y

iju i i. Aigeora . • ' ' ! ' • : < • ' ,

1 l 'j 4«— p for x* we must have An— 3r=p *-e- r 5—

.\ tReqd. coefficient «2wCj,,.lp

:' j ' 3

" ! •" 2tf ! ,

, EXAMPLES XIII (b) (on Page 147)

T H - i - 3 6 - r + l y 4 • •r r 11

:' ! _ 3 1 - r v 4 _ _ I 2 4 - 4 r :

Ans.

i r 11 l l r

If T U i > T r t h e n l 2 4 - - 4 r > l l r or i;24>15r .'. Greatest value of r can be 8 •Thusj the greatest term in the gi^m expansion will be

?th. It's value when x = l l and y~=4 is

,m I = :ll'^»°CB(A)^22T^r!X:*8-nM AnS' 1 We can write ( 2 x - 3;>)28' as (2*)!8 (1 - | . J-/*)a8

H e r ' T r t l _ 2 8 - r + l 3 4 ,

: ' _ 2 9 - r 2 _ 5 8 - 2 ( ' r ' 3 3r

If Tr+i>T r t h e n 5 8 - 2 r > 3 r ^ O A 58>5r

' !< W e s t value from it of r can fee 11. Hence the great-*Li ra the given expansion will be 12th. Ans.

s / b X1* 3. (2«+Z>)14 cad be written as (la)]* [}+-^J

Examples XHI (b) (on Page 14 /) 251

Now ^ = H = L + i x 4 = 7 S - ^ T r + i>T r then 75 — 5r>8r or 75>13/-

From it greatest value of r#can be 5 Hence the greatest term in the given expansion will be

6th. Ans. 4. We can write-(3-F2*)1* as (3)1S (I+£*)'*•'

Here ^ i = l A ± t l X A x A when *»5/2

80-5r 3r

IF- T r + 1 >T r then80-5r>3r or 80>8r Here we see that for all values of r up to 9 we have

T,+l>TV But when r=10 then T r + 1=T r and these are greatest. Hence the greatest term in the expansion are two 10th and 11th which are numerically equal. . . 5. (l+*)n when.r=-|, n = 6

„ , , TH-I n - r + 1 6- r - f l 2 14-2r We have -'-±1= —•—- . x= -±- . ^ - = ^ 5 —

If ? ^ ? > t or T r M >T r then 14-2r>3r *

or 14>5r. From it greatest value of r can be 2. Thus the greatest term is'3rd. Its value is

= e C a ( l ) 4 . ( £ ) 2 = Y = 6 | . Ans. 6. (a+x)n when x=£, «=9, a=l

We can write (<*+*)" as a" ( l - f — J

„ere %-*=*!. 4 _9-r+l 2 .=^^l f

r ' 3 3r Now when T,+1>T r then 20-2/->3r or 20>5r

252 i Algebra ! i "'

Hence for all values of r up to 3 we have T r + l>T r but when r=4, TH . l=T r and these are greatest; Hence the grea- y, test terms are two 4th and • 5th and are numerically equal. Their value is •

';. !<i)' • 8C4 (i)*-(|)«x J - i | ^ | x l , - . i Ans.

':, 7.1 In (l-f*)an, the middle term will be («+l)thterm 1 A Tn+l=»nCn (*)« .

2ft ' Coefficient of *n is 2 nC,= , , ...(A)

i n ixnl

In the expansion of (l-t-#)*B+l, the middle terms will be two1 nth and (n+l)th term.

and T ^ - ^ C (*)n

Sum of the coefficients of xn and x"-1. ««"-3Gfl-1+

8,1=1Cfi=SftC,, ' ...(B) ' (A);and (B) are the same. Hence Proved. 8. We know that

^•f-fl)n=*0+nCiXn- ia+f,C2xn '8as-r-nC3xn-3aa+ * ', +"€,*"-V-J- +a*

We can write it as

(x+a)"=(;cp+BCain-2aHnQxn-V+ ) -MBC1Jca-1a+uC3x

n-3a°+ ) =(M-B) (say) „.{{)

Again,similarly we will have ,(*— a)*=[x*+°Ci&-*a*+ )

- t»CJ*»-»a+»C^»-'-a»+ ) = {A-B) (ii)

Multiplying (i) and (ii) we get ( x + ^ - a ^ C A - r - B ) (A-B)

or ( * 2 - a V = A 8 - B 3 . Proved. .

Examples XIII (b) (on Page 147) 253

= (*B+"C2*n-* a*+ )a

-( f lC, x"-1 a+"CznCax

n-* cP+ )a Proved.

9. In the expansion of (x+y)n we have

2nd lerm=nC1 xn~x y=n x"^1 y

3rd term = "Ca »--* . « B J ^ p H *"-* . j - s

4th term="C3xn-3 . j>*=* 3-j • *n~3 . yz

From the given condition we have

nxn-l.y*=24Q ».(i)

«(n—1) (n—2) - , 3 1 r ton ,..., __-i ' i i .xn-3 _ -[,3—1080 ...(Hi)

'4-^-3 - ( A ) Dividing (ii) by (i) and (iii) by (ii) we get

" - I y _o 2 ' x

_ - . - - y ...(B)

Now dividing (B) by (A) we have

HzLx~-l~=2 o r 3 n - 3 = 4 n - * 2 rt—2

Solving for n we have «=5

Now from (A) 2yfx=3 or y=3x}2.

From(i) 5 . * 4 . y « 2 4 0

i.e. ^ = 3 2 of x=2 3* 3x2

Now from y=-£ w e n^ve .}>= —=— ~ 3

Hence x='2Jy—3, n=5 Ans. 10. For expansion <1 +2X-X2)* put 1 +2x=>'

254 \ Algebra .

Thus (y-x*)*~y* + *(^y\-x2)+*Ciy*<t-x*)2+iCM-xi)i + + *C4(-xa)«

= (l+2xy-iqix2 Cl+2^)3+1C2x* (l+2x)a

-4C3x° a+2je>+*C4;c8

=(14-4 . 2*+6 . 4s3+4 . 8x34-16.r*)-4.v2

(l-4-6x4-]2x2-h8x3J+6iV4 (I+4x+4xa) - 4 x 8 (l+2x) + x8

Simplifying it further we have ( l+2x-x 2 )* = l+8x+20;c2+8x3-26.*4-8x5+26xs

-8x 7 +a 8 Ans. 11. Regarding (3.v3—lax) .as a single term .we have

(3x*-2ax+ 3a3)3^ {(3*2*- 2a x) + 3a2}3

= (-3x3—2^)3-f3C1(3x3-2ax)2(3fl2)+3Ca(3,v2-2ax)(3fl2)2

+3C3 (3a2)3

= 27xs—54a.tB+36<22x4-8a3.v3+9a2 (9a;t4-12tfx3-f4a2x2) .*--f-27fl4 (3*a-2ax) + 27fl6

Simplifying it further we get =^27.tfl ~54flxG-|-117aV-U6a5Jc3+'l 17ff*Jt2~54fl5x

_ ** +27ae Ans. 12. rth lerm in any expansion of power n from the end

will be («4-2—r)th term from the beginning /. T m . ^ ,= n C„ + 1 . r x / - 1 Xfl n + 1 " '

« ! ° ~ t e - r + l j | ( r - - l T l ^ - 1 - « " - r + ' An*.

13. From the end (p+2)th term will be the same as (2/r-}-3— p—2)th term from the beginning.

= 2 n + 1 C 2 n _ I , X x :+" X ( - 1 /x)an-»

= 1--XJ-L: / ^ n j n - p v2p-2fli. l

. (2«-p) ! ( l+p ) ! ( ! ) ' * ' "

H_l)p _ ? P ± i L U > « • « Ans'

be ^ ~ = (n+l) th terra

Examples XIII (b) (on Page 147) 21

14. In the expansion of (1 -fx)43

T2H-1=43C3,x2 ' ' and T^^C^x'+i

.'. coefficient of x2r is equal to that of xr+1

4 S P I B P • y->2,— ^ H - I

or 2 r - f r + I = 43 i.e. r = 1 4 Ans.

15. In (l-j-x)n we have T 3 = 2 n C 3 r _ 1 . *»'-i and T H * = M C r + l * '+ 1

As given we have

or 3r—l'- | - r+l=2n or « = 2 r Ans.

16. In the" expansion of (l+3cj*nthe middle term u

. T _a n r „„_ (2/0 !

= 1 - 2 . 3 . 4 . 5 . 6 ( 2 f i - 2 ) ( 2 n - l ) . _ 2 « n !XH !

It can be written as _ > . 3 . 5 . 7 . 9 (2/2-1) 3 . 4 . 6...2H> •

l « + i - •: ST • n ! X *

1 . 3 . 5 . 7 . 9 ( 2 n - l ) v 2 n f l . 2 . 3./....«) v

ft I n I

= 1 _ . 3 . 5 . . - . ( 2 H - 1 ) H . l x > . « ! n\

J ^ l ^ ^ S n - z l L . 2«*» Proved. n !

17. Substituting the corresponding Values of C1} H have Cx + «C„+ aC a+*C i + f lC f l

„_L? « ! («- ! ) . ! . 3/i ( « - I ) ( n - 2 ) , =n+2 . — Y \ — + 31 r + «

61; \ ; • Algebra '.

';. U+^>+^»M + +, -• r n ( I + ( i r i ) + ( - l ^ ^ ) + ,j It is = * [1 + !]"->=/(. 2"-1 Proved.

: have

is. c0 + T-fT + - j — - — ^ Taking L. H. S. and putting the corresponding values ive,

= 1 -L-g.-i.-^fr1-1* , n(n-lHn—2) 1 -• 1 " t " 2 i r - 3 (2 if"1" 4(3!) "*" "*~ n-f-1

- ! '' TfB-i'n'i "•("+'> 4."( | ,- ' '>'0»+l) , n nTr «B+1>+ - r r - + — 3 i — + 1]

• ^ J ^ .[it is clearly (l + l)^1]

I9.-&+J&+S+ 3 -* - 0 * - l * - 2 , ' - ' f t - l

We have =**--=- =n

' C . . 2 1 1 ( 1 1 - 1 ) , „

^cr—inr-^"15

•' 3Ca^3«(n-l)Cn~2) 2! - _ c3, 3i X ^ 7 ^ i r - ^ * 2 '

*.<UIIUIMUB A m \D) (.on rage 147) 257

Adding these all we get

c « + Q + C i + + c U " = " + ( « - l ) + ( « - 2 ) + + 3 + 2 + 1 = 1+2+3+4+ !+(«-2) (fl-i) (ll)

^ y ^ + I ) Proved.

20. We have. C0+C1 & 1 + * which can be written as 1 - ' t 1 +g> . a 3 Q(I+«)

Similarly

C 1 + C 2 = « + ^ i ^ i ) Q ' K « ± l ) = = C l («+i) 2 2 2 ""

(V C t=n)

2! + 31 ^~

^^T^xf^+l^c^/z+i) 2' u ; r~

Multiplying these all together wc have ' • «Vf-Q) ( Q + C X Q + Q ) (C^+Cn)

~^x^x<i£±!!x ^ ^ ^CpdQCaC, C,_, . («+!)«

1 . 2 . 3 «

258 I' ., Algebra*,

• ,. / V C0=Crt=l)1 Ana:. 21. We have , ,, .„

J 2 C + ^ 4 - ^ + ? & r ' i 2"*1C«

5 ^ | ^ j ^ ^ ' : 2 « + +**]

-^2^S^

^2.- The expansion of fll^*)* is- : .. •

We&iowthat f G ^ . G ^ , ^ / *"• •:;> ,> -o?fCfl j'Ga^Gn-j-, GaP='Gn-a etc. fieHceWegeV , -. f \ t . , - • ,<• -

. Multiplying (1) add',(2):and' collecting* ^coefficient: of x* onty from" tlie product* we get'

•Now in ( l + x ) 2 * ^ occurs in (it'+l)th'teto , '-%, Tnn=?nGmi.-x$\iie.: coefficient-of*n'is-^G,,;' Equating it to (3) we get

Examples XII} (b) (on Page 147)# 259

CoHC 1«+Q'+. . .Cw»= 2 f lCM=~-^ 1 Ans.

23. We have a§ in Q. No. 22 (l+^r=C0+C1*+Csx

2+.. ;C rx, '4-...Cnxn ...(1)

or {\+xr=Cn+C^1x+C^txs+...Cn-tx'+...-+Cfix

n - (2) Multiplying (I) and (2) and collecting the coefficient of

*n+ ' from the product we get * ( C o a + Q C ^ r f C2C,+ s+.. .+C,_, . Q.) * t ,+r+ , But in (1+x)*n, xn+'' occurs in the («+r+l)th term.

.'. Tn+H.I==2"Ctl+rxn+''

Hence the coefficient of xn+r is a nC n^ Equating the coefficients of xn+r we have CflCf-f QC^i+CjCrtjf . . . +CB- r . Gfl=

2f,Ctl+r

<2n) 1 A n s

<iH-r)!x(Ji-r)l *

! CHAPTER XII

BINOMIAL THEOREM (any Index)

EXAMPLES XIV (a) (on Page 155)

1. We know that

Putting *^=i we have

2. Putting n « | in (A) we have

ii+xr'-iU'+^^r^* <4"1] j 1 — * *+••••. ' = 1 +|jC+|.xi_1Vxa Ans.

3L Putting x=»—* and n=$in(A)we have on simplification (l-*)* ,6«l-^*«—A*--*!*'* Ans.

^ 4i Putting *=jcj'and n=—2 in (A) we get

• + ( - 2 ) ( ^ l ) ( - 2 - 2 ) ( ; c a ) 3 + ^

1 . L W , • * * * • "2! 3 ! T""

«—I-2*"+3x 4 -4* < " Ans, 5. Putting JC=—3* and n= J in (A) we get

i^Hi^>(-3, ) a + , 3!

Examples XIV (a) (on Page 155)

=*1— x-~xl—5/3 JC3 Ans. 7. Putting x = — 3* and n= —\ in (A) we get

(-3x)-»*=l-H-|) ( - 3 J e ) + n i ^ l = U (-3*)*

= l + » + 2 x 8 + y x3+ ... Ans.

7. Putting x=2x and n= — J in (A) we get

= I -*+|x"-4x 8+ ... Ans. 5. Putting x=x}3 and n= — 3 in (A) we get

, - 3 ( -3-1) ( -3-2) / * \»

- l - . + y ^ - g j M - . . . A n s .

9. Putting * = ( y ) and «=-§ in (A) we get

« 1 + J C + T " 5 4 + ; " Ans.

10. (1 +k)-*=l +(-*) <«/2> + — " j r ? 1 ^ ' - ^ 2 )

11. (2+A:)-3 can be written as

rar"(1+fr-*.(1+-ff Now putting .'c=*/2 anc?/i= —3 in formula (A) we get

+ - 3 ( -3 -1 ) ( - 3 - 2 ) / ^

Ti V f) + -]

Ans.

= 4 [l^#*+**a-4*3+.-.J Ans. 12. Taking 9 common vye get 91'2 (1 +|jr)wa

Now putting x=-£;c and w=4 in (A) we get

13. (8+l2ay'3*=(8)a's(l+4<0*'a

^ t t - 0 « - 2 ) V 3 f l y , ^ -|

-4[l + . - ^ + - 2 ] An, 14. Taking 9 common we get

- ( 9 - 6 J T 3 ' 2 = ( 9 ) - - ; " 2 U—fc*r"WiWT (I—SxV8" Now expanding we get

' «A-[l+*+^+K*8+».l Ans.

Examples XIV (a) (on Page 155) 263

15. •(4a-8x)r*,««(4fl)-1'1 | ~ 1 - —l"1'*

H_(-*)(-£~-l)(-&-2) / - 2x y (

V * L " tf 2 ' a% "** 2 " a 8 + - J ADS.

16. We know that in the expansion of(l+x)", (r+l)th term is T,+1="C, . fJ1 (n~U \n~2) <»-*)-l*-r±l) ( j t),

-'. Sth term of (1 + 2*)-1'8 by putting x=2x and n = - J , r = 7

^( - i ) ( - i -0 ( - i -2 ) (-4-3)...(-j-7+l> „

- ( " 1 > ? ^ - n L 2 3 3 5 4 7 7 : 6 3 7 W ^

Jfi" x Ans.

17. 1 Ith term of (I -Ix*)*1'* will be T » - y ( V - » (V-2) rV-3>...(V- 10+j) ( _ J ) 1 0

( ^ W ** 1 0 . 9 . 8 . 7 . 6. 5 . 4T37T-1 <2> "**">

S S * " Ans. 18. 10th term of (\ + 3a2)18; * T30=

1"»CB (3a")«

>4 i' . ' Algebra ",

'• I - 1 6 . 1 3 . 1 0 ' . r , 4 . l . 2_j_._i 38. , 18.

J M . j P y t P I8 - 1040, ,B

" 19. 5th term of (3d—2b)-} will be

• ' t3a)» <*« [ ^ T l J

243*7* I 0 ° [_ 4 X 3 x 2 x l J 243d5 Ans.

20. '(r+Uth term of ( l - * y * will be

' T f + 1 - -*C v ( -x)* . , « ( L. i)- ( ^ - 2 (-3) (-4) ( -5) . , . ( - 2-j-r+l)

I T * *

. i* I Ans.

21: ( r + I)th term of (1 —jc)-1 will be .

= f - W ( ; 0 - 4 (-S)(-6)(-7).,.(-4+r+t) *» r •* ^ * r ! • 6

4 . 5 . 6 . 7 r(M-l)

rl -( ' - i v * ' 1 . 2 ,>3 . 4 ..5. 6 .7,..f. (r+1) (r+3)(r+3) ^ - l ) . x . ^ v T . 2 . 3 x r l

(r iH)fr+2)(r+3>Xi ' ! '. i 3 1 x r !

Examples XIV (a) (on Page 155) 265

(r+I)(f+2)(r+3) or". 31

_ (r+l)(r+2)|r+3) ^ x , 6 * Ans,

22. (r-fl)th term of (l-f*) i ;a will be

T r + l =» ! C r (*) ' _* ( i - n ft-2) ( j -31 ( W + l )

r !

, J ( - j H - f l - f j - i - H ) r !

<*r , - — . (*r

-1.3.5...(2r-3^ 1 u 2 ' x r ! • Ans.

23. (r-M)th term of (1 +x)«'3 will be

TH . ,« u"C r(x) ' '

- . V ( \ r a " l ) ( V - 2 ) ( V -3 ) . . . fV- r + l) , . ,

„ V - f r . i . i . ( - * ) ( - i ) . . . ( y - H - l ) -r ! • *

(-l)"-4 • 11.8.5.2.1.4..,(3r-14) , y . r I ' * Ans.

24. 14th term of (210-27*) is 'a will be"

T„=13 ; 'C13 (210)"'2-" (-27*)13

=(210)-13" (-27*)13 . 1J/aC13 •* *=(2)-" (-1)18 (201) (*)". ""C1 3

=(2)M . (—l),a (x13) . Vt V — !)('•? — 2)...(1T

a — 134-1)

- m « f rt»f«»> 13.11.9.7.5.3.1.1.3.5.7.9.U

.7.6.5.4.3.2.1

*6<f, ; Algebra

' i =(2)» (*») (~1 ) . 2 ^ ^ « - 2 » . . 11.7.3. *»

\ = - 8 . 11 . 7 . 3 . *«=—J848* Ans.

25.1 (384-64x)11" can be written a$ ' r" 2**1*1" (3")lu-«[l+^-j

•Now. 7th terro==(38)1I/4 .[ l i ; aCa ; ( | ? * ) J .

' =(3)»a. p««Ca] (4)" • (**) : ' â[y(v-i)(y^îâi)] | i .^

« _ ? ^ 11 -7 .3 .Kj^9 A 19712 ^ 32 ' 1 . 2 . 3 . 4 . 5, 6'; 3 Ans.

(EXAMPLES XIV (b)-(o*"?og« s6i) Formula :—',

(r+l)th term of (1-f x)a is given W ,

• TW^-C (*)' . .,.(A)

1. Putting n*=~\ we get (r-f l)tfc term of (1 +*)-Va

T -<-*X-4) (-•« -izt^±l>-y r\

^ j . 3 . 5 . 7 , g r - l ) ' (2)''. f I Ans.

2. (r+l)th terra of (1— x)-* by putting * = - * and n=i—5 ia .

bAaiiij/iba A I T yvj you rage 101; jLOf

* ' r ! __ t.2.3.4.5.6.7.8-...r.(r+l)...(f+4) r ~ + ' IXITxTl '* _f l ( i -+U(f+2)( f+3)( r4-4) .

4»Xi-I ' _ ( r+ l ) ( r+2 ) (r+3)(f+4)«;

24 Ans. 3. T r + 1of(l+3x)1 / 3

_ i ( * - I ) ( i - 2 ) - f ^ - T + l )

~< SJ * ! , X Ans. 4 T (-D("IX-T) (-W+1) .

2.5.8.11 (3r - l ) =*<—!) (s) • -j — . v Mis.

5. (/•+l)thtermof(l+.v2)-3

Muliiply above and below by 2 ! ( - 1 / . 1.2.3.4.5 ( r+l ) ( r+2) v

r ! x 2 ! * « / ,«- r_j_(r+l)(f+2)

'• Algebra ,1

6. T,+ iof(l-2x)-3 ' !wiHbe T . (-1? ( - 4 - D ( - 4 - 2 ) . . : . . . ( - f - r + j )

• ' 1 3.5.7.,....(2r+l) y .

= i ^ ^ ± l ) An, •

7* (r+l)th term of {a+bx)-1 will be . *

- - ^ (fat/ . -=l-^i2> ("3) t ( -4) . ( - 1 - f r - I )

o1 H-1)"--^-*' Ans-8- (r+l)th term of (2-*) - ' will be -*3C r(2)- '- '(-x)r „ ' _ I r m - n ( - 2 ) ( - 3 ) ( - 4 ) . . . ( - 2 - r + l )

(2)*+' l M } : v\ e j _ L . , m ( _ 1)2, t xr m 1.2,3.4 r . (r+1)

l2)"+3 v V '~ • r l , i l

=2r+r*r-(r+l) Ans.

9. (r+l)th term of ^>[(a3-*3)*] '• 'or of (a3-*3)2 '3 will be

Ld*-* (-1) ' . xw. at4--im-2)...(i-H-i)

L L f - « w aar 2,1,4... [3r-5)

= 2.1 .4 ( 3 r - 5 ) x*i y^r\ ' fl»'-i A n s '

Ia vc iW = = ( , + 2 J C ) " 1 / a

( r+ l ) th term of {\ + 2x)~^2 will be

T^J-*H-$).;....(--Wtl> (2jc/ ' = f _ i y 1.3 5 . . . . U / - - 0 j r „,

_(_lru^^-l)^ Ans

. t , 2.5.8 ( 3 r - l ) . ^ 1J ' 3 ' x r ! ' J ' * 2.5.8 ( 3 r ~ l ) , .

= —• ' xr. Ans.

12. ' =ffl"—njc^-u"

—.('-Sf r J(-4X4-'X-T-a>-K-^Q]

(-30' = L f ( - 0 ^ 1 . («+0 (2«+l) + ...(wr+l-ii)> «^£ '

a I • " if. r! 3 ' u"

te (/_H-1)(2«-H) { B f r -1) +1} jf_ • r ! ' t f n ' + i

\ 1

27ff. Algefra

v 13, ( l + x ^ w h e n x ^ ' . T , + 1 _ n + r - l

.« . „ 7 + r - l 4^.6-J-r 4_ \ r 15 r " 15

' .\ T r H<T J . i

• /;. 24+4r>15r or 24>l l r

_24+4r 15r

• Hence for all values up to two we have T r+1>TF

The greatest value of r consists with this is 2. J

,\ ' Greatest term is 3rd. Ans,

14 .. tm.w-r+I ,

.'. for the given expression we • T V ^ y - M - i „ v - i

T r • r r 23-2r

, 3r If TH-i>t r then 23—2r>3r or

-

have

—- • -s ; when x=%

2-3>5r The greatest value from it can be 4. Thus the greatest

term df the given expression is 5th. ' Ans.

- j r 7 • *

"TT the given expression (I—7*)~n'4t we have

28r+49 ~ 32r .

If T , t l > t , then 28r+49>32r OF 49>4r From it' we have 12 as the greatest value of r. , Hence

the greatest terra is 13th. Ans.

16-. - p * + 5 / j n = 2 » . * » f ( l + ^ "

Examples XIV (b) (on Page 161) 271

Tr+1 n - r + l 12—r+1 fSy\

1 2 - r + l / 5 x 3 \ , 0 -

_13-~r I5_195-15r. r 16** 16r IQ'S—15r

If T r41>T f t h e n ^ ; ^ > 1

or 195-15r>16 or 195>31r

From it we have 6 as the greatest value of r. Hence the greatest term is 7th. • Ans.

17; ( $ - 4 i ) - ' = 5 - 7 ( 1 - f . x ) - 7 - ^ il-ixy-i

-J? =—— —. x numerically

- ^ r - ! <-W

«7±£.ll („* . J) when A-^J

5r If T r + 1>T, then 12+2r> 5r or 12>3r

Hence for all values of/- up to 3 TV,|>T„ but Tr+1—Tr

when r=>4 and these are the greatest terms. Hence there will' be two greatest terms 4tH and 5th which are numerically equal and'greater than others. Ans.

4y3yn 18. l3jE"+j^--i3jra)-*(l ,+ ^ J

; . T A i - 5 + r r J . *? numerically T, r 3x2

Cli. I Algebra

_ 1 5 + r - j 4x8 .. ; r *3x8l ' '

when n=\5, x=9, v=-2 __448+32r ,

| ' 243r : . , If T r+1>T rthe448+32r>243r ', oir 448>243r--32>- or 448>2IIr

From it the greatest value of r can be 2. Thus the greatest term will be 3rd. ' Ans:'

19. V 9 8 = ( 1 0 0 - 2 ^ - a = l o ( l - A 2 y "

IU L 2 10*+ "2!^ I lTV + '""'J f l - i - i 2L+ I • 10

= ~Io~T*To3 + <*

-io-i--f+ ; = 10—-i—'0005 — =9'89949'tofive places of decimal. Ans.

;20. ^'(998)=(1000-2)1 '3=s(l03-2) l 'a

Expanding binomially we have

•-10V1-I l—±. >.-•+. 1

., 3 102 9 1 0 5 ^ " " " .'

examples x iv (\y) (on Page 161) 273

= 10-1 J . _ ± ^ 3 '100 9 " IOOOC(T

- t n _ ^ ? '000004-3 " 9

= 10--00666--000004. = 10--006664=9"993336 =9-99333 to five places of decimal. Ans.

21. Wehave^/(1003)=(1000f3)1,3-(103+3)u8

I+ioooy - , 0 (

= 10L1+iooo~io6oFoo+ "'J

- 1 0 + I O O - 1 0 0 W + - = 1 0 0 0 9 9 9 Ans.

22. •Wehave^<2400H(240i-l)1"

-^""'O-HOT)"1

/ 1 \ll* / 1 V'*

Expanding we have

- 7 * X 1 - 3 X l 4-

- 7 - ^ v » -1.4285 4 7 •

and l/72=-020407 =6'99927 to five places of decimal Ans. 23. We can write A - / 1 2 8)

a s ( !28)"1/a

Algebra

or; <128h*'»«(125+3H2,8='<SM-3)-i'» 1, i / 3 V 1 ' 9

» 5 V 3 1251 2!- . 6 + ' "J

5L M 2 5 + 5 6 + ""J !5 5*- 5' T «/2--0qi6+-0Q00256«-19842 Ans.

1 ^ ) can be written " ( l + ^ J j

^ 3 250^ 2 ! ^250^ + * " ; ' 4 I I 1 / ,4 \ 2 '

= 1 + T^4^<25Or"~TV4xi50j + *" . . , 4_ '_ _1_ 16

j rali"3 ,T*Too5 9 'looqoob' "• ] , , '004.; :OOOOI6 ;* I - i + - T i i 9— - ;: ^

I = l + *0O1333—000001 = 1-00133

To five places of decimal • ! • Ans.

|25. (630)^=(625+5>*«(5f+5)"4

- * ( 1 + F ) " '•!

JT Ll+T*3» 32 'y+fJ i'

Examples XIV (b) (on Page 161) 211

- l a + -73—3^»-. s4-+ . . .

= 125+ 75--00075 = 125*74925

But (630)-3'4 l l

(630)3'1 125*74925 =•00795 Ans.

26. We have ^312S=(3125+3)"8

f 3 \1 / 5

= 5 + 5 3 r + - " 5 + T ^ + -

=5+-OO096=5<00096 -Ans.

27. Expanding each binomially we have (l-7jr)" J(l+2x)- '«

t 23 . . 2 3 e l - T ^ " = 1 - r Ans.

28. Expanding each binomially we have

1 < , I

276 Algebra

;29. We have r , An 9.

(8+3x) , , ,1 (••+¥)'

• i /a

; -1*2

T( I + T X »~TX I + T ) , (neglecting terms of higher powers of fy

', 4 ~~ 2 "*" S.1"3 "8 "p . Ans.

I, 30; We have '

'. J _ [4+x)""

; ; _CI+f)>Cl'+T*)lir

HfO-^X'+^.-OO-^-)}

Examples XIV (b) (on Page 16V) 277

=-Lri_3*_-7 U~l=l/ ' i l0x\ 4 |_ "8 24 J 4 V 3 J 1 5x

T " 6 * Ans '

o-H"4+o-i*r

2_103

„,17je V2 20 * J ^ + W " '""30

= . _#I7x _ 103 103x17x? 60 40 * + ~40x~60

Algebra

, 17x.JG3 y ! l ~ 6 0 " 40"

(neglecting other terms of higher powers of x)

343 !l

• „ 1 ^ ( 8 + 3 * ) - - ^ « - * ) _ '•!

(I-rf*)"B-l-(4+f) ; *

_ 2(l+>)"3-(l--y)^_ij

„ _ 2(1+4. |y)-(l- j«). i

"(l+-5.W+2(l + f . ^ ) (neglecting terms of higher powers of x)

33. iletx* occurs .in the (r+l)th term of (l-4x)~1'*

Examples XIV (b) (on Page 161) 27?

_ 1-3.5 (2r- i) . „

Multiplying above and below by r ! we have

- ' " 3 - 5 ( 2 r - 1 ) . (2x)r. (1.2.3.4 5 r) , {fly

, 1.2-3.4.5. (2r -Q (M)2 • * • f2-4-6 ^

g=1.2,3.4.5......(2r--i)

4. Reqd. coefficient is j—.[• Ans.

. 34. Taking R. H. S. we have

2 LV i+* + ~ T 2 ~ U + V + J

==2V (-4~^==(l+*)n . Hence Proved.

35- ^ehave iiT3ow+4^=t i+3ei:8( i+4^ i /8

- ( 1 - 2 J C + 3 X , + . . . ) (1-2*4-—* C ~*~- . I f i ^ + . A

^ 1 -2*-f3s ,-2*+4*!!4-6;ra

(neglecting higher powers of x) . *=(l-4x-H3*«) Ans. '

' 280 . Algebra

36. We'have I l ± ^ ± V ( l ± 5 x )

, , 5 , 2V,2 ) „. \

;,' (14 2x+3x»+...)

' % • (I+2»+3*«)

- ( 2 + " * - "iy x a ) (l"+2xf 3^)

I ; (neglecting higher powers of x) ., \ • - . I3x 103 „ , . , 13 » , , , " ' ^ I' = 2 + 4 — x--f4x-f~jxVr6x3

\ •' V , 29 ^ , 297. 2 A i i;.-f=2+-j x +-32-*2 Ans. : • , -

-^. Ijfljbe expansion of (1— ,r)-°, /rth coefficient will occur in nth" term and (n-l)th., coefficient will qctiur inv

(H— l)th terra;

• ' . .• . T r t =- f l c f ] _ 1 (^ ) n - 1

1 • - « ( - » M ) i ( - n - 2 ) . ( - « 4 3 ) . . :

., L f - 1)*» * ( « + l H « + 2)(ii+3)+...<2n~2)

-.'.. ' nttilcoefficient is 1 /I ,»,.," (n+n(«-+2)-(2»-2) V

or

Examples XIV (c) (on Page 167) 28

n ( n + l ) ( n 4 - 2 ) (2n-2)

Also (/i—l)th coefficient will be

n ( « + 1 ) ( I I + 2 I . . ! . . . ( 2 H - 3 )

(n -2 ) !

nth coefficient (n-r-l)th coefficient

n(«+l)(/ i-f2). . . (2fl-2)

= ( « - D 1 ]_ n(«-H) ffi+2)...(2a—3I

_ f « - 2 ) ! x ( 2 n - 2 ) ( i i - 0 !

_ 2 ( / i - 2 ) ! ( » - ! ) (« - i ) (« -2 )r s 2 Proved-

EXAMPLES X I V (c) (on Page 167)

• ' " ( T ^ - ( 3 - S * ) ( l - . ) -

= (l-5x) (\ + 2x+3x*+4xS+ ^-IOOJC"

+ 101*"°+... On multiplying coefficient of x100 will be obtained b

multiplying 100, 101 by 3 and—5 respectively Adding-w. get - 5 0 0 + 3 0 3 = - 1 9 7 Ans.

2. We toe 4-^~f ~ t 4 + 2 a - o * ) ( l . + g ) ^

= ( 4 - f . 2 a - a a ) ( l - 3 a + . . . r 1 0 ^ 9 + r n f l " + ...rfla'', On multiplying these two factors we see that coefficien

ofa12 will be obtained from the product of P10, P n , PIZ will — 1, 2, 4 respectively.

. T « ( « - l ) ( « - 2 ) («~3). ( « - r + l ) „ . . lr+i=* _ p[— . a

Algebra

„ f - 3 ) i ( - 4 ) ( - 5 ) . i , . . . < - ^ M : i l

i * , - ,1

A p^c-jrx^^ee : ••(

M ' i

r i , '• ' ;'

A. Reqd. coefficient^—Ix66-f 2 (^7g).f4x9f I" I, =-66-156+354^142 Ans.

, ' „T ^ •' 3* ' -2 3 > - 2 ' I '• 3. We have —r-„ "-TT-I—L I *+*a *,'.(] -f x) . l /3x 1 -2V/"l \ „ -!

'; =(3x-2fx) (l-je+**-J«*+„.)V

Multiplying these two!; factors ancjjufcjifojy j&rjnsjtovnt' g A" together we'have the required coefftc,cnt

: 4(-i^1<3-2)=(-i)-> ;: Anfc

4. Coefficient of xn in the expansion 0f '' •2-fx+«' - , '


We have i±f+£!=(2-f-x+x») <!+*->)

=(2 +*+*«) (1 -3*+6x»+.. . +r„_2 . *"-' -+'*-i*n-'+'**n-f- )

Coefficient of *" will be obtained by multiplying 2,x and x2 of 1st factor to r„s°f rn^. s"~l and rn_8x*" respectively.

Collecting these together we have the coefficient of x" which is=(2 . r„+rn_ l+r„- l> ....(i)

Now w<-3><-4"~j> (-»-"+»,.

, (-l)w(n+1)(n+2) ° = 2! *

, ..( , i y - t a ( i ' + ' >

and , _ , . ( _ , » - . & = » « 2!

Substituting these values in (i) we get the required coefficient.

Simplifying it further we get __( 1 ) n (2H '+4W+4)

«(- l )"(« ,+2n+2) Ans. 5. Wc can write

1 _ 1 Lx 1 - 3 ±S-l^l l J . 1-3-5-7 1 2 * "2 "^14 " 2* 2.4.6 ' 2* i"2^4.6.8 * 2 * " ' "

: Algebra

I, 'i,.C-l) ( - I - D t - H * ) / i V +

6. l,+Hi-*+^+... i

We can write it as

a=(lpl)'3 '3-='(i)r8/3=st2)8 '3=(2!l)1 'a=v'8 A™-7. Takingl.R.H.S. we get |

, , f . n .J i ( i i+ l ) ,H(Bf l ) t« '+2) 1 I 1 3 3.6~^ 3Xr; '1"*"i

• i2"{l+(-n)(^i)+-n^|-1).;(^^ I, I' ; - n ( - n - l ) ( - n - 2 ) ,.3

j ^ 1.2.3 ! •' *} + ' "

• «2 f l { l -J}- n=2B [i\-n=2n [4]n=3n. ...(A) i

Now taking L. H. S. we get

,<=(llr3)-" = Q)-'>=3n . .:.(B) j •

We see that (A) is equal to (B) . \< Hence proved. • ,'

Examples XIV (c) (on Page 167) • 285

8. We can write L. H. S. as

. n ( n r l ) ( n - 2 ) ,a 1 + 1*7273 ( T ) + ' " j

which is clearly the expansion of 7" (1-K)" .". L . H . S . * 7 " ( 1 + 4 ) B ^ 7 B ( T ) " = 8 " ...(I)

R. H. S. can be written as

=4" {l-J}-»=4° . (J)-"=4"X2"=8" ...(2)

From (1) and (2) we see that L. H. S.«R. H. S.

Hence

' t 1 + 7 + 7 .14 + 7 . 1 4 . 2 1 ^ J

-4»{ l+$+"# + -} • Proved.

9. We can write

' • K ' - 4 0 " ( ' - T " ) "

286 '* ! Algebra '

•• :• 0 - f * + g * - - )

1 ' / / • ! 9 , 243 " A

A ^ 9 113 A / , i : 9 „ 243 a \

l + T 128 • " * 8

1 256

64 ** + 243 256

Proved. 1:10. Let«,be the integral part and m be the fractional

part of (5+2^6)" then ,' ( 'i ' n+m=(5 r+ n C, ( 5 T 1 . (2y'6)+f,Ca (5)n-8(2-v/6)s+.-(l)

Let'w's=(5-2V^6)W :

J" J =(5)h-»'fcs (5)*-' (2y 6)+"jC, <5)«-« (2v/6)2- . . . ( l ]

; Adding (1) and (2) we have jj

,7H-ro-fm'=2 { 5 " + " ^ (5)n72 (2V6)2+...}

;] —2 {all +ve terms} , ! ! =an even integer ' i:But

one.' m and m' are proper fractions, their sum must be

Hence n=an even integer —(m+m') ..'. . n=anodd integer. > ,

Examples XIV (c) (on Page 167) *

11 As above let n denotes the integral part and denotes the fractional part of (8+3 V7)" t n e n

n+m=(8+3-v/7)n

»8m+"C! (8)"-1 (3V7)+nCa (8)""; (3v'7)a + . . . Since (8—3V7) is positive and less than unity. Therefore (8 —3^/7)* will be a proper fraction. Let it be m'. Then -

m'=8n~nC1 (Hf-1 (3V7j-HnC2 (8)"-= (2V/7J2-...

Adding (A) and (BJ we get « + » + m ' « 2 {(8)*-(-"Ca (8;"-3(3V7)2+...}

*=2 {all -fveterms} = aneven integer

Y m and m' are proper fractions. Their sum will be .*. n=an even integer -(«i'-fm)

= an odd integer. Ans.

12. Coefficient of** in the expansion of (1—2x+3*a-4*5+. —n

• i we have l-2x+3x ,-4**+

2!

.'. <l-2x4-3x a-4x8+.. .)-n=[(l+s:)-2]^=(l+x)-2 E

In the expansion of (1+*)-*°, (w + l)th term will g the coefficient of xn.

.'. Reqd. coefficient^2"^ (V T„+1=2'iCM. „ (2«M

(ti) !x(n) ! * Ans. f 1 \ 4 «

J 3. Middle term of. f x+ ~- J will be

•—J— th<*(2n+l)'th term from the beginning

288 !',' ' Algebra' Tj

TSM+i.=<nCa-(x)fln-»n..fl7i)2n j

Coefficient of xa in the expansion of {(1— 4x)-<n+i/») w m be given by the (ji-f--i)th' term. ' i

.v TL^-VV^Cii (-4*)" * \ ;, "•' 1 - 2 n - i / _ 2 B - 3, \ / - 2 I J - 5 / — • 2 B - 3 \ / - 2 I J - 5 V

/ — 2 n - l • , \

(-4*) • n!

>(-DSQ. (ir • n , ^ " - (-4x)»

Multiplying above and below by (2n)! and simplifying

I „ I' ' . (4«)! '":

x 2 V xa ,' i (2/IJI . n! (n+,1) (n+2) (n+3)...2n.. 2"

L (4«)i... - , | (2n)!x(2n)J ,1.(2}

From(l) and ,{2) we see that the 'two results are the same. Hence proved. *

14.". (l-j-x)8n +3«x i[l-:.*)3n-9+...canbe written as il(l_xy.n-j.n . Ox) ( l - « ) (l^*)*°-»

| +nl^-fl> <3*}' (l-*)» (I-x)*-M-...

It is;clearly the expansion .of

=i{(l-*)8+3x (l-*)}» '! „ « [ l - 3 * + 3 x ^ * s + 3 * - 3 * » ] W(|~**;n

' l1 - Proved.


15, Coefficient of xn in the expansion of . — -~ 3

We have _ 1 (1-*)* ^ I j ^*

V+x+x* ( l - x ) ( l + x + " ^ ) {\-x*) - ( l - x X I - * 8 ) - 1

^[\— x) {l+» s+xQ+x° + rc3»>-3 +*3m+.v3m+3+...}

= 1—X+X3-X4 + X 8-X 7+ X3n'~3-Xa™-3+... We see that

when n=s3wi; coqff. of xn=\ when n = 3m + l; coeff of xn= — 1 when n = 3m—1; coeff, of *n=0

.*. n the expansion of rr—r — coeff. of »n are

1, — 1, 0 Proved.

16. See Art. no. 190, 191. The number of terms _ j H - « - l ) !

n ! ( r - l ) ! • Here we have r=3 ,n=8

A t. (3 + 8-1)! 10! Ae

. . reqd. number= g i i \ = 8 T 2 l (a+6+c)8=^8+8Cifl7 m (ft+,c) + 8C3^ (ft+c)2+6C3a

6 (6+c)s

+8C4 . A* (Z>+<7)*+BCsa3 ( A + c j H ^ a 2 (6+c)6+8C7« (b +c)T

+ (6+c)8

.". Reqd. sura of the coefficients^ 1 +aCt,. 2+8Cs (2+2) + S Q . <l-j-3+3 + l)+8C4 . C,Co+....4C4)+

8C5sCl+...«C5)1

V C 6 (8C0+...6C6) + 8C7 (

7C0+...,'C7) + 8C0+:..8C8)

= 1 + 16+112+448+70 (2<) + 56 (2*)+28 (28)+S (27)+28

= 1 + 16+112+448+1120+1792 + 1024+256=6561. Ans.

2901 !* Algebra

„.! I ^ + J ,„+H. l 1 1 ! (n-1) ! ' 3•! (n-3) I 5! (n^5)\

_ 1 f > ( n ~ l ) ( f t - 2 ) , a («*-!) Q:-2) (n-3) (n-4) , 1

—JH lCx+Cs+CB+ ]

=^T [*(Co+C1+C2+C3+ )]

18. (i) We know that ( l~x ) n =C 0 -C 1 x+C a x a -C 3 x a +. . . ( - i rC^ r +. . . and (1—x)- l«l+x+«1+x3+...x , ' :f... On multiplying these two series, we get

• ; { C B - c 1 + C k - c a + . : . i - i / c r } ^coefficient of x' in the product of the two series ^coefficient of *r in (1—x)nX(l— x)-1

coefficient of x ' in {l—xy^l^*-lCr (—\)r

* , , r (1-1) 1

(ii) L. Hf>S.«[Co--C1+Ca-CB + . . . ( - i rC ( J +[-Ci+2C--3Ca+...(-l l»flC J 1]

, -0-[C 1-2C 2+3C 3+.. . ( - l ) n«C t l] . (V Co-0] -JTn - 2ff(ft^}) 3n(n~l) (n~2) 1 •

• L 2!" "*" " 3! ~+ J i T, « - I , ( « - ! ) ( M - 2 ) ' 1

= 4-n [1 — l r - ^ O Ans. (iii) We know that

1. ( l+x) n =C 0 +C,xf C2x2+-'.C„x"

Examples XIV (c) (o Page 167) 291

«a+C«_ lx+C*- 1* 1 + ...CoXB ».(') (".' Equidistant coefficients from the beginning and

from the end are equal).

Also ( I - j : r = C 0 - C l j : + C 4 x a - . . . + (-l)"Cnxn ...(ii) Multiplying relations (i), (ii) both sides, we get

C 0> -C l HCV-C» + ...(-1)oCV

coefficient of x" in (1+«)»(!— *)" ^coefficient of x" in (1—x2)" « ( - l ) - ' 1 . -C i l . a f 0 .

according as n is even or odd. Proved.

Since when n is odd there is1 no terra containing xn,

n > or (~~\)nti —

according as n is odd or even.

Henee Proved.

19. (i) (l-x)-3=^l+3x+6t l+10x3-r-l5x*+... =S1+S ix+Sax

i !+S4x8+S5x1+...SBx' l-i#

' + Proved.

(i) We know that (l-x)-3=S1-t-Sa*.-f-S3«a+S4x3+...

Sn5:n-l+S«+1xn+...S2^1x»-2+SSw>:='H1 + ...

Again writing ( l -x ) - 3 =S 1 + Sax+S3xs+...Swx"-1

+ SB+1«n-!-.MSZ(;_1x

Bfl-»-r-....Strix*'*-*-f-... Multiplying these two series, we get 2 (S1San+S2Sa,_1+....S/,SBf i)

^coefficient of x8"-1 in (1— x)-3x(!-x)-» =coefficienl of*2™-1 in (1—x)"3.

292 Algebra,,

_ f - t f ) f -7) ( - 8 ) . . . ( ^ 6 - 2 « - H - f I) ' ( 2 B - 1 )

_ J 6 - 7 , 8 :.(2H+4) ^ 1 w . \2 / i - l ) !

• (taking—common from each)

_ 5 i ' . 6 . 7 . 8...{2«+4)_ * ( 2 H + 4 )

5 r . ( 2 n - l ) I , 5 ! . (2/1-1)!

20. (i) We know that

' Again repeating the same series,

( i ; -*)~ ] '2= 1H- fflx+ft*a+tfaxB+».ffii-i^I+?«*" •

- ; - " • +ft.+i*"+1

On multiplying .these two series and collecting the oefficieht of xSfl.+1, we get v

2 fe(,+i+ffift»-f?Bftii-i+..,-...?1.-i9»fa+...?fc?«+i] • | ^ c o e f f i c i e n t o f x ^ + M n l l - x r ^ x ^ - x ) - 1 ' 3

=coojn\;ient of x2"+1 in (1—x)-'1

_ _ ( - ! ) ( -21 ( - 3 ) - . . . ( - 2 n - l ) , , „ n + 1

(2n+l) !

! l *' (2/1+1)

.'• fe2n+l+gi?2«+02?2»-l + " - 3 n - l ? « + 2 + ? n ? * + l ] = £ A n s *

(ii)i -We know that

', . ...?2n-2'c2n-2 + ff2„-lx2n-1+?2„^+...

• and ( l + s ) - » 8 = l - g 1 x 4 ^ 2 + . . . ( - l ) n - 1 ? » - 1 ^ - 1

Examples XIV (c) (on Page T67) 293

• Multiplying these two series and collecting the coeffic ients of x2n , we get

2[ft«-ffj?a,-,>rf8n-.-l-...+(-l)n-1^-1?«+i] + ( - l ) , , ? na

coefficient of xin in {\~xf~Ui (1-t-x)-1'2

^coefficient of *2n in (I—x8)*-'1

^coefficient of s2n in

( - * ) ( - * ) ( - j ) - H * - " + 1 ) / y2y, 1 . 2 . 3 ,4 . . .n v '

( J ' 2.4.6...2« ^n

{ V ( - D 3 n = I }

21. We know that'(I-f-x)n =C0-f-C1*-j-C2x2+-"Cw.rn

Putting x = I in both sides.

(1+1)B = 2 " = C 0 + C a + C2 + C 3 f ...C„.

(c0+c1+c2+...gi=(C0J+c1HC2

2+...c^) + 2(C0Cl + C1Ca+...)

2(C 0 C 1 +.C l C a + . . . ) * ( C „ + C I + C a + ..CB)« - ( C 0

3 + C 12 i - C 2

2 + . . . C n2 )

22n (2n) '

• _ 2 « - ^ ' ( 2 " ) !

J n ! « ! " Ans. ' 22. Expanding binomially

(7+4-v/3)£ we have = 7"+" Q7"-1*(4v/3) + n C s

n 7 n " 2 C 4 V 3 ) a + - ( V 3 ) n

1

294 I Algebra .

; **OH-P) (say), where p is positive and p is a prcp*er fraction. *

We see that (7—4-\/3) js positive and is less than 1.

'' ;'. (7-4,V3)n=7"-C1.7?,-14v/3-fC2.7n-2(4V3)I!

", . + . . .(_1)«(4V3)B

-=(1-P). -".' (^+P)(l-P)=(7+4> /3)".(7-4V3r=C49-16x3)n

= \n=jc [V « ispositive]

_ Hence 0 + P ) ( l - P ) ~ l Ans.

'23. L e t S n = f l C I - - ^ + ^ 3 - - K - l ) " - 1 ^ -2 ' 3 ' v ' H (i) Putting n=/ i+l

By subtracting (i) from (ii) we have

s«+1-s„=(n+1C1-' ,co-M ,1+1c8-nc2)+...... „

v B«cr-ncr-

ncl:_1

=-^-[(«+i)-K»+1). f lc1+...+(-i)n. f t+1c f l+1]

V •"C,«(«/r).-iCr-J

1 Ci-{l-n + 1Ci+ i ,+1c a- . . .}] «+i

' Examples XIV (c) (on Page 167) * 295

or Sn+1=Sn+—-p

Now putting tt=0, 1, 2 (n—1), we have

S r - I . S ^ S j + H l + i

CHAPTER XV .,

MULTINOMIAL THEOREM

j EXAMPLES XV ( Pages 173-174)

1. The required coefficient is .

1 ^ " m ^ ' ' ^ 1 ^ - ' 2 ' 0 " A n ,

2#| Required coefficient is ', 1 8 ! , -,,, 8.7.6.5! '. . „ . ; = r r 5 " ! T ! ( - 1 ) = 2 T 5 i r = ~ 1 6 8 A°S"

3. Required coefficient is 7 ! : . 23x 3 = 3360 Ans.

3 ! 3 ! 4. Required coefficient is .

; = = Y T 3 9 ! ! 7 r f l 2 - ( - - ^ C , = - , 2 6 0 a W Ans.

5. 'For' coefficient of .v3 in the expansion of (1 + '3A--*2X2)3;

We hav'e by picking out the coefficients of x3 from the first few'terms of the expansion of {1+(3JC—2X2)}3 by the' Binomial theorem,' that is,, from

l+3(3x-2x a)+3 (3.v-2.v3)3+(3*-2A'a)3 ; we stop at this term as all the other terms involve powers of » higher than x3.

Hence the required coefficient=—36+27=—9 Ans.

6 The required coefficient is found by picking out the coefficients of x* from the first few terms of the expansion of {l-HZv+3**)10} i. e. from ' •

l + 10(2fl:+3jc2)+45(Zr-r-3x2Ja+12O(2x-f3x2J3-H210(2*-f3^J1;;

i

Examples XV (on Page 173-174) 297

we stop at this term because all the other terms involve powers of* higher than **.

Hence the required coefficient is =45x9 + 36x120+210x16^8085 Ans.

7. The required coefficient is found by picking out the coefficients of x8 from the terms of the expansion of {l + (2x-x2)}6

i.e. from 1 + 5 (2x-xa) + 10(2x-x2)8+10 (2x-»V + 5 (2X-A- 2 ) 4 +(2X-* 2 )5

.". The required coefficient= — 10+12iJ+(—80)=30 Ans.

8. From Art. 197, the general term of (l+&x+CA-a+<fc3+...)nis

n'(B.-l)(fi-2)...(i»-g + l) fcfi ryJ 0+2Y + 3B+...

where p+Y+8+ ...=/? ^ Hence the general term of (I— 2x+3x*— 4r*)4 is

4 ( 4 - i ) ( 4 - 2 ) ( 4 - 3 ) . . . ( 4 - / > + 1 ) fly ,4* r P + 2 Y + 3 S ~ p f y ! 5 !" —\r)^x

(I) Thus we have to obtain by trial, all the positive integral

values of p, y, 5 which satisfy the equation p+2y+38-8 and then p is found from the equation /> = P + y+S. The required coefficient will, be the sum of the corresponding values of the expression (1).

In finding p, y, 6 it will be better to commence by giving to 5 successive values beginning with greatest admissible.

The values will be found as follows : 8=2, y=l,f5=0,/>=3 S=2,Y=0,fS=2,p=4 5 % l , T - 2 , p = l , / > = 4 6=0,Y=4,(3=0,/?=4

298 i • Algebra " 9

Substituting the values in (1), the reqd. coefficient

. + 2 7 f 7 T ! <-2)3 <3>" <r4>'+ otfrTTT (~2)0 (3)* ( '4 ) 0

=576+864+384 + 81 = 1905, Ans.

9. Coefficient of *23-in the expansion of ( l - 2s ;+3^ a - x 4 - * 6 ) 5

. The, general term is S(5 - l ) ( 5 -2 ) . . . ' ( 5 - | , + l ) a p y ,.6 a+2P+4T+58

P ! y ! B ! a ! J ^ W V-M V W •

where a + p + Y + 6 ~ p and ce+2p+4Y+55=23

By trial, the positive integral values are S = 3 , Y = 2 ) P = = 0 , a - 0 ^ = 5

Substituting in the above expression for general t e rmor we have the required coefficient of x'*3

# =—10 ( 'Ans.

10. Coefficient of x* in the expansion of (1-2JT+3*1)-"*

The genera] term in she expanMon of (l — 2x-\-ixz)~112 is

^i!rll^4.) :^± = £ ± 1) (_2 )« (3 ) ? . / + 2 P . _ (1)

where <x+2p=5, a+p«/»

By trial all the positive integral values of «, (S, p are as follows:

p=2, a = l , ^ « 3 P=I,a=3,/>=*4 P = 0 , a = 5 , p = 5

Examples XV (b) (on Page 173-174) \ 299

» Substituting these values in (1), we have the required coefficient of a5

(~g) ( T) (~T) / ry\ (•»*• . f ~ g ) (—%> (~~T't 9 \ 3 -,

1 1 2 ! *• . . 3 ! 1 f k } •* 4- (~ *).-(-*) (-*-2)°

5 ! " 135 105 . 63 12- 3

11. Coefficient of x3 in the expansion of (1-2X+3JC E -4 ;C 3 ) 1 ' 2

The general term of this is

«-tX-«. . . ( - i -P + l) (_2) V ( ~ 4 ) T . »a+2?+3r

where * + P+Y=P» an(* a+23 + 3Y=3

By trial all the positive integral values of af$,y and /> are

. . Y = U P = 0 , a = 0 , p = l T «0,p«I ,«=l , j»* B 2 Y=-0,fi=l,a=3,p=3.

Substituting these values and adding, we get the» requirec coefficient of xz

=.—2+7—i=- l . Ans.

12. Coefficient of A8 in the expansion of '

We know that the general term of .the expansion is ( -2) (~3) . . . ( -2- /> + l ) r __„«,MPV 2^+4(3

where «.+?«/»» 2 K + 4 | 3 = 8

By trial all the positive integral values of txt [3 and/7 are

300 •' Algebra »

p = l j a = 2 , p = 3 , !;

% ' {5=0, a=4,/7=4 '•! Substituting, these values in (1), We have the required ;|

, ' g / n i (-24) 1 I j _ n o 1_ Coefficient of * - j f ( ^ ) + T 1 ~ T ' T 4 f 81

r A 27 *1 - 8 1 / 8»

13. V (2-4x+2x*)-*=L ( 1~2*4- | *V , we shall

first obtain the general1 term from (1—2x+4*9)~a-Now the general term is ; ^ H - ^ ^ f e ^ + a (_2/(|)P ,.+* ; (i)

where «+2p=4, «+(*=/?. » By trial all the positive integral values of a, [3 and p are

, p - 2 , a=0,/>=2 |i (3=l , a =2, ,p=3 fj=0,a=4,p=4 ,!

Substituting all these values of 4, ? and /> in (1), we have the required coefficient of x4 - ,

• 14. Find the coefficient of x° in the expansion of ', (l+4xs4-10x4+20*6)-3'4 "t

The general term is ti !

j-i) (~il..j^p±UAfl^m-t ,tf-+4p+6ri

vhere «+p4-Y—^i 2a-f 4J3+'6Y=6

By trial all the positive integral values of a, p, V, and /* xeas follows-1 r = l . ?—0, a = 0 , ^ = l !i

•i 7=0, p = i , a = i , p = 2 • ;;

k amples XV (on Page 173-174) 303

equidistant from- the beginning and from the end will be equal.

* .". {\+x+x*)n=a0+alx+aix* + a3xi+...aHxn *

Putting * = — x, we get

(l-x+x*)n=a0-a1x+a2x*-a5x*+„.+(-\r . «„*" - +...fl3x2n-3+flax3n-2-a1x2f l-1+o0x2n

Multiplying the two expressions on the right, we get the coefficient of xin

= 2 {fl0»-fl l"+fl,"-fl . I+.. .(-l) , ,-1flVi}-K-l)*fl"« and it will be equal to the coefficient of xZn in

(l+x+x*)" (l~-x+x*)n

or (l+*a+-v1) f l, that is it must be equal to an. ,

Hence 2 { f lo 8 -« iHff> , - t fa a +-»( - l ) , , ^V- i}

or V - « i M - f l aa - o 3

2 + . . . ( - i r - 1 . a V i = K { l - ( - l ) " a - } Ans.

21. Since ( l + A : + ^ 2 ) B = f f o + « i * + ^ - x H ^ ^ 2 + ^ * + ...aSRx>» ...(1)

Putting * = l on both sides, we have * •

or ((70+«B+'76+..-)+(^+ff4+O7+-)+(ffsftf54 :fl8+-) = 3 " ...(A)

Again in (1), putting * = G J , we get

or ( 0 i + a 3 + c l + — ) + « (<'i+«4+fl7+—) +<of (fls+0B+*B:f ....)=0. ...(B)

Also in (I), putting * = u2, we have u 0 +a jw3+o2<)i*+03w6 + • • •—0

«=US("l+«« + * 7 + - ) = 0 ...(C)

304\ Algebra i •} •

\ ,Let- ff0+fl3+(76-f-....be P, a2+a5-\-ae-\-... be Q ajid

flr+fl4+fl7"+.»be R ; then from (B), P+wR-f<o3Q==0 « ».(2)

and from (C)., , P + C J Q + M 2 R . = 0 ,

Adding, 2P+« (R+Q) + «* (R+Q)==0 \ • 2P+(R+Q) (w+w2)=0.

.\ *2P«R+Q. P=R/2+Q/2.

Substituting these values of P in (2), .we have

or " R+Q-f2R6i+.2Q6)a=P Or \ R(H-2cd)=-Q'(l+2u2) . Squaring both sides, we have • R2 (l,+ 4o+4to2)=Q2 (l+46>H4w).' [V ca4=o3co=fi>] ;

i.e. R=Q. ". But we have ^

2P = R + Q = R + R = 2 R ', ".'. l P = R = Q ,

But from (A), we have that P + Q + R = 3 n . • i.e. 3P=3n or P«3n/3«3"-1 .

/ . p=Q=R= =3"- l . Hence (a0+a3+aBj-...)=(a1+ai+a7+.„)

. =(^+a 5+fl 8+.- . )=3 n - a . '„ Proved. i'

Y

V-irtAtM £K. A V i

LOGARITHMS *

Examples XVI (a) (Page 178-179)

1. Let x be the rcqd. logarithm ; then by definition of log

we have (v

r2)*=I6, or 2*/8=24

/ . x/2=4 (equating the indices) .'. x—8 Aus.

Similarly (2\/3)*=1728 = (2\/3)6

.*. x=6. Ans. 2. Let * be the reqd. logs ; then

(5N/5)1!=I25=(5x/5j8 .*. x=2 Similarly (4)*=-25=J=(4)-1

.'. *=—1 ' Ans.

3. A s above (_V2)^ 25r= ( - 2 V 2 ' ? , 7 3 - ( 2 , / 2 , - »

.'. K=—V Ans. • Also 9I=-3 = J = l/(9)1/^9"1 '1

, .". *—(--$) Ans.

4. log of -0625 to base 2, and of 8 ODD to base 01

Suppose 2*='G625~TV=-'I/21 or 2*--=2-» .*. x= — 4 Ans.

Also (-01)*=: 2000= -rQTW = *01-3'3.

.*. J C = — | A::"s.

5- log of 'O00S to base -#0i, and "1 to basc9v '3

Ifjcbe therc<;d. log, tUn {-Q0I)*=-CG01*(-Q01.)«« ,\ . 5 - } Am.

^306 \

Algebra

*\

Also ( 9 V 3 ) ^ a = ^ ^ ^ 5 = ( 9 V 3 r " &

.*. s = —$ \ Ans.

'6. log of ^(a8")i 4 a " #(«""'*) to base a.

If x be the required log, then • A -.

« - i Ans.

Also ^ fl.^^^fl-1/a . .

and o!»=^/ta:i5'2)=(a-"i2JU3=fl-5j2

7. Let'logg 128=x ; then \ 8"=128«87 '3; / .

Also if log6 -SITT—*? then i

and if log2J -rr=x, then

A*——1 .Ans.

/ . Sc=—I Ans.

- I — 1 - K - .

«==£ ">Ans. v.

K= —3 Ans. \

'Ans.

\

27««^. or 3 ^ = ^ = 3 - " ; \

• .*. 3 x = - 4 or x=—4 Ans. and, if log343 49=x, then

343J=49 or 73*=7a orN 3* = 2; .\ *=•£

8. log[V(0268)]c=61ogflfc3'2

= 6 [log a-flog A3'3] \ = 61og<i-f6x£ log & N

^=6 log o+9 log b Ans, ' v\ 9. log(^f lsx-^6*)=log'(f l2 '3x68") v

=log # ' 3+Iog f f « S log fl-J-4 log fr -;, '• Ans.

10\ log [^(o-^B j l= log (fl-*/*x 6a'°)

\ , =lcgrt-*'°+log6V3

\ *=—4 l o g a + i log/t Ans.

Examples XVI (a) (o . Page 178-179) 307

11. log Wia-tyx^Cab-*)} = log.(a-W8x« l '36-M =log a-1* log bw+log «"3-f log 6"J

t=—log fl+| log 6-H log a—log b = — £ log a—J log 6 Ans.

12. log [ K f l - ' x ^ f i ^ V l f c V " ) ] =tog ( f l - ' - ' x ^ r ^ ' ^ f l 1 ! ' ) =log ( /r l ' "x6 s fl)-tog {bznxaM) =log a-1'3 flog 6s '6-log 6s'.1-tog a"1

« - j log a+# log b-\ log 6 - J log a = —•r'j log a—log 6 Ans.

13. We have log ffj^ffi '

~log Wfcfr -^f l - log (a-l6-*c-*)1" «dog <tf1*s£-"»<r*'8)-log (<rV»£-i «>*-«") =*Iog a"»+log &-1,a+log c"2 '3-log fl_1,G-log 6"1'3

—log c~£rt

« 4 log a—1 log 6 - 4 log c f J log «+4 tog 6+# log c =*4 log a. Ans. •

^ ** {(^r-(^)5ican'bc witten

^log {(Pc-^-ib-WFr =Iog {(6-J5c15)-f-(Z>-15cM)} = - 1 5 log 6+15 log c+15 log 6-20 log c — — 5 logc. Ans.

51'* x 2 i a o

15. We have log j ^ j s ^ r s

=log 5u*+log 2lfl0~log iSx'3—log 2l G

= i log 5+TV log 2-log 21/s-log 32/3-$ log 2 =-1 log 5-CA log 2-1 log 2 - 4 log 3 - J log 2 = | log 5-4 log 2 - 4 tog 3 Ans, ^

'/

as

308 Algebra

\ \ 16. We have log -^{729-^ (9"1 . 27"1'3)} - l o g <P{729-9-v?. 27-*")

\ * =log (729 1 ' 4 x9- , m x27- 1 ' a )

= i log 7 2 8 - ^ log 9—fr log 27

\ = i ^g 36 - IT log ,3 a-4 log 33

= | log 3 - £ log 3-r i log3=log 3 Ans. '17. , l o g * f - 2 " l o g 4 + i q g / f t

« log 75- log 16—2 (log 5 - log 9)+log 32- log 243 =log 25+log 3 - 4 log 2 - 2 log 5 + 4 log 3 + 5 log 2 - 5 log 3 = 2 1 o g 5 + I o g 3 - 4 1 o g 2 - 2 1 o g 5 + 4 1 o g 3 + 5 1 o g 2 - 5 1og 3 =k>g 2' Ans.

18; The equation is ax=cb" Taking logarithms of both sides, we have

x log a=x log &+log c or ' x (log a—log b)=\o% c ;

v log a—log b 19. Equation is a%s . b3x=cn

Taking log of both sides, we have l

* 2x . log a + 3 * log b=5 log c or x (2 log a + 5 log b) = 5 log c ;

• • r- 5 l o g C ' Ans •

, '• * ~ ( 2 I o g a + 3 l o g £ ) \

20. ^ •

Taking logarithms of both sides, we have ( JC+I ) log a—(*—1) log 2>=2JC Iog.c

or x (log a—log 5—2 log c)=*-Iog .a—log b ; • loga+logb • •

2 log c—log tf+log 5 21. a-* . 63v=m5 ; a9* .bsv=~m10 '' Taking logarithms of both sides in both the equations,

1 Examples XVI (a) (on Page 178-179) * 309

wabave 2x log c-f 3;> log b=5 log m ...(1) -*and 3*.Iog a+2y.\ogb=10 log m ...(2)

Multiplying equation (1) by 3 and (2) by 2 antS then subtracting, 5y log 6=—5Jog m ;

" y logb

Again multiplying (1) by 2 and (2) by 3 and subtracting xc= 4 log m

log« Ans.

22. •.' log (x*y*)=a and log (»/;•)=*

.'. 21ogx+3 log j = a ...(1)

and log JC—log y—b ...(2)

Multiplying (2) by 2 and subtracting from (1), we have

5 log y=a—2b

A logy= —£--

Again multiplying (2) by 3 and adding, 5 Iogx=a + 36 ;

/ . Jog # = — — . Ans.

23. Since a3+*.#*==a*+s.63* a3"a!_fc31

or 0-^-5=^ac-Jir o r a-te-r_£-£0

Taking log on both" sides, ( - 2 x - 2 ) log a^=—2x log *

or (*+l) log a=x log b or K log a+Iog tf=.v log 6 or log a=x (log 6—log '0 or l©g A=X log (f?|a). Ans. 24. fa*-2fl«>3 ,+/y)*-1-(a-6j** m+6)-*

310 • Algebra

„ i Taking logarithms of both sides, we have , (* - l ) log (a*-2a862+A')^2.v log(a-.6)-2 log <«+&)

i „br (*- l ) log (aa'-:6«)«ix log (<7-&)-2 lag (d-f-6) ^ or <2x-2) log (a?-6*)=2x log ( a - 6 ) ~ 2 log (a+6)

' or 2(jrf-l) {log (a*-b) -Hog (a+*)} ' = 2 * log ( a - £ ) - 2 log (a+W ' o r ^ - 2 log (tf-£)-f-(2x-2) log (*+fc)=~2 log (a+b)

or —iog(o-6)+x log(a-hft)-log(a-r-6) =—iog(o+*) i or jclog (a+6i=l6g(a—6);

"• log(fl+fc) ' >|AnS'

EXAMPLES XVI (b) (on Page i85)

M. Characteristic of logarithm 21735 will be 4. (V No digits is 5, subtracting 1 from, the number

of digits) ( . < Characteristic of logarithm 23*8 will be 1 Characteristic of logarithm. 350 will be 2 Characteristic of logarithm '035 will be 2~

(adding one to the number oT ciphers «l . , immediately after decimal).

Characteristic of logarithm *2 will be 1 Characteristic.of logarithm *87'will beT Characteristic of lpgarithm '875 will beT 2. Given the mantissa of 7623— 8821259

.'. toff 7-623='882i2S9 f ', log762'3=s2-8821259

Iog'-007623= F-882I259 ; log 762300=5-8821259 ; log "000007623=76 -8821259. Ans.

3. The numbers of digits in | the integral parts are " 5, 2,4,1 respectively. i, •

• ! • * fi

4. For T '7781513, the first significant figure is seeor 4ecimal place.

-* £or '6910815 tbc first significant figure is unit's place. For T'487!3S4,-the first significant figure k fif.

decimgj place, 5. log 64=log 25=6 log 2 (7 64=7.")

= 6 ('30103O0)=r8061800. Ans. 5 log84=log (22X3X7)

=21og2+log 3+log7 «-60206OG-p477!213+*845O9$0 = 1-9242793. Ans.

«log27~log(10)' = 7 log 2 - 3 log 10 *7('301O3O0)--3 =2-O72100=T'107210O. Ans.

8. log .0I25«log - ^

125X8 . 100 1 ~ log IcoooxT^108 TxTcoW = b g~la ^log 1 - log 10" log 8 •

M - 3 log 2=*log 1-iog 10~log(23)

= -l- '9030900=T'096910o. Ans. 5. log l4-4=log SV

=10^144—105 10=log (16x9)~log 10 =logT6-*-log9~loglO « 4 log 2+2 log 3 - 1 ' = 4 (30l0300)+2 (*47712I3)-i = 1'2041200+-9S42426-1 = M583626 Ar

" 10. logr4|=logV ~log 14-Iog 3=Iog (2x7)-.i0g 3

312 • * Algebra 7 \

= iog2-Mog 7-log3 #

; '~-3010300-f-8450930--47712I3 %

' % =-6690067, ( Ans. ; 11. log ^'12=log 12l/a=Iog (4X3)V3

=£ (log 4 +log 3) =4(2Iog2+log3) « i (-6020600'+ -4771213) • =•3597271. "• Ans.

; 1 2. log VI l°*i Pog 35-log 27]. =^[Iog(7x5)-log'(3)3] W [ I o g 7 + l o g 5 - 3 1dg3] «H'84509804>log V - 3 (-4771213)] =4 [-8450980+loglO-log 2-T4313639] =2 [-8450980+1--301030O—1-4313639] =•0563520 Ans.

13. log-e/COt05)=logl-Ol05)1'^ i i 105 . f , 21X10 "1

"*Iog foooo = * L tog 1005x2-J : = i log T - ^ "•.,

# = 1 [log 2l-log(l000x2i] = i [log 7+log 3-log 1G00 -log 2] = i [-8450980+-477I213-3--30103001

< =1-5052973 Ans. 14. Let (00324)1" be equal to ,v, ie. (00324) 1 I ?«JC.

Taking logarithms of birth sides, " :• |[Iog-00324]=log.-c

or log *=4 [ log - T ^ - ] = * [ log (!Lj<* )] =£[log81+Iog4+51og,10] - f [4 log 3+2 log 2T-5] = 4 [4 (-4771213+2) (-30I0300)-5] ^

I' «4[I-9084852+'6020600-5]

\ Examples XVI (b) (on Page 185) 312

=•?- [2-5105452-51-4- [1-5105452] =4[T+4-5l0452] ; log x= I '6443436;

/ . x=Antilog ( T - '6443636) = '44092388 Ans. 15. Suppose {(39-2)2}1 '«=A-

Taking logarithms of both sides, log *=log 09-2)* '"=^ log (39-2)

=TV log Vo-»=-& {hS 392-log 10] =,^[Iog49x8-loglO] = i r log [log 49+iog 8—log 101 = A [2 log 7+3 log 2-11

. =Tr[2(*8450980+-9030900-l] =T

9T [1-690I960+-9030900-1)^-2896S83

A x^Antilog (•2896883) = 1"94844S Ans. 16. Suppose x=37-203x3'7203x-0037203x372030 Taking logarithms of both sides, we have

og x=log 37'203+log 3'7203+log -0037203+log 372030 ' = 1 -5705780+-5705780+3~5705780 +5*5705780 = 5*2823120. #

/ . x^ Antilog (5-2323120) = 191563 • 1. Ans.

»* jm yf 354/3

~ "2^0 :* I of i 3+^ log 5 - i log 2 ~ * C4771213)+| (log 10-lDg 2J-K-301030C

( V 5 - V « $ (-4771213)+$-$ (-3010300;-J (-30103OC „-9542426_^_1.204l200_^ . 0 5 0 I 7 1 6

3 r a 3 ='3l80808 + l-3333333-'4013733--050I716 •=1-1998692 Ans.

18. log(^48xl08 l ' 4X 1V6)

=log (481!\x 108"*)- log 61,M

=log (I61 '"* 31,8)+i log I 0 8 - T V log (2X3) - 1 ^ g 2«+§ log,3+J [log 27+log 4]

\ • • ' "f —&{log2+log3} - I log 2+1 C4771213)-H log 3-H log 2 '

'. "* ' - — ^ J o g 2 - A I o g 3 •\ - t t + l ^ f t ) log 2 + ( i + # - A ) ^ g 3

- 1 C36l0300)+C477I213> , . . 3 J S » 1 0 0 + . 4 7 7 | 2 1 3 • *' •

\ «-5268O25+"4771213«VG03923R.. Ans.

[given log 2,' log 3, log 7 ; also 1 log 9076-226==3'9579053]

/ 294x125 \»'a

\ • " J , 42X32,\J Taking logarithms of both aides, we,have •

log a=£ [log 294+log 125-log 42-^log 321 ;

= 4 [log (45>x6)+log 5»-log (7x6)-|:log 26] = i [2log ,7+log2+Iog 3+3 log 5-Iog 7-log 2-log 3

= # [log 7 - 5 log 2+3 log V) v ' = 4 [log 7—5 (log 2)+3 log 10-3 log 2] =#[•8450980-8 ('30I0300)+3] * •' (V log 10=1) ~ | [3-8450980-2 4082400]='95?9053.'-

,\ :X=Antil6gC95790S3)=9O7(5226. Ans.

20. Suppose x=(330+49)«-^(22x 70). Taking logarithms of both sides, we have,

log 3-.=l0g{(33O-r49)*~^(22x7O)} ' =log (330-r49)4-log (22X70)1'8

= 4 [log 330-log 49 ] -* [log 22+log 70],

\ Examples XVI (b) (on Page 185) • 31 i

= 4 [log 11+Iog 10+Iog 3—2 log 7] ' - 4 [log t l + log 2+log 7+log 10

=(4--J) log U+(4 -I) log 10+4 log 3 - ( 8 + i ) I o g 7 - i l o g 2

= V (1B0413927) + V+'*'4771213 {-V ("8450980)} - $ (-3010300

^ l l - 4 5 5 3 1 9 7 + n i . 9 0 8 4 8 5 2 _ 2 ^ 7 4 5 0 0 _ . ^ ^ 3 ' 3 ' ' " - " " - 3

«3-8184399+3"6666666+l-9084852 '-7-O424833--100343

« 2 2507651.

.'. x=Anti!og (2-2507651)~l 78i415I6 • Ans.

21.- Suppose x= 3 " x28 .

Taking logarithms of both sides, we have log*=121og 3+8 log 2

= 12('4771213)+8 ('3010300) ^5-72545564-2*4082400 =8 1336956

Since the characteristic of log JC=8

.'. number of digits in x will be 9 Ans. 22. Suppose (££)10*=;t / . log *=100 [log ir—log 20] =100 [log (7X3)

- log (2x10 = 100 [log 7+log 3-log 2-log It)] = 100&-8450980+"4701213--'-3010300-l] = 100 [1-32293-1-3010300] «100L'0211893]=2-I1893

,\ sr=Antu*og (2*11893) =three digits in the integral part and decimi

part. But hundred (IQO) contains only 3 digits and no decira

part. •

316 l • Algebra'; . i-

.\ (4i)ioo>ioo : 23. Suppose x== (J)1000

; • / . log x=1000 [log 1—log 2] , Wl00O[O-'30lO30O]

\ =~301-Cf300=302'9600 Since the characteristic is ~3"02~J so the number of

ciphers==30i. Ans." 24; , log.(3l-a)=log('5)

or (JC—2) log 3=log 5

'• . KX l \ log,3 • r„" log 5 , Jog 10/2

, •' " * - log3^"T5gT +

_log 10-log 2 . ? 1--3010300 -i o g l T | '4771213 "*"

* , 6 9 8 9 7 0 ° -.2«3-46 •4771213 -[•*—-« A o S t

25. V 5*=103

Taking log of both sides, we get % x log 5=3 log 10 ,

or * log — - 3 log 10

-[log 10-log 2] = 3, log 10 3 log 10

x— log 10-log 2 3.1 =4-29 Ans.

1—'30.10300 "6989700 26. 58-3*^2*+*" Taking log of both sides, we have

. (5-3*)log5=(.v+2)log2 (5-3x) {log 10-log 2}=(*+2) log 2

* [log 24-3 (log 10-log2)]=5 (log 10-log 2 ) -2 log 2

I Examples XVI (b) (on Page 185) 317

• or x ['3010300+3-3 ("3010300)1=5 (1--3010300) - 2 ('3010300)

or x [3--6020600] = 5 ('6989700)-*6020600 3-49485OO--602O60O . -Aj- .

» * = — 3 ^ 2 0 6 0 1 ) — ^ 1 2 ° 6 A n S ' 27. 2l*«=2**+l.5* Taking log of both sides,

x log 21=(2x+l) log 2+s log 5 or x [log 21—2 log 2—log 5]"—log 2 (taking x terms tc one side) or x [log 7+log 3 - 2 log 2-log 10+log 2]=log 2 or x [•8450980+-4771213--'6O20600-l+'301O3003="0l0300 or * [1*6232493-1-6020620]=*3010300

•3010300 , , . . , .

•• x-*nm-lA™ Ans-28. 2*x6s-'=52a!x7<l*-*> Taking log of both sides, log (2*x6*-*)=Iog (5*x71-")

x log 2+(x-2) log 6 - 2 * Iog-5+(l-*) log 7 or x log 2+JC log 6 - 2 * log 5 + x log 7=2 log $+Jog 7 or x [log 2+lqg 2+log 3 -2 log 10+log 2+log 7]

= 2 (log 2+log?3)+log 7 or x [3 log 2+log 3-24-log 7]=2 log 2+2 log 3+Iog 7 or .T--2( |oe2+loB,3)+log7

3 1og2 + log3-2+log7 te 2(;3O1030O + -4771213)+j45O98O _ . •

3 C3l)]030D)+r47712l3)-2+"84509SO~ 5 6 , 29. 2 ^ = 6 * . 3°«3x2i'+I, taking log of both sides

we have (*+>') log 2=y log 2+^ log 3 or x log 2—;• log 3=0 and" x log 3—log 3 + (y+l) log 2 or .v log 3-y log 2=log 3+Iog 2

.x log 2 - y log 3=0 ....(A of * log 3—^ Jog 2=Jog 3 + log 2 ...(B

Algebra \ 1

Putting in (B), x*=y .5Lj , we have

', ^ . ^ 8 - > - ] o g V 2 = : l o g ; 3 f l o g 2

or 1 y {(log 3)8r(Iog 2)a}-log 2 {log 3+log 2}

nr l V ^ W 2 (log 3+log 2) ) •

" V (log3)*-(log2)\ \ log,2{log3+log2}' •',_ log 2 . ,

(log 3+log 2) (log3-lbg2) Iog3-log2 '•,

••• Ffom<A)^!ofi ;; _ log 2 y l o g j ^ log 3 "(log 3-log 2} log, 2 (log 3-log 2)

30. a1-*"*—4-«, i^^ym— Taking'log in 1st equation, ': ',•

' ( l-Ar-^)Iog3 = -2^1og2;;. or . x log 3-f-r (log 3 -2 log 2)=Iog 3 ...(1) and from 2nd we have * ' \

* ' (2*- l ) log 2=(3>>--x) log 3 or lx (Jog 3+2 log 2)-3y log 3=lpg 2 ...(2) Suppose log 2=o, log 3=ft ; then we have

xb+y(b-2ci)=±b * »..(3> and x (b+2a)—3ybr<* -\ —(4) Multiplying (3rby 3b and (4) by (6—2a)i'and adding,

1 *(361+i*-4a*)«=36»-+a6—2a1

or , 4* (6+0) (&-a)=(36-2a) (b'+a) ;

4(2>-c) 4 (log 3-log 2),-• „ . . . 3b—2a . „ . ;

Substituting- «—4 /fa_fl\ 'i(3)> we gel ,', >

v= '- 6 - ' 'og '1 A n s

' 4.0-a) 4(lbg-3-log2) <1 Ans-

Examples XVI (b) (on Page 185) 319

31. Given logia 2="30l03t find loga5 200 Iogi6 200==Iog10 200x.log20 10

=log10 200x ~ - 2 j <v Io&* I 0X l o&° 2 5 = i >

=(log10 2+2 log10 10) X j ^ i ^

_ '30103 + 2 _ 2 30103 log10 100-log10 4 2 - 2 (-30103)

2-30103 _ 2-30103 . . . , . , " 2 ^ 6 0 2 0 6 " F39794 = 1 6 4 6 0 1

32. Given log]0 2='301O3, log10 7="84509

We have log, (V2)=logl0V2+log7 10 (change base)

=logl0V2 x 'log10 7

= i C 3 0 1 0 3 ) X ^ l ^

= -15051_. ^84509" 1 7 8 1

Similarly-, log ,27=logj0 7xIog 2 | 0

E l ° 7 XlogM V2 - i !ogM2

log10 2 '30103

CHAPTER XVII

EXPONENTIAL AND LOGARITHMIC SERIES

EXAMPLES XVII (on Page 195)

1. p u t t i n g J t = l o n b o t h s ide s o f ,

w e h a v e

1 ^ 1 1 _. 1 1 . 1 •;* .

log62=l-T+y-T+y-y+.. . 2. We know that •[

Putting *=$ w e get

log, 4=Iogd 3 -Jog, 2 Ans.

3. Since log, (1 +x)—log (1 -x)

- I * ( 1 + ^ - 1 0 8 . ( 1 - 1 . )

' ^ ^ - t o g t i - i o g h i t t f x - i )

Examples XVII (on Page 195) 321

4. We have Xz X3 Y* «5 *

y * 2 r 3 4 ^ 5 -*

<| 1 4

But we know that log ( l + s ) = x - ^ - + ^ - - — +-... j£ J *T

or (1 +*)=ev (by definition of log)

But fi»»i+y+£!+£.+...

A 1+^=1+3;+^-+^+...

u 2 y 3

or * = J ' + T - , + 3~, + .. . Ans.

5. Show that a-+T(^)+T(~) + v

Since

logiCi-xH-x—j—g—...; #

=-[,+f4+...] or ,+^ . + y+...«-Iog(l-.»)

^ • a - *

Putting * = - — »

In the above result we get

K-_(log 6—log a)«log a—log 6 Ans.

Algebra

6i We have i'l

l0Se "99J = 1 ° 8 # 0 + 9 9 9 ) » V; IO%{\+X)=X-X*I2+XV3+L

B 999 ^999 -2 ^999 J T 3 ^999/ ^ Putting their values, we have =;0020000006666670 • Ans. ,;

1 , • •

7. > i o v B t h . t e - " - 2 ( l + / I + i + . . . )

^=•00200. . .

(9i)3=-oooo.

Let nth terra of the given series beuTw ; then 2n+l

; L(2»+I) U (2n+l)! (M+l) ! ( 2 B + 1 ) ! » T--ian"eiJi)i[v ^'f^+ntwi Now putting n = l , 2, 3,..., we have'.'

( l l _ 2 ! 3 !

l a ~ 4 ! 5 !

T s ~ 6 ! 7~1 'J Adding all these terms we get th£,sum<to infinity

/ . sum=(T\ -pT s+T3+... co)

r ^ 2 J •',, ' g+e" l -2 -e+e - ' -+2 „2e-1^i^_1 Proved*

Examples XVir(on Page 195) * 323

, 8. Prove that

> log ( 1 + ^ a"«)"-2(^+3^+^+. . . ) .

Here L. H. S.=log {(l+x)1+V (1-JC)1"1}

=log (1 +x)1+*+log (1 -x)1-* " «(1+*) log (1+*)+(!-*) log(l-je)

... .( x2 x2 x* x5 \

Simplifying it further we get

+{(—T-T-7-T-")

Now collecting same powers of x together we get

Thus L.H.S-R.H.S. Ans.

We have on collecting x terms separately and y terms separately

324 ' Algebra

', It can be written as *

V F 2 ! "*" 3 ! " t " 7 U ! "*" 2 ! "•" 3 ! *"')

=e —e Ans.

10. Before finding the values of log 7, log 11, log 13,

we shall find logc 3 with the-help of the following series :—

log, (n+l)-logc «=2 [ ^ + — ^ 3 + - - ^ ^ + . . . ]

(see result § 226), ,

TfweputM*=2, we get ' j A. — -2 Io&,3—lo& 2 "\ 5

i-04

= 2 r 2 + ^ 8 + ' 0 0 0 0 6 4 J - 5*

, + • 0000018 + ... J 5*

«2[-2+'0026666+-000064+-OOOOOi;R] ' , ^ = = ' 0 0 0 3 2

rr=*008

='0016

= 2[*20273141. .\ log]0 3-Iog10 2==-4054628xn

= (•4054628) (-43429448) : .'. log10 3 = logI0 2+04054628) (-43429448) Z ; « 0 + - 1 7 6 0 9 1 3 j .,' ^^ -0000018

^5=-000064

- - = •0000128

Now in order to find the value of log" 7, put n=6 in the above series; then '•.

Examples XVII (on Page 195) • 325

. Iogc7-logc6 ( ! ^ = . 0 7 6 9 2 3 0

* ~ 2 [ l 3 + 3"T3^+5T13^+ ' " J I . = 2 [ - 0 7 6 9 2 3 9 + i ° 5 ^ 5 i ] . ] ^='0059171

- 2 [076923O+-000I517] ! 1 _ . m n , , S I

= 2f0770747) j l P " ~ N * 1 ™ 1

log# 7-loge 2-Ioge 3='1541494

or Iog10 7-log I0 2-logJ0 3=( 1541494) X fi ; .'. log10 7=log10 2+Iog10 3+C1541494)x(-43429448)..

= •8450980 Ans. .

(ii) For log 11, putting n=10 in logI0 (n+1)—Iog10 n

,we get log10 ll-Iog1010=f, ( 1 - ^ - + 3 ^ , - ..-)

= (-043429448-*002l71472+-000144765)

.'. log10 H=IogI0 10+-043429448--00317I472 +•000144765

= 1-0413927 Ans. •

(iii) For log 13, putting «=1000 in

Iog10 (n+l)-log1 0 n - a ( 7 — y ^ r + 3-^3 - • • • )

m getlog10 1001-lofco 1 0 0 e - | i [ 1 ^ - 3 X ( / 0 0 ( y + - ]

ogio 13+log10 77 -3 logl0 10="00043429448--000000217I... og10 13=3 —IogI0 7-log10 11+'00,043429448-,'O0O00O217I.

Putting these values of Iogl0 7 and Iog10 10 and ' Jimplifying, we get logl0 13 = 1-1139434. Ans. ,

11. If ax2 and ajx2 are each less than unity, we are toi/ prove that •'''

326 Algebra

=Iog(l+^+fl*'.4^J-)

; We have R.H.S.=Iog (x+ax*+a* + J - )

> ' =log | l (l+<Jx2)-H£(l + ax*) j ; (factorizing)

\ «Iog{(l+d*")(l+£)}

; [V log (m«)=iogm-j-log«] By logarithmic expansion we have

r „• a2*4 a3*0;', aixs ...)

T \ x * 2x* P 3 . ^ 6 4K 8 H " V * 6 4 A'8

Collecting coefficients of a, a2, a3'etc. we get

Thus L. ric'S.=R. H. S. Ans/j

12.; We have L. H. S.=log (1^3x+2*2). It can be written as'log (l+*+2x+2x2) \

.'. 'fog 0 +-3x+2x*)=}og\(l +x) (1 +2*}] =Iog(I+x)+Iog(l+2x) '",

/ X2 , X3 JC4 . \ '.'

- l-A*—4*' - 8* 3 16*4

2 ' 3 _ Sx*9x* IIJC' . I '

" 3 *"" r 2~+~y Q—f- „T Ans.

\

!£+...)

Examples XVII (on Page 195) , " 327

Clearly the general term of the series will be equal to tfce sum of general terms of the two above geometric series,

Hence T , - ± ^ - • + J r I E l * l * . r r

= (~1J"1- .{?+\).xr Ans.

13. We can write

log {~2~**-log (l + 3jr)-log (1-2*) .

( , 9A-2 , 2 7 X 3 81 x* ,

V 2 3 4 " V . 5*2 , 35*a 65** , = 5 .v - - r +- T - - 4 -+ . . . Ans.

General term=sum of the general terms of the above two series

14. We have

15. Ke*r+e-fa] '• • '

« i iYi J.1* +(fje)2 +<*>* + . a x ) 4 4. >

~*L^1+n+T7 +Tr+™4T + - J , / » . ( - to ) * . (TMC)* , ( - / x ) » , \ 1

328 1 * , Algebra fi

fiLv !•' 2! 3 r 4 ! + , , J , A ix x2 ix3 , x4. \-i

: ••• H l*"rr2T+r!+4-!+"OJ

16,, We have

• -f h* + *' '+ j ! _ . 1

.-. Now R.H.S.=21og e(*+/0--]og e 'x+Iog e(^±^J

^2"log, (x+/i)-Iog, x+Iog, x (*+2fc)-Iog. (x-i-A)2

=log3 .(x4:A)B—fog, ."c+logc x-Kog* (x+2/0

=logc(x+2//) Ans. ^J

17. V ;«, P are the roots of x2—px+q=0

.". a+$=p, aP=9 (puttingp=a+p, ? = a »

We have logc (1 +i>x+<7As)

-logeO+te+P) *+«?*•}, - l o g e i l + c c x + ^ + a M ^ l o g ^ l (l+a*)+Px(l+aur)}

=lo-ge {(l+*x) •(l + Px)}=loge (I+flweJ+lege (1+fc)

Examples XVII (on Page 195) • - 329

/ atx* , a3*3 a V \ - ( « - -7T+-3—r + • • • ;

collecting the coefficients of x, x3, X3 etc. we get

Proved. 18. 4xa+#:e8-f-£**+|.xB + ... can be written as

/ S 3 X3 Xi X* \

(adding and subtracting x to both the factors) We get =(x+x2+x34-x*+x5+...cc)

(V x+x9+x* + ..:o: in a G. P.) X -flog (1-A-)

19. R. H. S. 1 1

= 1 - 2 (n-f-1) 2.3 («+l)2" 3.4 (n+I)3 . ""*

=i_r_j__—J_vf ~J - L i - j _ ! i__ r

13 (n-fl)3 4 ( j j+ l f j -

\ 1 + 2(n + iy+3T^U3+"4^'+lT3~+ '"J __J 1 1

330 " , ; Algebra \

• f M + 1 ) ( t«+l)"- 2(n + l)2 3(»+l!)» "')•

; +( L; 1 _L_L.;> <, ^ v B + I 2(n+n» 3(ii+i)a | )

— <n+1> { to* C1 ^ - V ) ] +Ioe< fr-aTr) (n-HO loga-r-rr+loSo n+1 n+lv

: - n ) 0 g - ^ T - l 0 g — f + l 0 g ^ T

' = —nlog - n

r_jn Y 1 B + l

. . ^ ( s t i ) : . =«[log(l + l/n)]-Iog(l+l/n)«

Thus R. H. S.=L. H.S. !" Proved. ,',

SO. We have ';

l0ge ( 1-HHW* ) : \

"lo8i [ ( i+^mi+^)]i0 gnT^+ l 0 f c(wJ =ioge (i+*H+io&a+*«r»:: ft

log-(l+«)™log-(l+»") «

- / v a ^ 4 - * " • ' fiffl+fiQ!^- ^ V* 2~*"3 '".2r- " r 2 r + l '"V

- ( •

'Examples XVII (on Page 195) 331

+ - L + . . . . + . . . ) + H + ^ - 4 + . .

^ 2r 2r+l "J Now if n be odd, the coefficient of xn is —lfx from the

1st series and from second it is zero. But if n be of the form (4r-f-2), the coefficient of*"

from the 1st series is 1/n and—2/n from the second, that is the coefficient is — 2/n-f \\n — -^\\n.

But if n be of the form 4«, the coefficient of xn is +l/n from the 1st Series, together with 2/n from the second series, that is, the coefficient is l/n+2/n=3/n, Ans.

21. Let the nth term of the given series be TB ; ther we have

T ^ " 3 - "a - "3 - n ' - l + l "„ n ! n ( n - l ) ! (n-1) I (n-1) P

l j ? ! r L + i •_(«-!? («+i», i («-1) I^Cn- l ) ! (n-1) ! ^ ( t t - l j

^ («+l) • 1 _ n - 2 + 3 . _ 1 (n -2 ) ! ' ( n - 1 ) ! (n-2)! ' ( n - 1 ) !

( n - 2 j r ( n - 2 ) ! ^(n-1) . ' 1_ , 3 1

T « - ( „ -3 ) i "** (n-2> I + (n-1) I

Putting K = 1, 2, 3 etc, we have

l r ~ (-2)1 ( -1)1 0 !

• * ~ : ( - l ) ! 01 r l !

32 • Algebra. •

3 o Fir 21 alwa\s remember that 7— : —r-Vr~0

' (any negative number) ! J T*~n*r!+r!

•

Adding, required sum t, «CTx+Ta+ta+...oo) i

"(oVnf27 +- °°) +3 (£+n+ri+;l- «)

=e-f3e+e«5e Proved. 22. Prove that ',, *

21ogc n-lofeC+U-log, Oi-I)« ^ - + 1 + - ! + 4

L. H. S.=2 log, n-loge («+l)—loge (n-1) \ % = 2 l0gc K-{l0gfi (ft-f-I)-fJoge («+!)} •»

=2 log, n-logc {(n-1) (n+1)} * «log, n2—Ioga in%-\) c=-p0ge(nB-l)-Iogen

a]

= r_!-_lL~j 1 L n2 2n* 3n8 "*J

n* T 2n* T 3n6 ^ "" Ans; 23. ' We know that

1 / , N * * * 3 * 4 •

Examples XVII (sn Page 195) 333

i /i ^ r x2 x* x* i •

Now if we put x = , •-\

on both sides we get

01 ^i+¥Wi?+3_d+ip"1""-=-l0g(sTT) =—logfl+log(n+l)

•**(?*) = l o g ( l + ± )

\n 2n2"t'3«3 4£*+"7

+ I)a 3 (H+1)3"**"" \n 2«2_r3«3 "V " #t+I ' 2 (/improved.

^ w i 9 <. i 2 4 I. t 8 0

24. If Iogc JQ = - » . ' ^ ^ ^ '*°S c 8T,==C'

we are to show that 7a—2ft+3c=Ioge 2, lla-36+5<;=Ioge 3, loge5=l6a-4ft+7c.

(i) 7a-26 + 3c =7 log -^--2 loge 24 + 31oge gQ

( V <7~]oge~,& = 1 0 g e | ^

334 " • Algebra I!

= 7 [log, 2+Iogc 5 - 2 Ioge 3 ] -2 [2 Iogc 5 - 3 Ioge 2-Ioge 3] • ' ;! +-3 [4 loge 3~4lloge 2—log,, 5]

=(7+6-12) loge 2+(7-4-:3) loge 5+(..-14+2r|-12) loge 3 =logc 2. i . : • jj Proved.

(ii) lla-3fe+5c , h - 1 1 l0ge V - 3 logT'll + SiogeW J = 11 [loge-2+1oge 5 -2 Ioge 3 ] -3 [2 loge1'5

-.3 logc 2-Ioge 31+5 (4 Ioge 3 - 4 loge'!2-Ioge 5) =(11+9-20) loge 2+(-22+3+20) lo£ 3

; + ( l l - 6 - 5 ) l o g e 5 =loge 3. I' Proved.

(iii) 16a—46+7c ,j = 16 We V - 4 loge + 7 lOge H jj = l6;[Ioge 2+logc 5 - 2 loge 3] -4 [2 loge|5

- 3 loge 2-log* 3]+7 [4 loge 3 - 4 loge'2-loge 5] =(16+12-28) loge 2+(-32+4+28) logc 3

+((16|-8-7)loge5 = Iogc 5, ' ' I! Proved.

I pw V 0=- log c (A) = -IogeUr-:rV) jj „ 1 J, 1 , 1 " , 1 , 1 f

! ' i •;' 6 x i o a ^ " '

= -l + -005 + -00033333+'OOOp25+-000002' + Jli+'0D000016.

==•105360516 !'; ,...(A) Again 6,= -loge -|t = -lpg e (1—TV)

4 \ -—tofc(i—X55) p

« = ± i _.16"_ , _64_ j 5 6 _ , 1024 <j_ 10 2 +2xl0 1 _ h3xl0° 4xl08 i~5Xl010]T"'

='O4 + 'O008+'O0O02133 + ... + '00000064 jj [f-000000024

Examples XVII (on Page 195) 335

= •040821995 . ...(B)

Similarly c=lo& ^ = = l o | ( 1 4 ^ )

25 , 125 400 2X160UUO 3x64000000

_ J525 4x25600000000

Simplifying, =-012422520 ...(L) V loge 2=7a—2b+3c

= 7 (-105360516)—2 (-040121995)4-3 (-012422520) = •69314718

[putting values of ay bt c from (A), (B), (L)] Again log* 3 = l l a -36+5c

= 11 (-105360516)-3 (040821995) + 5 (-012422520)

= 1*09861229. Similarly log* 5 = 16a--4&-f 7c •

= 16 (-105360516)—4 (040821995) H-7 (-012422520)

= 1;60943792.

',1

50

CHAPTER XVIII ii

INTEREST & ANNUITIES

. EXAMPLES XVIII (a) (on Page 203)

1. p = £ 100, n=50 years, :, 5 'l

' ,,£=Too ^r 'l& *n t e r e s t o n £ ')•

/ . amount M=P (l+r)n=100 ( I + J^Q)' 'i

Taking log of both sides we have I or log M=log 100+50 log ^- 8

'• =ldg 100+50 (log 21-log 20) =2+50 [log (7X3)-log (2x fp)l

11 = 2 log 10+50 [log 3+log 7- tag 2-log IUJ =2+50[4771213+-8450980 I',

f-3010300-11 V =2+50 [1-3222193-1-3010300]

=2+50 [-0211893]=2+l-0594'65 =3*0594650. {

log M=log 1146*74; •" \ ','. M=1146"74=£ 1146. 14 s; 10 d. \ Ans.

•i

2. Let the sum be P, time n years, rate of interest for £ 1 and discount be D. ll

/ . Simple interest of the sum ', P for n years=P«r=£ 90. —CO^

Discount=r-i—=£ 80.

From (i) we have

Examples XVIII (b) (on Page 206) 341

Hence ,amount= 100 (—-• % — j

_100 (1-4JJ7) - 045

n 141 '11 _ 141170 = 0*45 45 = £ 3137; 2 s. 2 | d. Ans.

3. Required rent will be=interest of £ 2750 at 4% 2750x4x1 .

one year= <QQ—=£ 110 Ans. 4. If the rate of interest is r per cent; then we have

1 2 0 = 4 0 0 0 x ^ x 1 or 120=40r:

.-. r = 3 % Ans. 5. Let the number of years be n; then we have

P=P*255Xn

[It be the amount of freehold estate] 20Q 7 „=-^=28£ Ans.

6. Rate per cent= ^ =4

Amount=n A+ -~7t— rA

= 2 x 6 2 5 + ^ x ^ x 6 2 5

[v A-625(B-2,r-120] = 1250+25=^1275 Ans.

7. Rate per«cent= r- = 5

••• '-i4"-°5-QaBdR-(1+s)

342 Algebra

If the annuity be A, then

i4» •JST)_A L r ^ ^ 5 — A tiff)3

2522-A ^ g - j A 2 r , — 2 0 :\

. r926l-r8000T A R.1261X201 " A l_T926l)7^ J~ A |j;~926i " J !

2522x9261 =4; yzo.'^ s. ans

J

A = T 2 6 1 ^ 2 0 = £ 9 2 6 ' 2 S' AnS--8. If M is the sum to be paid then

• ., M J O O X O ^ ; . . . (VU-1-04,

Taking log of both sides we have ' .*. logM=log {10000 x (1-04)-10} l\

^Iog 10000-10 log (1-04) 'A =41ogl0-10C0170333) • =,4-*I70333=3'8296670 \ =Tog 6755*65 ; \\

M=6755 65^£ 6755,, 13 s. Ans. J 9. If the sum] is P ; then

• Amount—P e*r («=number of years, ft r=rate of interest of £\ \

• .-. 500=P e50* <2/i«» or 500=Pe ; - (see §\234) 500 ''< i\

or F=-^=500xe-1=500x(3678), \ ^ 1 — •

.'. P=183*90=£ 183, 18 s. • Ans. ^ 10. Since 30 years* purchase is paid for an annuityUb

continue for In years and 25 years purchase is paid for an Annuity to continue for « years, then with the usual assumed notations, we have '\

2 5 = 1 P ^ T m - ^

Mr 2 5 » g ^ y T ) or 2 5 R n ( R - D - R n - l

or

%or

,or

or

or

or

or

Examples XVIII (b) (on Page 2Vb)

R"(25R-25~1)=-1 Rn(26-25R) = l

•' R « - l

26-25R 1— R-BB

Abo 30= - R ^ ~ Dividing (B) by (A) we get

6 _ J - R - 2 n _ R 2 n - l R" _ R n + l 5 1—R"n R2n * R " - I Rn

- l ^ 1 ^ 3 ^ 2 6 - 2 5 1 1

ii = 27^25R

I35-jl25R=6; .'. R = J*? /. , , 129 4 ( l + r ) = - I 2 T ; .'. r = T f 5

JH3

-.(I)

...(B)

From (i)

•

A v 1 on Hence the rate=100r=--~^=3»<yo ^

11. Here r='Q4t R=l*04, n=*10, V=5000 ; i J)— n

Hence from V=A 0 — - , we have

1—n-041-10

5 O O O = A - L _ L L ^ 1 _ (1-04)10—1 •

5000 C04) = A U( l . 0 4 r

A _ 5000 C04) X O ^ ) 1 0

'"" A ~~ -(l":04)ll)-l snnrw™ 10 ' Suppose (I•«)»•=.* 5000(04) g.1 7 5 5 6 5 ' / . log x=10 log (V041

A = = , =10 (.0170333) ~ " A . 7 ^ - 1 I ='170333

6 775565 =1—-829667 = 5000x4x10x6-775565 I =log 10-log 6-75565-

100x6-7755'65(IO-6T775565)j . _ 10 ^ 2000 _ 4000000Q " x~ lo2 6-77J56T

3-2443"5 64887 . 10

x= :£616,9s,Mid. Ans, , " 6*775565

u g t j u i a • ' •' " 1!

12. Since rate is 5% and the" capital is. OOOO, hence the amount obtained in 17 years will be 20000 (££)"

As he spends £ 1800 every year, so it may! be regarded* as the annuity. *' i

Therefore the man for his expenses must require the .otal amount for 17 years given ^y ]i

^ r 1 ii =36000 ( ^ ) 1 7 - 36000

Now we have't^show-that 36000 f#l)"—36000 must be jreater than 20000 ( » ) " .. e. 36000 (4$)*7-36000 >20000 (^gp \ x 16000 (U)l7> 36000 f, • v 16000 (44)"-36000 > 0 • 1;

• Taking log, we have • ;j log 2d+3 log tO+17 (log 21~log 2Q)-2 log 2 -

\; - 2 1 o g 3 - 3 1 o g l 0 > 0 r 4 tog 24 3+17 (log 3+log 7-log 2-I6g 10) ']

- 2 C3010300)-2 (-4771213)—3>0 r 4(-3010306)+3+17 047712134--84S0980-'3010300-1>

-•6020600--9542426-3>0 : 4-20412004^602181—4-5563026 must be positive r 4*5640381-r4-5563026=a positive quantity. ? i|

Since the amount he. would need is greater, than that ; would 3raw 'in 17 years, hence he shall be ruined before ie end of 17th year. Ans. ji

. C - . _ 500(1-06)-13_500 (1-06)-20

fine= •06

Now if x=(V06)-»;

log * = - 1 3 ('0253059) = -•3289767 =T*6710233

=log (-4688385); :. JC=*46883^5 ;

' ° 6 G [See Art. 242 of the book]

and )>=(-06)-ao fi .*.' logj'==-20C0253b59)

==—•5061180] = T -4938820 =log (-3118042);

.y^'3118042

II

Examples XVtfl (a) (on Page 202) * 337

9 0 =80 1+nr

or 9=8+8/V; nr=$. P«r=90, *

,\ P=90x 8=^720 Ans. 3. Let the sum be P. / . amount (M)=2P and suppose the time is n years.

••• *-* ('+&)" •"-(§)" Taking log of both sides we get

or log 2=n (log 21-log1 20) ;

log 21 —log 20 _, log_2

log3+log7—log 2—log 10 ^ -3010300

•4771213 -K843O98O--3O1O300-1 ^ . -3010300 ^ 1 3222193-1-3010300' •

•3010300 , , . " ^ n 8 9 J - , * 2 y M n Ans.

A. Let P be sum given, V the present value ; tBSh *p«V (l+r)B [r is interest on £ 1]

or 10000=V ( l + f Q V > V ( | ) a

Taking log of both sides we get

log 10000^1og V+8 (log 21 - log 20),

o r 4^1og V+8 (log 3+log 7-log 2-log 10) =logV+8(-4771213-f-8450980~'3010300-l) =logV+81[l'3222l93-l-3O10300)

338 Algebra

=Iog V+8 (0-2U893) \, or logV=4--1695!44 '\ •"' =3*8304856=log 6768-3%

V=6768-394=£ 6768, 7 s. 10J d. APS. 5. *If n be the number of years; then from n=P (1+r)" we have *, f

'2500=1000 ( l + ^ o ) " or S - ^ ) " ;

Taking log of both sides V>

,\ log 5=log 2+n (log 11 —log 10) " or log 10- log 2=log 2+n (log 11 - log 1,0)

' {;: sJ% \ or n (log 11—log 10}==log I0-2Iog 2 ; \ n (1-0413927-1)«1 -2 ('3010300) . 'rt

1-2 (-3010300) _ 1-'6020600 \ " 10413927-1 , "0413927 \

•3979400 ' I 0 1 n = s T 0 4 l 3 9 2 7 = 9 ' 6 y e a r 8 ; Ani 6. Suppose P be the sum due, r the rate of-interest on

£ 1. and n the number of years.-• A

.*. -interest on sum due^Pnr \-<•• i

.". harmonic mean of P and P/ir \\ =2 ( P x P y f „ 2P*iir.* ^ IVnr \

P+Pnr P ( l + w ) \ + nr i P»r "

.". i of harmonic means=~-r——discount. » 1 + w

Hence Proved. \\

7. Let the sum be P and amount M ; then .100 M- >(1+mF-r*$y

E*xamples XVIII (a) (on Page 202) 339

P \20j

taking log of both sides we get

o r l0g ±£=100 (log 21 -log 20)

= 100 (log 3-flog7-log 10-log 2) s=l00(*4771213+*845O98O~l-*3OlO3O0) =100(-0211893)«=2*11893

M , .*. p-=100-f-decimal part«more than a hundred.

Hence Proved. 8. If the sum is P ; then we have 1000=P (1 +Ta^)1 ' or log 1000=Ic*; P+12 (log 106-log 100),

3=IogP+12 (2-0253059-2), 3=log P+12 C0253059); .'. IogP=3-'3036708

_ ^ _ _ =2'6963292=log 496"97 ; .*. P«£496 97 &T £496, i9 s. 4# d. Ans, 9. Let n be the number of ye^rs. *

Since the bill is renewed every half.year, so the number of half years=2n. Then

6000=600 (« + , & • ) "

taking log of both sides Jog 10=2« (log 118—log 100),

l - 2 n (2-071882-2), l=2/i (-071882);

2X(-071882) -143764 «=about 7 years. Ans.

Algebra

1 " / 10. 1 f a r t h i n g s j ^ ^ j ^ ^ - p o u n a U ^ . , 1 *

(VJ 4 farthings=1 d.)

100/

"4x12x20 u

If the amount be M ; then M = l , ( i f

.".' log Ni=200 (log 106-k>g 100) \ =200(2-0253059-2) «5-0611800 \ =log II5127*0 \

M«ll5127ferthwies

Examples XVIII (b) (onVage 206)

1. Amount=nA+^-^—- /'A (formula) 2

.'. 6 7 2 = 5 x l 2 0 + ^ ~ X " 1 2 a x r

•or ' 72=1200r; -' 72 6

1200 100

100 fate percental00.-T=^iX 100=6% Ans|

2. Amount<=*'A * - l \\

Here R, the amount of £ 1 for one ycar= I-045 ^

".-. a m o u n t i ^ o W ^ ' ^

Suppose ' (l-045)20=x ' ,\ Iogx=20 log (1*045)

•" =20 (-0191163), • ''.= ,3823260=log 2-4117;

*=2*4117

Examples XVUI \b) (on Page 206) •

500(4-4688385) 500 (-3118042)

= * * [-1570343]

= 500X100X . 1 5 7 0^ 3

o

6

14. We have

5X10000X1-570343

£1308, 12 s. l | d .

6 5X1570343

Similarly

1-R-"

b=sk~r I -R- 3 "

and c= R - l Squaring-(A)-and (B) we get

R"^R* 1 r»n i o B n

a (R-l)2

i-^2 +-L

(R- l ) 3

Multiplying (A) and (B) we get

- 1 - 1 - l + '

« * — (R=TF~ Multiplying (A) and (C) we get

346 Algebra*

Now <P+b*^ab will be

f 1 R" + R*« + l R»» + R4« ~ * + R* + R7« ~" R3«-

.— n» (Rc-D

i_L—L+.i_ =ac

R" _R3^

(P—ab^b^ac Hence 15. With the help of Art. 240 we may say

present worth due in i year of £ 10 is 10 (J.-05)-1 and worth of £ 20 due in 2,yearsis 20 (I-OS)-'.

Similarly present worth of £ 30 due in 3 year 30 (1-05)-3.

Adding all these to infinity, we get the prese of a perpetual annuity.

==10 (1*05)^+20 (l-05)-3+30T(l-05ra+.»'» = 10 (^)+20 (^)3+30xC4^)3+.T7«r^-" «10(#f ) [ l+2 . (**) + 3 . <^) a+. . .«]

• «10 (£•?•) U— IT!""8 fcy binomial theo know that (l-rA

,)~8=l+2x+3;t2+...co]

200 X(21)2=200x2I

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