Bahasa
Halaman
Hukum
jee main
1. Of the students attending a school party, 60% ofthe students are girls, 40% of the students like to sing. After these students are joined by 20 more boy students, all of whom like to sing, the party is now 58% girls. How many students now at the party like to sing?(a) 248 (b) 250 (c) 252 (d) 256
2. P is a point on the ellipse xa
yb
2
2
2
2 1+ = with foci S
and S1 and eccentricity e. The locus of the incentre of triangle PSS1 is a conic of eccentricity
(a) 11−+
ee (b) 1
1−+
ee
(c) e
e1+ (d) 21
ee+
3. In triangle ABC, if a, b, c are in A.P. and C A− = π2 ,
then cos B =
(a) 12 (b)
23 (c) 3
4 (d) 45
4. The area bounded by the curve x2/3 + y2/3 = 1 is
(a) 32π (b) 3
4π (c) 3
8π (d) 5
16π
5. If Crr =
30, then C r
r4
0
7
=∑ =
(a) 228 (b) 229 (c) 230 (d) 215
jee advanced
6. Let a i j k= − +2 ^ ^ ^,
b i j k= + −^ ^ ^2 and c i j k= + −^ ^ ^ .2A vector in the plane of
b cand whose projection on a
is of magnitude 23
is
Maths Musing was started in January 2003 issue of with the suggestion of Shri Mahabir Singh. The aim of Maths Musing is to augment the chances of bright students seeking admission into IITs with additional study material.
During the last 10 years there have been several changes in JEE pattern. To suit these changes Maths Musing also adopted the new pattern by changing the style of problems. Some of the Maths Musing problems have been adapted in JEE benefitting thousand of our readers. It is heartening that we receive solutions of Maths Musing problems from all over India.Maths Musing has been receiving tremendous response from candidates preparing for JEE and teachers coaching them. We do hope that students will continue to use Maths Musing to boost up their ranks in JEE Main and Advanced.
Set 152
(a) 2 3 3i j k^ ^ ^+ − (b) 2 3 3i j k^ ^ ^+ +(c) − − +2 5i j k^ ^ ^ (d) 2 5i j k^ ^ ^+ +
comprehension
Consider the odd integers 1, 3, 5, 7, 9,....
7. If S = 13 – 33 + 53 – 73 + ... + 213, then S is divisibleby(a) 5 (b) 7 (c) 11 (d) 13
8. If S = 14 – 34 + 54 – 74 + ... + 254, then the sum ofdigits of S is(a) 15 (b) 20 (c) 25 (d) 30
integer match
9. If 2x+y = 6y and 2x = 3 × 2y+1, then x = loga3⋅logb6,where ab is
matching list
10. Column II contains curves satisfying the conditionsin Column I.
Column-I Column-IIP. Subtangent is constant 1. parabolaQ. Subnormal is constant 2. ellipse
R. Subtangent = twice the abscissa 3. hyperbola
S. Subnormal = twice the abscissa 4. exponential
curveP Q R S
(a) 4 1 1 3 (b) 2 3 3 1 (c) 1 4 4 2 (d) 1 2 3 4
See Solution set of Maths Musing 151 on page no. 80
| august ‘1512
MEASUREMENT OF AN ANGLE
Sexagesimal orBritish system
Centesimal orfrench system
Circularsystem
1. Sexagesimal or British System : In this system ofmeasurement, a right angle is divided into 90 equalparts called degrees.Each degree is then divided into 60 equal parts calledminutes and each minute is further divided into 60equal parts called seconds.Thus, 1 right angle = 90°, 1° = 60′ and 1′ = 60′′
2. Centesimal or French System: In this system ofmeasurement, a right angle is divided into 100 equalparts called grades. Each grade is then divided into100 equal parts called minutes and each minuteis further divided into 100 equal parts calledseconds.Thus, 1 right angle = 100g, 1g = 100′ and 1′ = 100′′
3. Circular System : The angle subtended at the centre ofa circle by an arc whose length is equal to the radius ofthe circle is called a radian and is denoted by 1c.1 right angle = (p/2) radians.
y
TRIGONOMETRIC FUNCTIONS AND PRINCIPLEOF MATHEMATICAL INDUCTION
Series3
The length l of an arc PQ of a circle of radius r is given by l = rq, where q is the angle subtended by arc PQ at the centre of the circle in terms of radians.
Relation between DegRee anD RaDian
Radian measure of an angle = ×p180
degree measure of the angle
Degree measure of an angle = ×180p
radian measureof the angle.
Thus, if the measure of an angle in degrees and radians
be D and C respectively, then D C
180=
p
tRigonometRic FunctionsTrigonometric ratios for an acute angle is defined as the ratio of sides of a right angled triangle. The extension of trigonometric ratios to any angle in terms of radian measure are called trigonometric functions.
sin q = yr
, cosq = xr
, tan ,q =yx
cosec , sec , cotq q q= = =ry
rx
xy
P
Ql
r
r y
xO
X
YP x y( , )
signs oF tRigonometRic Functions in FouR DiFFeRent QuaDRants
Quadrant Signs of various T-functions
I All T-functions are positive.
II sinq and cosecq are positive. All others are negative.
III tanq and cotq are positive. All others are negative.
IV cosq and secq are positve. All others are negative.
sin and cosec are +veII
All are +veI
tan and cot are +veIII
cos and sec are +veIV
XX
Y
Y
O
| August ‘15 13
Domain and Range of Trigonometric functions
f(x) = sin xDf = R
Rf = [–1, 1]
f(x) = cos xDf = R
Rf = [–1, 1]f(x) = tan x
Df = R – {(2n + 1)p/2 ; n ∈ Z} Rf = R
f(x) = sec xDf = R – {(2n + 1)p/2 ; n ∈ Z}
Rf = R – (–1, 1)
f(x) = cosec xDf = R – {np ; n ∈ Z}
Rf = R – (–1, 1)
f(x) = cot xDf = R – {np ; n ∈ Z}
Rf = R
gRaphs oF tRigonometRic Functions
4π3π
2π π π2
π 2π 3π4π
Y
Y′
X′ X1
O
1
y = sin x y = cos x
O
y = tan x y = cot x
Y
y = sec x
Y
y = cosec x
| August ‘1514
Values oF t-Functions oF some paRticulaR angles
q 0p6
p4
p3
p2
p32p
2p
sin q 012
12
32
1 0 –1 0
cos q 1 32
12
12
0 –1 0 1
tan q 013
1 3not
defined 0 not defined 0
Note : sin (• q + 2np) = sin q, cos (q + 2np) = cos qtan (q + np) = tan q, cot (q + np) = cot qsec (q + 2np) = sec q, cosec (q + 2np) = cosec qIn case of –• q, p ± q, 2p ± q, t-ratio remains the
same and in case ofp q p q2
32
± ±, , t-ratio changes
as follows : sin → cos, cos → sin, tan → cot,cot → tan, sec → cosec, cosec → sec.
sin (–• q) = – sinq, cos (–q) = cos qtan (–q) = –tanq, cot (–q) = – cot qsec (–q) = secq, cosec (–q) = – cosec q
some impoRtant FoRmulae
sin(• A + B) = sinA cosB + cosA sinBsin(• A – B) = sinA cosB – cosA sinBcos(• A + B) = cosA cosB – sinA sinBcos (• A – B) = cosA cosB + sinA sinB
tan( ) tan tantan tan
A B A BA B
+ = +−1 •
tan( ) tan tantan tan
A B A BA B
− = −+1
•
cot( ) cot cotcot cot
A B A BA B
+ = −+
1 •
cot( ) cot cotcot cot
A B A BB A
− = +−
1 •
sin sin cos tantan
2 2 21 2A A A A
A= =
+ •
cos2• A = cos2A – sin2A = 1 – 2sin2A
= 2cos2A–1 = −+
11
2
2tantan
AA
tan tantan
2 21 2A A
A=
− •
sin3• A = 3sinA – 4sin3Acos3• A = 4cos3A – 3cosA
tan tan tantan
3 31 3
3
2A A AA
= −−
•
sin sin sin cosA B A B A B+ = +
−
2
2 2 •
sin sin cos sinA B A B A B− = +
−
22 2
•
cos cos cos cosA B A B A B+ = +
−
2
2 2 •
cos cos sin sinA B A B B A− = +
−
2
2 2 •
2si• nAcosB = sin (A + B) + sin(A – B)
2cos• A sinB = sin(A + B) – sin (A – B)
2cos• AcosB = cos(A + B) + cos (A – B)2sin• AsinB = cos (A – B) – cos (A + B)
sin cos
,
,
A A
A
A21
2
2
2
= ± −
+
−
+ ve if lies in I or IIquadrants
ve if lies in IIII or IV
quadrants
•
cos cos
,
,
A A
A
A21
2
2
2
= ± +
+
−
ve if lies in I or IVquadrants
ve if lies in II oor IIIquadrants
•
| August ‘15 15
tan coscos
A AA
A
A211
2
2
= ± −+
+
−
ve, if lies in I or IIIquadrants
ve, if liees in II or IVquadrants
•
tRigonometRic eQuations
1. Trigonometric equation : Equation involving one ormore trigonometric functions (ratios) of unknownangle is called a trigonometric equation.
2. Solution or root of a trigonometric equation : Avalue of the unknown angle which satisfies the giventrigonometric equation is called a solution or root ofthe equation.
3. Principal solution of a trigonometric equation :The solutions of a trigonometric equation lying inthe interval [0, 2p) are called principal solutions.
4. General solutions of some trigonometric equations:The solution consisting of all possible solutionsof a trigonometric equation is called its generalsolution.sin • q = 0 ⇒ q = np, n ∈ Z
cos • q = 0 ⇒ q p= + ∈( ) ,2 12
n n Z
tan • q = 0 ⇒ q = np, n ∈ Z
cot • q = 0 ⇒ q p= + ∈( ) ,2 12
n n Z
sin • q = sin a ⇒ q = np + (–1)n a, n ∈ Zcos • q = cos a ⇒ q = 2np ± a, n ∈ Ztan • q = tan a ⇒ q = np + a, n ∈ Z
pRinciple oF mathematical inDuctionMathematical statements : Statements involving mathematical relations are known as the mathematical statements. We shall be using notation P(n) or P1(n) or P2(n) etc. to denote such statements.First principle of mathematical induction :Let P(n) be a statement involving the natural number n such that(I) P(1) is true i.e. P(n) is true for n = 1 and(II) P(m + 1) is true, whenever P(m) is true i.e. P(m) is true ⇒ P(m + 1) is true.Then, P(n) is true for all natural numbers n.Second principle of mathematical induction :Let P(n) be a statement involving the natural number n such that(I) P(1) is true i.e. P(n) is true for n =1 and
(II) P(m + 1) is true, whenever P(n) is true for all n, where 1 ≤ n ≤ m.Then, P(n) is true for all natural numbers.
Very short answer type
1. Find the radian measure of –37°30′.
2. Find the value of cosec (–1410°).
3. Prove that : sin
coscot
21 2
xx
x−
=
4. Find the general solution of cosec q = 2.
5. If P(n) is the statement "n(n + 1) (n + 2) is divisibleby 12", prove that the statements P(3) and P(4) aretrue, but P(5) is not true.
short answer type
6. If tan a = x + 1, tan b = x – 1, then show that2cot (a – b) = x2.
7. If a + b = 90°, find the maximum value of sin a sin b.
8. Solve the equation sin x + sin 3x + sin 5x = 0.
9. Using the principle of mathematical induction,prove that n < 2n for all n ∈ N.
10. If sin andq p q p= − < <45
32
, find the values of all
the other five trigonometric functions.Long answer type
11. (i) Solve : 2 sin2 x – 5 sin x cos x – 8 cos2 x = –2
(ii) Show that cot cotp p4 4
1+
⋅ −
=x x
12. Prove that cos cos cos cos4 4 4 48
38
58
78
32
p p p p+ + + = ,
13. Ifsin( )cos( )
q aq a
+−
= −+
11
mm
, then prove that
tan . tanp q p a4 4
−
−
= m
14. If tan( ) tan( ) tan( )q a q b q g+
=+
=+
a b c, prove that
a ba b
b cb c
+−
− + +−
−sin ( ) sin ( )2 2a b b g
+ +−
− =c ac a
sin ( )2 0g a
15. If A + B + C = p, then prove that
cos cos cos sin sin sin .2 2 22 2 2
2 12 2 2
A B C A B C+ + = +
| August ‘1516
soLutions
1. we know that, 180 1180
° = ⇒ ° =
p pcc
Now, − ° ′ = × −
37 30180
752
p c= −
524
p c
2. cosec(–1410°) = – cosec 1410°= – cosec (4 × 360° – 30°) = cosec 30° = 2
3. L.H.S.sin
cos=
−2
1 2x
x=
2
2 2sin cos
sin
x x
x = cotx = R.H.S.
4. We have, cosec q =2
⇒ =sinq 12
⇒ =sin sinq p6
⇒ = + − ∈q p pn n Zn( ) ,16
5. P(3) is the statement "3(3 + 1) (3 + 2) = 60 isdivisible by 12", which is true.P(4) is the statement "4(4 + 1) (4 + 2) = 120 is divisibleby 12", which is also true.P(5) is the statement "5(5 + 1) (5 + 2) = 210 is divisibleby 12". Clearly it is not true.
6. We have, tana = x + 1 ⇒ cota = 11x +
... (i)
tanb = x – 1 ⇒ cotb = 1
1x −... (ii)
L.H.S. = − =+
−2
2 1cot( )
(cot . cot )cot cot
a ba bb a
= + −+
−−
+
2 11
11
1
11
11
x x
x x
.[From (i) and (ii)]
= −+
+ − −
−
=2 1
11
1 1
1
22
2
2
2xx x
x
x( ) ( )
. = =x2 R .H.S
7. Let y = sin a sin b
⇒ y = = − − +12
2 12
( sin sin ) [cos( ) cos( )]a b a b a b
= − − ° = −12
90 12
[cos( ) cos ] cos( )a b a b
Since maximum value of cos(a – b) is 1
\ Maximum value of y = 12
8. Given equation is sin x + sin3x + sin 5x = 0⇒ (sin 5x + sin x) + sin 3x = 0⇒ 2 sin 3x cos 2x + sin 3x = 0⇒ sin 3x (2 cos 2x + 1) = 0⇒ sin 3x = 0 or 2 cos 2x + 1 = 0Now sin 3x = 0 ⇒ 3x = np, n ∈ Z
⇒ = ∈xn
n Zp
3, ...(i)
Also, 2 cos 2x + 1 = 0
⇒ = − ⇒ =cos cos cos2 12
2 23
x x p
⇒ 2 2 23
x n n Z= ± ∈p p ,
⇒ x n n Z= ± ∈( ) ,3 13p ...(ii)
But solution in (ii) are induced in solution (i), because (3n ± 1) is also an integer. Hence, required
solution is x n n Z= ∈p3
,
9. Let P(n) : n < 2n ∀n∈NWhen n = 1, L.H.S. = 1 and R.H.S. = 21 = 2Clearly, 1 < 2Thus, P(n) is true for n = 1, i.e., P(1) is true.Let P(k) be true. Then, P(k) : k < 2k ∀ k ∈ NNow, k < 2k
⇒ 2k < 2k + 1 ⇒ (k + k) < 2k + 1
⇒ (k + 1) ≤ (k + k) < 2k + 1 [Q 1 ≤ k]⇒ (k + 1) < 2k + 1
\ P(k + 1) : (k + 1) < 2k + 1
⇒ P(k + 1) is true, whenever P(k) is true.Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.
10. Clearly, q lies in the third quadrant in which tanqand cotq are positive and all the other trigonometricfunctions are negative.
We have, sin , cosecsin
q qq
= − = = −45
1 54
Now, cot (cosec )2 2 1 2516
1 916
q q= − = −
=
⇒ = =cot ,q 916
34
tancot
= =1 43
Also, cos sinq q= − − = − − = −1 1 1625
35
2
\ = = −seccos
1 53
| August ‘15 17
Hence, cos ; tan ; cot ;q q q= − = =35
43
34
sec and cosecq q= − = −53
54
11. (i) Given equation is2 sin2x – 5 sin x cos x – 8 cos2x = –2Dividing both sides by cos2x, we get2 tan2x – 5 tan x – 8 = –2 sec2 x
⇒ 2 tan2 x – 5 tan x – 8 + 2 (1 + tan2x) = 0⇒ 4 tan2 x – 5 tan x – 6 = 0⇒ 4 tan2 x – 8 tan x + 3 tan x – 6 = 0⇒ 4 tan x (tan x – 2) + 3 (tan x – 2) = 0⇒ (tan x – 2) (4 tan x + 3) = 0\ Either tan x – 2 = 0 ⇒ tan x = 2 = tan a (say)⇒ x = n p + a, where tan a = 2, n ∈Z
or 4 tan x + 3 = 0 ⇒ = − =tan tan ( )x 34
b say
⇒ x = np + b, where tan b = − 34
, n ∈Z
(ii) L.H.S. =+
⋅ −
+
−
cos cos
sin sin
p p
p p4 4
4 4
x x
x x
=−
−=
−
−=
cos sin
sin sin
sin
sin
2 2
2 2
2
24
4
1212
1
p
p
x
x
x
x
12. We have, 78 8
58
38
p p p p p p= − = −and
⇒ = − = −cos cos cos cos78 8
58
38
p p p pand
⇒ = =cos cos cos cos4 4 4 478 8
58
38
p p p pand
\ = +L.H.S 28
2 38
4 4cos cosp p
=
+
2
838
22
22
cos cosp p
= +
+
+
2 14
2
1 34
2
2 2cos cosp p
= +
+ +
24
14
1 34
2 2cos cosp p
= +
+ −
=1
21 1
21 1
232
2 2 = RHS
13. Given,sin( )cos( )
q aq a
+−
= −+
11
mm
⇒+ + −+ − −
= − + +− − −
sin( ) cos( )sin( ) cos( )
q a q aq a q a
1 11 1
m mm m
[By componendo and dividendo rule]
⇒+ + − −
+ − − −
=−
sin( ) sin ( )
sin( ) sin ( )
q a p q a
q a p q a
2
2
22m
== − 1m
⇒
+ + − + + − + −
+ + − + + − + −
2 22
22
2 22
2
sin cos
cos sin
q a p q a q a p q a
q a p q a q a p q a
22
1= −m
⇒+
−
+
−
= −sin . cos
cos . sin
p a q p
p a q p4 4
4 4
1mm
⇒ +
−
= −tan . cotp a q p4 4
1m
⇒ − +
−
= −tan cotp a p q4 4
1m
⇒ = +
−
m cot tanp a p q4 4
= − +
−
tan tanp p a p q2 4 4
= −
−
tan . tanp q p a4 4
14. We have,tan( ) tan( )q a q b+
=+
a b
⇒ =++
ab
tan( )tan( )
q aq b
⇒ +−
=+ + ++ − +
a ba b
tan( ) tan( )tan( ) tan( )
q a q bq a q b
⇒ +−
=+ +
−a ba b
sin( )sin( )
2q a ba b
| August ‘1518
⇒ +−
− = + + −a ba b
sin ( ) sin( )sin( )2 2a b q a b a b
⇒ +−
−a ba b
sin ( )2 a b
= + + −12
2 2{ sin( )sin( )}q a b a b
⇒ +−
−a ba b
sin ( )2 a b
= + − +12
2 2 2 2{cos( ) cos( )}q b q a ... (i)
Similarly, we haveb cb c
+−
− = + − +sin ( ) { cos( ) cos( )}2 12
2 2 2 2b g q g q b... (ii)
and, sin ( )c ac a
+−
−2 g a
= + − +12
2 2 2 2{cos( ) cos( )}q a q g ... (iii)
Adding (i), (ii) and (iii), we geta ba b
b cb c
+−
− + +−
−sin ( ) sin ( )2 2a b b g
+ +−
−c ac a
sin ( )2 g a
= 12
{cos(2q + 2b) – cos(2q + 2a) + cos(2q + 2g)
– cos(2q + 2b) + cos(2q + 2a) – cos(2q + 2g)}
= × =12
0 0
15. L.H.S = cos cos cos2 2 22 2 2A B C+ +
= + + + + +12
1 12
1 12
1( cos ) ( cos ) ( cos )A B C
= + + +32
12
(cos cos cos )A B C
= + +
−
+
32
12
22 2
cos cos cosA B A B C
= + −
−
+ −
32 2 2 2
12
1 22
2cos cos sinp C A B C
= + −
−2
2 2 22sin cos sinC A B C
= + −
−
22 2 2
sin cos sinC A B C
= + −
− − +
2
2 2 2 2sin cos sinC A B A Bp
= + −
− +
2
2 2 2sin cos cosC A B A B
= +
2
22
2 2sin sin sinC A B
= +
=2 1
2 2 2sin sin sinA B C R.H.S.
nn
| August ‘15 19
| august ‘1520
1. It is given that the polynomial P(x) = x3 + ax2 + bx + chas three distinct positive integer roots and P(14)= 231. Let Q(x) = x2 – 2x + 232. It is also given thatthe polynomial P(Q(x)) has no real zeros. Then|a| =(a) 21 (b) 43 (c) 67 (d) 91
2. Evaluate the infinite series
S = − + − +1 21
32
43
3 3 3
! ! !....
(a) 1e
(b) −1e
(c) 2e
(d) −2e
3. The smallest possible sum of a1b1 + a2b2 +.......+ a8b8 where a1, a2, ...., a8 and b1, b2, ..., b8 are rearrangements of the binomial coefficients (7C0, 7C1, 7C2, ...., 7C7) is (a) 14C3 (b) 2·14C3
(c) 12
147⋅ C (d) (14C7)2
4. In the sequence 1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8,..., the number occupying the 100th position is(a) 26 (b) 210 (c) 2100 (d) 2101
5. Evaluatea
a a
a
a a
a
a a13
1 12
23
2 22
1013
101 10121 3 3 1 3 3 1 3 3− +
+− +
+ +− +
...
where each a nn =
101(a) 50 (b) 51 (c) 100 (d) 102
6. The number of positive real numbers xsatisfying the equation (8x – 5x)(7x – 2x)(6x – 4x)+ (9x – 4x)(8x – 3x)(5x – 2x) = (105)x is k then
k10
= , ([·] denotes G.I.F.)
(a) 0 (b) 1 (c) 2 (d) 37. A gambler tosses a fair coin and scores one point
for each head that turns up and two points for each
tail. The probability of the gambler scoring exactly 10 points is
(a) 20493072
(b) 30123048
(c) 8731024
(d) 4191512
8. If three distinct numbers a, b, c are roots of theequation x3 – x2 – x – 1 = 0 then a possible value ofa b
a bb c
b cc a
c a
a a a a a a−−
+ −−
+ −−
(a > 21, a ∈ N) is
(a) 2 (b) 2 (c) 310
(d) − 13
9. Let set S = {a1, a2, ..., an}. The G.M. of the G.M.'s ofall non-empty subsets of S is
(a) aii
n n
=∏
1
1/
(b) aii
n
=∏
1
(c) 1
1nai
i
n
=∏
(d) 1
1nai
i
n
=∑
10. A set of n n( )+12
distinct numbers are arranged at
random in a triangular arrangement as shown. LetTk be the largest number in the kth row from thetop. The probability that T1 < T2 < T3 < ... < Tn is(n ≥ 3)
(a) 2
1
n
n( )!+ (b) 2
1
1n
n
+
+( )!(c) n n
+12
(d) nn2 1+
11. Two persons Mr. A and Mr. B randomly select aninteger from 1 to 9, independently of each other.The most likely value(s) for the units digits of thesum of their numbers is(a) 0 (b) 3 (c) 7 (d) 9
| august ‘15 21
12. The reduced fractionpq
with q ∈ [8, 99] is closest
to 37
, then q =
(a) 23 (b) 45 (c) 83 (d) 9613. The number of real values of p for which the solutions
a, b of the equation x2 – 2px + p2 – 2p – 1 = 0
have the property that ( )[( ) ]a ba b
− −+ +
2
22
2 2 is an integer
is/are(a) zero (b) 2 (c) 4 (d) ∞
14. The bottom part of a network of roads is shownbelow. It extends upto level 99 in a similar way. 2100
people leave point A (level0). At every intersection halfgo in L direction and half inR direction. The number ofpeople ending up at the next-to-left position at level 99 is(a) 100 (b) 198 (c) 210 (d) 29
15. Suppose an + 1 = 2(an + bn), bn + 1 = 2bn for alln ∈ N. If an + 3 = tan + kbn and bn + 3 = yan + gbn then t + k + y + g =(a) 40 (b) 45 (c) –40 (d) –45
16. The value of 4 3 129 72 3+ − =(a) 9 (b) 9 3(c) 9 3+ (d) 9 3−
17. A set of positive integers has the Y-property if it hasthree distinct elements that are the lengths of thesides of a triangle whose area is positive. Considerthe set {4, 5, 6, ..., t} of consecutive integers, all ofwhose 10 element subsets have the Y-property. Thelargest possible value of t is(a) 251 (b) 252 (c) 253 (d) 254
18. The number of integers for which 8 3 42 1
3 2( )n nn
− +−
is an integer is(a) 4 (b) 8 (c) 11 (d) 28
19. Let E = 1! 2! 3! ... 100!. The factor that must beerased from E so that E becomes a perfect squareis(a) 25! (b) 50! (c) 99! (d) 100!
20. Mr. X wrote 21 numbers on a circle. Each numberwas half the sum of its two neighbours. If one of thenumber is 3, the sum of all the numbers written onthe circle is(a) 21 (b) 63 (c) 42 (d) 84
21. The number of common terms in the two equationsA : 1, 1, 3, 5, 11, 21, ....B : 1, 7, 17, 55, 161, 487, ... is/are
(a) 1 (b) 2 (c) 10 (d) 21
22. A natural number n for which 39 + 312 + 315 + 3n isa perfect cube of an integer is(a) 13 (b) 14 (c) 17 (d) 19
23. The largest 3-digit prime factor of the integer2000C1000 is(a) 559 (b) 661 (c) 743 (d) 893
solutions
1. (a) : Q(x) = x2 – 2x + 1 + 231 = (x – 1)2 + 231⇒ Q(x) takes all values from [231, ∞)Let P(x) = (x – x1)(x – x2)(x – x3)P(14) = 231 = (14 – x1)(14 – x2)(14 – x3) = 3 × 7 × 11So, 14 – x1 = 3, 14 – x2 = 7, 14 – x3 = 11
a = –(x1 + x2 + x3) = –(11 + 7 + 3)⇒ |a| = 21
2. (b) : (n + 1)th term Tn
nnn
n+ = − ⋅
+= ∞1
31
10( )
( )!
, to
= − − − + − + +( )!
[ ( )( ) ( ) ]1 1 2 6 1 7 1n
nn n n n n n
= −−
+ −−
+ −−
+ −( )( )!
( )( )!
( )( )!
( )!
13
6 12
7 11
1n n n n
n n n n
S Ten
n= = −
∑ +=
∞1
0
1
3. (b) : By rearrangement inequality, we know thatSa1b1 is minimized when one of the two sequences (a1, a2, ...) and (b1, b2, ...) is increasing and other is decreasing. Since, the binomial coefficients are increasing from (7C0 to 7C3) and decreasing from (7C4
to 7C7), the minimum sum is 2 7 74
0
3( ) ( )C Ck k
k+
=∑
= ∑ =−=
2 27 73
0
3 143( ) ( ) [ ]C C Ck k
k
4. (a) : an = 2k, where k is the largest power of 2appearing in the expansion of n when n is expressed as sum of powers of 2.
100 = 22 + 25 + 26, so a100 = 26
5. (b) : Ta
a a
a
a an
n
n n
n
n n=
− +=
+ −
3
2
3
3 31 3 3 1( )
=
+ −n
n n
3
3 3101( )⇒ Tn + T101 – n = 1
| august ‘1522
So T1 + T100 = 1T2 + T99 = 1------------------------------
T50 + T51 = 1⇒ S100 = 50
S100 + T101 = 516. (a) : Notice that x = 1 is a solution.Dividing throughout by (105)x, we have
85
1 1 27
2 43
−
−
−
xx x
xx
x
+
−
−
−
97
47
83
1 1 25
x x xx x
=x
1
Now, using the concept that if a > b > 0, then ax – bx is monotonically increasing. For x ≥ 0, we see that the above equation has no solution other than x = 1.So, only one solution exist.7. (a) : Let Pn be the required probability then
P P Pn n n= ⋅ +− −12
121 2 with P1
12
= and
P212
12
12
34
= + ⋅ =
Forming the character equation of the above equation
and solving it, we have Pn
n= + −
13
2 12
and so, P1020493072
=
8. (b) : Let r a ba b
s b cb c
t c ac aa
a a
a
a a
a
a a= −
−= −
−= −
−, ,
Now, ( )
( )( )
a a a a
b b b br r r r
n
n
n n n n
3 2
3 2
3 2 1
1 0
1 00
− − − = ×
− − − = ×+
− − − =
↓+ + +
Similarly, for s and t. Now an easy induction proves that rn + sn + tn is an integer for all n ∈ N.9. (a) : Consider a particular element ai. It occurs inn – 1Ck – 1, k element subsets of S with exponent 1
kin
each of the corresponding G.M.'s. Hence the exponent of ai in the G.M. of the G.M.'s is given by
12 1
1 12 1
111
1 1nn
kk
n
nn
kk
nC
kC
n−⋅ ⋅∑ =
−⋅∑−
−= =
= ⋅−
⋅ =∑=
1 12 1
11n
Cnn
nk
k
n
which is same as that in the G.M. of S.
10. (a) : Let Pn be the probability when there are nrows. Clearly P1 = 1, P2 = 2/3. In general, the largest number should go in the last row, the probability for
which is nn n n( )+
=+1
2
21
. There is no restriction on the
remaining numbers of the last row. The probability that the numbers of the previous (n – 1) rows are suitably
placed is Pn – 1. Hence, Pn
Pn n=+
⋅ −2
1 1
So, Pnn
n=
+2
1( )!11. (a) : Out of 81 equally likely pairs, lets find thenumber of ways each sum can occur.2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 181 2 3 4 5 6 7 8 9 8 7 6 5 4 3 2 1
So, last digit 0 occurs in 9 ways and other last digits occurs in 8 ways.
12. (d) : Consider37
3 77
− =−p
qq p
q| |
we find q as large as possible with |3q – 7p| = 1. We look for multiples of 7 which are slightly less than 300 and are not divisible by 3. The largest is 287 = 41 × 7 which differs from 3 × 96 by 1. So reduced fraction pq
is 4196
.
13. (b) : a + b = 2p, ab = p2 – 2p – 1⇒ (a – b)2 = 8p + 4
np
pI p=
++
∈ ⇒ = −4 1
4 21
42and
p n nn
= ± − −1 1 2 12
( )
⇒ = =n p1 12
, .so
14. (b) : This is like Pascal's triangle. At level n, the
fraction of people in horizontal position k is n
kn
C
2and so the number of people in position at level 99 is 99
199
100
22 198
C× =
15. (a) :ab
Aab
A =n
n
n
n
+
+
=
1
12
1 10 1
where
So, ab
Aab
n
n
n
n
+
+
=
3
3
38
⇒ t = 8, k = 24, y = 0, g = 8
16. (a) : 4 3 9 4 3 4 3 9 4 3 92 ( ) | |
17. (c) : Let us find a 10-element set S that does nothave Y-property. Starting from the smallest possible values S = {4, 5, (4 + 5), 5 + (4 + 5), ....}S {4, 5, 9, 14, 23, 37, 60, 97, 157, 254}. So the largest set having the Y-property is
{4, 5, 6, ...., 253}So, tmax = 253
18. (b) : Note that8 3 4
2 14 10 5 27
2 1
3 22( )
( )n n
nn n
n
So, 2n – 1 = ±27, ±9, ±3, ±1
19. (b) : E = 1! 2! 3! ... 100!= (1!)2 · 2 ·(3!)2 · 4 .... (99!)2 · 100= (1! 3! 5! ... 99!)2 × 250 × [1 2 ... 50]= (1! 3! 5! ... 99!)2 × 250 × 50!
20. (b) : Let ai be the maximum of the numbers writtenthen ai ai + 1 and ai ai – 1
a
a ai
i i1 12
but according to question, aa a
ii i
1 12
So, ai = ai + 1 and ai = ai – 1 All numbers are same = 21. Sum of numbers = 3 × 21 = 63. 21. (a) : mod 8 : A : 1, 1, 3, 5, 3, 5, ....
B : 1, 7, 1, 7, 1, 7, ....An easy induction reveals that A and B are periodic and 1 is the only common term.22. (b) : (33)3[1 + 3 · 32 + 36 + 3n – 9] 3n – 9 – 35 = 0So, n = 14
23. (b) : 20001000
2000 1999 10011000 999 3 2 1
C
.........
The greatest 3-digit prime factor is the one which occurs once in denominator and atleast twice in numerator.
It must be less than 13
2000 and near to it.
1. Let A = [aij]m × n be a matrix such that aij = 1 for alli, j. Then(a) Rank (A) > 1 (b) Rank (A) = 1(c) Rank (A) = m (d) Rank (A) = n2. The equation
( ) ( )x y x y− + + + + =2 2 42 2 2 2 represents(a) a circle (b) a parabola(c) a pair of lines (d) an ellipse
3. lim ( !) /
n
nnn→ ∞
1 equals
(a) e (b) e–1 (c) 1 (d) None of these
4. Given thatx
x a x b x cdx
2
2 2 2 2 2 20 ( )( )( )+ + +
∞∫
=+ + + + +
∞
∫p
2 4 92 20( ) ( ) ( ), then
( )( )a b b c c adx
x x is
(a) p60 (b)
p20 (c)
p40
(d) p80
5. For all real x x x, sin cos4 42 2
+ is (a) ≥ 4 (b) > 4(c) < 4 (d) None of these6. The length of the perpendicular drawn from
(1, 2, 3) to the line x y z− = − = −
−6
37
27
2 is
(a) 4 (b) 5 (c) 6 (d) 77. The number of equivalence relations on the set{1, 2, 3} containing (1, 2) and (2, 1) is(a) 3 (b) 1 (c) 2 (d) None of these
8. Let A x R x B x R x= ∈ ≥ 1
= ∈ ≥
: and :2
34
If f : A → B is defined as f(x) = x2 – x + 1, then the solution set of the equation f(x) = f –1(x) is
(a) {1} (b) {2} (c) 12
(d) None of these
9. The complex number z = x + iy which satisfy the
equation z iz i
−+
=55
1 lie on
(a) the x-axis (b) the straight line y = 5(c) a circle passing through origin (d) None of these10. The equation of the curve satisfying thedifferential equation y(x + y3)dx = x(y3 – x)dy and passing through the point (1, 1) is (a) y3 – 2x + 3x2y = 0 (b) y3 + 2x + 3x2y = 0(c) y3 + 2x – 3x2y = 0 (d) None of these
11. If lim ,x a
x a
x aa xx a→
−−
= − 1 then a is equal to
(a) 0 (b) e(c) 2 (d) None of these12. Let f(x) = sin x, g(x) = x2 and h(x) = loge x.If G(x) = hogof(x), then G′′ (x) is equal to (a) 2 cosec2x cotx (b) –2 cosec2x cotx(c) 2 cosec2x (d) –2 cosec2x13. For any integer n, the integral
e n x dxxcos cos ( )2 3
02 1+∫
p has the value
(a) 1 (b) p (c) 2p (d) None of these
14. The area defined by 1 ≤ |x – 2| + |y + 1| ≤ 2 is(a) 2 (b) 4 (c) 6 (d) None of these15. C0, C1, C2 ,...... , C15 are the binomial coefficients inthe expansion of (1 + x)15, then CC
CC
CC
CC
1
0
2
1
3
2
15
14
2 3 15+ + + +..... is
(a) 32 (b) 64(c) 128 (d) None of these16. Let the function, f, g, h are defined from Rto R such that f x x g x x( ) , ( )= − = +2 21 1 and
h xx
x x( )
ifif
=<≥
0 00
, then (ho(fog))(x) is
(a) x (b) x2 (c) 0 (d) None of these
AMUENGG.
| august ‘1524
17. AB is a diameter of a circle and C is any point onthe circumference of the circle. Then(a) the perimeter of DABC is maximum when it is
isosceles(b) the area of DABC is minimum when it is isosceles(c) the area of DABC is maximum when it is isosceles(d) None of these
18. If
a a a
b b b
c c c
2 3
2 3
2 3
1
1
1
+
+
+
= 0 and vectors (1, a, a2), (1, b, b2)
and (1, c, c2) are non coplanar, then the product abc equals (a) 2 (b) –1 (c) 1 (d) 0
19. If
a i j b j k= + = +4 6 3 4and then the vectorform of component of
a balong is
(a) 18
10 33 4( )j k + (b)
1825
3 4( )j k +
(c) 18
33 4( )j k + (d)
1825
4 6( )i j +
20. Let f(x) = sinx – tan x, x ∈ (0, p/2), then tangentdrawn to the curve y = f(x) at any point will (a) lie above the curve (b) lie below the curve(c) nothing can be said (d) be parallel to a fixed line21. If [x] denotes the greatest integer ≤ x, then thevalue of
lim | |[cos ]
x
xx→ 0
is
(a) 0 (b) 1(c) –1 (d) does not exists22. The differential equation of all conics whose centrelies at origin is of order (a) 2 (b) 3 (c) 4 (d) None of these
23. Let
a b c, , be three vectors such that
a b c× =and
c a b× = , then(a)
a b c. | |= 2 (b)
c a b. | |= 2
(c)
b c a. | |= 2 (d)
a b c| | ( )×24. If cos–1x > sin–1x, then x belongs to the interval(a) (– ∞, 0) (b) (–1, 0)
(c) 0 12
,
(d) −
1 12
,
25. The number of non-zero diagonal matrices of order 4
satisfying A2 = A is(a) 2 (b) 4 (c) 16 (d) 1526. Find the number of ways in which 52 cards can bedivided into 4 sets, three of them having 17 cards each and the fourth one having just one card.
(a) 52
17 3( ) (b) 52
17 33( ) (c)
51
17 3( ) (d)
51
17 33( )27. If R is a relation on a finite set having n elements,then the number of relations on A is
(a) 2n (b) 22n (c) n2 (d) nn
28. The area bounded by the curve y = f(x), x-axis andthe ordinates x = l and x = b is (b – 1) sin(3b + 4). Then f(x) is (a) (x – 1)cos (3x + 4) (b) sin(3x + 4)(c) sin(3x + 4) + 3(x – 1)cos(3x + 4) (d) None of these29. If the angles of a triangle are in the ratio 4 : 1 : 1,then the ratio of the longest side and perimeter is (a) 3 2 3: ( )+ (b) 1 : 6(c) 1 2 3: ( )+ (d) 2 : 630. If |z – 1| + |z + 3| = 8, then the range of values of|z – 4| is(a) (0, 8) (b) [0, 8] (c) [1, 9] (d) [5, 9]31. If a, b, c are the roots of equation x3 –px2 + qx – r = 0,
then the value of 1 1 12 2 2a b c
+ + is
(a) p qr
r
2
22−
(b) q pr
r
2
22−
(c) r pq
q
2
22−
(d) r pq
p
2
22−
32. If f x dx F x x f x dx( ) ( ), then ( )= ∫∫ 3 2 is equal to
(a) 12
2 2 2 2{ ( ) ( ) }x F x F x dx− ∫
(b) 12
2 2 2{ ( ) ( ) }x F x F x dx− ∫(c) 1
212
2 2x F x F x dx( ) ( )−
∫
(d) None of these33. In three dimensional space x2 – 5x + 6 = 0represents(a) two points (b) two parallel planes(c) two parallel lines (d) a pair of non parallel lines
34. If f(x) = logx (logx), then d
dxf x( ( )) is
(a) 1e
(b) e
(c) e + 1 (d) None of these
| august ‘15 25
35. Let f x x x x
b x
xx
( )
,
, if .( )
= + − +
=
≠
−
3 2 16 20
2
22 2
if
If f(x) is continuous for all x, then b is equal to(a) 7 (b) 3 (c) 2 (d) 536. The locus of the middle points of the chords of thecircle x2 + y2 = a2 which subtend a right angle at the centre is
(a) x y a2 22
2+ = (b) x2 + y2 = 2a2
(c) x y a2 22
4+ = (d) None of these
37. An integrating factor of the differential equation
y y dxdy
x ylog log+ − = 0 is
(a) log(log y) (b) log y
(c) 1
log y (d) 1
log(log )y38. A fair die is rolled. Consider the events A = {1, 3, 5},B = {2, 3} and C = {2, 3, 4, 5}. Then the conditional probability P((A ∪ B) | C) is
(a) 14
(b) 54
(c) 12
(d) 34
39. India play two matches each with West Indies andAustralia. In any match, the probabilities of India getting 0, 1 and 2 points are 0.45, 0.05 and 0.50 respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points is
(a) 0.0875 (b) 1
16(c) 0.1125 (d) None of these40. The region represented by the inequation systemx, y ≥ 0, y ≤ 6, x + y ≤ 3 is (a) unbounded in first quadrant (b) unbounded in first and second quadrant(c) bounded in first quadrant (d) None of these41. If x2 + px + 1 is a factor of ax3 + bx + c, then(a) a2 + c2 = ab (b) a2 + c2 = –ab(c) a2 – c2 = ab (d) None of these
42. The set of values of x for which tan tantan tan
3 21 3 2
1x xx x−
+ ⋅=
is
(a) f (b) p4
(c) n np p+ =
4
1 2 3, , , , ........
(d) 24
1 2 3n np p+ =
, , , , ........
43. If y = a log x + bx2 + x has its extremum values atx = –1 and x = 2, then
(a) a = 2, b = –1 (b) a b= =2 12
,
(c) a b= − =2 12
, (d) a b= = −2 12
,
44. The angle through which the axes must be rotated,without translation, in anti-clockwise sense so that the expression ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 does not contain the mixed product xy, is given by
(a) tan−−
1 2ha b
(b) 12
21tan−−
hb a
(c) 12
21tan−−
ha b (d)
12
1tan−−
ha b
45. If cos cos sinA C B+ = 42
2 , then the sides a, b, c of
the triangle are in (a) A.P. (b) G.P.(c) H.P. (d) None of these46. The two consecutive terms in the expansion of(3 + 2x)74 whose coefficients are equal are(a) 30th and 31st terms (b) 31st and 32nd terms(c) 29th and 30th terms (d) None of these
47. If g f x x f g x x( ( )) | sin |, ( ( )) (sin ) , then= = 2
(a) f(x) = sin x, g(x) = |x| (b) f x x g x x( ) , ( ) sin= =2
(c) f and g can not be determined (d) f x x g x x( ) sin , ( )= =2
48. Let * be a binary operation on the set Q+ of all
+ve rational numbers defined by a b ab* =
100for all
a, b ∈ Q+. The inverse of 0.1 under opeation * is (a) 105 (b) 106
(c) 104 (d) None of these49. If a, b, c, d and p are distinct real numbers such that(a2 + b2 + c2 ) p2 – 2 (ab + bc + cd) p + (b2 + c2 + d2) ≤ 0, then a, b, c, d are in (a) A.P (b) H.P (c) ab = cd (d) G.P50. If the position vectors of three points are
a b c− +2 3 , 2 3 4 7 10
a b c a c+ − − +and , then the threepoints are(a) collinear (b) non-collinear(c) coplanar (d) None of these
solutions
1. (b) : We have, A
m n
=
×
1 1 11 1 1
1 1 1
| august ‘1526
Applying Rk → Rk – R1 (where k = 2, 3, ...., m), we get
A =
1 1 10 0 0
0 0 0
m×n\ Matrix A has only one non-zero row.Hence, rank of matrix A = 12. (c) : We have,
( ) ( )x y x y− + + + + =2 2 42 2 2 2
⇒ + − + + + + + =x x y x x y2 2 2 24 4 4 4 4
By hit and trial method, we getx = ±2, y = 0
1 2 3–3 –2 –1
1
2
3
0x
y
x = ±1, y = 0and x = 0, y = 0Thus, given equation represents a pair of lines (as shown in figure).
3. (b) : Let z nn
nnn
n
n n
n= =
→ ∞ → ∞
lim ( !) lim !/ /1 1
⇒ =
→ ∞
zn n n
nnn
nlim . . . .....
/1 2 3 1
Taking log on both sides, we get
log lim logzn
rnn r
n=
→ ∞ =
∑1
1
⇒ = ∫log logz x dx0
1
⇒ log z = –1 ⇒ z = e–1
4. (a) : Let I dxx x
=+ +
∞∫ ( )( )2 20 4 9
⇒ =+ + +
∞∫I
x dx
x x x
2
2 2 2 2 2 20 2 3 0( )( ) ( )
⇒ =+ + +
I p2 2 3 3 0 0 2( ) ( ) ( )
= p60
5. (a) : Since A.M. ≥ G.M. " real x
⇒+
≥ +4 42
42 2
2 2sin cossin cos
x xx x
⇒ + ≥4 4 2 42 2sin cosx x
⇒ + ≥4 4 42 2sin cosx x
6. (d)7. (c) : Equivalence relation of the set {1, 2, 3}
containing (1, 2) and (2, 1) areA1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}A2 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 1),
(3, 2), (1, 3)}So, only two equivalence relations exist.8. (a)
9. (a) : We have z iz i
−+
=55
1 ⇒ |z – 5i| = |z + 5i|
⇒ + − = + +( ) ( ) ( )x y x y2 2 2 25 5⇒ x2 + y2 + 25 – 10y = x2 + y2 + 25 + 10y⇒ y = 0 ⇒ z lies on x-axis.10. (c) : We have, y(x + y3)dx = x(y3 – x)dy⇒ y3(y dx – x dy) + x(y dx + x dy) = 0
⇒−
+ =x yy dx x dy
xxd xy2 3
2 0( )
⇒ −
+ =yx
dyx
d xy
x y
( )2 2 0
On integrating, we get
−
− + ′ =
yx
xyc
2
21 0 ⇒ + + =
y
x xyc
2
221 0
⇒ y3 + 2x + 2cx2y = 0Since it passes through the point (1, 1), so
1 + 2 + 2c = 0 ⇒ = −c 32
\ The curve is y3 + 2x – 3x2y = 0
11. (d) : We have, lim formx a
x a
x aa xx a→
−−
= −
1 00
⇒−
+= −
→
−lim
log
( log )x a
x a
xa a ax
x x
1
11
⇒ −+
= −−a a aa
a a
a a
alog
( log )
1
11
⇒ log a – 1 = –log a – 1 ⇒ log a = 0 ⇒ a = 112. (d) : f(x) = sinx, g(x) = x2 and h(x) = logexG(x) = hogof(x) = hog(sin x) = h(sinx)2
⇒ G(x) = loge(sinx)2 = 2logesin xDifferentiating both sides, we get
G xx
x′ = × ×( )sin
cos2 1 = 2cotx
Again differentiating, we getG′′ (x) = –2cosec2x
13. (d) : Let I e n x dxx= +∫ cos cos ( )2
0
3 2 1p
...(i)
⇒ I e n x dxx= + −−∫ cos ( ) cos ( )( )2 3
02 1p
pp
| august ‘15 27
⇒ = + − +∫I e n n x dxxcos cos (( ) ( ) )2 3
02 1 2 1p
p
= − +∫ e n x dxxcos cos ( )2 3
02 1
p...(ii)
(Q cos(np – q) = –cos q if n ∈ odd integer)Adding (i) and (ii), we get2I = 0 ⇒ I = 014. (c)15. (d) : CC
CC
CC
CC
1
0
2
1
3
2
15
14
2 3 15+ + + +......
= + × +
+ +
15 2 7 3 133
15 115
( ) .....
= 15 + 14 + 13 + ...... +1 = +15 15 12
( ) = 120
16. (b) : (ho(fog))x = hof(g(x)) = +hof x( )2 1
= +
−
= =h x h x x22
2 21 1 ( )
17. (c) : AB is the diameter of thecircle and C is any point on the circumference.So, ∠ACB = 90° A B
C
\ DACB is a right angle triangle and area of right angled triangle is maximum when it is isosceles.
18. (b) : We have,
a a a
b b b
c c c
2 3
2 3
2 3
1
1
1
0
+
+
+
=
⇒
a a
b b
c c
a a a
b b b
c c c
2
2
2
2 3
2 3
2 3
1
1
1
0+ =
⇒a a
b b
c c
abc
a a
b b
c c
2
2
2
2
2
2
1
1
1
1
1
1
0+ =
⇒ + =( )abc
a a
b b
c c
1
1
1
1
0
2
2
2
(1, a, a2) (1, b, b2) and (1, c, c2) are non-coplanar.⇒ abc + 1 = 0 ⇒ abc = –1
19. (b) : Vector form of component of
a balong
=
a bb
b.
| |
^
=+ +
+
+
+
( ).( )4 6 3 4
3 4
3 4
3 42 2 2 2
i j j k j k
= + +
+
0 18 05
3 45
( )j k
= +1825
3 4( )j k
20. (a) : We have, f(x) = sin x – tan xDifferentiating both sides, we getf ′(x) = cos x – sec2x ...(i)f ′(x) ≥ 0 " x ∈ (0, p/2)Now, again differentiating (i), we getf ′′(x) = –sin x – 2sec x (sec x tan x)
= –(sin x + 2sec2x tan x)f ′′ (x) < 0 " x ∈ (0, p/2)⇒ f(x) is concave downward.Hence tangent will lie above the curve.
21. (b) : We have, lim | |[cos ]
x
xx→ 0
...(i)
When x → 0, then 0 ≤ cos x ≤ 1⇒ [cos x] = 0 when x ≠ 0From (i), we have lim | | lim
x xx
→ →= =
0
0
01 1
22. (b) : Consider the general equation of all conics isax2 +2hxy + by2 + 2gx + 2fy + c = 0 ...(i)As centre of the conic is (0, 0) ⇒ g = 0, f = 0\ (i) becomes ax2 + 2hxy + by2 = –c
or ac
x hc
xy bc
y−
+−
+−
=2 22 1
or Ax2 + 2Hxy + By2 = 1 is the general equation of all conics whose centre lie at origin.Since, it has three arbitrary constants.Hence, order of the differential equation is 3.23. (d) : We have,
a b c× = ⇒ ⊥ ⊥
a c b cand ...(i)Also,
c a b× = ⇒ ⊥ ⊥
c b a band ...(ii)From (i) and (ii), we get
a c a b⊥ ⊥and⇒ ×
a b c| | ( )24. (d) : We know that,
0 < cos–1x < p and − < <−p p2 2
1sin x
Also, cos–1x is decreasing function and sin–1x is increasing function.And sin cos− −= =1 1
4x x p
\ cos–1x > sin–1x when
cos ,− ∈
14
x p p ⇒ ∈ −
x 1 12
,
25. (d)
| august ‘1528
26. (b) : We know that if 3m things are divided intothree equal groups, then the number of divisions
= 33
mm m m
Using this result, 52 cards are divided into 3 equal sets (each containing 17 cards) and 1 set (containing only
one card) represented as 52
17 17 17 3 152
17 33=( )
27. (b) : Number of elements in A = nNumber of elements in A × A = n × n = n2
\ Number of relations R in A are subsets of A × A.
⇒ Number of relations = 22n
28. (c) : The area bounded by the curve y = f(x) withordinates x = l and x = b is
f x dx b bl
b( ) ( )sin( )= − +∫ 1 3 4
Differentiating w.r.t. b, we getf(b) = sin(3b + 4) + (b – 1) cos(3b + 4) × 3⇒ f(x) = sin(3x + 4) + 3(x – 1) cos(3x + 4)29. (a) : Given that angles of the triangle are in the ratio4 : 1 : 1\ Angles are 120°, 30° and 30°.By sine rule, we havesin sin sin
(say)120 30 30°
=°
=°
=a b c
g
⇒ =°
=asin120 3
2g g, b =
°=
sin 30 12g g
,
c =°
=sin 30 1
2g g
\ Perimeter = + + = +
3
21
21
23 22g g g g
Now, required ratio =+
= +32
3 22
3 3 2g g
:( )
: ( )
30. (c) : Given that |z – 1| + |z + 3| = 8⇒ z l ies on the ellipsewhose foci are (1, 0) and (–3, 0)\ Vertices of the ellipse are (3, 0) and (–5, 0)Hence, the maximum and minimum values of |z – 4| is 9 and 1 respectively.\ Range of |z – 4| is [1, 9].31. (b)32. (a) : Let I = x f x dx3 2( )∫
Put x2 = t ⇒ 2x dx = dt
⇒ I = =∫ ∫t f t dt t f t dt( ) ( )
212
⇒ I = − ∫12
t F t F t dt( ) ( )
(1, 0)(–3, 0)O(–5, 0) (3, 0)
(4, 0)
= −
∫
12
2 2 2 2x F x F x dx( ) ( )
33. (b) : We have, x2 – 5x + 6 = 0⇒ (x – 3) (x – 2) = 0 ⇒ x – 3 = 0 or x – 2 = 0In three dimensional space, these two equations represent two parallel planes.34. (Question is not correct)
35. (a) : f xx x x
xx
b x
( ) ( ), if
, if
=+ − +
−≠
=
3 2
216 202
2
2
\ lim ( ) lim( )x x
f x x x xx→ →
= + − +−2 2
3 2
216 202
=− + −
−→lim
( )( )
( )x
x x x
x2
2
22 3 10
2
=− + −
−→lim
( )( ) ( )
( )x
x x x
x2 22 5 2
2= + = + =
→lim ( )
xx
25 2 5 7
f(x) is continuous for all x.\ =
→f f x
x( ) lim ( )2
2⇒ b = 7
36. (a) : Circle x2 + y2 = a2 having centre at origin andradius a. Let AB be the chord of the circle which subtend a right angle at the centre and C(h, k) be the mid-point of the chord.Now, in DACO, sin 45° = OC
OA
⇒ =12
OCa
⇒ =OC a2 A B
(0, 0)Oa
C h k( , )⇒ − + − =( ) ( )0 0
22 2h k a
⇒ + =h k a2 22
2Hence, locus of C(h, k) is x2 + y2 =
a2
237. (b)38. (d) : A = {1, 3, 5}, B = {2, 3}, C = {2, 3, 4, 5}⇒ A ∪ B = {1, 2, 3, 5}.and (A ∪ B) ∩ C = {2, 3, 5}
Now, P[(A ∪ B)|C] =∪ ∩ n A B Cn C
( )( )
= 34
39. (a) : India will get atleast 7 points.\ India get 2 points in all 4 matches or 2 points in 3 matches and 1 point in one match.Thus, probability of getting at least 7 points= (0.5)4 + 4C3(0.5)3 × 0.05 = 0.0875
| august ‘15 29
40. (c) :
1 2 3 4 5 6–6 –5 –4 –3 –2 –1
1
2
3
4
5
6
6
5
4
3
2
1 xy+
=3
y = 6
0
Hence, area of the given inequality is bounded in first quadrant.41. (c) : x2 + px + 1 is the factor of ax3 + bx + c = 0then let a, b be the roots of x2 + px + 1 = 0.⇒ a + b = –p ...(i)⇒ ab = 1 ...(ii)Now, a, b, g be the roots of ax3 + bx + c = 0. ⇒ a + b + g = 0 ...(iii)
ab bg ga+ + = ba
...(iv)
abg =− ca
...(v)
From (i) and (iii), we get g = p
From (ii) and (v), we get g = − ca
Now, (iv) becomes
1 + + =b ap p ba
⇒ + = −p ba
( )a b 1
⇒ − = −p p b aa
( ) ⇒ − = −p b aa
2
⇒ − = −g2 b aa
⇒ −−
= −ca
b aa
2
⇒−
= −c
ab a
a
2
2 ⇒ a2 – c2 = ab
42. (a) : We have,tan tan
tan tan3 2
1 3 21x x
x x−
+ ⋅=
⇒ tan(3x – 2x) = 1 ⇒ = =
tan tanx 1
4p
⇒ = + ∈x n n Zp p4
, . But if x n= +p p4
, then
tan tan2 24
x n= +
p p = +
tan 22
np p=
= ∞tan p2
Thus, there is no value exist for which given equation holds. Hence ,solution set = f.43. (d) : Given, y = a log x + bx2 + x
⇒
dydx
ax
bx= + +2 1
y has extremum values at x = –1 and x = 2.
Hence, dydx
−1
= 0 ⇒ –a – 2b + 1 = 0 ...(i)
anddydx
=2
0 ⇒ a + 8b + 2 = 0 ...(ii)
Solving (i) and (ii), we get a b= = −2 12
and44. (c) 45. (a)46. (a) : Let rth and (r + 1)th terms of (3 + 2x)74 hasequal coefficient.⇒ Coefficient of Tr = Coefficient of Tr + 1⇒ 74Cr – 1 (3)75 – r 2r – 1 = 74Cr (3)74 – r 2r
⇒ = ⇒ − + =−
74
741
32
74 1 32
C
Cr
rr
r⇒ 5r = 150 ⇒ r = 30Hence, required terms are 30th and 31st.47. (d) : From option (d), we getg f x g x x x( ( )) (sin ) sin | sin |= = =2 2
and ( ( )) ( ) sin (sin )f g x f x x x= = =2 2
48. (a) : We have, a b ab* =
100Let e be the identity element, thene * a = a * e = a
⇒ =a ea100
⇒ e = 100
Now, let inverse of 0.1 be x, then
0.1 * x = e
⇒ 0 1100
100. x
= ⇒ x = 100000 = 105
49. (d) : We have(a2 + b2 + c2)p2 – 2(ab + bc + cd) p + (b2 + c2 + d2) ≤ 0⇒ (a2p2 –2abp + b2) + (b2p2 – 2bcp + c2) +
(c2p2– 2cdp + d2) ≤ 0⇒ (ap – b)2 + (bp – c)2 + (cp – d)2 ≤ 0This is possible only if ap – b = bp – c = cp –d = 0
⇒ = = =ba
cb
dc
p ⇒ a, b, c, d are in G.P.
50. (a,c) : Let l1, l2 and l3 be the three points suchthat( ) ( )
a b c a b c− + + + − +2 3 2 3 41 2l l( )− + =7 10 03 a c l
⇒ + − + − + +( ) ( )l l l l l1 2 3 1 22 7 2 3
a b( )3 4 10 01 2 3l l l− + =c
Equating the coefficients, we getl1 + 2l2 – 7l3 = 0 ...(i)–2l1 + 3l2 = 0 ...(ii)3l1 – 4l2 + 10l3 = 0 ...(iii)On solving (i), (ii) and (iii) we getl1 + l2 + l3 = 0Hence, given vectors are collinear and coplanar.
nn
| august ‘1530
| AUGUST ‘15 31
introductionFor the complex number z = a + ib, a is called the real part, denoted by Re(z) and b is called the imaginary part denoted by Im(z). A complex number z is said to be real if Im(z) = 0 and is said to be purely imaginary if Re(z) = 0. Thus the complex number 0 = 0 + 0i, is both real and purely imaginary. a + ib is called an imaginary number if b ≠ 0.Two c ompl e x nu mb e rs z 1 = a + ib and z2 = c + id are equal if and only if a = c and b = d. However there is no order relation in the set of complex numbers i.e., the expressions of the form a + ib < c + id and a + ib > c + id are meaningless unless b = d = 0.
AlgebrA of complex numbersLet a, b, c, d, l ∈ R, then
(i) Addition : (a + ib) + (c + id) = (a + c) + i(b + d)(ii) Subtraction : (a + ib) – (c + id) = (a – c) + i (b – d)(iii) Multiplication :
(a + ib)(c + id) = (ac – bd) + i(ad + bc) (iv) Division :
For c id a ibc id
ac bdc d
i bc adc d
+ ≠ ++
= ++
+ −+
0 2 2 2 2,
(v) Multiplication by a real number : l(a + ib) = la + i(lb)
tHe conJugAte of A complex number Let z = a + ib be a complex number. The conjugate of z, denoted as z , is the complex number a – ib, i.e., z a ib= − .
Thus a z z z= = +Re( )2
and b z z zi
= ( ) = −Im2
Let z, z1 and z2 be complex numbers. The following results can easily be proved. (i) z = z ⇔ z is a real number. (ii) z = – z ⇔ z is a purely imaginary number.
(iii) z z z z1 2 1 2+ = + (iv) z z z z1 2 1 2− = −
(v) z z z z1 2 1 2= ⋅ (vi) zz
zz
z1
2
1
22 0
= ≠,
(vii) z z( ) = (viii) ( ) ( ) ,z z n In n= ∈
(ix) f z f z( ) ( )= , f being a polynomial with real coefficients. (x) Let z = x + iy, then zz x y= +2 2. Thus zz is a non-negative real number for any complex number.
note : The imaginary roots of a polynomial equation over reals, occur in conjugate pairs and hence any polynomial equation of odd degree over reals has at least one real root. Obviously this is not the case for the polynomial equations over C, the set of complex numbers.
tHe modulus of A complex number Let z = a + ib be a complex number. The modulus of z, denoted by |z|, is defined to be a non-negative real number a b i e z a b2 2 2 2+ = +, . . | | .
This column is aimed at Class XI students so that they can prepare for competitive exams such as JEE Main/Advanced, etc. and be also in command of what is being covered in their school as part of NCERT syllabus. The problems here are a happy blend of the straight and the twisted, the simple and the difficult and the easy and the challenging.
*
| AUGUST ‘1532
Alternate definition :Let z be a complex number then |z| can be defined as ( ) /zz 1 2 .Let z, z1 and z2 be complex numbers, then(i) | | | |z z= ≥ 0 and equality holds if and only if
z = 0. (ii) Re(z) ≤ |z| and equality holds if and only if
Im(z) = 0 and Re(z) ≥ 0.(iii) Im(z) ≤ |z| and equality holds if and only if
Re(z) = 0 and Im(z) ≥ 0.
(iv) zz zz
zz
z= = ≠| || |
,22
1 0and hence
(v) | | | | | | Re( )z z z z z z1 22
12
22
1 22+ = + +
(vi) | | | | | | Re( )z z z z z z1 22
12
22
1 22− = + −
(vii) | | | | (| | | | )z z z z z z1 22
1 22
12
222+ + − = +
(viii) |z1z2| = |z1|·|z2|
(ix) zz
zz
z1
2
1
22 0= ≠,
geometricAl representAtion of complex numbers
P a, b( )
Q a, –b( )
O
y
x
a b+ 22
A complex number a + ib corresponds to the ordered pair (a, b), which can be represented geometrically as the unique point (a, b) in the xy plane and vice-versa. The xy plane having a complex number assigned to each of its points is called the complex plane or the Argand plane. Let z = a + ib. Then z is represented as the point P(a, b) on the argand plane. Obviously
OP a b z= + =2 2 | | . Hence |z| is the distance ofz from origin. Further if Q(a, –b) is the mirror image of P(a, b) in the x-axis (also called as real axis), then Q is the geometrical representation of z . Also note that a complex number is real if and only if it lies on the real axis and a complex number is purely imaginary if and only if it lies on the imaginary axis. For any complex number z1
and z2, it is easy to see that |z1 – z2| is the distance of z1 from z2, when z1 and z2 are represented as points on the argand plane. note : The locus of z satisfying |z| = r, r > 0 is the circle having centre at origin and radius r. Similarly, locus of z satisfying |z – z1| = r, r > 0 is the circle having centre at z1 and radius r.the geometrical representation of z1 + z2 and z1 – z2 : Let P, Q represent two complex numbers x1 + iy1 and x2 + iy2 respectively in the Argand plane. Join the origin O with the points P and Q and complete the parallelogram OPRQ. It is clear that the coordinates of the point R are (x1 + x2, y1 + y2) and hence the point R represents the complex number (x1 + x2) + i(y1 + y2) = z1 + z2.I n t r i a n g l e O P R , O R = d i s t a n c e o f R from O = |z1 + z2|. Similarly OP = |z1| and PR = OQ = |z2|. Using the triangle inequality in triangle OPR, we get OR < OP + PR ⇒ |z1 + z2| < |z1| + |z2|. However if O, P and Q are collinear (P and Q being on the same side of O), then OR = OP + PR, i.e. |z1 + z2| = |z1| + |z2|. Hence for any complex numbers z1, z2; |z1 + z2| ≤ |z1| + |z2|. This is called the triangle inequality. |z1 + z2|2 + |z1 – z2|2 = 2(|z1|2 + |z2|2) Its geometrical interpretation can now be explained as OR2 + PQ2 = OP2 + PR2 + RQ2 + QO2, in a parallelogram the sum of the squares of the diagonals is equal to the sum of the squares of the four sides.
Argument of A complex number Let z be a non-zero complex number, which corresponds to the point P on the Argand plane. If |z| = r, then q satisfies x = r cosq, y = rsinq, where z = x + iy, and q is called the argument of the given complex number. The angle of the line segment OP with the positive direction of x-axis is q. The angle can also be taken as 2np + q, n ∈ I. This angle is called argument or amplitude of the complex number z denoted by arg(z) or amp(z). Value of q having minimum modulus value is called the principal argument of z. Thus for any non-zero z ∈ C, principal arg(z) ∈ (–p, p]. Obviously a non-zero complex number z is real if and only if principal argument of z is 0 or p and it is purely imaginary if and only if
principal argument of z is ± p2
.
| AUGUST ‘15 33
some properties of Argument Let z be a non-zero complex number and n ∈ I (i) arg(–z) = arg(z) + (2n + 1)p (ii) arg( ) ,z z n+ = p z not being purely imaginary.
(iii) arg( ) ( )z z n− = ±4 12p , z not being purely
real.
(iv) arg arg argz zzz z
n=
=
+1 2 p
(v) arg argz z n+ = 2 p
method of finding principal argument of a non-zero complex numberLet z = x + iy be a given non-zero complex number, whose principal argument is to be found.
Let a = −tan .1 yx
Now a, p – a, – p + a, – a
becomes the principal argument of z according as the point P(z) lies in first, second, third, fourth quadrant respectively.
polAr form of A complex number Let z be a given non-zero complex number, which corresponds to the point P in the Argand plane. Let z = x + iy. Then OM = x, PM = y. Let r = |z|, then OP = r. Let q = arg(z), then ∠POM = q. Now x = rcosq and y = r sinq.Hence z = x + iy = r (cosq + i sinq). This form is called polar form of z, sometimes written in short as rcisq. Note that (r, q) are polar coordinates of z. note : It is a known fact that
cos ......q q q q= − + − +12 4 6
2 4 6
and sin ......q q q q q= − + − +3 5 7
3 5 7
Hence cos sin ( ) ( ) ......q q q q q+ = + + + +i i i i11 2 3
2 3
Also we know that e x x xx = + + + +11 2 3
2 3......
Hence, eiq = cosq + isinqsome consequences of euler formula
Let z z be a non-zero complex number and let reiq be its polar form. As arg(z) = q, so arg(z) = 2np + q, n ∈ I. Hence z rei n= +( ).q p2
Now log log log ( ), .( )z re r i n n Ii n= = + + ∈+q p q p2 2
Thus logarithm of an imaginary number is not unique.
cos sin cos( ) sin( )q q q qq q+ = ⇒ − + − = −i e i ei i z
Thus, andcos sinq qq q q q
= + = −− −e e e ei
i i i i
2 2de moiVre’s tHeorem
We define eiq = cosq + i sinq.Let n be any integer, then(cos sin ) ( ) cos sinq q q qq q+ = = = +i e e n i nn i n in
Thus (cosq + isinq)n = cos nq + isin nq, n ∈ IThis result is called De Moivre’s Theorem for an integer. However if n p
qp q p q= ∈ ≠ >, , , ,I 0 1and
then(cos sin ) (cos sin ) cos sin ,/q q q q
q q+ = + = +i i
pq
ipq
n p q
which is one of the values of (cosq + i sinq)p/q. To make this point more clear, consider (–1)1/2 i.e., all the complex numbers whose square is –1,which are i and –i, for –1 = cosp + i sinp. Hence
(–1)1/2 = (cos p + i sin p)1/2 = + =cos sin ,p p2 2
i i
which is one of the value of (–1)1/2.For the other value, we write –1 = cos 3p + i sin3p
⇒ − = + = −( ) cos sin/1 32
32
1 2 p pi i
note : De Moivre’s Theorem is true for irrational values of n also. Applications of de moivre’s theorem to trigonometry(i) Let n be a positive integer. Then
(cos q + i sin q)n = cos nq + i sinnq. Also expanding (cosq + isinq)n by using Binomial Theorem and comparing the real and imaginary parts, we get
cos nq = cosnq – nC2 cosn–2q sin2q + nC4 cosn–4q sin4q – ....
sin nq = nC1 cosn–1q sinq – nC3 cosn–3q sin3q +....
Hence, tantan tan ......
tan tan ...n
C C
C C
n n
n nqq q
q q=
− +
− + −1 3
3
22
441 ....
(ii) Let z1 = r1(cosq1 + isinq1) and z2 = r2 (cosq2 + isinq2), then
z z r e r e r r er r i
i i i1 2 1 2 1 2
1 2 1 2 1
1 2 1 2= ⋅ == + + +
+q q q q
q q q q
( )
(cos( ) sin( 22))
Hence |z1z2| = r1r2 = |z1| |z2| and arg(z1z2) = q1 + q2 = arg(z1) + arg(z2)note : (i) Infact, arg (z1z2) = arg z1 + arg z2
+ 2np, for some n ∈ I Similarly,
for somearg arg arg ,
zz
z z nn I
1
21 2 2
= − +∈
p
and arg (zn) = n arg(z) + 2kp, k ∈ I
| AUGUST ‘1534
(ii) Let z rei= a. Then ze rei iq a q= +( ). Thus thecomplex number zeiq (in the Argand plane) can be obtained from z by rotating z about origin by an angle q.
geometricAl representAtion of z1z2 And z1/z2
Let P and Q represent two complex numbers z1 and z2 respectively in the Argand plane. Join the origin O with the points P and Q. Now we are in search of a point R(z1z2) in the Argand plane. For that let I be the point representing the number 1. Now consider a triangle OQR, which is similar to the triangle OIP.
R z(z )1 2
P z( )1
Q z( )2
O
y
x
12 1 I(1)
Now arg u m e nt o f t h e c omp l e x nu mb e r corresponding to R is q1 + q2 = arg(z1z2)FurtherOROP
OQOI
ORz z
= ⇒ =⋅| | | |1 21
Thus the complex number corresponding to R is |z1|.|z2|. (cos(q1 + q2) + isin(q1 + q2))= |z1|(cosq1 + i sinq1) |z2|(cosq2 + i sinq2) = z1z2For the representation of z
z1
2 in the Argand plane
once again, let P and Q represent the complex numbers z1 and z2 in the Argand plane (as shown in the figure). Now construct a triangle OPR similar to triangle OQI. Then argument of the complex number corresponding to R is
q q1 21
2− =
argzz
Further OROP
OIOQ
OR OPOQ
zz
zz
= ⇒ = = =| || |
1
2
1
2
Hence the complex number corresponding to R is
| || |
(cos( ) sin( ))zz
i1
21 2 1 2q q q q− + −
=++
=| |(cos sin )| |(cos sin )
z iz i
zz
1 1 1
2 2 2
1
2
q qq q
concept of rotAtion Let z1, z2 and z3 be the vertices of a triangle ABC described in the anticlockwise sense. AB = |z2 – z1|, AC = |z3 – z1| and BC = |z3 – z2| = |(z3 – z1) – (z2 – z1)|
Now arg arg( ) arg( )z zz z
z z z z3 1
2 13 1 2 1
−−
= − − − = a
Alsoz zz z
ACAB
3 1
2 1
−−
=
Therefore the polar form of z zz z
3 1
2 1
−−
isACAB
i(cos sin )a a+
if z z z z z z z z z12
22
32
1 2 2 3 3 1+ + = + + , then the triangle ABC is equilateral.
condition for tHe four points to be concyclic
Let z1, z2, z3 and z4 be four concyclic points, represented by A, B, C and D in the Argand plane. As the points A, B, C and D are concyclic,
∠ADB = ∠ACB ⇒−−
=−−
arg argz zz z
z zz z
2 4
1 4
2 3
1 3(using the concept of rotation)
⇒−−
×−−
=argz zz z
z zz z
2 4
1 4
1 3
2 30
⇒− −− −
( )( )( )( )z z z zz z z z
2 4 1 3
1 4 2 3is a positive real number.
note : The converse of the above result is also true, i.e., if a complex number of the form ( )( )( )( )z z z zz z z z
2 4 1 3
1 4 2 3
− −− −
is a positive real number, then
the points represented by z1, z2, z3 and z4 on the Argand plane are concyclic.
tHe section formulA Let P(z1) and Q(z2) be two given points. Let R(z) be the point which divides the join of P and Q internally in l : 1 as shown in the figure.
| AUGUST ‘15 35
Using the concept of rotation at the point R, we get
z zz z
ei−−
=1
2l p ⇒ =
++
zz z1 2
1ll
If R divides PQ externally in l : 1, then it can be
shown that zz z
=−−
1 21
ll
.
tHe nth roots of unity Consider the polynomial equation zn = 1. The roots of this equation are n in number and are called nth roots of unity. In order to find the roots of zn = 1, we write the polar form of 1, i.e.,
zn = cos 2kp + i sin2kp, k ∈ I
⇒ = + = ∈z kn
i kn
e k Ik incos sin ,2 2
2p p
p
Now if we give different integral values to k, then we always get roots of zn = 1, but some of them may be repeated. However, if we take k = 0, 1, 2, 3, ..., n – 1, then we get (all) distinct roots of zn = 1. Thus n, nth roots of unity are
ek in
2 p, k = 0, 1, 2, ......, n – 1.
If we take ap
= ei
n2
, then the nth roots of unity are 1, a, a2, ....., an–1.
properties of n, nth roots of unity (i) If n , nth roots of unity are represented on the
argand plane, then they form the vertices of a regular n-gon having centre at origin and circumcircle of this n-gon is |z| = 1, as modulus value of every root is 1.
(ii) 1 + a + a2 + ..... + an – 1 = 0 (iii) 1 + ap + (ap)2 + .... + (ap)n – 1 = 0, if g.c.d.
(p, n) = 1.(iv) 1⋅a⋅a2 ...... an – 1 = (–1)n–1 (v) zn – 1 ≡ (z – 1)(z – a)(z – a)2 ..... (z – an–1)(vi) The imaginary roots form conjugate pair.
eQuAtion of strAigHt line in complex form
(i) Equation of line through the points z1 and z2 in parametric form is given by
z = z2 + l(z1 – z2), l ∈ R(ii) Equation of line through z1 and z2 is given by
z zz z
z zz z
z zz zz z
−−
=−−
=1
2 1
1
2 11 1
2 2
111
0or
(iii) General equation of a line is given by az az b+ + = 0, where a is a complex number and b is a real number. Slope of this line is given by − Re( )
Im( )aa
and its distance from a point
z1 is given by az az ba
1 12+ +
| |.
complex slope of A line If a line passes through the points z1 and z2 then its
complex slope is defined as z zz z
1 2
1 2
−−
. The complex
slope of the line az az+ is given by − aa
.
eQuAtion of A circle in complex form (i) The equation of the circle having centre at zo
and radius r is given by |z – zo| = r (ii) The general equation of circle is
zz az az b+ + + = 0, where a is a complex number and b is a real number. The centre
of this circle is ‘–a’ and radius is | | .a b2 −(iii) The equation of the circle described on the
line segment joining z1 and z2 as diameter is given by ( )( ) ( )( ) .z z z z z z z z− − + − − =1 2 2 1 0
(iv) z zz z
−−
= > ≠1
20 1l l l, , represents a circle
having diameter AB, when A and B divide the join of z1 and z2 in l : 1 internally and externally respectively.
(v) arg ,z zz z
−−
=1
2q represents an arc of the circle
through z1 and z2. (vi) |z – z1|2 + |z – z2|2 = k, represents a circle
provided k z z> −12 1 2
2| | .
eQuAtion of conic section in complex form
(i) Equation of parabola with focus at z1 and directrix
az az b+ + = 0 is | || |
.z zaz az b
a− =
+ +1 2
(ii) Equation of ellipse with foci at z1 and z2 and length of major axis l, is |z – z1| + |z – z2| = l.
(iii) Equation of hyperbola with foci at z1 and z2 and length of transverse axis l, is ||z – z1| – |z – z2|| = l.
| AUGUST ‘1536
ProblemsSection-i
Single correct Answer type
1. If n ≥ 3 and 1, a1, a2, a3 , ..... , an–1 are n roots of
unity, then value of a ai ji j n1 1≤ < ≤ −∑ is
(a) 0 (b) 1 (c) –1 (d) (–1)n
2. Let z = cosq + isinq. Then, the value of Im( )z m
m
2 1
1
15−
=∑
at q = 2° is
(a) 12°
(b) 13 2sin °
(c) 12 2sin °
(d) 1
4 2sin °3. If the complex number z satisfying z + |z| = 2 + 8i then value of |z| =(a) 8 (b) 17 (c) 15 (d) 244. If |z + 2 – i| = 5 then maximum value of|3z + 9 – 7i| = (a) 20 (b) 15 (c) 5 (d) 165. If l ∈ R and non real roots of 2z2 + 2z + l = 0 and(0, 0) forms vertices of an equilateral triangle, then l =
(a) 1 (b) 12
(c) 13
(d) 23
6. If z and w are two complex numbers such thatz iw+ = 0 and arg (zw)= p, then arg (z) =
(a) p4
(b) p2
(c) 34p (d) 5
4p
7. If A(z1), B(z2), C(z3) are vertices of a triangle such
that zz iz
iz z3
2 11 21
3 4=−−
= =and , and
z iz z z2 1 1 2+ = + , then area of triangle ABC is (a) 5
2 (b) 0 (c) 25
2(d) 25
4
8. The radius of the circle given by argz iz i
− ++ −
=5 43 2 4
p
is
(a) 5 2 (b) 5 (c) 52
(d) 2
9. If f(x) = 2x3 + 2x2 – 7x + 72 then f i3 52−
=
(a) 1 (b) 2 (c) 3 (d) 410. If cos a + cos b + cos g = 0 = sin a + sin b + sin g then sin2a + sin2b + sin2g = (a) 1
2 (b) 3
2(c) 4 (d) 1
11. If z1 and z2 are two complex numbers such thatz1
2 + z22 ∈ R and z1(z1
2 – 3z22) = 2, z2(3z1
2 – z22) = 11, then
z12 + z2
2 =(a) 5 (b) 125 (c) 25 (d) 15
12. If z is a non-real complex number, then the
minimum value of ImIm
zz
5
5 is
(a) –1 (b) –2 (c) –4 (d) –5
13. Let zr(1 ≤ r ≤ 4) be complex numbers such thatz rr = +1 and |30z1 + 20z2 + 15z3 + 12z4|
= k|z1z2z3 + z2z3z4 + z3z4z1 + z4z1z2|Then the value of k equals(a) |z1z2z3| (b) |z2z3z4|(c) |z4z1z2| (d) None of these
14. If P and Q are represented by the complex numbers
z1 and z2 such that 1 1 1 1
2 1 2 1z z z z+ = − , then the
circumcentre of DOPQ, (where O is the origin) is
(a) z z1 22− (b) z z1 2
2+ (c) z z1 2
3+ (d) z1 + z2
15. If a is non real root of x7 = 1, then 1 + 3a + 5a2 +7a3 + ... + 13a6 is equal to
(a) 0 (b) 14
1 − a(c) 14
1a − (d) none of these
16. If z1, z2 are two complex numbers satisfying the
equation z zz z
zz
1 2
1 2
1
21
+−
= , then is a number which is
(a) Positive real (b) Negative real (c) Zero (d) Lying on imaginary axis
17. If z1, z2 and z3 are the vertices of DABC, whichis not right angled triangle taken in anti-clockwise direction and z0 is the circumcentre, then
z zz z
AB
z zz z
CB
0 1
0 2
0 3
0 2
22
22
−−
+−−
sinsin
sinsin
is equual to
(a) 0 (b) 1 (c) –1 (d) 2
18. If 'a' is a complex number such that |a| = 1, thenthe values of 'a' so that equation az2 + z + 1 = 0 has one purely imaginary root is
(a) a i= + =−
−cos sin , cosq q q 1 5 12
(b) a i= + =−
−sin cos , cosq q q 1 5 12
(c) a i= + =−
−cos sin , cosq q q 1 5 14
(d) a i= + =−
−sin cos , cosq q q 1 5 14
| AUGUST ‘15 37
19. Let A(z1), B(z2), C(z3) be the vertices of a triangleoriented in anti-clockwise direction. If BC : CA : AB
= 2 : 2 3 1: ,+ then the imaginary part of z zz z
3 1
2 1
4−−
is
(a) 0 (b) − +7 2 6 (c) 7 2 6− (d) cannot be determined
20. If z is a complex number having least absolute valueand |z – 2 + 2i| = 1, then z =
(a) 2 12
1−
−( )i (b) 2 12
1−
+( )i
(c) 2 12
1+
−( )i (d) 2 12
1+
+( )i
21. Sum of common roots of the equationz3 + 2z2 + 2z + 1 = 0 and z1985 + z100 + 1 = 0 is(a) –1 (b) 1 (c) 0 (d) 1
22. Let z be a complex number having the argument
q, 02
< <q p and satisfying the equation, |z – 3i| = 3.
Then cot q − =6z
(a) i (b) –i (c) 2i (d) –2i
23. Let z1 and z2 be any two complex numbers then
z z z z z z1 12
22
1 12
22+ − + − − is equal to
(a) z z z z12
22
12
22− + + (b) z z z z1 2 1
222− + +
(c) z z z z1 2 12
22+ + + (d) z z z z1 2 1 2+ + −
24. Both the roots of the equation z2 + az + b = 0 are ofunit modulus if(a) |a| ≤ 2, |b| = 1, arg b = 2 arg a (b) |a| ≤ 2, |b| = 1, arg b = arg a (c) |a| ≥ 2, |b| = 2, arg b = 2 arg a (d) |a| ≥ 2, |b| = 2, arg b = arg a
25. If |z – i| = 1 and arg(z) = q where q p∈
02
, , then
cot q − 2z
equals
(a) 2i (b) 3i (c) i (d) –i
Section-ii
Multiple correct Answer type
26. If z1 = a + ib and z2 = c + id are complex numberssuch that |z1| = |z2| = 1 and Re( )z z1 2 0= then the pair of complex numbers w1 = a + ic and w2 = b + id satisfies(a) |w1| = 1 (b) |w2| = 1 (c) Re( )w w1 2 0= (d) w w1 2 0=
27. If points A and B are represented by the non-zerocomplex numbers z1 and z2 on the Argand plane such that |z1 + z2| = |z1 – z2| and O is the origin, then(a) orthocentre of DOAB lies at O
(b) circumcentre of DOAB is z z1 22+
(c) Argzz
1
2 2
= ± p
(d) DOAB is isosceles
28. The adjacent vertices of a regular polygon of nsides whose centre is at origin are given by ( , ),1 2 1+ ( , ).1 2 1+ − Then the value of n is (a) 8 (b) 4 (c) 12 (d) 6
29. If a is a variable complex number such that |a| > 1
and z = +aa1
lies on a conic then
(a) Eccentricity of the conic is 2
1 2a
a+(b) Distance between foci is 4
(c) Length of latus rectum is 2 1
1
2
2( )a
a
−
+
(d) Distance between directrix is aa
+
12
30. If thenz i= + +1 65
65
cos sinp p
(a) z = 2 35
cos p(b) z = 2 2
5cos p
(c) arg z = 35p (d) arg z = − 2
5p
31. Let z1, z2, z3, ...., zn are the complex numbers such
that |z1| = |z2| = ... = |zn| =1. If z zzk
k
n
kk
n=
= =∑ ∑
1 1
1 then
(a) z is purely imaginary (b) z is real (c) 0 < z ≤ n2 (d) z is a complex number of the form a + ib
32. If a, b, c are non – zero complex numbers of equalmoduli and satisfy az2 + bz + c = 0 then
(a)
min z = −5 12
(b) min |z| = 0
(c) min |z| does not exist (d) max z = +5 12
33. If |z – 3| = min{|z – 1|, |z – 5|} then Re(z) =
(a) 2 (b) 52
(c) 72
(d) 4
34. The complex slope m of a line containing thepoints z1 and z2 in the complex plane is defined as
| AUGUST ‘1538
z zz z
1 2
1 21 2
−−
. ,If m m are the complex slopes of two lines
L1 and L2, then(a) L1 and L2 are perpendicular if m1 + m2 = 0 (b) L1 and L2 are parallel if m1 + m2 = 0(c) L1 and L2 are perpendicular if m1m2 = –1 (d) L1 and L2 are parallel if m1 = m2
Section-iiicomprehension type
paragraph for Question no. 35 to 37Let z1 be a complex number of magnitude unity and z2 be a complex number given by z2 = z1
2 – z1.
35. If arg z1 = q, then |z2| is equal to
(a) 22
sin q(b) 2
2cos q
(c) 22
sin q(d) 2
2cos q
36. If arg z1 = q and 4np < q < (4n + 2)p (n is aninteger), then arg z2 is equal to
(a) 32q (b) p q− 3
2
(c) p q+ 32
(d) p q+2
37. If arg z1 = q and (4n + 2) p < q < (4n + 4)p (n is aninteger), then arg z2 is equal to
(a) p q2
3+ (b) 32
32
p q+
(c) 32
3p q+ (d) p q2
32
+
paragraph for Question no. 38 to 40Let A1, A2, A3, ..., An be a regular polygon of 'n' sides whose centre is origin O. Let the complex numbers representing vertices A1, A2, A3, ..., An be z1, z2, z3, ..., zn respectively. Let OA1 = OA2 = ...... = OAn = 1
38. The value of |A1A2|2 + |A1A3|2 + ...... + |A1An|2 =(a) n (b) 2n (c) 2(n – 1) (d) 2(n + 1)
39. The distances A1Aj (j = 2, 3, ..., n) must be equal to
(a) sin jnp (b) 2sin
( )jn− 1 p
(c) cos jnp
(d) 2sin( )j
n+ 1 p
40. The value of |A1A2| |A1A3| ..... |A1An| must be equalto (a) 1 (b) n (c) n (d) n2
paragraph for Question no. 41 to 43
Consider a complex number wz iz
=−+2 1
where z = x + iy and x, y ∈ R
41. If the complex number w is purely imaginary thenlocus of z is(a) a straight line
(b) a circle with centre −
14
12
, and radius 54
(c) a circle with centre 14
12
, −
and passing through origin
(d) neither a circle nor a straight line
42. If the complex number w is purely real, then locusof z is(a) a straight line passing through origin(b) a straight line with gradient 3 and y intercept
(–1) (c) a straight line with gradient 2 and y intercept 1(d) a circle
43. If |w| = 1, then locus of z is(a) a point circle (b) an imaginary circle(c) a real circle (d) not a circle
paragraph for Question no. 44 to 46If z = x + iy and (x, y) is a point represented by the complex number 'z' in the argand plane. |z1 – z2| denotes distance between z1 and z2 in the argand plane.
44. The complex number z = x + iy which satisfies the
equation z iz i
−+
=55
1 lies on the
(a) x-axis (b) y-axis(c) circle with radius 5 and centre at origin (d) line y = 5
45. If |z| = 5 then the points representing the complex
number − +iz
15 lie on the circle with
(a) centre (0, 1) and radius = 3 (b) centre (0, –1) and radius = 3 (c) centre (1, 0) and radius = 5 (d) centre (–1, 0) and radius = 15
46. If |z – i| ≤ 2 and z1 = 3 + 4i, then the maximumvalue of |iz + z1| is(a) 20 2− (b) 9 (c) 20 2+ (d) 8
| AUGUST ‘15 39
Section-iV
Matrix-Match type
47. Consider complex number
z i= + < <cos sin ,a a a p06
. Then the argument of
column-i column-ii
(A) 1 + z3 = (p) 22
a p−
(B) 1 – z4 = (q) p a2 2
+
(C) 11
3
4+−
zz
= (r) 32a
(D) zz
4
311
−+
= (s) p a2 2
−
48. The complex numbers z1, z2, ....., zn represent thevertices of a regular polygon of n sides, inscribed in a circle of unit radius and z z Az Az xn3 1 2+ = + , [ ] be the greatest integer ≤ x. Then
When n equals to [|A|] equals to
(A) 4 (p) 0
(B) 6 (q) 1(C) 8 (r) 2
(D) 12 (s) 3
49. Match the following.
column-i column-ii
(A) The number of integral solutions of the equation (1 – i)n = 2n is
(p) 4
(B) The number of common roots of the equations x3 + 2x2 + 2x + 1 = 0 and x2000 + x2002 + 1 = 0 is
(q) 3
(C) The number of all non-zero complex numbers ‘z’ satisfying z iz= 2 is
(r) 2
(D) If z is a complex number, then the number of solutions of z2 + |z| = 0 is
(s) 1
section-V
integer Answer type
50. If 35
2
1
izz
is purely real, then find 53 73 7
1 2
1 2
z zz z
+−
.
51. If a complex number z satisfies |z – 8 – 4i| + |z – 14 – 4i|
= 10, then the maximum value of arg( ) tan ,zk
= −1 113
find k.52. If ‘a’ and ‘b’ are complex numbers. One of the rootsof the equation x2 + ax + b = 0 is purely real and the other is purely imaginary then a a kb2 2− = , find k.53. The sum of the real parts of the complex
numbers satisfying the equations z iz i
−−
=42
1 and z i
z ik− +
+=8 3
335 5
is , find k.
54. If the equation z4 + a1z3 + a2z2 + a3z + a4 = 0 wherea1, a2, a3, a4 are real coefficients different from zero has purely imaginary roots then find the value of the
expression aa a
a aa a
3
1 2
1 4
2 3+ .
55. 1
1
1 121
1
−
= − = −=
−
∑e
nk
k iijnj
n
p , . ( )find
56. If z1, z2, z3 ,...., zn are in G.P with first term as unitysuch that z1 + z2 + z3 + .... + zn = 0. Now if z1, z2, z3,....,zn represents the n vertices of a polygon, then the distance between incentre and circumcentre of the polygon is represented by 4k. Find k.57. Let l, z0 be two complex numbers. A(z1), B(z2),C(z3) be the vertices of a triangle such that z1 = z0 + l,
z z e z z e ABC ki i2 0
43 0
7 11 322
= + = + ∠ =l l pp p/ /, andthen the value of k is
58. The argument of ( )( )z a z b− − is equal to that of( )( )3 1 3
1+ +
+i i
i, where a, b are real numbers. If locus
of z is a circle with centre 32+ i , then find (a + b).
59. If z i= −12
3( ) then the least positive integral value of 'n' such that (z101 + i109)106 = zn is 'k', then 25
k =
solutions
1. (b) : xn – 1= (x – 1)(x – a1)(x – a2).....(x – an–1) = xn – xn–1 (1 + a1 + ..... + an–1)
+ + + +
+ − =−−
≤ < ≤ −∑xn
i j ni j n
21 2 1
1 11 0a a a a a..... ......
⇒ + + + =≤ < ≤ −
∑ a a a a ai j ni j n
1 21 1
0.....
| AUGUST ‘1540
a ai ji j n
=≤ < ≤ −
∑ 11 1
2. (d) : Given that z = cosq + isinq = eiq
\ =−
=
−
=∑ ∑Im( ) Im( )z em
m
i m
m
2 1
1
152 1
1
15q = −( )
=∑ Imei m
m
2 1
1
15q
= sinq + sin 3q + sin 5q + ...... + sin 29q
=
+
×
sin sin
sin
q q q
q
292
15 22
22
= ( ) ( ) =°
sin sinsin sin
15 15 14 2
q qq
3. (b) : Let z = a + ib
⇒ + + + = +a ib a b i2 2 2 8
⇒ = + + =b a a8 64 22,
⇒ a2 + 64 = a2 – 4a + 4⇒ –4a = 60 ⇒ a = –15
\ = + = + = =z a b2 2 225 64 289 174. (a) : |3z + 9 – 7i|= |3z + 6 – 3i + 3 – 4i| ≤ |3(z + 2 – i)| + |3 – 4i|
= + − + +3 2 3 42 2z i = 3(5) + 5 = 205. (d) : Let z1, z2 be roots of 2z2 + 2z + l = 0
⇒ z z z z1 2 1 212
+ = − =, l
When origin, z1, z2 forms equilateral triangle, we have z1
2 + z22 = z1z2
(z1 + z2)2 = 3z1z2
1 32
23
= ⋅ ⇒ =l l
6. (c) : z iw z iw z iw+ ⇒ − = ⇒ =0Arg (zw) = p ⇒ arg(z) + arg(w) = p ⇒ arg(iw) + arg w = p ⇒ arg i + 2 arg w = p
p p2
2+ =arg w ⇒ 22
arg w = p
⇒ arg arg( )w z= ⇒ =p p4
34
7. (d) : |z2 + iz1| = |z1| + |z2| ⇒ z2, iz1, 0 are collinear.\ arg(iz1) = arg z2 ⇒ arg i + arg z1 = arg z2
⇒ − =arg argz z2 1 2p
Now, zz iz
i32 11
=−−
⇒ (1 – i)z3 = z2 – iz1⇒ z3 – z2 = i(z3 – z1)
z zz z
iz zz z
z z z z3 2
3 1
3 2
3 13 2 3 12
−−
= ⇒−−
= − = −arg p and
\ AC = BC \ AB2 = AC2 + BC2
⇒ 25 = 2AC2 ⇒ =AC 52
Required area = sq. units12
52
52
254
× × =
8. (a) : A(5, –4), B(–3, 2) subtends an angle p4
at
C(z) on the circle. Hence p2
at centre
OM AB AM AB^ \ =2
= + = =64 362
102
5
Radius = + = =25 25 50 5 2
9. (d) : Let x i= −3 52
⇒ 2x = 3 – 5i
⇒ (2x – 3)2 = (–5i)2 ⇒ 4x2 – 12x + 9 = 25i2
⇒ 4x2 – 12x + 34 = 0 ⇒ 2x2 – 6x + 17 = 0
)2 6 17 2 2 7 724
2 6 17
2 3 2
3 2
x x x x xx
x x x
− + + − ++
− +− + −( ) ( ) ( )____________________
( ) ( ) ( )____________________
8 24 72
8 24 68
4
2
2
x x
x x
− +
− +− + −
10. (b) : Let x = cis a, y = cis b and z = cis g
Clearly x y zx y z
+ + = + + =0 1 1 1 0,
x y z x y z xyzx y z
2 2 2 2 2 1 1 1 0+ + = + +( ) − + +
=
⇒ cis2a + cis2b + cis2g = 0⇒ cos2a + cos2b + cos2g = 0⇒ 1 – 2sin2a + 1 – 2sin2b + 1 – 2sin2g = 0
⇒ + + =sin sin sin2 2 2 32
a b g
11. (a) : We have, z z z1 12
223 2( )− =
⇒ z z z z z12
14
24
12
229 6 4( )+ − =
( )z z z z z12 3
12
24
14
229 6 4+ − = ...(i)
Also, z z z22
12
22 23 121( )− =
⇒ + − =( )z z z z z22 3
22
14
12
249 6 121 ...(ii)
(i) + (ii) ⇒ + =( )z z12
22 3 125 ⇒ z z1
222 5+ =
| AUGUST ‘15 41
12. (c) : Let z = a + ib, b ≠ 0, where Im z = bz a ib a C a bi C a b i
C a b i C ab i i b
5 5 5 51
4 52
3 2 2
53
2 3 3 54
4 4 5 5
= + = + +
+ + +
( )
Im z5 = 5a4b – 10a2b3 + b5
y zz
ab
ab
= =
−
+ImIm
5
5
4 25 10 1
Let sayab
x x R
= ∈ +2
( ),
y = 5x2 – 10x + 1 = 5[x2 – 2x] + 1 = 5[(x – 1)2] – 4Hence ymin = –413. (c) : We havez z z z k z z z z
z z z z1 2 3 4
1 2 3 41 2 3 42 3 4 5 60
1 1 1 1+ + + = + + +
Now, andz z z z z z z z1 1 2 2 3 3 4 42 3 4 5= = = =, ,
So, kz z z z
z z z= = = =60 602 3 4 5
301 2 3 4
4 1 2
14. (b) : 1 1 1 1
2 1 2 1z z z z+ = − ⇒ |z1 + z2| = |z1 – z2|
⇒ + =z z z z1 2 2 1 0 ⇒zz
1
2is purely imaginary.
⇒
= ±argzz
1
2 2p ⇒ ∠ =POQ p
2Circumcentre of DPOQ is the midpoint of PQ.15. (c) : Let A = 1 + 3a + 5a2 + 7a3 + ... + 11a5 + 13a6
aA = a + 3a2 + 5a3 + 7a4 + ... + 11a6 + 13a7 Now, (1 – a)A = 1 + 2a + 2a2 + 2a3 + ... + 2a6 – 13a7
= –12 + 2[a + a2 + ... + a6] = –14 ⇒ A = −−14
1 a
16. (d) : z zz z
z zz z
i1 2
1 2
1 2
1 21
+−
= ⇒+−
= +cos sina a
where a is the argument of z zz z
1 2
1 2
+−
.
Applying componendo and dividendo, we get zz
ii
1
2
11
= + +− + +
cos sincos sin
a aa a
=
+
22 2 2
22
cos cos sin
sin co
a a a
a
i
i ss sincot
a aa
2 22
+
= −
ii
Purely imaginary in nature.17. (c) : Taking rotation at 'O'
z zz z
C i C0 1
0 22 2
−−
= −cos sin
A z( )1
C z( )3B z( )2O z( )0
z zz z
A i A0 3
0 22 2
−−
= +cos sin
Nowz zz z
AB
z zz z
CB
0 1
0 2
0 3
0 2
22
22
−−
+−−
sinsin
sinsin
=
− ++
sin cos sin sin cos sinsin sin
sin
2 2 2 2 2 22 2
2
A C i A C A Ci A C
B
=+
= −sin( )
sin2 2
21
A CB
18. (a) : az2 + z + 1 = 0 ... (i) Taking conjugate of both sides,
az z a z z2 21 0 1 0+ + = ⇒ + + =( )az z2 1 0− + = ...(ii)(since z z= − as 'z' is purely imaginary)Eliminating 'z' from both the equations, we get ( ) ( )a a a a− + + =2 2 0
Let a = cosq + isinq (since |a| = 1) so that (–2isinq)2 + 2(2cosq) = 0
⇒ =− ± +
cos q1 1 4
2Only feasible value of cosq is 5 1
2−
Hence a = cosq + isinq, where q =−
−cos 1 5 12
19. (a) : cos A A= ⇒ =12 4
p
\−−
=−−
( )z zz z
z zz z
3 1
2 1
3 1
2 14cis p /
⇒−−
=+
⇒−−
= − −
z zz z
e
z zz z
i3 1
2 1
4 4
3 1
2 1
4
23 1
3 12
p
4
20. (a) : OP = OC – CP= −2 2 1
\ OP : PC = −( ) :2 2 1 1 OP
C(2, –2)
coordinate of P =2 2 2 1
2 22 2 2 1
2 2( ) , ( )− − −
= −
− −
2 12
2 12
,
21. (a) : (z + 1)(z2 + z + 1) ⇒ z = –1, w, w2
Let f(z) = z1985 + z100 + 1 f(–1) ≠ 0, f(w) = f(w2) = 0 \ w + w2 = –1
| AUGUST ‘1542
22. (a) : r = OA sinq = 6sinqz = 6sinq (cosq + isinq) ⇒ − =cot q 6
zi
A
P(z)
(0, 3)
O
C
y
x
23. (d) : If andz z z u z z z v1 12
22
1 12
22+ − = − − =
We have
u v u v u v2 2 2 212
12
+ = + + − = + −2 212
12
22z z z
And so u v z z z z+( ) = + − +{ }21
212
22
222
= + + − + −z z z z z z1 22
1 22
12
222
= + + −( )z z z z1 2 1 22
24. (a) : Let z1 = cosf1 + isinf1 andz2 = cosf2 + i sinf2 be the roots of z2 + az + b = 0z1 + z2 = (–a) and z1z2 = b
−−
+
+
+
=22 2 2
1 2 1 2 1 2cos cos sinf f f f f f
i a
⇒ =+
arg af f1 2
2arg b = f1 + f2 \ arg b = 2 argaAlso |z1z2| = |b| = 1 and |a| ≤ 2
25. (c) : ∠ = −AOP p q2
cos sinp q q2 2
−
= =z (0, 2)
(0, 1)
O x
y
P z( )
A
C
–Also 2
2
iz
OAOP
i
zi
= +( )
= +( )
sin cos
sin cos
q q
q q
= 1 + icotq ⇒ = − +2z
i cot q ⇒ − =cot q 2z
i
26. (a, b, c) : |z1| = |z2| = 1⇒ a2 + b2 = c2 + d2 = 1 ...(i)And Re( ) Re{( )( )}z z a ib c id1 2 0 0= ⇒ + − =
⇒ ac + bd = 0 ...(ii)Now from (i) and (ii), we get
a b a a cd
a d2 2 22 2
22 21 1+ = ⇒ + = ⇒ = ...(iii)
Also c d c a cb
b c2 2 22 2
22 21 1+ = ⇒ + = ⇒ = ...(iv)
| | [w12 2 2 2 1= + = + =a c a b From (i) and (iv)]
andFrom (i) and (iv)]
| |[
w22 2 2 2 1= + = + =b d c d
FurtherFrom (ii) and (iv
Re( ) Re{( )( )}[
w w1 2 0= + − = + =a ic b id ab cd))]
Also Im( )w w1 2 1= − = ±bc ad | | , | | Re( )w w w w1 2 1 21 1 0= = = and
27. (a, b, c) : |z1 + z2| = |z1 – z2|⇒ + + = − −( )( ) ( )( )z z z z z z z z1 2 1 2 1 2 1 2
⇒ z z z zzz
zz1 2 2 1
1
2
1
20+ = ⇒ = −
⇒zz
1
2is purely imaginary.
Also, |z1 – z2|2 = |z1|2 + |z2|2DOAB is a right angled triangle, right angled at O.
So, circumcentre =+z z1 22
28. (a) : z i z i= + + = + −( ) , ( )1 2 1 2For adjacent vertices
z zn n
ii
i
i= ⇒ = + −
− −=
+
−=cis cis cis2 2 1 2 1
1 2 1
18
18
28
p pp
pp( )
( )
tan
tan⇒ n = 8
29. (a, b, d) : Let |a| = r > 1 and a = rcisq then
z x iy rr
= + = + = + −aa
q q1 cis cis( )
⇒ = +
= −
x rr
y rr
1 1cos sinq qand
Eliminating givesq x
rr
y
rr
2
2
2
21 11
+
+
−
= ,
which is an ellipse.
a rr
= + 1 , b rr
r a b= − = > ⇒ >1 1( )a
\ = − =+
= =e ba r
r
ae1 21 2 4
2
2 , distance between foci
Distance between directrix = 2ae
30. (b, d) : z i= +
2 35
35
35
cos cos sinp p p
= −
2 25
25
25
cos cos sinp p pi
31. (b, c) : z z z zz z zn
n= + + + + + +
( ... ) ...1 21 2
1 1 1
| AUGUST ‘15 43
⇒ |z1 + z2 + .... + zn|2 → which is real
≤ + + + + =z z z z nn12
22
32 2 2...
32. (a, d) : |a| = |b| = |c| = r|c| = |–az2 – bz| ≤ r|z|2 + r|z|⇒ |z|2 + |z| – 1 ≥ 0 ...(i)and az2 = – (bz + c) |a||z|2 ≤ r|z| + r⇒ |z|2 – |z| – 1 ≤ 0 ...(ii)Solve (i) and (ii) to conclude that
min | | max | |z z= − = +5 12
5 12
and
33. (a,d) : If |z – 1| ≤ |z – 5|
Then z z z z z z z z2 21 5 5 25 3 1− − + ≤ − − + − = −& |
⇒ + ≤ − − + = − − +4 24 3 3 9 1( )z z zz z z zz z zand⇒ z z+ ≤ 6 and 2 8( )z z+ = ⇒ z z+ = 4 ⇒ Re(z) = 2⇒ Re(z) ≤ 3Similarly, for |z – 1| ≥ |z – 5|, we get Re(z) = 4
34. (a, d) : We observe that if z0 is a non-zero complexnumber and c is a real number, then the equation z z z z c0 0 0+ + = represents a straight line with complex
slope −zz
0
0.
Let andL z z c L z z d1 20 0: :a a b b+ + = + + =where a = (a, b) and b = (p, q) are non zero complex numbers. Then their cartesian equations are
ax by c px qy d+ + = + + =2
02
0and
\ L1 ^ L2 ⇔ ap + bq = 0 ⇒ + =ab ab 0
⇔ + = ⇔ + =aa
bb
m m0 01 2 ,
where andm aa
m bb1 2= − = − are the complex slopes
of L1 and L2 respectively.L L aq bp1 2 0 0 ⇔ − = ⇔ − =ab ab
⇔ = ⇔ =aa
bb
m m1 2
35. (a) : z1 = cosq + isinqz2 = cos2q + isin2q – (cosq + isinq)= (cos2q – cosq) + i(sin2q – sinq)|z2|2 = (cos2q – cosq)2 + (sin2q – sinq)2
= 2 – 2(cos2qcosq + sin2q sinq) = 2 – 2cosq = 42
2sin q
36. (c) : z2 = (cos2q – cosq) + i(sin2q – sinq)
= − +2 32 2
2 32 2
sin sin cos sinq q q qi
= +
22
32
32
i isin cos sinq q q
arg arg sin arg cos sinz i i2 22
32
32
=
+ +
q q q
= +p q2
32
4 4 2 22
2 1n n n np q p p q p< < +( ) ⇒ < < +( )
⇒ >sin q2
0
37. (b) : (4n + 2)p < q < (4n + 4)p
⇒ + < < + ⇒ <( ) ( ) sin2 12
2 22
0n np q p q
\ = +arg z232
32
p q
38. (b) : 2 2 2 2 2 4−
+ −
cos cosp p
n n+ + − −
...... cos ( )2 2 1 2n
np
= − − + + + −
2 1 2 2 4 1 2( ) cos cos ... cos ( )n
n nn
np p p
= 2(n – 1) – 2(–1) = 2n39. (b) : Consider the triangle OA1AjLet angle A1OAj = a Then (A1Aj)
2 = 12 + 12 – 2(1)(1) cos a
= − = ⋅2 1 2 22
2( cos ) sina a
⇒ = = −A A an
jj1 22
2 1sin ( )a pbut
⇒ =−
A Aj
nj1 21
sin( )p
40. (b) : 2 2 11nn n
nn
n− −=sin sin .......sin
( )p p p
41. (b) : wz iz
x x y y i y x
x y=
−+
=+ + − + − −
+ +2 12 2 2 2 1
2 1 4
2 2
2 2( ) ( )
( )
Re( )w x y x y= ⇒ + + − =02
02 2
42. (c) : Im(w) = 0 ⇒ y = 2x + 1
43. (c) : | |w ww x y x y= ⇒ = ⇒ + + + =1 1 43
23
02 2
44. (a) : Locus of 'z' is perpendicular bisector of theline segment joining (0, 5) and (0, –5). Its equation is y = 0 (i.e., x-axis)
45. (b) : Let w iz
w iz
= − + ⇒ + =15 15
⇒ + = =| || |
w iz
15 3
Locus of 'w' is a circle with centre (0, –1) and radius 3.
| AUGUST ‘1544
46. (c) : |z – i| ≤ 2 represents a disc with centre at(0, 1) and radius 2.|iz + z1| = |i| |z – iz1| = |z – i (3 + 4i)| = − + − ≤ − + − ≤ +|( ) ( ) | | | | |z i i z i i4 2 4 2 2 20
Maximum value of | |iz z+ = +1 20 247. A → r; b → p; c → s; d → q(A) 1 + z3 = 1 + cos3a + isin3a
=
+
2 32
32
32
cos cos sina a ai
⇒ + =arg( )1 32
3z a
(b) 1 – z4 = 1 – cos4a – i sin4a
= −
+ −
2 2 22
22
sin cos sina a p a pi
⇒ arg( )1 22
4− = −z a p
(c) 11
32
2 2 2 2 2
3
4+−
= −
+ −
zz
icos
sincos sin
a
ap a p a
⇒ +−
= −arg 1
1 2 2
3
4zz
p a
(d) zz
i4
311
232
2 2 2 2−+
=
+
+ +
sin
coscos sina
aa p a p
arg zz
4
311 2 2
−+
= +p a
48. A → r; b → r; c → q; d → qLet a be an interior angle of the polygonz z z z e z e z e zn
in
i i− = − ⇒ = − +1 2 1 1 21( ) ( )a a a
z z z z e z e z e zi i i3 2 1 2 3 1 21− = − ⇒ = + −− − −( ) ( )a a a
z3 + zn = (1 – 2i sina) z1 + (1 + 2i sina)z2A = 1 – 2i sina
An
= − = −
3 2 2 3 2 4cos cosa p
When n A= ⇒ =4 5n = 6 ⇒ |A| = 2n A= ⇒ =8 3
n A= ⇒ =12 2
49. A → s; b → r; c → q; d → q(A) |1 – i|n = 2n ⇒ n/2 = 0 ⇒ n = 0 (b) x3 + 2x2 + 2x + 1 = 0 ⇒ x = – 1, w, w2 But x = w, w2 will only satisfy x2000 + x2002 + 1 = 0(c) x + 2xy = 0 and x2 – y2 + y = 0
⇒ − − −i i i, ,32 2
32 2
(d) x y x y xy z i i2 2 2 2 0 2 0 0− + + = = ⇒ = −and , ,
50. (5) : Let (real)35
2
1
izz
K= then zz
Ki
2
1
53
=
53 7
3 75
3 353
3 353
2
12
1
+
−=
+
−
zzzz
KiKi
= 5 35 935 9
5K iK i
+−
=
51. (4) : Locus of z is an ellipse( ) ( )x y− + − =11
254
161
2 2
Equation of tangent is y – 4 = m(x – 11) + c ⇒ c = 11m – 4As c2 = a2m2 + b2 for standard ellipse
⇒ − = + ⇒ = =( )11 4 25 16 0 1112
2 2m m m mor
\ = ⇒ = −tan tanq q1112
1112
1
52. (4) : Let a and ib, a, b ∈ R are roots ofx2 + ax + b = 0 ⇒ a + ib = – a, iab = b
a b− = −i a⇒ = − + = − −2 2a b( ) ( )a a i a aand
\ = − ⇒ = −4 42 2 2 2i a a b a aab
53. (5) : z iz i
z x i−−
= ⇒ = +42
1 3 , using this in
z iz i
x i x i− ++
= ⇒ − + = +8 33
35
5 8 6 3 6
⇒ x = 8, 17Two complex numbers are 8 + 3i, 17 + 3iSum of real parts = 8 + 17 =2554. (1) : Let z = iy⇒ − − + + =y a y i a y ia y a4
13
22
3 4 0
⇒ − + =y a y a42
24 0 ....( )i and − + =a y a y1
33 0
⇒ = =y yaa
0 2 3
1or (ii)....
From (i) and (ii), we get aa
a aa
a32
12
2 3
14 0− + = ⇒ + =
aa a
a aa a
3
1 2
1 4
2 31
55. (2) : Let ei
n2p
a= then 1
1
11
11
112
1
1
2 1
−
=−
+−
+ +−=
−
−∑e n
ijj
n
np a a a.......
| AUGUST ‘1546
where a is a nth root of unity. (a, a2, a3 ,...., an–1) are the
roots of xx
x x xn
n−−
= − − − −11
2 1( )( ).......( )a a a
Taking log on both sides
log log( ) log( ) ....... log( )xx
x x xn
n−−
= − + − + + − −11
2 1a a a
Differentiating w.r.t. x and use limx→1
⇒ − =−
+−
+ +− −
nn
12
11
11
112 1a a a
........
56. (0) : Let vertices be 1, a, a2, ......, an–1
Given 1 + a + a2 + ........ an–1 = 0 ⇒ an – 1 = 0 ⇒ z1, z2, z3 ,......., zn are roots of an = 1which form regular polygon. So, distance is zero.
57. (5) : |z1 – z0| = |z2 – z0| = |z3 – z0| = |l|
z zz z
ee
ei
ii3 0
2 0
7 11
417 44−
−= =
p
pp
/
//
⇒ ∠ = ⇒ ∠ =BSC BAC1744
1788
p p
Similarlyz zz z
e ACBi2 0
1 0
48
−−
= ⇒ ∠ =p p/
\ ∠ = − − =ABC p p p p8
1788
1522
58. (3) : tan( )
( )− −
+ − + +=1
2 2 4a b y
x y a b x abp
⇒ x2 + y2 – (a + b)x – (a – b)y + ab = 0
Centre =3
23
+⇒ + =
ia b
59. (4) : z i i i= − + =12
1 3 2( ) wz101 = iw(z101 + i109)106 = (–iw2)106 = – w2
–w2 = (iw2)n = inw2n or w2n–2 in = – 1This is possible only when n = 4r + 2 and 2n – 2 is multiple of 3 i.e., 2 (4r + 2) – 2 is a multiple of 3.i.e., 8r + 2 is a multiple of 3 ⇒ r = 2\ =
\ =
n
k
1025
4
nn
| AUGUST ‘15 47
Definition of Matrix A system of m × n numbers arranged in the form of an ordered set of m rows and n columns is called an m × n matrix. It can be read as m by n matrix. It is represented as A = [aij]m × n and can be written in expanded form as
A
a a aa a a
a a a
n
n
m m mn
=
11 12 1
21 22 2
1 2
. . .
. . .
. . .
Different types of Matrices (i) rectangular matrix and square matrix : If in
an m × n matrix m ≠ n, then it is a rectangular matrixBut if m = n, it is a square matrix.
(ii) column Matrix : A matrix which has only one column and m rows is called a column matrix of length m.
(iii) row Matrix : A matrix which has only one row and n columns is called a row matrix of length n.
(iv) Diagonal Matrix : A square matrix of any order with zero elements everywhere, except on the main diagonal, is called a diagonal matrix.
(v) scalar matrix : A matrix whose diagonal elements are all equal and other entries are zero, is called a scalar matrix.
(vi) identity or Unit Matrix : A square matrix in which all the elements along the main diagonal (elements of the form aii) are unity is called an identity matrix or a unit matrix. An identity matrix of order n is denoted by In.
(vii) null or Zero Matrix : The matrix whose all elements are zero is called null matrix or zero matrix. It is usually denoted by O.
(viii)triangular Matrix : A square matrix whose elements above the main diagonal or below the main diagonal are all zero is called a triangular matrix. note : (i) [aij]n × n is said to be upper triangular matrix if i > j ⇒ aij = 0, (ii) [aij]n × n is said to be lower triangular matrix if i < j ⇒ aij = 0.
(ix) sub Matrix : A matrix obtained by omitting some rows or some columns or both of a given matrix A is called a sub matrix of A.
(x) Horizontal Matrix : Any matrix in which the number of columns is more than the number of rows is called a horizontal matrix.
(xi) Vertical Matrix : Any matrix in which the number of rows is more than the number of columns is called vertical matrix.
eqUality of two Matrices Two matrices A = [aij] and B = [bij] are said to be equal if they are of the same order and their corresponding elements are equal. If two matrices A and B are equal, we write A = B.
operation on Matrices addition of Matrices
If A and B are two matrices of the same order m × n, then their sum is defined to be the matrix of order m × n obtained by adding the corresponding elements of A and B.
This column is aimed at Class XII students so that they can prepare for competitive exams such as JEE Main/Advanced, etc. and be also in command of what is being covered in their school as part of NCERT syllabus. The problems here are a happy blend of the straight and the twisted, the simple and the difficult and the easy and the challenging.
*
| AUGUST ‘1548
subtraction of two MatricesIf A and B are two matrices of the same order then the sum A +(–B) , written as A – B, is the matrix obtained by subtracting B from A.
properties of Matrix addition If A, B, C are three matrices of the same order, then(i) A + B = B + A (commutative law of addition)(ii) A + (B + C) = (A + B) + C (associative law
of addition)(iii) A + O = O + A = A(iv) A + B = A + C ⇒ B = C (v) A + (–A) = (–A) + A = O.
Multiplication of a Matrix by a scalarIf A = [aij]m × n and l is a scalar, then lA = [laij]m × n
Multiplication of two Matrices For two matrices A and B, the product matrix AB can be obtained if the number of columns in A = the number of rows in B.
properties of Matrix Multiplication (i) Matrix multiplication is associative i.e., (AB)C = A(BC), A, B and C are m × n,
n × p and p × q matrices respectively. (ii) Multiplication of matrices is distributive over
addition of matrices i.e., A(B + C) = AB + AC (iii) Existence of multiplicative identity of square
matrices. If A is a square matrix of order n and In is the identity matrix of order n, then
AIn = InA = A. (iv) Whenever AB and BA both exist, it is not
necessary that AB = BA. (v) The product of two matrices can be a zero
matrix while neither of them is a zero matrix.
(vi) In the case of matrix multiplication of AB = 0, then it doesn’t necessarily imply that A = 0 or B = 0 or BA = 0.
trace of a Matrix Let A be a square matrix of order n. The sum of the diagonal elements of A is called the trace of A.
\ = = + + +=∑Trace ( ) ... .A a a a aii nni
n
11 221
transpose of a Matrix The matrix obtained from any given matrix A, by interchanging rows and columns, is called the transpose of A and is denoted by A′ or AT.
properties of transpose of a matrix(i) (A′)′ = A (ii) (A + B)′ = A′ + B′ (iii) (aA)′ = aA′, a being any scalar. (iv) (AB)′ = B′A′ special Matrices
symmetric Matrix z
A matrix which is unchanged by transposition is called a symmetric matrix. Such a matrix is necessarily square i.e., if A = [aij]m × n is a symmetric matrix then m = n, aij = aji i.e., A′ = A. skew symmetric Matrix z
A square matrix A = [aij] is said to be skew symmetric, if aij = – aji for all i and j.Thus if A = [aij]m × n is a skew symmetric matrix, then m = n, aij = – aji i.e., A′ = –A. Obviously diagonal elements of a square matrix are zero.note : Every square matrix can be uniquely expressed as the sum of symmetric and skew symmetric matrix.
i e A A A A A. ., ( ) ( )= + ′ + − ′12
12
where and12
12
( ) ( )A A A A+ ′ − ′
are symmetric and skew symmetric parts of A.orthogonal Matrix z
A square matrix A is said to be orthogonal, if AA′ = A′A = I, where I is a unit matrix.note : (i) If A is orthogonal, then A′ is also orthogonal. (ii) If A and B are orthogonal matrices then AB
and BA are also orthogonal matrices. idempotent Matrix : z A square matrix A is called idempotent provided if it satisfies the relation A2 = A.involutory Matrix z
A square matrix A is called involutory matrix if A2 = Iperiodic Matrix z
A square matrix A is called periodic, if Ak + 1 = A, where k is a positive integer. If k is the least positive integer for which Ak + 1 = A, then k is said to be period of A. For k = 1, we get A2 = A and we called it to be idempotent matrix.nilpotent Matrix z
A square matrix A is called a nilpotent matrix, if there exists a positive integer m such that Am = O. If m is the least positive integer such that Am = O, then m is called the index of the nilpotent matrix A or nilpotency of matrix.
| AUGUST ‘15 49
DeterMinant Equations a1x + b1y = 0 and a2x + b2y = 0 in x and y have a unique solution if and only if
a1b2 – a2b1 ≠ 0. We write a1b2 – a2b1 as a ba b
1 1
2 2and call it a determinant of order 2. Similarly the equations a1x + b1y + c1z = 0, a2x + b2y + c2z = 0 and a3x + b3y + c3z = 0 have a unique solution if a1(b2c3 – b3c2) + b1 (a3c2 – a2c3) + c1 (a2b3 – a3b2) ≠ 0
i ea b ca b ca b c
. ., D = ≠1 1 1
2 2 2
3 3 3
0
The number ai, bi, ci (i = 1, 2, 3) are called the elements of the determinant. The determinant obtained by deleting the ith row and jth column is called the minor of the element at the ith row and jth column. We shall denote it by Mij. The cofactor of this element is (–1)i+j Mij, denoted by Cij.Let A = [aij]3 × 3 be a matrix, then the corresponding determinant (denoted by det A or |A|) is
a a aa a aa a a
11 12 13
21 22 23
31 32 33It is easy to see that |A| = a11C11+ a12C12 + a13C13 (we say that we have expanded the determinant |A| along first row). Infact value of |A| can be obtained by expanding it along any row or along any column. Further note that if elements of a row (column) are multiplied to the cofactors of other row (column) and then added, then the result is zero : a11C21 + a12C22 + a13C23 = 0.
properties of DeterMinants The value of a determinant remains unaltered, if its z
rows are changed into columns and the columns into rows. If all the elements of a row (or column) of a z
determinant are zero, then the value of the determinant is zero. If any two rows (or columns) of a determinant z
are identical, then the value of the determinant is zero. The interchange of any two rows (columns) of a z
determinant results in change of its sign.If all the elements of a row (column) of a determinant z
are multiplied by a non-zero constant, then the determinant gets multiplied by that constant.
If each element of a row (or column) of a determinant z
is a sum of two terms, then determinant can be written as sum of two determinant in the following way: a b c da b c da b c d
a b ca b ca b c
a b da b d
1 1 1 1
2 2 2 2
3 3 3 3
1 1 1
2 2 2
3 3 3
1 1 1
2 2
+++
= + 22
3 3 3a b dThe value of a determinant remains unaltered under z
a column operation of the form Ci → Ci + aCj + bCk(j, k ≠ i) or a row operation of the formRi → Ri + aRj + bRk (j, k ≠ i)product of two determinants a b ca b ca b c
m m mn n n
1 1 1
2 2 2
3 3 3
1 2 3
1 2 3
1 2 3
×l l l
=+ + + + + ++ +
a b m c n a b m c n a b m c na b m c n a
1 1 1 1 1 1 1 2 1 2 1 2 1 3 1 3 1 3
2 1 2 1 2 1
l l ll 22 2 2 2 2 2 2 3 2 3 2 3
3 1 3 1 3 1 3 2 3 2 3 2 3
l ll l
+ + + ++ + + +
b m c n a b m c na b m c n a b m c n a ll3 3 3 3 3+ +b m c n
(row by column multiplication)
=+ + + + + ++ +
a b c a m b m c m a n b n c na b c a
1 1 1 2 1 3 1 1 1 2 1 3 1 1 1 2 1 3
2 1 2 2 2 3
l l ll l l 22 1 2 2 2 3 2 1 2 2 2 3
3 1 3 2 3 3 3 1 3 2 3 3 3
m b m c m a n b n c na b c a m b m c m a
+ + + ++ + + +l l l nn b n c n1 3 2 3 3+ +
(row by row multiplication) We can also multiply determinants column by row or column by column. limit of a determinant
Let D( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )
,xf x g x h x
x m x n xu x v x w x
= l
then lim ( )
lim ( ) lim ( ) lim ( )
lim ( ) limx a
x a x a x a
x ax
f x g x h x
l x→
→ → →
→=D
xx a x a
x a x a x a
m x n x
u x v x w x→ →
→ → →
( ) lim ( )
lim ( ) lim ( ) lim ( )
,
provided each of nine limiting values exist finitely. Differentiation of a determinant
Let D( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )
,xf x g x h xl x m x n xu x v x w x
=
then ′ =′ ′ ′
+D ( )( ) ( ) ( )
( ) ( ) ( )( ) ( ) ( )
( ) ( )x
f x g x h xx m x n x
u x v x w x
f x g xl
hh xx m x n x
u x v x w x
f x g x h xx m x n x
( )( ) ( ) ( )( ) ( ) ( )
( ) ( ) ( )( ) ( ) ( )
′ ′ ′
+′
l
luu x v x w x( ) ( ) ( )′ ′
| AUGUST ‘1550
integration of a Determinant
Let D( )( ) ( ) ( )
,xf x g x h xa b cl m n
=
where a, b, c, l, m and n are constants,
then D( )
( ) ( ) ( )
x dx
f x dx g x dx h x dx
a b cm n
a
b
a
b
a
b
a
b=
∫ ∫ ∫
∫l
Note that if more than one row (column) of D(x)
are variable, then in order to find D( ) .x dxa
b
∫ First
we evaluate the determinant D(x) by using the properties of determinants and then we integrate it.
special DeterMinants skew symmetric Determinant A determinant of a skew symmetric matrix of odd order is always zero.circulant Determinant A determinant is called circulant if its rows (columns) are cyclic shifts of the first row (columns).
e.g.,a b cb c ac a b
. It can be shown that its value is
– (a3 + b3 + c3 – 3abc) .1 1 1
2 2 2
a b c
a b c
a b b c c a= − − −( )( )( )
1 1 1
3 3 3
a b c
a b c
a b b c c a a b c= − − − + +( )( )( )( )
1 1 12 2 2
3 3 3
a b c
a b c
a b b c c a ab bc ca= − − − + +( )( )( ) ( )
aDjoint of a sqUare Matrix Let A = [aij]n × n be an n × n matrix. The transpose B′ of the matrix B = [Aij]n × n, where Aij denotes the cofactor of the elements aij in the |A|, is called the adjoint of the matrix A and is denoted by the symbol adj A.
Thus, the adjoint of a matrix A is the transpose of the matrix formed by the cofactors of A. properties of adjoint MatrixIf A, B are square matrices of order n and In is corresponding unit matrix, then(i) A(adj A) = |A| In = (adj A) A
(Thus A(adj A) is always a scalar matrix)(ii) |adj A|= |A|n–1
(iii) adj (adj A) = |A|n–2 A (iv) |adj (adj A)|= |A|(n–1)2
(v) adj (AT) = (adj A)T
(vi) adj (AB) = (adj B)⋅(adj A)(vii) adj (Am) = (adj A)m, m ∈ N(viii) adj (kA) = kn–1 (adj A), k ∈ R(ix) adj (In) = In (x) adj 0 = 0A is symmetric ⇒ adj A is also symmetric.A is diagonal ⇒ adj A is also diagonal.A is triangular ⇒ adj A is also triangular.A is singular ⇒ |adj A| = 0
inVerse of a sqUare Matrix Let A be any n–rowed square matrix. Then a matrix B, if exists, such that AB = BA = In, is called the inverse of A. Inverse of A is usually denoted by A–1 (if exists). We have, |A|In = A(adjA) |A| A–1 = (adj A). Thus the necessary and sufficient condition for a square matrix A to possess the inverse is that |A| ≠ 0 and then A–1 =
adj ( )| |
.A
AA square matrix A is called non-singular if |A| ≠ 0. Hence, a square matrix A is invertible if and only if A is non-singular. properties of inverse of a Matrix (i) (A–1)–1 = A(ii) (AT)–1 = (A–1)T
(iii) (AB)–1 = B–1A–1 (iv) (An)–1 = (A–1)n , n ∈ N(v) adj (A–1) = (adj A)–1 A = diag(a1, a2, ...., an). A is symmetric ⇒ A–1 is also symmetric (|A| ≠ 0).A is diagonal matrix and |A| ≠ 0 ⇒ A–1 is also diagonal matrix.A is scalar matrix ⇒ A–1 is also a scalar matrix.A is triangular matrix and |A| ≠ 0 ⇒ A–1 is also triangular matrix.
| AUGUST ‘15 51
systeM of linear siMUltaneoUs eqUations Consider the system of linear non-homogeneous simultaneous equations in three unknowns x, y and z, given by a1x + b1y + c1z = d1, a2x + b2y + c2z = d2 and a3x + b3y + c3z = d3
Let Aa b ca b ca b c
Xxyz
Bddd
=
=
=1 1 1
2 2 2
3 3 3
1
2
3
, ,
,
Let | | , ,Aa b ca b ca b c
d b cd b cd b c
x= = =D D1 1 1
2 2 2
3 3 3
1 1 1
2 2 2
3 3 3obtained on replacing first column of D by B.
Similarly, let andD Dy z
a d ca d ca d c
a b da b da b d
= =1 1 1
2 2 2
3 3 3
1 1 1
2 2 2
3 3 3It can be shown that AX = B, xD = Dx, yD = Dy, zD = Dz. Determinant Method of solution We have the following two cases :case iIf D ≠ 0, then the given system of equations has unique solution, given by x = Dx / D, y = Dy / D and z = Dz /D. case iiIf D = 0, then two sub cases arise:(a) at least one of Dx, Dy and Dz is non-zero, say
Dx ≠ 0. Now in x⋅D = Dx, L.H.S. is zero and R.H.S. is not equal to zero. Thus we have, no value of x satisfying x⋅D = Dx. Hence given system of equations has no solution.
(b) Dx = Dy = Dz = 0. In the case the given equations are dependent. Delete one or two equation from the given system (as the case may be) to obtain independent equation(s). The remaining equation(s) may have no solution or infinitely many solution(s).
Matrix Method of solution z
(a) D ≠ 0, then A–1 exists and hence AX = B ⇒ A–1(AX) = A–1B ⇒ X = A–1 B and therefore unique values of x, y and z are obtained.
(b) We have AX = B ⇒ ((adj A)A)X = (adj A)B ⇒ DX= (adj A)B.If D = 0, then DX = O3 × 1, zero matrix of order 3 × 1. Now if (adj A)B = 0, then the system AX = B has infinitely many solutions, else no
solution. note : A system of equation is called consistent if it has atleast one solution. If the system has no solution, then it is called inconsistent.
systeM of linear HoMogeneoUs siMUltaneoUs eqUations
Consider the system of linear homogeneous simultaneous equations in three unknowns x, y and z, given by a1x + b1y + c1z = 0, a2x + b2y + c2z = 0 and a3x + b3y + c3z = 0. In this case, system of equations is always consistent as x = y = z = 0 is always a solution. If the system has unique solution (the case when coefficient determinant ≠ 0), then x = y = z = 0 is the only solution (called trivial solution). However if the system has coefficient determinant = 0, then the system has infinitely many solutions. Hence in this case we get solutions other than trivial solution also and we say that we have non-trivial solutions.
ProblemsSection-i
Single correct Answer type
1. A and B are two non singular matrices so thatA6 = I and AB2 = BA(B ≠ I). A value of K so that BK = I is(a) 31 (b) 32 (c) 63 (d) 64
2. For each real number x such that –1 < x < 1, let
A(x) be the matrix ( ) .11
1 11−
−−
= +
+−x
xx
z x yxy
andThen,(a) A(z) = A(x) + A(y) (b) A(z) = A(x) [A(y)]–1
(c) A(z) = A(x) A(y) (d) A(z) = A(x) – A(y)
3. If Ai ii i
B=−
−
=
−−
and
1 11 1
, then A8 equals to
(a) 4B (b) 128B (c) –128B (d) –64B
4. If a – 2b + c = 1, then the value ofx x x ax x x bx x x c
+ + ++ + ++ + +
1 22 33 4
is
(a) x (b) –x (c) –1 (d) 1
5. If A B C ABAT=−
=
−
=
cos sinsin cos
, , ,q qq q
1 01 1
then ATCnA equals to (n ∈ I+)
| AUGUST ‘1552
(a) −
n 11 0
(b) 10 1
−
n (c)
0 11 −
n (d)
1 01−
n
6. If p + q + r = 0 andpa qb rcqc ra pbrb pc qa
ka b cc a bb c a
= ,
then k = (a) 0 (b) abc(c) pqr (d) a + b + c
7. x
rx
rx
ry
ry
ry
rz
rz
rz
r
xr
xr
xr
yC C C
C C C
C C C
C C C+ +
+ +
+ +
++
++
−1 2
1 2
1 2
11
22
CC C C
C C Cr
yr
yr
zr
zr
zr
++
++
++
++
=11
22
11
22
(a) 0 (b) 2n
(c) x+y+zCr (d) x+y+zCr+2
8. If a, b, c, d > 0 ; x ∈ R and(a2 + b2 + c2)x2 – 2(ab + bc + cd) x + b2 + c2 + d 2 ≤ 0,
then 33 1465 2797 40
logloglog
abc
=
(a) 1 (b) – 1 (c) 0 (d) none of these
9. If A and B are two non singular matrices and bothare symmetric and commute each other then (a) Both A–1B and A–1B–1 are symmetric.(b) A–1B is symmetric but A–1B–1 is not symmetric.(c) A–1B–1 is symmetric but A–1B is not symmetric.(d) Neither A–1B nor A–1B–1 are symmetric.
10. If f(x) = ax2 + bx + c, a, b, c ∈ R and the equationf(x) – x = 0 has imaginary roots a and b and g and d be
the roots of f(f(x)) – x = 0, then 2
01
a db ag b
is
(a) 0 (b) purely real (c) purely imaginary (d) none of these
11. Suppose a matrix A satisfies A2 – 5A + 7I = 0. IfA5 = aA + bI, then the value of 2a + b is(a) –87 (b) –105 (c) 1453 (d) 1155
12. If ‘a’ is a root of x4 = 1 with negative principalargument, then the principal argument of D(a) where
D( )a a a a
a a
= + +
+
1 1 1
1 1 0
1 3
1
n n n
n n
is
(a) 514π (b) − 3
4π (c) π
4(d) − π
4
13. If z is a complex number and l1, l2, l3, m1, m2, m3
are all real, then l l ll l ll l
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3
z m z m z z m zz m z m z z m zz m z m z z m z
+ + ++ + ++ + + ll3
is
equal to (a) |z|2 (b) 3(c) (l1l2l3 + m1m2m3)2|z|2 (d) 0
14. Let x > 0, y > 0, z > 0 are respectively the 2nd, 3rd, 4th
terms of a G.P and D = = − −
+ +
+ +
+ +
x x x
y y y
z z z
rr
k k k
k k k
k k k
1 2
1 2
1 2
221 1 1( )
(where r is the common ratio) then(a) k = –1 (b) k = 1 (c) k = 0 (d) None of these
15. Let a, b, c be positive real numbers. Then thefollowing system of equations in x, y, z xa
yb
zc
xa
yb
zc
xa
yb
zc
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
21 1 1+ − = − + = − + + =, ,
has(a) no solution (b) unique solution (c) infinite solution (d) finitely many solution
16. Ifa ab c ca b
ab c b bc a
ac b bc a c
c bc a
b a
2 2
2 2
2 2
+ + −
− + +
+ − +
−−
−=
l l l
l l l
l l l
ll
l(( )1 2 2 2 3+ + +a b c
then l is equal to (a) 0 (b) 1 (c) –1 (d) ±2
17. The system of equations x + ky + 3z = 0,3x + ky – 2z = 0, 2x + 3y – 4z = 0 possess a non-trivial solution over the set of rationals, then 2k is an integral element of the interval (a) [10, 20] (b) (20, 30) (c) [30, 40] (d) (40, 50)
18. If A and B are two square matrices of order 3 × 3which satisfy AB = A and BA = B then (A + B)7 is (a) 7(A + B) (b) 7I3×3 (c) 64(A + B) (d) 128 I3×3
19. A and B are square matrices and A is non-singularmatrix, (A–1BA)n, n ∈ I+, is equal to (a) A–nBnAn (b) AnBnA–n (c) A–1BnA (d) A–nBAn
| AUGUST ‘15 53
Section-ii
Multiple correct Answer type
20. If thenD =
− +− +
− − −+ +
=
a c c a a cc b b c b c
a b b c a cx y z x y
01
0,
(a) a, b, c are in A.P. (b) b, a, c are in A.P. (c) a, c, b are in G.P. (d) a, c, b are in H.P.
21. If ‘A’ is a matrix of size n × n such that A2 + A + 2I= 0, then(a) A is non-singular (b) A is symmetric
(c) |A| ≠ 0 (d) A A I− = − +1 12
( )
22. Let thenA =
1 2 22 1 22 2 1
,
(a) A2 – 4A – 5I3 = 0 (b) A A I− = −13
15
4( )
(c) A3 is not invertible (d) A2 is invertible
23. If the sum of two idempotent matrices isidempotent then(a) AB + BA = 0 (b) AB = BA = I(c) AB ≠ BA (d) AB – BA = 0
24. If a square matrix A = [aij], aij = i2 – j2 is of evenorder, then(a) A is a skew symmetric matrix (b) |A| is a perfect square(c) A is a symmetric and |A| = 0 (d) A is neither symmetric nor skew symmetric
25. If
D( )x
x x x
x x x
x x x
ax bx c=
+ − +
+ − +
− + −
= + +
2
2
2
3 2
4 3 2 4 13
2 5 9 4 5 26
8 6 1 16 6 104
xx d+ ,
then(a) a = 3 (b) a = 0 (c) c = 0 (d) b = 2
26. If f x
x x
x x
x x
a b
a b
b a
( )
( ) ( )
( ) ( )
( ) ( )
,=
+ +
+ +
+ +
1 1 2 1
1 1 1 2
1 2 1 1
where
a, b being positive integers, then(a) constant term of f(x) is 0 (b) coefficient of x in f(x) is 0(c) constant term in f(x) is (a – b) (d) coefficient of x in f(x) is (a – b)
27. The value of the determinantcos( ) sin( ) cos
sin cos sincos sin cos
q a q a aq q aq q l a
+ − +
−
2is
(a) independent of q for all l ∈ R (b) independent of q and a when l = 1(c) independent of q and a when l = –1 (d) independent of l for all q
28. Suppose that a, b, c are real numbers such that
a + b + c = 1. If the matrix Aa b cb c ac a b
=
be an orthogonal
matrix, then (a) A is an involutory matrix (b) |A| is negative(c) a3 + b3 + c3 – 3abc = 1 (d) atleast one of a, b, c is negative
29. If A is a square matrix such that
A A⋅ =
( )adj then4 0 00 4 00 0 4
(a) |A| = 4 (b) |adjA| = 16
(c) adj adjadj
AA
( )= 16 (d) |adj 2A| = 128
30. System of equations x + 3y + 2z = 6, x + ly + 2z = 7and x + 3y + 2z = m has (a) unique solution if l = 2, m ≠ 6 (b) infinitely many solution, if l = 4, m = 6 (c) no solution if l = 5, m = 7 (d) no solution if l = 3, m = 5
31. If A =
1 1 11 1 11 1 1
, then
(a) A3 = 9A (b) A3 = 27A (c) A + A = A2 (d) A–1 does not exist
Section-iiicomprehension type
paragraph for question no. 32 to 34A Pythagorean triplet is triplet of positive integers (a, b, c) such that a2 + b2 = c2. Define the matrices P, Q and R by
P Q R=
= − − −
=− − −1 2 3
2 1 22 2 3
1 2 22 1 2
2 2 3
1 2 22, and 11 22 2 3
| AUGUST ‘1554
32. If we write Pythagorean triplets (a, b, c) in matrix
form as abc
then which of the following matrix product
is a Pythagorean triplet ?
(a) Q345
(b) P345
(c) R345
(d) none of these
33. Which one of the following does not hold good ?(a) P–1 = adj.P (b) (PQ)–1 = adj.(PQ) (c) (QR)–1 = adj.(QR) (d) (PQR)–1 ≠ adj.(PQR)
34. Tr(P + QT + 2R) equals(a) 17 (b) 15 (c) 14 (d) 18
paragraph for question no. 35 to 37If A is 3 × 3 matrix then a non trivial solution X = (x y z)T such that AX = lX(l ∈ R) yields 3 values of l say l1, l2, l3. For any such matrix A, l’s are called eigen values and corresponding X’s are called eigen vectors. It is known that, for any 3 × 3 matrix Tr(A) = l1 + l2 + l3, det A = l1l2l3. Answer the
following questions for matrix A =
1 1 22 2 12 1 2
35. Tr(A–1) =(a) 1
3(b) 1
2(c) − 1
2(d) − 1
336. Tr(A3) =(a) 149 (b) 101 (c) 128 (d) 133
37. Which of the following is false?(a) $ a non trivial solution X such that
AX X= +( )2 7
(b) $ a non trivial solution X such that AX = X (c) $ a non trivial solution X such that
A X X− = −1 2 7( )
(d) The total number of non trivial solutions X such that AX = lX is 3.
paragraph for question no. 38 to 40Let D ≠ 0 and Dc denotes the determinant of cofactors, then Dc = Dn–1, where n (> 0) is the order of D.On the basis of above information, answer the following questions:
38. If a, b, c are the roots of the equation x3 – px2 + r = 0,then the value of
bc a ca b ab c
ca b ab c bc a
ab c bc a ca b
− − −
− − −
− − −
2 2 2
2 2 2
2 2 2
is
(a) p2 (b) p4 (c) p6 (d) p9
39. If a, b, c are the roots of the equationx3 – 3x2 + 3x + 7 = 0, then the value of
2
2
2
2 2 2
2 2 2
2 2 2
bc a c b
c ac b a
b a ab c
−
−
−
is
(a) 9 (b) 27 (c) 81 (d) 0
40. Suppose a, b, c ∈ R, a + b + c > 0, A = bc – a2,B = ca – b2 and C = ab – c2 and A B CB C AC A B
a b cb c ac a b
= 49, then equals
(a) – 7 (b) 7(c) – 2401 (d) 2401
Section-iV
Matrix-Match type
41. Let p( )sin coscos sinsin cos
,qq qq qq q
=−
− −
211
q( )sincosq
qq=
−−−
2 1 12 4 3
2 7 5; r q
q q qq q qq q q
( ) = −− −
cos sin cossin cos sincos sin cos
and cosecs( )
sec
cos cos
cos cot
q
q
q q q
q q
=
2
2 2 2
2 2
1 1
1
Match the functions on the left with their range on the right.
column i column ii
(A) p(q) (p) [0, 1]
(B) q(q) (q) [ , ]0 2 2
(C) r(q) (r) [–2, 2]
(D) s(q) (s) [ , ]− − −5 2 5 2
| AUGUST ‘1556
42. column i column ii
(A)A is a matrix such that A2 = A. If (I + A)8 = I + lA, then l + 1 is equal to
(p) 64
(B)If A is a square matrix of order 3 such that |A| = 2, then (adjA–1)–1 is equal to
(q) 1
(C)
L e t | | | |A aij= ≠×3 3 0 . E a c h element aij is multiplied by li– j . Let |B | is the resulting determinant, where |A| = l |B|, then l is equal to
(r) 256
(D)
If A is a diagonal matrix of order 3 × 3 is commutative with every square matrix of order 3 × 3 under multiplication and trace (A) = 12, then |A| =
(s) 4
43. A, B are two matrices of 3 × 3 order such that
column i column ii(A) A is non singular and
AB = 0 then(p) A = 0
(B) B is non singular and AB = 0 then
(q) B = 0
(C) A & B both are non zero matrix and AB = 0 then
(r) det A = 0
(D) An = 0 for some n ≥ 2 then (s) det B = 0
Section-V
integer Answer type
44. Let An, (n ∈ N) be a matrix of order(2n – 1) × (2n – 1), such that aij = 0 " i ≠ j and aij = n2 + i + 1 – 2n " i = j where aij denotes the element of ith row and jth column of An. Let Tn = (–1)n × (sum of all the elements of An). Find
the value of Tn
n =∑
1
102
520200, where [.] represents the greatest
integer function.
45. Find the value of f π6
, where
f ( )
cos cos sin sin
cos sin sin cossin cos
q
q q q q
q q q qq q
=
−
−
2
2
0
46. det ,Pa b cc a bb c a
= where 'P' is an orthogonal matrix.
Then the value of |a + b + c| is
47. If A =−−
3 41 1
, then det(A2005) equals to
48. If x, y, z are cube roots of unity and
D
x y z z
x y z x
y y z x
=
+
+
+
2 2 2 2
2 2 2 2
2 2 2 2
,
then the real part of D is
49. Find the coefficient of x in the determinant( ) ( ) ( )
( ) ( ) ( )
(
1 1 1
1 1 1
1
1 1 1 2 1 3
2 1 2 2 2 3
+ + +
+ + +
+
x x x
x x x
x
a b a b a b
a b a b a b
)) ( ) ( )
,a b a b a bx x3 1 3 2 3 31 1+ +
where ai, bi ∈ N (i = 1, 2, 3)
50. If f xx x xx x x( )
cos( ) cos( ) cos( )sin( ) sin( ) sin( )sin
=+ + ++ + +
a b ga b g
(( ) sin( ) sin( )b g g a a b− − −
and then find f f rr
( ) , ( ) / .2 6 301
25=
=∑
51. If f(x) satisfies the equationf x f x f x( ) ( ) ( )+ + +
− =1 8 1
1 2 52 3
0l
for all real x. If f is periodic with period 7, then find the value of |l|.
52. If A =−
−
1 1 10 2 32 1 0
, B = (adj A) and C = 5A, then
find the value of adj B
C.
53. If A =−
1 01 7 and A2 = 8A + kI2, then find the
value of |k|.solutions
1. (c) : A5(AB2) = A5BA⇒ B2 = A5BA⇒ B4 = (A5BA)(A5BA) = A5B2A = A5(A5BA)A⇒ B4 = A4BA2
| AUGUST ‘15 57
⇒ B8 = (A4BA2)(A4BA2) = A4B2A2 = A4(A5BA)A2
⇒ B8 = A3BA3
⇒ B16 = (A3BA3)(A3BA3) = A3B2A3 = A3(A5BA)A3
= A2BA4
⇒ B32 = (A2BA4)(A2BA4) = A2B2A4 = A2(A5BA)A4
= ABA5
⇒ B64 = (ABA5)(ABA5) = AB2A5 = A(A5BA)A5 = B⇒ B63 = I
2. (c) : A z Ax y
xy( ) =
++
1
=+
− −
−+
+
−+
+
11 1
11
11
xyx y
x yxy
x yxy
( )( )
\ A(x) A(y) = A(z)3. (b) : We have, A = iB⇒ A2 = (iB)2 = i2B2 = –B2 = −
−−
= −
2 22 2
2B
⇒ A4 = (–2B)2 = 4B2 = 4(2B) = 8B⇒ (A4)2 = (8B)2 ⇒ A8 = 64B2 = 128B
4. (c) :x x x ax x x bx x x c
a b ca b c b
+ + ++ + ++ + +
− + =− + − =
1 22 33 4
2 11
sinceor ( ) ( )
Apply the operation,R1 → R1 – 2R2 + R3R3 → R3 – R2, the determinant reduces to
0 0 12 3
1 11x x x b
c b+ + +
−= −
5. (d) : A =−
cos sinsin cos
q qq q
AAT = I ... (i)Now, C = ABAT ⇒ ATC = BAT ... (ii)Now, ATCnA = ATC·Cn – 1A = BATCn – 1A (from (ii)) = BATC Cn – 2A = B2ATCn – 2A = .......
= Bn – 1AT CA = Bn – 1BATA = Bn =−
1 01n
6. (c) : p + q + r = 0 ⇒ p3 + q3 + r3 = 3pqrpa qb rcqc ra pbrb pc qa
pqr a b c abc= + + −( )3 3 3 3
= ⇒ =pqra b cc a bb c a
k pqr
7. (a) :
xr
xr
xr
yr
yr
yr
zr
zr
zr
C C C
C C C
C C C
+ +
+ +
+ +
1 2
1 2
1 2
=
++
++
++
++
++
++
xr
xr
xr
yr
yr
yr
zr
zr
zr
C C C
C C C
C C C
11
12
11
12
11
12
By applying C2 → C2 + C1, C3 → C3 + C2Now apply C3 → C3 – C2, we get
xr
xr
xr
yr
yr
yr
zr
zr
zr
C C C
C C C
C C C
++
++
++
++
++
++
=
11
22
11
22
11
22
0
8. (c) : We have(a2 + b2 + c2)x2 – 2(ab + bc + cd)x + b2 + c2 + d2 ≤ 0 ⇒ (ax – b)2 + (bx – c)2 + (cx – d)2 ≤ 0⇒ (ax – b)2 + (bx – c)2 + (cx – d)2 = 0
⇒ba
cb
dc
x= = =
⇒ b2 = ac or 2logb = loga + logc,
Now, 33 1465 2797 40
130 5465 2797 40
logloglog
log logloglog
abc
a cbc
=+
[Apply R1 → R1 + R3]
⇒ =0 0 0
65 2797 40
0loglog
bc
[Apply R1 → R1 – 2R2]
9. (a) : AB = BAPre & post multiplying both sides by A–1.A–1(AB)A–1 = A–1(BA)A–1
(A–1A)(BA–1) = A–1B(AA–1) ⇒ (BA–1)′ = (A–1B)′ = (A–1)′B′ (reversal law)
= A–1B (as B = B′ and (A–1)′ = A–1) ⇒ A–1B is symmetric.Similarly for A–1B–1.10. (b) : f (x) – x > 0 or, f (x) – x < 0 " x ∈ Rf (f (x)) – f (x) > 0 or, f (f (x)) – f (x) < 0 Adding, f (f (x)) – x > 0 or, f (f (x)) – x < 0⇒ Roots of f (f (x)) – x = 0 are imaginary.
Let z =2
01
a db ag b
⇒ z z= = =2
0
1
20
1
a db ag b
b ga bd a
11. (a) : A3 = AA2 = A(5A – 7I)= 5A2 – 7A = 5(5A – 7I) – 7A = 18A – 35I
| AUGUST ‘1558
A4 = A · A3 = A(18A – 35I) = 18(5A – 7I) – 35A = 55A – 126I
A5 = A⋅A4 = A(55A – 126I) = 55(5A –7I) – 126A = 149A – 385I
⇒ a = 149, b = –38512. (b) : Clearly a = –i, where i2 = –1
So D( )a aa
a a
a
= = −nn i i
i
11 1 1
11 1 0
1 1 11
1 0
3
= 1(–i) + 1(i2) + (1 + i2) = –1 – i
So, principal argument of D(a) is − 34π .
13. (d) :l ml ml m
z zz z z
z
1 1
2 2
3 3
000
1
1 00× =
14. (a) : x y z
ar a r
ar a r
ar a r
k k k
1
1
1
2 2
2 2 4
3 2 6
= a3(k + 1) · r3(2k + 1)[(r – 1)(r4 – 1) – (r2 – 1)2]⇒ k = –1
15. (d) : Let xa
Xy
bY z
cZ
2
2
2
2
2
2= = =, ,
X + Y – Z = 1, X – Y + Z = 1, –X + Y + Z = 1On solving X = Y = Z = 1⇒ x = ± a, y = ±b, z = ±c ⇒ 8 solutions
16. (b) : If D =−
−−
ll
l
c bc a
b a, other determinant (say
D′) is the cofactor determinant.DD′ = D3 (for 3rd order det.)D = l(l2 + a2 + b2 + c2), by comparing l = 117. (c) : For the given system to have a non-trivialsolution, we must have
1 33 22 3 4
01 30 2 110 3 2 10
0kk
kkk
−−
= ⇒ − −− −
=
[Applying R2 → R2 – 3R1, R3 → R3 – 2R1]
⇒ 20k + 11(3 – 2k) = 0 ⇒ k = 332
18. (c) : AB = A, BA = B ⇒ A2 = A and B2 = B(A + B)2 = A2 + B2 + AB + BA
= A + B + A + B = 2(A + B)
(A + B)3 = (A + B)2(A + B) = 2(A + B)2 = 22(A + B) \ (A + B)7 = 26(A + B) = 64(A + B) 19. (c) : For n = 2 ⇒ (A–1BA)2 = (A–1BA)(A–1BA)
= A–1B2AAlso (A–1BA)3 = (A–1B3A) and so on Thus (A–1BA)n = A–1BnA20. (a, c) : C3 → C3 – (C2 – C1)
C4 → C4 – (C1 + C2)a cc b
a b b c a c bx y z x y
a c b ab c
0 00 0
2 01
2 2− − + −
+ −
= + − −( )( )
21. (a, c, d) : A(A + I) = –2I|A(A + I)| = |–2I| |A| |A + I| = (–2)n ≠ 0 22. (a, b, d) : A2 – 4A – 5I3
=
−
1 2 22 1 22 2 1
1 2 22 1 22 2 1
41 2 22 1 22 2 1
−
51 0 00 1 00 0 1
=
+− − −− − −− − −
+−
−9 8 88 9 88 8 9
4 8 88 4 88 8 4
5 0 00 5 000 0 5−
=
=0 0 00 0 00 0 0
0
\ A2 – 4A – 5I3 = 0 or A–1A2 – 4A–1A – 5A–1I3 = 0 or (A–1A)A – 4I3 – 5A–1 = 0 or IA – 4I3 – 5A–1 = 0
\ = −−A A I13
15
4( )
Also, | |A29 8 88 9 88 8 9
=
= 9(81 – 64) – 8(72 – 64) + 8(64 – 72)= (9 × 17) – (8 × 8) + 8 × (–8)= 25 ≠ 0 \ A2 is invertible. and A3 = A·A2 = A(4A + 5I3) = 4A2 + 5A
=
+
=36 32 3232 36 3232 32 36
5 10 1010 5 1010 10 5
41 442 4242 41 4242 42 41
\ |A3| ≠ 0 \ A3 is invertible.
| AUGUST ‘15 59
23. (a, d) : (A + B)2 = A2 + AB + BA + B2 = A + B⇒ AB + BA = 0A2B + ABA = 0 ⇒ A2B + (–BA)A = 0⇒ AB – BA = 0
24. (a, b) : Let A =a aa a
11 12
21 22
0 33 0
=
−
′ =−
= −
−
= −A A
0 33 0
0 33 0
A is skew-symmetric matrix and | |A =−
=0 33 0
9 is a perfect square.
25. (b, c) : ′ =+ ++ +− −
D ( )xx xx xx x
2 4 2 4 134 5 4 5 26
16 6 16 6 104
+
+ −
+ −
− +
x x
x x
x x
2
2
2
4 3 2 13
2 5 9 4 26
8 6 1 16 104
= 0 + 2 × 13 × (0) = 0⇒ D(x) = constant ⇒ a = 0, b = 0, c = 026. (a, b) : Let
( ) ( )
( ) ( )
( ) ( )
...
1 1 2 1
1 1 1 2
1 2 1 1
2
+ +
+ +
+ +
= + + +
x x
x x
x x
A Bx Cx
a b
a b
b a
Putting x = 0, we get A = =1 1 11 1 11 1 1
0
Now differentiating both sides w.r.t. x and putting x = 0, we get
Ba b
a bb a
= + + =2 0
1 1 11 1 1
1 1 10 21 1 1
1 1 11 1 1
2 00
Hence coefficient of x is 0.27. (a, c) : We have,
cos( ) sin( ) cossin cos sincos sin cos
q a q a aq q aq q l a
+ − +
−
2
=
+ − +
−
12
2sin cos
cos( ) sin( ) cos
sin sin cos sin sin
cosa a
q a q a a
q a q a a
qccos sin cos cosa q a l a2
[Multiplying R2 and R3 by sina and cosa, respectively]
=
+
+
−
1
0 02 2
2
2sin cos
cos sin
cos
sin sin cos sin sin
cos coa a
a a
l a
q a q a a
q ss sin cos cosa q a l a2
[Applying R1 → R1 + R2 + R3]
= + +⋅ −
cos sin cossin cos
sin sin cos sincos cos sin c
2 2 2a a l aa a
q a q aq a q oos a
= +−
(cos cos )sin coscos sin
2 2a l aq qq q
= (1 + l)cos2a Therefore, the given determinant is independent of q for all real values of l. Also, l = –1, then it is independent of q and a.28. (a, b, c, d) : AAT = ATA = I.Also AT = A, so A2 = I ⇒ A is involutory matrix.⇒ |A2| = |A|2 = 1 or, |A| = ± 1.
But | | ( )Aa b cb c ac a b
a b cb cc aa b
= = + +111
= (a + b + c)(ab + bc + ca – a2 – b2 – c2)|A| = ab + bc + ca – a2 – b2 – c2 (Q a + b + c = 1)\ a2 + b2 + c2 – ab – bc – ca ≥ 0So |A| = –1. Hence a3 + b3 + c3 – 3abc = 1. Again a2 + b2 + c2 – ab – bc – ca = 1⇒ 1 – 3(ab + bc + ca) = 1, so ab + bc + ca = 0,⇒ atleast one of a, b, and c is negative.29. (a, b, c) : A(adjA) = |A|I \ |A| = 4|adjA| = |A|n – 1 = 16 ; |adj(adjA)| = |A|(n – 1)2 = |A|4 = 256 adj(KA) = Kn – 1 (adjA) \ |adj(KA)| = (Kn – 1)n |adjA|\ |adj (2A)| = 26·1630. (b, c, d) : x + 3y + 2z = 6 …(i)
x + ly + 2z = 7 …(ii)x + 3y + 2z = m …(iii)
(a) If l = 2, then D = 0, therefore unique solution is not possible. (b) If l = 4, m = 6
x + 3y = 6 – 2z x + 4y = 7 – 2zy = 1 and x = 3 – 2z
Substituting in equation (iii) 3 – 2z + 3 + 2z = 6 is satisfied \ Infinite solutions.(c) l = 5, m = 7
| AUGUST ‘1560
Consider equation (ii) and (iii) x + 5y = 7 – 2z x + 3y = 7 – 2z y = 0, x = 7 – 2z are solutions.
Substitute in (i) 7 – 2z + 2z = 6, which does not satisfy. \ No solution. (d) If l = 3, m = 5 then equation (i) and (ii) have no solution. \ No solution .
31. (a, d) : A A23 3 33 3 33 3 3
3=
=
A3 = A2A = 3A · A = 3A2 = 3(3A) = 9A and |A| = 0 \ A–1 does not exist.
32. (d) :1 2 32 1 22 2 3
345
262029
=
⇒ 292 ≠ 202 + 262
Similarly, P Q345
345
and
do not form pythagorean triplet33. (d) : detP = 1, detQ = 1, detR = 1det(PQ) = 1, det(QR) = 1, det(RP) = 1 det(PQR) = 1 34. (c) : Tr(P + QT + 2R) = 1435. (d) 36. (b) 37. (c)
For the matrix A =
1 1 22 2 12 1 2
,
|A – lI| = 0 ⇒ l3 – 5l2 + l + 3 = 0 ... (1)whose roots are l1, l2, l3l1 + l2 + l3 = 5, l1l2 + l2l3 + l3l1 = 1, l1l2l3 = –3 ⇒ Tr(A) = 5, det(A) = –3
Also, AX = lX ⇒ =−A X X1 1l
⇒ = + + = ∑ = −−Tr( )A 1
1 2 3
1 2
1 2 3
1 1 1 13l l l
l ll l l
AX = lX ⇒ A3X = l3X ⇒ Tr(A3) = l13 + l2
3 + l33
= (l1 + l2 + l3)(l12 + l2
2 + l32 – l1l2 – l2l3 – l3l1)
+ 3l1l2l3 = 5(25 – 3(1)) – 9 =101Solving (1) gives l1 = 1, l2 2 7= + , l3 2 7= − , which by theory yield non trivial solutions. In particular for
A–1, the values of l yieldings non trivial solutions are
1 12 7
12 7
, ,+ −
i.e. 1 2 73
2 73
, ,+−
−−
.
Hence (c) is false.38. (c) : a + b + c = p, ab + bc + ca = 0\ a2 + b2 + c2 = (a + b + c)2 – 2(ab + bc + ca)
= p2 – 0 = p2
If D =a b cb c ac a b
\ =
− − −
− − −
− − −
Dc
bc a ca b ab c
ca b ab c bc a
ab c bc a ca b
2 2 2
2 2 2
2 2 2
= D3–1
Now, D2
2
= = ×a b cb c ac a b
a b cb c ac a b
a b cb c ac a b
=
+ + + + + +
+ + + + + +
+ +
a b c ab bc ca ab bc ca
ab bc ca a b c ab bc ca
ab bc ca a
2 2 2
2 2 2
bb bc ca a b c+ + + +2 2 2
= =
p
p
p
p
2
2
2
6
0 0
0 0
0 039. (d) : Q x3 – 3x2 + 3x + 7 = 0⇒ (x – 1)3 + 8 = 0 ⇒ (x – 1)3 = (–2)3
⇒ −−
= ⇒ −−
= =x x12
1 12
1 13
1 3 2( ) , ,/ w w
⇒ x – 1 = –2, –2w, –2w2
or x = –1, 1 – 2w, 1 – 2w2
\ a = –1, b = 1 – 2w, c = 1 – 2w2
Q
a b cb c ac a b
a b cb c ac a b
a b cb c ac a b
2
= ×
=−−−
×a c bb a cc b a
a b cb c ac a b
(row by row)
=
−
−
−
=
2
2
2
2 2 2
2 2 2
2 2 2
2bc a c b
c ac b a
b a ab c
a b cb c ac a b
= (a3 + b3 + c3 – 3abc)2
| AUGUST ‘15 61
= {(a + b + c)(a2 + b2 + c2 – ab – bc – ca)}2
= + + − + − + −{ }14
2 2 2 2 2( ) ( ) ( ) ( )a b c a b b c c a
= − + + =94
8 1 1 02 2{ ( ) ( )}w w w
40. (a) :A B CB C AC A B
bc a ca b ab c
ca b ab c bc a
ab c bc a ca b
=
− − −
− − −
− − −
2 2 2
2 2 2
2 2 2
= =a b cb c ac a b
2
49 \ = −a b cb c ac a b
7
41. a → q; B → s; c → r; D → p(a) Expand along C1 to obtain p( ) ( )( ) ( )( sin cos )
( )(sin cos )
q q q
q q
= − − + − −
+ − −
2 1 1 2
1 2 2
= + + = + +
2 2 2 2 2 24
sin cos sinq q q π
\ Range of is p( ) ,q 0 2 2
(B) Applying R2 → R2 + 4R1, R3 → R3 + 7R1, we get
q( )sin
cos sinsin
q qq
=−
++
2 1 12 4 2 0 1
2 7 2 0 2= 2 cos 2q + 8 sin2q – 2 – 7 sin2q = 2 cos 2q + sin2q – 2
As 2 cos 2q + sin 2q lies between − 5 5to , we get range of q(q) is [ , ]− − −5 2 5 2(c) Using C1 → C1 + C3, we get
r( ) cossin coscos sinsin cos
cosq qq qq qq q
q=−
=2100
2
\ Range of r(q) is [–2, 2] (D) Taking sec2q common from R1, we get
s( ) sec
cos cos
cos cos
cos cot
q q
q q
q q q
q q
= 2
2 2
2 2 2
2 2
1
1
cosec
R3 → R3 – R1, we get
s( ) sec
cos cos
cos cos
cot cos
q q
q q
q q q
q q
=
−
2
2 2
2 2 2
2 2
1
0 0
cosec
= sec2q (cot2q – cos2q)(cos2q – cos4q)= (cot2q – cos2q)sin2q = cos2q – cos2qsin2q = cos4q\ Range of s(q) is [0, 1]
42. a → r; B → s; c → q; D → p(a) (I + A)8 = 8C0I + 8C1IA + 8C2IA2 + ......+ 8C8IA8
= 8C0I + 8C1A + 8C2A + ...... + 8C8A= I + A(8C1 + 8C2 + ..... + 8C8) = I + A(28 – 1) ⇒ l = 28 – 1
(B) adj( ) | || |
A AA
− −= =1 1 22
1
( ( ))| |
| |adjadj
AA
A− −−= = = =1 1
12 21 2 4
(c) | |Aa a aa a aa a a
=11 12 13
21 22 23
31 32 33
⇒ =
− −
−| |B
a a a
a a a
a a a
111
122
13
21 221
232
31 32 33
l l
l l
l l
= =13
211 12 13
221 22 23
231 32 33
l
l l
l l
l l
a a a
a a a
a a a
A| |
Hence, |A| = |B| ⇒ l = 1. (D) A diagonal matrix is commutative with every square matrix, if it is a scalar matrix. So, every diagonal element is 4. \ |A| = 64.
43. a → q, s; B → p, r; c → r, s; D → r(a) Q |A| ≠ 0 ⇒ A–1 exist. \ AB = 0 ⇒ A–1AB = 0 ⇒ IB = 0 ⇒ B = 0(B) AB = 0, |B| ≠ 0 ⇒ B–1 exist. ABB–1 = 0 ⇒ AI = 0 ⇒ A = 0(c) AB =0 ⇒ |A| |B| = 0 ⇒ three cases (i) either |A| = 0, |B| ≠ 0 or (ii) |A| ≠ 0, |B| = 0 or (iii) |A| = 0, |B| = 0 (i) |B| ≠ 0, AB = 0 ⇒ A = 0 but in question A ≠ 0 ⇒ Ist case is not possible.(ii) case is also not possible, hence both |A|, |B| should be zero.(D) An = 0 ⇒ |An| = 0 ⇒ |A| = 0 44. (2) : aij = 0 " i ≠ j and aij = (n – 1)2 + i " i = j
Sum of all the elements of A n ini
n= − +
=
−
∑ [( ) ]1 2
1
2 1
| AUGUST ‘1562
= (2n – 1) (n – 1)2 + (2n – 1)n = 2n3 – 3n2 + 3n – 1 = n3 + (n – 1)3
So, Tn = (–1)n [n3 + (n – 1)3] = (–1)n n3 – (–1)n–1(n – 1)3 = Vn – Vn–1
⇒ T V V V Vnn
n nn=
−=
∑ ∑= − = − =1
102
1 102 03
1
102102( ) ( )
Tnn=∑
=1
102
5202002
45. (1) : Applying C1 → C1 – sinq C3 andC2 → C2 + cosq C3, we get
f ( )sin
cossin cos
qq q
=−
−
1 00 1
0
Applying R3 → R3 – sinq R1 + cosq R2, we get
f ( )sin
cos
sin cos
q
q q
=−
+
=1 00 1
0 0
12 2
Thus, f π6
1
=
46. (1) : PPa b cc a bb c a
a c bb a cc b a
T =
=
1 0 00 1 00 0 1
⇒
+ + + + + +
+ + + + + +
+ +
a b c ab bc ca ab bc ca
ab bc ca a b c ab bc ca
ab bc ca a
2 2 2
2 2 2
bb bc ca a b c+ + + +
2 2 2
=
1 0 00 1 00 0 1
47. (1) : A =−−
3 41 1
A2 3 41 1
3 41 1
5 82 3
=−−
−−
=
−−
A A A3 2 5 82 3
3 41 1
7 123 5
= =−−
−−
=
−−
Observing A, A2, A3 we can conclude that
An n
n nn =
+ −−
1 2 41 2
det( )An n
n nn nn =
+ −−
= − + =1 2 4
1 21 4 4 12 2
det (A2005) = 1
48. (4) : Applying R1 → R1 – R2 – R3, we get
D
y x
x y z x
y y z x
=
− −
+
+
0 2 22 2
2 2 2 2
2 2 2 2
Applying R2 → R2 – R1 and R3 → R3 – R1, we get
= −2
0
0
0
2 2
2 2
2 2
y x
x z
y z
= 4x2y2z2 = 4 (1⋅w⋅w2)2 = 4
49. (0) : Let
( ) ( ) ( )
( ) ( ) ( )
(
1 1 1
1 1 1
1 1 1 2 1 3
2 1 2 2 2 3
+ + +
+ + +
x x x
x x x
a b a b a b
a b a b a b
11 1 13 1 3 2 3 3
0 1 22
33
+ + +
= + + + +
x x x
x x x
a b a b a b) ( ) ( )
...l l l lFor l1, differentiate w.r.t. x and put x = 0 So l1 = 050. (5) : Clearly f ′(x) = 0\ f(x) = c = 6
\ = == =∑ ∑f rr r
( )1
25
1
256 150
51. (4) : On solving, we get(2l + 15) f(x + 1) – (l + 10) f(x + 8) – f(x + 1) = 0(2l + 14) f(x + 1) = (l + 10) f(x + 8) Since f is periodic with period 7\ f(x + 1) = f(x + 8) ⇒ 2l + 14 = l + 10 ⇒ |l| = 4
52. (1) : adj adj adjBC
AA
AA
A= = =
−( )( )5 5 125
3 1
3
32
Now |A| = 5
\ =adj B
C1
53. (7) : Here |A –lI| = 0
⇒1 0
1 70
−− −
=l
l
⇒ (1 – l)(7 – l) = 0⇒ l2 – 8l + 7 = 0⇒ A2 – 8A + 7I2 = 0⇒ A2 = 8A – 7I2⇒ k = – 7 ⇒ |k| = 7
nn
introductionA set of mn numbers arranged in a rectangular array of m horizontal lines and n vertical lines, enclosed by [ ] or ( ), is called a matrix of order m by n, written as m × n matrix.The numbers which form a matrix are called entries or elements of the matrix. The horizontal lines in a matrix are called rows and the vertical lines are called columns of the matrix.The element in the ith row and jth column of a matrix A is denoted by aij. The first suffix indicates the row and the second suffix indicates the column to which that element belongs. With this notation a matrix A of order m × n can be written as:
A
a a a a
a a a a
a a a a
n
n
m m m mn
=
11 12 13 1
21 22 23 2
1 2 3
or, A = [aij]m × n ; i = 1, 2, ..., m ; j = 1, 2, ..., n or simply A = [aij]m × n
Equality of matricEsTwo matrices A = [aij]m×n and B = [bij] m×n are said to be equal if(i) they have the same order, i.e., the number of rows
and columns in both the matrices are same.(ii) their corresponding elements are equal, i.e.,
aij = bij for each i and j.Thus, two matrices are equal if and only if one is a duplicate of the other.typEs of matricEs1. Row matrix : A matrix having exactly one row and
any number of columns is called a row matrix.2. Column matrix : A matrix having exactly one
column and any number of rows is called a columnmatrix.
MATRICES AND DETERMINANTS
Series3
3. Square matrix : A matrix having same number ofrows and columns is called a square matrix.
4. Diagonal matrix : A square matrix is called adiagonal matrix if all its non-diagonal elementsare zero.i.e., the square matrix [aij]m×n is a diagonal matrixif aij = 0 for i ≠ j.
5. Scalar matrix : A square matrix is a called a scalarmatrix if its non-diagonal elements are zero anddiagonal elements are equal.i.e., the square matrix [aij]m×n is a scalar matrix if
ak i j
i jij =
=
≠
, for
, for0
6. Unit matrix or identity matrix : A square matrixis called a unit matrix if its non-diagonal elementsare zero and diagonal elements are all equal to one(unity).i.e., the square matrix [aij]m×n is a unit matrix if
ai j
i jij =
=
≠
1
0
, for
, for
7. Null (or zero) matrix : A matrix having all theelements zero is called a null matrix or zero matrix.The zero matrix of order m × n is denoted by thesymbol Om × n.
8. Upper-triangular matrix : A square matrix iscalled an upper-triangular matrix if all the elementsbelow the principal diagonal are zero.i.e., the square matrix [aij]m×n is an upper-triangularmatrix if aij = 0 for all i > j.
9. Lower-triangular matrix : A square matrix iscalled a lower-triangular matrix if all the elementsabove the principal diagonal are zero.i.e., the square matrix [aij]m×n is a lower-triangularmatrix if aij = 0 for all i < j.
| August ‘15 63
10. Triangular matrix : A square matrix is called atriangular matrix if it is either an upper-triangularmatrix or a lower-triangular matrix.i.e., the square matrix [aij]m×n is a triangular matrixif aij = 0 for all i > j (or all i < j).
11. Comparable matrix : Two matrix A and B are saidto be comparable if they are of the same order, i.e.,they have the same number of rows and the samenumber of columns.
algEbra of matricEs1. Addition of matrices : Let A and B be two comparable
matrices, each of order (m × n). Then, their sum(A + B) is a matrix of order (m × n), obtained byadding the corresponding elements of A and B.i.e., if A = [aij]m × n and B = [bij]m × n thenA + B = [aij + bij]m × n.
2. Negative of a matrix : Let A = [aij]m × n. Then,the negative of A is the matrix (–A) = [–aij]m × n ,obtained by replacing each element of A with itscorresponding additive inverse. (–A) is called theadditive inverse of A.
3. Subtraction of matrices : If A and B are twocomparable matrices, then we define (A – B) =A + (–B).
4. Scalar multiplication : Let A be a given matrixand k be a number. Then, the matrix obtained bymultiplying each element of A by k is called thescalar multiple of A by k, to be denoted by kA.If A is an (m × n) matrix, then kA is also an (m × n)matrix.If A = [aij]m × n , then kA = [k · aij]m × n.
5. Multiplication of matrices : Let A = [aij]m × n andB = [bjk]n × p be two matrices, then the productAB = C = [cik]m × p is called multiplication of twomatrices where cik = ai1 b1k + ai2 b2k + ... + ain bnk.
p r o p E r t i E s o f m at r i x a d d i t i o n a n d multiplication
S.No. Properties of matrixAddition and Multiplication
1. Commutative If A and B are two matrices then A + B = B + A and AB ≠ BA
2. Associative If A, B and C are three matrices, then(A + B) + C = A + (B + C) and (AB)C = A(BC)
3. Distributive If A, B & C are three matrices, thenA(B + C) = AB + AC and (A + B)C = AC + BC
4. Existence of identity
The nul l matrix is the identity element for matrix addition i.e., A + O = A = O + A. IfA is a square matrix, then identity element for matrix multiplication is I, i.e., IA = A = AI, of same order.
5. Existence of inverse
For every matrix A = [aij]m × n, there exists a matrix [– aij]m × n denoted by – A, such that A + (– A) = O = (– A) + A
matrix polynomialLet f(x) = a0xm + a1xm–1 + … + am–1x + am be apolynomial in x and A be a square matrix of order n, then f(A) = a0Am + a1Am–1 + … + am–1A + amIn is calleda matrix polynomial in A. Thus, to obtain f(A), replace x by A in f(x) and the constant term is multiplied by the identity matrix of the order equal to that of A.The polynomial equation f(x) = 0 is said to be satisfied by the matrix A iff f(A) = O.
transposE of a matrixLet A = [aij] be an m × n matrix. Then, the transpose of A, denoted by AT or A′, is an n × m matrix such thatAT= [aji] for all i = 1, 2, ...., m ; j = 1, 2, ..., n.
Thus, AT is obtained from A by changing its rows into
columns and its columns into rows.For example, if
A AT=
=
1 2 3 42 3 4 13 2 1 4
1 2 32 3 23 4 14 1 4
, then
Properties of transpose For any two matrices A and B of the same order,(i) (AT)T = A(ii) (A + B)T = AT + BT
(iii) (kA)T = k(AT), where k is a scalar.(iv) (AB)T = BTAT
somE spEcial matricEsSymmetric matrix : z A square matrix A = [aij]m×n is called a symmetric matrix, if A = AT i.e., aij = ajifor all i, jSkew-symmetric matrix : z A square matrix A = [aij] m×n is a skew-symmetric matrix if AT = –A i.e.,
aij = – aji for all i, j z Orthogonal matrix : A square matrix A of m × n order is called an orthogonal matrix if AAT = ATA = I.
| August ‘1564
ElEmEntary opEration (transformation) of a matrixThe following three operations applied on the rows (columns) of a matrix are called elementary row (column) operations:(i) Interchange of any two rows (columns) : If ith row
(column) is interchanged with jth row (column), we write Ri ↔ Rj (Ci ↔ Cj).
(ii) Multiplying the elements of a row (column) by a non-zero scalar : If the elements of ith row (column) are multiplied by a non-zero scalar k, we write Ri → kRi (Ci → kCi).
(iii) Adding to the elements of a row (column), the constant times the corresponding elements of another row (column) : If k times the elements of jth row (column) are added to the corresponding elements of the ith row (column), we write Ri → Ri + kRj (Ci → Ci + kCj).
Invertible matrices Let A be a square matrix of order n. Then a square matrix B of order n, if it exists, is called an inverse of A if AB = BA = I. In that case A is said to be invertible. The inverse of an invertible matrix A is denoted by A–1. Uniqueness of inverse Inverse of a square matrix, if it exists, is unique.Inverse of a matrix by elementary operations Let A is a matrix of order n. To find A–1 using elementary row operations, write A = IA and apply a sequence of row operations on A = IA till we get I = BA. The matrix B will be the inverse of A. Similarly, if we wish to find A–1 using column operations, then write A = AI and apply a sequence of column operations on A = AI till we get I = AB.dEtErminantsA determinant is a pure number associated with a square matrix, denoted by det A or |A|, written as
| | detA A
a a aa a a
a a a
n
n
n n nn
= =
11 12 1
21 22 2
1 2
A determinant having n rows and n columns is called a determinant of order n.Determinant of a square matrix of order 1Let A = [a11] be a 1 × 1 matrix, then the determinant of A is the number a11 itself, i.e., |a11| = a11.Determinant of a square matrix of order 2
Let Aa aa a
=
11 12
21 22 be a 2 × 2 matrix, then
| |Aa aa a
a a a a= = −11 12
21 2211 22 12 21
i.e. the determinant of a 2 × 2 matrix is obtained by taking the product of the entries on the main diagonal and subtracting from it the product of the entries in the other diagonal.Notes :
det z In = 1, where In is unit/identity matrix of order n.det z On = 0, O is null matrix (square matrix of n order).A z = [aij]n × n, then |A| = |A′| {Reflection Property}.det ( z AB) = det A⋅det B, where A & B are matrices of same order.det ( z kA) = kn det A, if A is of order n × n.det ( z An) = (det A)n if n ∈ I+.
minors and cofactorsThe minor of an element z aij in |A| is defined as the value of the determinant obtained by deleting the ith row and jth column of |A|, and it is denoted by Mij.The cofactor z Cij of an element aij is defined as Cij = (–1)i + j⋅ Mij.
propErtiEs of dEtErminantsThe value of a determinant remains unaltered if the z
rows and columns are interchanged.If two rows (or columns) of a determinant are z
interchanged, the value of the determinant is multiplied by –1.If any two rows (or columns) of a determinant are z
identical, then the value of determinant is zero.If the elements of a row (or column) of a determinant z
are multiplied by a scalar, then the value of the new determinant is equal to same scalar times the value of the original determinant.If each element of any row (or column) of a z
determinant is the sum of two numbers, then the determinant is expressible as the sum of two determinants of the same order.
Evaluation of determinants using elementary operationsWe shall use the following notations to evaluate determinant1. The operation of interchanging the ith row and jth
row will be denoted by Ri → Rj.2. The operation of multiplying each element of the ith
row by a number k will be denoted by Ri → kRi.3. The operation of adding to each element of the ith
row, k times the corresponding element of the jth row(j ≠ i) will be denoted by Ri → Ri + kRj.
| August ‘15 65
Similarly notations are used for operations on columns replacing R by C.
application of dEtErminantsArea of triangle with vertices z A(x1, y1), B(x2, y2) and
C(x3, y3) is given by 12
111
1 1
2 2
3 3
x yx yx y
.
If area is zero, then three points are collinear. z
adjoint of a matrixLet A = [aij] be a square matrix of order n and let Cij denote the cofactors of aij in |A|. Then, the adjoint of A, denoted by adj A, is defined as adj A = [Cji]n × n.Singular and non-singular matricesA square matrix A is said to be(i) singular if |A| = 0, (ii) non-singular if |A| ≠ 0.propErtiEs of adjoint of a matrix
For every square matrix of order z n A(adj A) = (adj A)A = |A|·I
If | z A| = 0, matrix A is called singular else non singular.If z A and B are non singular matrices of the same order, then AB and BA are also non singular matrices of same order.(adj z AB) = (adj B)·(adj A)(adj z A)′ = adj A′If z A be n × n matrix, then(i) |adj A| = |A|n–1
(ii) adj(adj A) = |A|n–2A(iii) |adj(adj A)| = |A|(n–1)2
invErsE of a squarE matrixIf A be any n-rowed square matrix, then a matrix B, if it exists such that AB = In = BA, is called inverse of A. The inverse of A is denoted by A–1.
AA
A− =1 1| |
(adj )
Properties of the inverse of a matrix1. A square matrix is invertible if and only if it is non-
singular.2. The inverse of the inverse is the original matrix itself,
i.e., (A–1)–1 = A.3. The inverse of the transpose of a matrix is the
transpose of its inverse, i.e. (A′)–1 = (A–1)′.4. (Reversal law) If A and B are invertible matrices
of the same order, then AB is also invertible and(AB)–1 = B–1A–1.
5. If A, B, C are invertible matrices of same order, then(ABC)–1 = C–1 B–1 A–1.
solution of systEm of EquationsLet z AX = B be the given system of equations :(i) If |A| ≠ 0, the system is consistent and has one unique solution.(ii) If |A| = 0 and (adj A)B ≠ O, then the system is inconsistent so it has no solution.(iii) If |A| = 0 and (adj A)B = O, then the system may be either consistent or inconsistent according as the system name either infinitely many solutions or no solution.
Very short answer type1. If a matrix has 12 elements, what are the possible
orders it can have?2. Simplify :
coscos sinsin cos
sinsin coscos sin
θθ θθ θ
θθ θθ θ−
+
−
3. Show thatb c c a a bc a a b b ca b b c c a
− − −− − −− − −
= 0
4. Find matrix X, if X +
=
− −
−34
76
53
86 .
5. If A =
2 51 3
, find adj. A.
short answer type6. Using properties of determinants, prove that
x y x xx y x xx y x x
x+++
=5 4 4 210 8 8 3
3
7. Prove that :cos cos cos sin sin
sin cossin cos sin sin cos
a β a β aβ β
a β a β a
−− =0 1
8. Using elementary transformations, find the inverse
of 2 61 2
−−
.
9. Verify that A =
2 31 2 satisfies the equation
A3 – 4A2 + A = 0.
10. If f xx xx x( )
cos sinsin cos=
−
00
0 0 1,
show that [f(x)]–1 = f(–x).
| August ‘1566
Long answer type11. A trust fund has ` 30,000 that must be invested in
two different types of bonds. The first bond pays5% interest per year, and the second bond pays7% interest per year. Using matrix multiplication,determine how to divide ` 30,000 among the twotypes of bonds if the trust fund must obtain anannual total interest of (i) ` 1800 (ii) ` 2000.
12. Let A =
1 2 22 1 22 2 1
. Prove that A2 – 4A – 5I = O.
and also find A–1.
13. If x ≠ y ≠ z andx x x
y y y
z z z
2 3
2 3
2 3
1
1
1
+
+
+
= 0 , then prove that xyz = – 1.
14. Find the matrix P satisfying the matrix equation2 13 2
3 25 3
1 22 1
−−
=
−
P
15. Discuss the consistency of the following system ofequations:
x + y + z = 12x + 2y + 2z = 23x + 3y + 3z = 4.
soLUtioNs1. If a matrix has mn elements, then its order is m × n
or n × m.Here, mn = 12 = (1) (12) or (2) (6) or (3) (4).Possible orders of the matrix are : 1 × 12, 12 × 1, 2 × 6,6 × 2, 3 × 4, 4 × 3.
2. We have,
coscos sinsin cos
sinsin coscos sin
θθ θθ θ
θθ θθ θ−
+
−
=−
cos sin cos
sin cos cos
2
2
θ θ θ
θ θ θ
+−
sin sin cos
sin cos sin
2
2
θ θ θ
θ θ θ
=+ −
− + +
cos sin sin cos sin cos
sin cos sin cos cos sin
2 2
2 2
θ θ θ θ θ θ
θ θ θ θ θ θ
=
1 00 1
3. Let D =− − −− − −− − −
b c c a a bc a a b b ca b b c c a
Applying C1 → C1 + C2 + C3, we get
D =− −− −− −
=000
0c a a ba b b cb c c a
[Q C1 consists of all zeros]
4. Let A B=
=
− −
−34
76
53
86and .
Then, the given matrix equation is X + A = B⇒ X = B + (– A)
=
+
−
− −−−
53
86
34
76 .
=
=
+ − + −
+ − − + −−
− −5 3 8 73 4 6 6
8 151 12
( )( ) ( )
Hence, X =
−
− −8 151 12
5. The cofactors of the elements of |A| are given byA11 = 3, A12 = –1, A21 = –5, A22 = 2.
∴ =−
−
=
−−
adj A
T3 15 2
3 51 2
6. Let the given determinant be D. Then,
D =+++
x y x xx y x xx y x x
5 4 4 210 8 8 3
Applying R R R R R R2 2 1 3 3 12 3→ − → −&
D =+++
x y x xx y xx y x
3 2 2 07 5 5 0
Expanding along C3, we get
D =++
xx y xx y x
. 3 2 27 5 5
Taking x common from C2
D =++
xx yx y
2 3 2 27 5 5
.
= x2 · [5(3x + 2y) –2(7x + 5y)] = (x2 · x) = x3
Hence, D = x3
7. Let the given determinant be D. Then,
D =−
−cos cos cos sin sin
sin cossin cos sin sin cos
a β a β aβ β
a β a β a0
| August ‘15 67
Multiplying R1 by sina, R3 by cosa and dividing D by sin a cos a
D = 1sin cos
.a a
sin cos cos sin cos sin sinsin cos
sin cos cos sin co
a a β a a β aβ β
a a β a
−−
2
0
ss sin cosa β a2
Applying R1 → R1 – R3
D = 1sin cos
.a a
0 0 102
−− sin cos
sin cos cos sin cos sin cos
β β
a a β a a β a
Expanding along R1
D = − − −1 2 2sin cos
. [ sin cos sin sin cos cos ]a a
a a β a a β
= − − + =1 12 2sin cos
. ( sin cos ) [sin cos ]a a
a a β β
Hence Proved.8. In order to use elementary row operations we write
A = IA, where A = 2 61 2
1 00 1
−−
=
and I
We have,2 61 2
1 00 1
−−
=
A
⇒−−
=
−
→ −
1 41 2
1 10 1 1 1 2A R R R, applying
⇒−
=
−−
→ −
1 40 2
1 11 2 2 2 1A R R R, applying
⇒−
=
−
−
→
1 40 1
1 112
1122 2A R R, applying
⇒
=
−
−
→ +1 00 1
1 312
141 1 2A R R R, applying
Thus, A− =−
−
11 312
1
9. We have, A =
2 31 2
A2 2 31 2
2 31 2
4 3 6 62 2 3 4
7 124 7
=
=
+ ++ +
=
A A A3 2 7 124 7
2 31 2
= =
=+ ++ +
=
14 12 21 248 7 12 14
26 4515 26
∴ − + =
−
+
A A A3 24
26 4515 26
47 124 7
2 31 2
=
−
+
26 4515 26
28 4816 28
2 31 2
=− + − +− + − +
=
26 28 2 45 48 315 16 1 26 28 2
0 00 0
Hence, A3 – 4A2 + A = 0
10. We have, f xx xx x( )
cos sinsin cos=
−
00
0 0 1
⇒ − =− − −− −
f xx xx x( )
cos( ) sin( )sin( ) cos( )
00
0 0 1
= −
cos sinsin cos
x xx x
00
0 0 1
...(i)
Let f(x) = A, then
| |cos sinsin cosA
x xx x=
−
00
0 0 1
= (cos2x + sin2 x) = 1 [Expanding along C3]
∴ =−
−adjcos sinsin cos
cos sinsin cosA
x xx x
x xx x
T00
0 0 1
00
00 0 1
∴ = = = −
− −[ ( )]adj| |
cos sinsin cosf x A
AA
x xx x1 1
00
0 0 1
...(ii)
From (i) and (ii), we get [f(x)]–1 = f(–x).11. Let the investment in the first bond = ` x.
Then, the investment in the second bond= ` (30,000 – x).
Rate of interest on first bond = 5%Rate of interest on second bond = 7%Let the investments represent the row matrix; [ x 30,000 – x]
| August ‘1568
And rate of interest represent the column matrix5
1007
100
The total interest may be represented by the product
[ , ]x x30 000
5100
7100
−
(i) Total interest = ` 1800
∴ −
=[ , ]x x30 000
5100
7100
1800
⇒ +−
=5100
30 000 7100
1800x x( , )
⇒ + − =5100
2100 7100
1800x x
⇒ − = − ⇒ =2100
300 2 30 000x x ,
⇒ x = 15,000Hence, investment in the first bond = ` 15000and investment in the second bond
= ` (30,000 – 15,000) = ` 15, 000(ii) Total interest = ` 2000
∴ −
=[ , ]x x30 000
5100
7100
2000
⇒ +−
=5100
30 000 7100
2000x x( , )
⇒ + − =5100
2100 7100
2000x x
⇒ − = −2100
100x
⇒ x = 5000Hence, investment in the first bond = ` 5000and investment in the second bond
= ` (30,000 – 5,000) = ` 25,00012. We have
A AA21 2 22 1 22 2 1
1 2 22 1 22 2 1
= =
9 8 88 9 88 8 9
∴ − − =
−
A A I2 4 59 8 88 9 88 8 9
41 2 22 1 22 2 1
−
51 0 00 1 00 0 1
=
−
−
9 8 88 9 88 8 9
4 8 88 4 88 8 4
5 0 00 5 00 0 5
=
=0 0 00 0 00 0 0
O
Since A2 – 4A – 5I = OMultiplying both sides by A–1, we get
A–1 A2 – 4A–1A – 5A–1I = A–1O⇒ (A–1A) A – 4I – 5A–1 = O⇒ IA – 4I = 0 + 5A–1
⇒ = − =
−
−5 41 2 22 1 22 2 1
41 0 00 1 00 0 1
1A A I
=
−
=−
−−
1 2 22 1 22 2 1
4 0 00 4 00 0 4
3 2 22 3 22 2 3
∴ =−
−−
−A 1 15
3 2 22 3 22 2 3
13. Let the given determinant be D. Then
D =
+
+
+
= +
x x x
y y y
z z z
x x
y y
z z
x x x
y y y
z z z
2 3
2 3
2 3
2
2
2
2 3
2 3
2 3
1
1
1
1
1
1
= + ⋅
x x
y y
z z
xyz
x x
y y
z z
2
2
2
3
3
3
1
1
1
1
1
1
( )
= + − ⋅
x x
y y
z z
xyz
x x
y y
z z
2
2
2
2
2
2
2
1
1
1
1
1
1
1
( )( )
[interchanging the columns of 2nd det. twice]
= +( )1
1
1
1
2
2
2xyz
x x
y y
z z
Applying R R R R R R2 2 1 3 3 1→ − → −,
| August ‘15 69
= − −
− −
x x
y x y x
z x z x
2
2 2
2 2
1
0
0
( ) ( )
( ) ( )
= + − − ++
( )( )( ) ( )( )
1100
2
xyz y x z xx xy y xz z x
Expanding along C3
= + − − ⋅ ⋅++
( )( )( )1 111
xyz y x z xy xz x
= (1 + xyz) (y – x) (z – x) (z – y).Q D = 0 ⇒ (1 + xyz) (y – x) (z – x) (z – y) = 0⇒ (1 + xyz) = 0 [Q (y – x) ≠ 0, (z – x) ≠ 0, (z – y)≠ 0, ]Hence, xyz = – 1
14. Let and , thenA B=
=
−−
2 13 2
3 25 3
| |A = = − = ≠2 13 2
4 3 1 0
and | |B =−
−= − = − ≠
3 25 3
9 10 1 0
So, A and B are non-singular and invertible matrices.
∴ =−
−
adj A
2 13 2
and| |
adjAA
A− = =−
−
1 1 2 13 2 [Q |A| = 1] ... (i)
∴ =− −− −
adj B
3 25 3
and| |
adj( )
BB
B− = =−
− −− −
=
1 1 11
3 25 3
3 25 3
...(ii)
[Q |B| = –1]The given matrix equation is
2 13 2
3 25 3
1 22 1
−−
=
−
P i e APB. ., =
−
1 22 1
⇒ =−
− −A APB A1 1 1 22 1
⇒ =−
−
−
−( ) ( )A A PB1 2 13 2
1 22 1
[Using (i)]
⇒ =− +
− + − −
I PB( )
2 2 4 13 4 6 2
⇒ =−
PB
0 51 8
⇒ =−
− −( )PB B B1 10 51 8
⇒ =−
−P BB( )1 0 51 8
3 25 3
[Using (ii)]
⇒ =+ +− −
PI
0 25 0 153 40 2 24
⇒ =− −
P
25 1537 22
15. The given system of equations can be written as
1 1 12 2 23 3 3
124
=
xyz
, i.e., AX = B, where
1 1 12 2 23 3 3
124
=
=
, Xxyz
Band
Here, det A = 1 1 12 2 23 3 3
0
=
∴ Nothing can be said about consistency. We proceed to find adj A.
Cofactors of elements of first row are
( ) ,( ) , ( ) , . ., , , .− − −+ + +12 23 3
12 23 3
12 23 3
0 0 01 1 1 2 1 3 i e
Cofactors of elements of second row are
( ) ,( ) ,( ) , . ., , , .− − −+ + +11 13 3
11 13 3
11 13 3
0 0 02 1 2 2 2 3 i e
Cofactors of elements of third row are
( ) ,( ) , ( ) , . ., , , .− − −+ + +11 12 2
11 12 2
11 12 2
0 0 03 1 3 2 3 3 i e
∴ =
=
adj A
T0 0 00 0 00 0 0
0 0 00 0 00 0 0
⇒ =
=
=(adj )A B0 0 00 0 00 0 0
124
000
0
∴ The given system of equations may have infinitely many solutions. From the last two equations, we observe that
2x + 2y = 2 – 2z ⇔ x + y = 1 – z
and 3x + 3y = 4 – 3z ⇔ x + y = 43 – z
⇒ 1 43
1 43
− = − ⇒ =z z , which is absurd.
Hence, the given system of equations is inconsistent and no set of values of x, y, z can satisfy all the equations.
nn
| August ‘1570
| AUGUST ‘15 71
cateGoRy - i
For each correct answer one mark will be awarded, whereas, for each wrong answer, 25% of total marks (1/4) will be deducted. If candidates mark more than one answer, negative marking will be done.
1. The value of
16
13
1 23
1 76
+
+
+
+
cos cos cos cosp p p pis
(a) 3/16 (b) 3/8 (c) 3/4 (d) 1/22. If 0 ≤ x ≤ p, 0 ≤ y ≤ p and sinx + siny = 2, then thevalue of cos(x + y) is (a) –1 (b) 0 (c) 1 (d) none of these
3. If 02
< < <x y p , then
(a) x – cosx > y – cosy (b) x – cosx < y – cosy(c) y – sinx < x – siny (d) none of these
4. If q and f are acute angles satisfying sin ,q = 12
cos ,f = 13
then
(a) p q f p3 2
< + ≤ (b) p q f p2
23
< + <
(c) 23
56
p q f p< + < (d) 56p q f p< + <
5. If tanq = 5 tanf, then the maximum value oftan2(q – f) is (a) 4/5 (b) 4/9 (c) 16/25 (d) 1
6. If sinx + siny = 3(cos cos )y x− , then sin3x + sin3yis equal to (a) 1 (b) 0 (c) 2sin3x (d) –1
7. If cosa + cos(q + a) = cos p7
and
sina + sin(q + a) = sin p7
, then the value of q is
(a) p3
(b) 23p
(c) 27p (d)
56p
8. If sin cosx y+ = 13
and cos sinx y+ = 12
, then
tanx y−
=2
(a) –5 (b) –1/5 (c) 1/5 (d) 5
9. In a triangle PQR, ∠ =R p2 , if tan P Q
2 2
and tan
are the roots of the equation ax2 + bx + c = 0, a ≠ 0 then(a) a = b + c (b) c = a + b(c) b = c (d) b = a + c
10. The value of cos cos sin15 7 12
7 12
° ° ° is
(a) 1/2 (b) 1/8 (c) 1/4 (d) 1/16
11. If P = +12
13
2 2sin cos ,q q then
(a) 13
12
≤ ≤P (b) P ≥ 12
(c) 2 ≤ P ≤ 3 (d) − ≤ ≤136
136
P
12. Simplest form of 2
2 2 2 2 4+ + + cos x is
(a) sec x2 (b) sec x (c) cosec x (d) 1
13. If ( ) tan1 1 1 1+ + = + −a a a , then sin4a =(a) 1 (b) a (c) 2a (d) 4a14. If p = sin6x + cos6x, then
(a) 34
1≤ ≤p (b) 14
1≤ ≤p
(c) 38
12
≤ ≤p (d) none of these
The entire syllabus of Mathematics of WB-JEE is being divided in to six units, on each unit there will be a Mock Test Paper (MTP) which will be published in the subsequent issue.
UNIT-II : TrIgoNomeTry
| AUGUST ‘1572
15. 2 − −−
=sin cossin cos
x xx x
(a) tan x2 8
−
p(b) tan x
2 8+
p
(c) tan x2 4
−
p(d) tan x
2 4+
p
16. The value of cos12° + cos84° + cos156° + cos228°is(a) –1/2 (b) 1 (c) 1/2 (d) 1/8
17. The value of tana + 2tan(2a) + 4tan(4a) + ........+ 2n – 1 tan(2n – 1a) + 2ncot(2na) is (a) cot(2na) (b) 2n tan(2na)(c) 0 (d) cota
18. The general solution of tanmx = cotnx is
(a) ( )
( )2 1
2rm n
++
p(b)
2rm n
p( )+
(c) ( )( )2 1rm n
++
p(d)
2rm n
p( )−
[where r is any integer]
19. The smallest positive value of p for which theequation cos(psinx) = sin(pcosx) has a solution in 0 ≤ x ≤ 2p is
(a) p2
(b) p2
(c) p2 2
(d) p4
20. If log0.5sinx = 1 – log0.5cosx, then the number ofvalues of x in –2p ≤ x ≤ 2p, is (a) 4 (b) 2 (c) 3 (d) 1
21. The principal value of sin tan− −
1 54p is
(a) p4
(b) − p4
(c) p2
(d) − p2
22. The trigonometric equation sin–1x = 2sin–1a, has asolution for
(a) | |a ≥ 12
(b) 12
12
≤ ≤| |a
(c) all real values of a (d) | |a ≤ 12
23. sin cot12
34
1− −
is equal to
(a) 15
(b) 25
(c) − 25
(d) − 15
24. If sin–1x < cos–1x, then
(a) − ≤ <1 12
x (b) 0 12
≤ <x
(c) 12
1< ≤x (d) –1 < x < 0
25. Number of solutions of
tan sin− −+ + + + =1 2 1 2 12
x x x x p is
(a) 0 (b) 1(c) 2 (d) infinite
26. (sin ) (cos ) ,− −3
+ =1 3 1 332
x x p if x is equal to
(a) 12
(b) 12
(c) 32
(d) 1
27. If the sides of a triangle are 3x + 4y, 4x + 3y and5x + 5y, where x > 0, y > 0, the triangle will be (a) right-angled (b) obtuse-angled(c) equilateral (d) none of these
28. In a triangle ABC, if ∠ = ∠ =B Cp p3 4
, and D
divides BC internally in the ratio 1 : 3, then sinsin
∠∠
BADCAD
equals
(a) 13
(b) 13
(c) 16
(d) 23
29. The perpendicular AD to the base of a triangledivides it into segments such that BD, CD and AD are in the ratio 2 : 3 : 6. The angle A of the triangle is (a) 90° (b) 60° (c) 45° (d) 30°
30. In a DABC, if a = 2b and A B− = p3
, then ∠C is
equal to (a) p/6 (b) p/4 (c) p/3 (d) p/2
cateGoRy-ii
Every correct answer will yield 2 marks. For incorrect response, 25% of full mark (1/2) would be deducted. If candidates mark more than one answer, negative marking will be done.
31. The value of 3 32
3 22
4 4 6 6sin sin ( ) sin sinp a p a p a p−
+ +
− +
+ −
3 32
3 22
54 4 6 6sin sin ( ) sin sin (p a p a p a p−
+ +
− +
+ − aa)
is equal to (a) 2 (b) 1 (c) 0 (d) 4
| AUGUST ‘15 73
32. The number of integral values of k for which theequation 7cosx + 5sinx = 2k + 1 has a solution, is(a) 8 (b) 6 (c) 7 (d) 9
33. The value of sin sin sin sinp p p p14
314
514
714
is
(a) 1/8 (b) 1/4 (c) 1/2 (d) 1
34. The base angles of a triangle are 22 12
°and
112 12
°. If b is the base and h is the height of the
triangle, then (a) b = 2h (b) b = 3h(c) b h= +( )1 3 (d) 2b = 3h
35. If cos cos cos ,− − −+ − + − =1 1 11 1 34
p p q p
then the value of q is
(a) 12
(b) 1 (c) 12
(d) 13
cateGoRy-iii
In this section more than 1 answer can be correct. Candidates will have to mark all the correct answers, for which 2 marks will be awarded. If, candidates marks one correct and one incorrect answer then no marks will be awarded. But if, candidate makes only correct, without marking any incorrect, formula below will be used to allot marks.2 × (no. of correct response/total no. of correct options)
36. If 2 3cos cos cosA B B= + and2 3sin sin sinA B B= − ,
then the value of sin(A – B) will be (a) 1/2 (b) 1/3 (c) –1/2 (d) –1/3
37. The value ofcos cossin sin
sin sincos cos
a βa β
a βa β
+−
+ +−
n n
(where n is a whole number) is equal to
(a) 0 (b) 22
tann a β−
(c) 22
cotn a β−(d) 2
2cotn a β+
38. If A = tan6° tan42° and B = cot66° cot78° then,(a) A = B (b) A2 – 2AB + B2 = 0(c) A + B = 0 (d) A2 + 2AB + B2 = 0
39. The solution of the equation
16 16 10 02
2 2cos sinx x x+ = < <
p is
(a) p6
(b) p3
(c) 23p
(d) 53p
40. In DABC, if cos ,A b cc2 2
= + then
(a) Area of the triangle is 12
ab
(b) Circumradius is equal to 12
c
(c) Area of triangle is 12
bc
(d) Circumradius is equal to 12
a
solutions
1. (a) : 16
13
1 23
1 76
+
+
+
+
cos cos cos cosp p p p
= +
+
+ −
+ +16
13
13
16
cos cos cos cosp p p p p p
= +
+
−
−
16
13
13
16
cos cos cos cosp p p p
= −
−
16
13
2 2cos cosp p
= −
−
= × =1 3
21 1
214
34
316
2 2
2. (a) : Q sinx ≤ 1, siny ≤ 1 and sinx + siny = 2\ sinx = 1 and siny = 1 \ x = 90° = yNow, cos(x + y) = cos180° = –13. (b) : x < y \ cosx > cosy
[ cosq is a decreasing function]⇒ –cosx < –cosy⇒ x – cosx < y – cosy
4. (b) : sinq q p= ⇒ =12 6
cos , cosp p3
12 2
0 13
12
= = < and 0 <
\ < <cos cos cosp f p2 3
or is decreasing in 0p f p f f p3 2 2
< < ≤ ≤
cos
\ + < + < +q p q f q p3 2
or p p q f p p6 3 6 2
+ < + < + or p q f p2
23
< + <
| AUGUST ‘1574
5. (a) : Given that tanq = 5tanf
Now, tan( ) tantan
q f ff
− =+4
1 5 2
⇒ tan( )cot tan
q ff f
− =+45
\ − =− +
tan ( )(cot tan ) cot tan
22
165 20
q ff f f f
The least value of the denominator on R.H.S. is 20.
\ Greatest value of tan ( )2 1620
45
q f− = =
6. (b) : sin sin (cos cos )x y y x+ = −3
⇒ + = −3 3cos sin cos sinx x y y
⇒ + = −32
12
32
12
cos sin cos sinx x y y
⇒ −
= +
cos cosx yp p6 6
⇒ − = + ⇒ − = +x y x yp p p p6 6
32 2
3
⇒ 3x = p + 3yNow, sin3x + sin3y = sin(p + 3y) + sin3y = 0
7. (b) : cosa + cos(q + a) = cos p7
... (i)
and sina + sin(q + a) = sin p7
... (ii)
\ Squaring and adding (i) and (ii), we get [cosa + cos(q + a)]2 + [sina + sin(q + a)]2
= +cos sin2 27 7p p
⇒ cos2a + cos2(q + a) + 2cosa cos(q + a)+ sin2a + sin2(q + a) + 2sina sin(q + a) = 1
⇒ (cos2a + sin2a) + {cos2(q + a) + sin2(q + a)}+ 2[cosa cos(q + a) + sina sin(q + a)] = 1
⇒ 2 + 2cosq = 1 ⇒ 2cosq = –1 ⇒ cos q = − 12
\ =
⇒ =cos cosq p q p23
23
8. (b) : sin cosx y+ = 13
and cos sinx y+ = 12
Now, sin cos cos sinx y x y+ + + = +13
12
⇒ + + + =(sin sin ) (cos cos )x y x y 56
... (i)
Also, sin cos cos sinx y x y+ − − = −13
12
⇒ − + − = −(sin sin ) (cos cos )x y y x 16
... (ii)
From (i), we get
22 2
22 2
56
sin cos cos cosx y x y x y x y+ −
++ −
=
⇒−
++
+
=22 2 2
56
cos sin cosx y x y x y ... (iii)
From (ii), we get
22 2
22 2
16
cos sin sin sinx y x y y x x y+ −
++ −
= −
⇒− +
++
= −22 2 2
16
sin cos sinx y x y x y ... (iv)
Now, (iv) ÷ (iii) gives tan //
x y−= − = −
21 6
5 615
9. (b) : Given that, tan P Q2 2
and tan are the roots
of ax2 + bx + c = 0, a ≠ 0.
\
+
= −tan tanP Q ba2 2
and tan tanP Q ca2 2
=
But ∠P + ∠Q + ∠R = 180°⇒ ∠P + ∠Q = 180° – 90° [Q ∠R = 90°]⇒ ∠P + ∠Q = 90°
⇒ ∠ + ∠ = °12
12
45P Q ⇒ tan tanP Q2 2
45+
= °
⇒
+
−
=tan tan
tan tan
P Q
P Q2 2
12 2
1
⇒−
−= ⇒ − = − ⇒ − =
baca
ba
ca
ca
ba1
1 1 1
⇒ c – b = a ⇒ c = a + b
10. (b) : cos cos sin15 7 12
7 12
° ° °
= ° ° °
12
15 2 7 12
7 12
cos sin cos
= ° ° = ° °12
15 15 14
2 15 15cos sin ( sin cos )
= ° = × =14
30 14
12
18
sin
| AUGUST ‘15 75
11. (a) : P = +12
13
2 2sin cosq q
= − + +14
1 2 16
1 2( cos ) ( cos )q q
= + − −
14
16
14
16
2cos q = −512
112
2cos q
\ ≤ ≤ − ≤ ≤13
12
1 2 1P [ cos ] q
12. (a) : 2
2 2 2 2 4+ + + cos x
=+ + +
2
2 2 2 1 4( cos )x
=+ +
=+ +
2
2 2 4 2
2
2 2 2 22cos cosx x
=+ ⋅
=+
2
2 2 2
22 22cos cosx x
=+
=⋅
=22 1
2
2 22
1
22( cos ) cos cosx x x = sec x
2
13. (b) : ( )tan1 1 1 1+ + = + −a a a
\ = + −+ +
tana aa
1 11 1
⇒ = + −+ +
tan sinsin
a qq
1 1 21 1 2
[Putting a = sin2q]
= + −+ +
=−
+
11
22
22 2
22
22
2
2
cos sincos sin
cos sin cos
cos sin c
q qq q
q q q
q q oos q2
=−
+
22
12
22
12
2
2
cos tan
cos tan
q q
q q= −
tan p q4 2
\ = − ⇒ = −a p q a p q4 2
4 2
\ sin4a = sin(p – 2q) = sin2q = a14. (b) : p = sin6x + cos6x= (sin2x + cos2x)3 – 3sin2xcos2x(sin2x + cos2x)
= −1 34
4 2 2( sin cos )x x = −1 34
22sin x
p will be maximum when sin2x = 0.\ pmax = 1
p will be minimum when sin2x = 1
\ = − =pmin 1 34
14 .
Thus, 14
1≤ ≤p
15. (a) : 2 − −−
sin cossin cos
x xx x
=− +
−
1 12
12
12
12
cos sin
sin cos
x x
x x
=− −
−
=−
−
14
4
2 12 4
2 12 4
2cos
sin
sin
sin
x
x
x
x
p
p
p
p
−
cos 12 4
x p
=−
−
= −
sin
costan
x
xx2 8
2 82 8
p
pp
16. (a) : cos12° + cos 84° + cos156° + cos228°= (cos228° + cos12°) + (cos156° + cos84°)= 2cos120° cos108° + 2cos120° cos36°= 2cos120°(cos108° + cos36°)= 2cos120°(–sin18° + cos36°)
= −
+ − −
= −2 1
25 14
5 14
12
17. (d) : tana + 2tan(2a) + 4tan(4a) + ....... + 2n – 1tan(2n – 1a) + 2ncot(2na)
= cota – cota + tana + 2tan(2a) + 4tan(4a) + .... ... + 2n – 1 tan(2n – 1a) + 2ncot(2na)
= cota – (cota – tana) + 2tan(2a) + .... ... + 2n – 1tan(2n – 1a) + 2ncot(2na)
= cota – 2cot2a + 2tan2a + ... + 2n – 1tan(2n – 1a) + 2ncot(2na) [Q cota – tana = 2cot2a]
.................................................= cota – 2n – 1cot(2n – 1a) + 2n – 1tan(2n – 1a)
+ 2ncot(2na)= cota – 2ncot(2na) + 2ncot(2na) = cota
18. (a) : tanmx = cotnx ⇒ =sincos
cossin
mxmx
nxnx
⇒ cosmx cosnx = sinmx sinnx ⇒ cos(m + n)x = 0
⇒ + = +( ) ( )m n x r2 12p
\ = ++
x rm n
( )( )2 1
2p , where r is any integer.
19. (c) : cos(psinx) = sin(pcosx)
⇒ cos( sin ) cos cosp x p x= −
p2
\ = −p x p xsin cosp2
⇒ + =p x x(sin cos ) p2
| AUGUST ‘1576
⇒ +
=p x x12
12 2 2
cos sin p
⇒ −
=p xcos p p4 2 2
\ =−
px
pp2 24
cos
Clearly, the least value of p is p2 2
.
20. (b) : log0.5sinx + log0.5cosx = 1⇒ log0.5(sinx cosx) = 1
⇒ = =sin cos .x x 0 5 12
⇒ 2sinx cosx = 1 ⇒ 1 – sin2x = 0⇒ (cosx – sinx)2 = 0 ⇒ cosx – sinx = 0⇒ tanx = 1
\ = − + > >x x xp p p4 4
0 0 and 2 [ sin , cos ]
21. (d) : sin tan sin tan− −−
= −
1 154
54
p p
= − +
= −
− −sin tan sin tan1 14 4
p p p
= sin–1(–1) = −
= −−sin sin12 2p p
22. (d) : Given, sin–1x = 2sin–1a
We know, − ≤ ≤−p p2 2
1sin x
⇒ − ≤ ≤ ⇒ − ≤ ≤− −p p p p2
22 4 4
1 1sin sina a
⇒ −
≤ ≤sin sinp p4 4
a
⇒ − ≤ ≤ ⇒ ≤12
12
12
a a| |
23. (b) : sin cot sin cot12
34
12
34
1 1− −−
= −
p
=
=−cos cot12
34
25
1
24. (a) : sin–1x < cos–1x (given)
We know that − ≤ ≤ \ − ≤ ≤−p p2 2
1 11sin ;x x
But sin–1x < cos–1x ⇒ 2sin–1x < cos–1x + sin–1x
⇒ < ⇒ <− −22 4
1 1sin sinx xp p
⇒ <
−sin(sin ) sin14
x p ⇒ <x 12
\ − ≤ <1 12
x
25. (c) : sin tan− −+ + = − +1 2 1 212
x x x xp
⇒ + + = +− −sin cot1 2 1 21x x x x
⇒ + + =+ +
− −sin sin1 2 12
1 1
1x x
x x
⇒ + + =+ +
x xx x
22
1 1
1⇒ x2 + x + 1 = 1 ⇒ x2 + x = 0⇒ x(x + 1) = 0 ⇒ x = 0, x = –126. (b) : (sin–1x)3 + (cos–1x)3
= + −
=−a a x a33
12p [Let sin ]
= +
−
+
−a a a a33 2
2 32
32
32
p p p
= −
+32 2 8
23p p pa a
= − ⋅ ⋅ +
−
+32
24 4 4 8
22 2 3p p p p p( )a a
= −
−
+32 4 16 8
2 2 3p p p pa
= −
+ −32 4 8
332
2 3 3p p p pa = −
+32 4 32
2 3p p pa
\ + = −
+− −(sin ) (cos )1 3 1 32 33
2 4 32x x ap p p
(sin ) (cos )− −+ =1 3 1 33
32x x p
\ a i.e. x= =−p p4 4
1 sin ⇒ =x 12
27. (b) : Clearly, the greatest side of the triangle is5x + 5yLet q be the greatest angle of the triangle
\ =+ + + − +
+ +cos
( ) ( ) ( )( )( )
q3 4 4 3 5 5
2 3 4 4 3
2 2 2x y x y x yx y x y
= −
+ +2
2 3 4 4 3xy
x y x y( )( ) < 0
\ q is an obtuse angle.\ The triangle is obtuse-angled.
28. (c) : Given that ∠ = ∠ =B Cp p3 4
,
and BD : DC = 1 : 3
| AUGUST ‘15 77
From DABD, we have
BD
BADAD
Bsin sin∠=
From DCAD, we have
A
B D C/3 /4CD
CADAD
Csin sin∠=
Q BD : CD = 1 : 3
\ ∠ ∠ =AD BADB
AD CADC
sinsin
: sinsin
:1 3
⇒ ∠∠
⋅ =sinsin
sinsin
BADCAD
CB
13
⇒ ∠∠
= ⋅sinsin
sinsin
BADCAD
BC
13
⇒ ∠∠
= ⋅sinsin
//
BADCAD
13
3 21 2
⇒ ∠∠
= × ×sinsin
BADCAD
13
32
21
⇒ ∠∠
=sinsin
BADCAD
16
29. (c) : Let BD = 2x, CD = 3xand AD = 6x and ∠BAD = qThen ∠CAD = A – q
A
B D C
Now, tanq = = =BDAD
xx
26
13
and tan( )A CDAD
xx
− = = =q 36
12
Also, tanA = tan{(A – q) + q}
= − +− −
=+
− ⋅= =tan( ) tan
tan( )tan//
AA
q qq q1
12
13
1 13
12
5 65 6
1
\ tanA = 1 ⇒ A = 45°30. (c) : In DABC, given that a = 2b and A – B = 60°
Now, tan cotA B a ba b
C− = −+2 2
⇒ ° = −+
tan cot602
22 2b bb b
C
⇒ = \ = = °13
13 2 2
3 30cot cot cotC C
\ = ° ⇒ = °C C2
30 60
31. (b) : 3 32
34 4sin sin ( )p a p a−
+ +
− +
+ −
2
256 6sin sin ( )p a p a
= 3[cos4a + sin4a] – 2[cos6a + sin6a]= 3[1 – 2cos2a sin2a] – 2[1 – 3cos2a sin2a]= 3 – 2 = 132. (a) : 7cosx + 5sinx = 2k + 1 ... (1)
⇒+
++
= +
+
7
7 5
5
7 5
2 1
7 52 2 2 2 2 2cos sinx x k
Clearly (1) will have a solution if 2 1
7 51 2 1 74 8 6
2 2
k k+
+≤ ⇒ + ≤ =| | . (approx.)
\ –8.6 ≤ 2k + 1 ≤ 8.6⇒ –9.6 ≤ 2k ≤ 7.6 ⇒ –4.8 ≤ k ≤ 3.8 ... (2)Clearly, integral values of k satisfying the inequality (2) are –4, –3, –2, –1, 0, 1, 2, 3i.e. the number of integral values of k is 8.
33. (a) : We have, sin sin sin sinp p p p14
314
514
714
Let p q q p q p14
14 72
= = ⇒ =, then
\ The given expression becomessinq sin3q sin5q sin7q
= −
−
−
=
sin sin sin sinp q p q p q p q p2
62
42
22
72
= cos6q cos4q cos2q
= ⋅ ⋅ −12 2
2 2 2 4 8sin
sin cos cos cos( )q
q q q p q[ 6q = 14q – 8q = p – 8q]
= ⋅ −12 2
2 4 4 82 sinsin cos ( cos )
qq q q
= − ⋅ = −12 2
2 8 8 12 2
163 3sinsin cos
sinsin
qq q
= − + = − +18 2
14 2 18 2
2sin
sin( )sin
sin( )q
q qq
p q = 18
34. (a) : From the triangle ABC, we geta b
sin sin22 12
45°
=°
⇒ =°
a b 2 22 12
sin
45°
B
h
D C A
a
b22
1
2
°1121
2
°
1121
2
°180° –
= 90° – 221
2
°
Again, from DBCD, we get ha
= ° − °
sin 90 22 12
⇒ =°
h a cos22 12
⇒ =° °
h b 2 22 12
22 12
sin cos
⇒ = ⋅ ×°
h b2
2 2 22 12
sin = ⋅ °b2
2 45sin
⇒ h b b h= \ =2
2
| AUGUST ‘1578
35. (c) : Let cos− − =1 1 p q ... (i)
\ = − ⇒ = −cos cosq q1 12p p⇒ 1 – cos2q = p ⇒ sin2q = p
\ = = −sin sinq qp pand 1 ... (ii)
From (i) and (ii), we get
cos sin− −− =1 11 p p ... (iii)
Now, the given equation is
cos cos cos− − −+ − + − =1 1 11 1 34
p p q p
⇒ + + − =− − −cos sin cos1 1 1 1 34
p p q p [From (iii)]
⇒ − = − = − =−cos 1 1 34 2
3 24 4
q p p p p p
\ − = ⇒ − =−cos 1 14
1 12
q qp
⇒ − = \ =1 12
12
q q
36. (b, d) : We have, 2 3cos cos cosA B B= + ... (i)and 2 3sin sin sinA B B= − ... (ii)Squaring and adding (i) and (ii), we get 2(cos2A + sin2A) = cos2B + cos6B + 2cos4B
+ sin2B + sin6B – 2sin4B⇒ 2 = 1 + (cos2B + sin2B)3 – 3cos2Bsin2B
+ 2(cos2B + sin2B)(cos2B – sin2B)⇒ 0 = –3cos2B(1 – cos2B) + 2(2cos2B – 1)⇒ 0 = –3cos2B + 3cos4B + 4cos2B – 2⇒ 3cos4B + cos2B – 2 = 0⇒ 3cos4B + 3cos2B – 2cos2B – 2 = 0⇒ (3cos2B – 2)(cos2B + 1) = 0
\ = ≠ −cos [ cos ]2 223
1B B
Now, sin2 1 23
13
B = − =
We have, cos cos ( cos )A B B= + =± +
1
2
23
1 23
2
2
cos A = ± 53 3
\ = ±sin A 23 3
Now, sin2(A – B) = (sinA cosB – cosA sinB)2
= sin2A cos2B + cos2A sin2B – 2sinAcosA sinBcosB
= × + × − × × × ×227
23
2527
13
2 23 3
53 3
23
13
= + − × ×× ×
= + − = =4 2581
2 2 59 3 3
4 25 2081
981
19
\ − = ±sin( )A B 13
37. (a, c) : cos cossin sin
sin sincos cos
a βa β
a βa β
+−
+ +−
n n
=
+ ⋅ −
+ ⋅ −
+
+ ⋅22 2
22 2
22
cos cos
cos sin
sin cosa β a β
a β a β
a β an −−
− + ⋅ −
β
a β a β2
22 2
sin sin
n
= − + − −cot ( ) cotn n na β a β2
12
= −
0
22
,
cot ,
when is odd
when is even
n
n n a β
38. (a, b) : Given that A = tan6° tan42° andB = cot66° cot78°
Now, AB
=° °° °
tan tancot cot
6 4266 78
= tan6° tan42° tan66° tan78°
= ° ° ° °° ° ° °
( sin sin )( sin sin )( cos cos )( cos cos
2 6 66 2 42 782 6 66 2 42 78 ))
= ° − ° ° − °° + ° ° +
(cos cos )(cos cos )(cos cos )(cos cos
60 72 36 12060 72 36 1120°)
=− °
° +
+ °
° −
12
18 36 12
12
18 36 12
sin cos
sin cos
=− −
+ +
+ −
+ −
= −12
5 14
5 14
12
12
5 14
5 14
12
3 5( ))( )( )( )
3 55 1 5 1
++ −
= −−
= =9 55 1
44
1 \ = ⇒ =AB
A B1
Q A = B \ A2 – 2AB + B2 = A2 – 2A2 + A2 = 0
39. (a, b, c, d) : 16 16 102 2sin cosx x+ =
⇒ + =−16 16 102 21sin sinx x
⇒ + =aa
ax16 10 [Putting 16 = ]sin2
⇒ a2 – 10a + 16 = 0⇒ (a – 8)(a – 2) = 0 \ a = 8 and a = 2
\ = ± ±sin ,x 32
12
40. (a, b)nn
Solution Set-151
1. (a) : S rsr s
= + + + + −<∑19 1 2 3 20 22 2 2 2( ... )
= 20(12 + 22 + 32 +... + 202) – (1 + 2 + 3 + ... + 20)2
= ⋅ ⋅ − ⋅ =20 21 416
20 214 13300
2 2 2
2. (b) : I x x dx x t= − =∫ ( ) ,1 2 9 9 2
0
1
I t t dt= − =∫ ( )! !
!1 212
9 414
9 4
0
1
\ 1 2 149 4
2 14 13 12 11 1024 20020I =
⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ =!! !
3. (c) : t3 + 2t = 20t + 8, t3 – 18t – 8 = 0t1, t2, t3 are roots ⇒ St1t2 = –18, t1 t2 t3 = 8Circle x2 + y2 + 2gx + 2fy + c = 0, x = t2, y = 2tt4 + (4 + 2g)t2 + 4ft + c = 0St1 t2 = 4 + 2g ⇒ 2g = –22,t1 t2 t3 = 8 ⇒ 2f = –4,t1 + t2 + t3 = 0, t4 = 0, c = 0\ Circle is x2 + y2 – 22x – 4y = 0, radius = 5 5
4. (c) : P(x1, y1) ⇒ chord of contact is xx1 + yy1 = a2
y xy x a
yxa
yb
= −
+ + =1
1
2
1
2
2
2
2 1touches
\ Locus of P is xb
ya
ab
2
2
2
2
2
2+ =
5. (d) : P(A) = 0.7, P B( ) .= 0 6P(A ∩ B) = P(A) – 0.5 = 0.2
P B A B( / ( )) .. . .∪ = + − =0 2
0 7 0 6 0 514
6. (a, b, c, d) : 10 < x ≤ 24⇒ 242 < x2 + 100, x2 ≥ 476, x = 22, 23, 24x > 24 > 10 ⇒ x2 < 242 + 100, x2 < 676\ x ≤ 25 \ x = 22, 23, 24, 25
7. (c) : ABC is an equilateral triangle of side 6
Distance of A from BC = altitude = =32 6 3
2
8. (b) : The plane OBC isx y z2 3 13 1 2
0=
⇒ 5x – y – 7z = 0
Distance of A = =1875
6 35
9. (5)(1 – x)10 = C0 – C1x + C2x2 – ... + C10x10
(1 – x)10 x4 = C0x4 – C1x5 + C2x6 – ... + C10 x14
S x x dx= − ⋅ =∫ ( )! !
!110 4
1510 4
0
1
\ 1 15 14 13 12 1124 3 5 7 11 13S = ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ (5 primes)
10. (c) : Letters CCLLUUAS Words Number of words
P.
2 132
41
32 36
1 1 153
3 6096
+ ⇒⋅ =
+ + ⇒⋅ =
!!
!woords
Q.
2 232
42
18
2 1 131
42
42 216
1 1
2+ ⇒⋅ =
+ + ⇒⋅ =
+ +
!( !)
!!
11 154
4 120
354
+ ⇒⋅ =
!
words
R. 2 2 132
31
52 2 270
2 1 1 131
43
+ + ⇒⋅⋅ =
+ + + ⇒⋅
!! !
⋅ =
+ + + + ⇒ =
52 720
1 1 1 1 1 5 120
1110!!
!
words
S. 2 2 2 62
90
2 2 1 132
32
64 1620
2 1 1 1
3+ + ⇒ =
+ + + ⇒⋅⋅ =
+ + + +
!( !)
!
1131
44
62 1080
2790
⇒⋅⋅ =
!
words
nn
| AuGuST ‘1580
| august ‘15 81
Section-1This section contains EIGHT questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. Marking scheme : +4 If the bubble corresponding to the answer is darkened, 0 In all other cases.
1. Let a1, a2, a3, ..., a11 be real numbers satisfyinga1 = 15, 27 – 2a2 > 0 and ak = 2ak – 1 – ak – 2 for k =
3, 4, ..., 11. If a a a12
22
112
1190
+ + +=
.... , then the value
of a a a1 2 1111
+ + +.... is equal to
2. Let a + ar1 + ar12 + ... + ∞ and a + ar2 + ar2
2 + ... + ∞be two infinite series of positive numbers withthe same first terms. The sum of the first series isr1 and the sum of second series is r2. The value of(r1 + r2) is
3. If a, b, c are in H.P. and ifa ba b
c bc b
+−
+ +−
≥ ∞2 2
l l l.... , then
the maximum integral value of l is
4. Let ‘m’ denotes the number of ways in which 4different balls of green colour and 4 different ballsof red colour can be distributed equally among 4persons if each person has balls of the same colourand ‘n’ be corresponding figure when all thefour persons have balls of different colour. Find( ) .m n+
1325. The polynomial p(x) = 1 – x + x2 – x3 + … + x16 – x17
can be written as a polynomial in y where y = x + 1,
then let coefficient of y2 be k then k400
(where [.]
denote greatest integer function) is
6. Number of ways in which 3A’s and 4B’s can bearranged in a row which reads the same backwardsand forwards, is
7. The number of solutions for the equationlog4(2x2 + x + 1) – log2(2x – 1) = 1, is
8. Consider the equation x3 – ax2 + bx – c = 0, wherea, b, c ∈ Q ; (a ≠ 1). It is given that x1, x2 and x1x2are the real roots of the equation. If b + c = 2(a + 1)
then x x ab c1 2
1++
is equal to
Section-2This section contains TEN questions. Each question has FOUR options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four option(s) is (are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. Marking scheme : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened, 0 If none of the bubbles is darkened, –2 In all other cases.
9. The system of equations6x + 5y + lz = 03x – y + 4z = 0x + 2y – 3z = 0 has(a) only a trivial solution for l ∈ R(b) exactly one nontrivial solution for some real l(c) infinite number of nontrivial solutions for one
value of l(d) only one solution for l ≠ –5
10. If a, b are the roots of the equation 8x2 – 3x + 27 = 0,
then the value of ab
ba
2 1 3 2 1 3
+
/ /
is
(a) 1/3 (b) 1/4 (c) 7/2 (d) 4
11. If a, b are the real roots of ax2 + bx + c = 0 anda4, b4 are the roots of lx2 + mx + n = 0, then the rootsof the equation a2lx2 – 4aclx + 2c2l + a2m = 0 are
PaPer-1
| august ‘1582
(a) real (b) imaginary(c) opposite in sign (d) equal
12. Let S1, S2, ….. be squares such that for eachn ≥ 1, the length of a side of Sn equals the lengthof a diagonal of Sn + 1. If the length of a side of S1 is10 cm, then for which of the following values of n isthe area of Sn less than 1 square cm?(a) 7 (b) 8 (c) 9 (d) 10
13. Let z1, z2, z3 be non-zero complex numberssatisfying the equation z iz2 = . Which of thefollowing statement(s) is/are CORRECT?(a) The complex number having least positive
argument is 32
12
, .
(b) Amp( )zkk=∑ =
1
3 52π
(c) Incentre of the triangle formed by z1, z2 and z3 is (0, 0).
(d) Area of the triangle formed by z1, z2 and z3 is 3
4.
14. If a, b, c are real numbers and z is a complex numbersuch that a2 + b2 + c2 = 1 and b + ic = (1 + a)z, then11
+−
iziz
equals
(a) b icia
−−1
(b) a ibc
++1
(c) 1−−
ca ib
(d) 1++
ab ic
15. If the equations x2 + bx + c = 0 and bx2 + cx + 1 = 0have a common root then(a) b + c + 1 = 0(b) b2 + c2 + 1 = bc(c) (b – c)2 + (b – 1)2 + (c – 1)2 = 0(d) b + c + 1 = bc
16. The independent events, the chance that only the firstoccurs is ‘a’, the chance that only the second occursis ‘b’ and the chance of only third is ‘c’. If the chances
of three events are respectively aa x
bb x
cc x+ + +
, , ,
where x is the root of the equation(a) (a – x)(b – x)(c – x) = x2
(b) (a + x)(b + x)(c + x) = x2
(c) (x – a)(x – b)(x – c) = x2
(d) None of these
17. If roots of ax2 + 2bx + c = 0 (a ≠ 0) are non-realcomplex and a + c < 2b, then(a) c > 0 (b) c < 0(c) 4a + c < 4b (d) 4a + c > 4b
18. 1111
is equal to k
j
j
i
i
n
===∑∑∑
(a) n n n( )( )+ +1 26
(b) n n n( )( )− −1 26
(c) nC3 (d) n + 2C3
Section-3This section contains TWO questions. Each question contains two columns, Column I and Column II. Column I has four entries (A), (B), (C) and (D), Column II has four entries (P), (Q), (R) and (S) . Match the entries in Column I with the entries in Column II. One or more entries in Column I may match with one or more entries in Column II. The ORS contains a 4 × 4 matrix whose layout will be similar to the one shown below.
(A)
(B)
(C)
(D)
P Q R S
P Q R S
P Q R S
P Q R S
For each entry in Column I, darken the bubbles of all the matching entries. For example, if entry (A) in Column I matches with entries (Q), (R) and (S), then darken these three bubbles in the ORS. Similarly, for entries (B), (C) and (D). Marking scheme : For each entry in Column I. +2 If only the bubble(s) corresponding to all the correct match(es) is (are) darkened, 0 If none of the bubbles is darkened, –1 In all other cases.
19. A is a set containing n elements. A subset P of set Ais chosen at random. The set A is reconstructed byreplacing the elements of the subset P. A subset Qof A is again chosen at random the probability that
Column I Column II
(A) P ∩ Q = f P. n(3n – 1)/4n
(B) P ∩ Q is a singleton Q. (3/4)n
(C) P ∩ Q contain 2 elements R. 2nCn / 4n
(D) |P| = |Q|, where|X| = number of elements in X
S. 3 12 4
2n
nn n− −⋅( )
| august ‘1584
20. Match the following:
Column I Column II
(A) Mr is defined as : Mr
r
r
r =−
−
1 1
1 11 2( )
and |Mr| is the corresponding determinant value
of Mr, then lim (| | | | ... | |)n nM M M
→∞+ + + =2 3
P. 0
(B If 1
11
00
0
cos coscos coscos cos
cos coscos coscos cos
a ba gb g
a ba gb g
= , then sin2a + sin2b + sin2g = Q. 4
(C) If A =−
3 11 1
and a matrix C is defined as C = (BAB–1)(B–1ATB), where |B| ≠ 0 and
|C| = k2, (where k ∈ N), then k =
R. 2
(D) If A =−
1 11 1
and A4 = –lI, then l equals S. 1
PaPer-2
Section-1This section contains EIGHT questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. Marking scheme : +4 If the bubble corresponding to the answer is darkened, 0 In all other cases.
1. Let ‘n’ ordinary fair dice be rolled once. If theprobability that at least one of the dice shows an
odd number is 3132
, then the value of ‘n’ is equal
to2. Two whole numbers are randomly selected and
multiplied. If the probability that the units place intheir product is “Even” is p and the probability thatthe units place in their product is “odd” is q, then
pq
+1 is equal to
3. If (y2 – 5y + 3)(x2 + x + 1) < 2x for all x ∈ R, thennumber of integral values of y will be
4. If in the expansion of (1 + x)m (1 – x)n the coefficientsof x and x2 are 3 and –6 respectively, then ‘n’ is
5. What is the value of ‘n’ if the ratio of the 6th termfrom the beginning to the 6th term from the end of
the binomial 3 2 32
+( )n is ?
6. The positive integer just greater than(1 + 0.00001)100000 is
7. Sum of all the real values of x for which
2 42 3
2− +− +
=xx
is
8. If x1, x2, x3 are such that, x1 + x2 + x3 = 2
x x x12
22
32 6+ + =
x x x13
23
33 8+ + =
find the value of –(x2 – x3)(x3 – x1)(x1 – x2).Section-2
This section contains EIGHT questions. Each question has FOUR options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four option(s) is(are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. Marking scheme : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened, 0 If none of the bubbles is darkened, 2 In all other cases
9. The probability that a student passes in Mathematics,Physics and Chemistry are m, p and c respectively.Of these subjects, a student has a 75% chance ofpassing in at least one, a 50% chance of passing inat least two and a 40% chance of passing in exactlytwo subjects. Which of the following relation aretrue?
| august ‘15 85
(a) p m c+ + = 1920
(b) p m c+ + = 2720
(c) pmc = 110
(d) pmc = 14
10. If loge2, loge(2x – 1) and loge(2x + 3) are in A.P.,then(a) 2x is rational (b) x is irrational
(c) ( )2 x is irrational (d) 2x2 is rational11. If a, b are the roots of the quadratic equation
ax2 + bx + c = 0 then which of the following expressionwill be the symmetric function of roots?
(a) ln ab
(b) a2b5 + b2a5
(c) tan(a – b) (d) ln (ln )1
+a
b2
2
12. If one of the root of the equation 4x2 – 15x + 4p = 0is the square of the other then the p is
(a) 12564
(b) − 278
(c) − 1258
(d) 278
13. If |z1 + z2|2 = |z1|2 + |z2|2 then
(a)zz
1
2 is purely real
(b) zz
1
2 is purely imaginary
(c) z z z z1 2 2 1 0+ =
(d) amp may be equal to zz
1
2 2
π
14. Let D = + ++
a a
a b a ba b
2
2
0
1 20 1 2 3
( ) , then
(a) a + b is a factor of D (b) a + 2b is a factor of D(c) 2a + 3b is a factor of D (d) a2 is a factor of D
15. Given, z1 + z2 + z3 = A, z1 + z2w + z3w2 = B,z1 + z2w2 + z3w = C, where w is a cube root of unity,then
(a) z A B C1
2
3= + +w w
(b) z z z A B C1 2 3
3 3 3
3= + +
(c) z z z A B C1 2 3
3 3 3
27= + +
(d) z A B C2
2
3= + +w w
16. If the conjugate of (x + iy) (1 – 2i) be 1 + i, then
(a) x = 15
(b) x iy i+ = +15
3( )
(c) x iy i− = +15
3( ) (d) x iy ii
+ = −−
11 2
Section-3This section contains TWO paragraphs. Based on each paragraph, there will be TWO questions. Each question has FOUR options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four option(s) is(are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. Marking scheme : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened, 0 If none of the bubbles is darkened, –2 In all other cases.
Paragraph for Q. No. 17 & 18Suppose p is the first of n (n > 1) AM’s between two real distinct numbers a and b; q is the first of n HM’s between the same two numbers.17. The value of p is
(a) na bn
++1
(b) na bn
++2 1
(c) nb an
++1
(d) nb a
n n++( )1
18. The value of q is
(a) ab nb an
( )++
1 (b) ab na bn
( )++
1
(c) ab nb an( )2 1++
(d) ab n na bn( ( ))+
+1
Paragraph for Q. No. 19 & 20There are three pots and four coins. All these coins are to be distributed into these pots where any pot can contain any number of coins.19. In how many ways all these coins can be distributed
if all coins are identical but all pots are different?(a) 15 (b) 16 (c) 17 (d) 81
20. In how many ways all these coins can be distributedif all coins are different but all pots are identical ?(a) 14 (b) 21 (c) 27 (d) none of these
anSwerS keyPAPER-1
1. (0) 2. (1) 3. (4) 4. (6) 5. (2)6. (3) 7. (1) 8. (1) 9. (c, d) 10. (b)11. (a, c) 12. (b, c, d) 13. (a, c) 14. (b, c)15. (a, c) 16. (b) 17. (b, c) 18. (a, d)19. (A) → Q; (B) → P; (C) → S; (D) → R20. (A) → S ; (B) → R ; (C) → Q ; (D) → Q
PAPER-21. (5) 2. (4) 3. (2) 4. (9) 5. (9)6. (3) 7. (1) 8. (6) 9. (b, c) 10. (a, b, c)11. (a, b, d) 12. (c, d) 13. (b, c, d)14. (a, b) 15. (c, d) 16. (b, d) 17. (a) 18. (b)19. (a) 20. (a)
nn
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