Relations and Functions - EduGorilla Study Material
69
| JUNE ‘17 12 CLASS XI Series 2 Relations and Functions X If A and B be two non-empty sets, then cartesian product of sets is A × B = {(x i , y i ) : x i ∈A, y i ∈B} X (x, y) = (p, q) ⇔ x = p, y = q X A × B = B × A ⇒ A = B X If n(A) = p, n(B) = q, then n(A × B) = pq X A × A × A = {(a, b, c) : a, b, c ∈ A} is called ordered triplet. X A × B = f ⇔ A = f or B = f X Subset of X × Y is called a relation from X to Y. X If n(X) = p and n(Y) = q then the total number of relations from X to Y is 2 pq . X e set of all first elements of the ordered pairs in relation R from set X to set Y is called the domain of the relation R. X e set of all second elements of the ordered pairs in relation R from set X to Y is called the range of the relation R. X The whole set B is called the co-domain of the relation R. h Range ⊆ Co-domain X If R = {(a, b) : a, b ∈ R}, then R –1 = {(b, a) : b, a ∈ R} X A subset f of X × Y is called a function (or map or mapping) from X to Y iff for each x ∈ X, there exists a unique y ∈Y such that (x, y) ∈ f. It is written as f : X → Y. h Set X is called domain and Set Y is called co-domain of the function f. h e set of elements of Y, which are assigned to the elements of X is called range of f. X Algebra of real functions If f : X → R and g : X → R, then h (f + g) (x) = f(x) + g(x), ∀ x ∈X h (f – g) (x) = f(x) – g(x), ∀ x ∈X h (af )(x) = a f (x), ∀ x ∈X h (f g)(x) = f(x)g(x), ∀ x ∈ X h f g x fx gx gx x X = ≠0 ∈ () () () , () , X Function ⊆ Relation ⊆ Cartesian Product X If A and B be two non-empty sets having n elements in common, then A × B and B × A have n 2 elements in common. IMPORTANT FORMULAE
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Transcript of Relations and Functions - EduGorilla Study Material
| JUNE ‘1712
CLASS XI Series 2
Relations and Functions
X If A and B be two non-empty sets, then cartesian product of sets is A × B = {(xi , yi) : xi ∈A, yi ∈B}
X (x, y) = (p, q) ⇔ x = p, y = q
X A × B = B × A ⇒ A = B
X If n(A) = p, n(B) = q, then n(A × B) = pq
X A × A × A = {(a, b, c) : a, b, c ∈ A} is called ordered triplet.
X A × B = f ⇔ A = f or B = f
X Subset of X × Y is called a relation from X to Y.
X If n(X) = p and n(Y) = q then the total number of
relations from X to Y is 2pq
.
X �e set of all �rst elements of the ordered pairs in relation R from set X to set Y is called the domain of the relation R.
X �e set of all second elements of the ordered pairs in relation R from set X to Y is called the range of the relation R.
X The whole set B is called the co-domain of the relation R.
h Range ⊆ Co-domain
X If R = {(a, b) : a, b ∈ R}, then
R–1 = {(b, a) : b, a ∈ R}
X A subset f of X × Y is called a function (or map or mapping) from X to Y iff for each x ∈ X, there exists a unique y ∈Y such that (x, y) ∈ f. It is written as f : X → Y.
h Set X is called domain and Set Y is called co-domain of the function f.
h �e set of elements of Y, which are assigned to the elements of X is called range of f.
X Algebra of real functionsIf f : X → R and g : X → R, then
h (f + g) (x) = f(x) + g(x), ∀ x ∈X
h (f – g) (x) = f(x) – g(x), ∀ x ∈X
h (af )(x) = a f (x), ∀ x ∈X
h (f g)(x) = f(x)g(x), ∀ x ∈ X
hfg
xf xg x
g x x X
= ≠ 0 ∈( )( )( )
, ( ) ,
X Function ⊆ Relation ⊆ Cartesian Product
X If A and B be two non-empty sets having n elements in common, then A × B and B × A have n2 elements in common.
IMPORTANT FORMULAE
| JUNE ‘1714
WORK IT OUT
VERY SHORT ANSWER TYPE
1. Let A = {1, 2} and B = {3, 4, 5}. Find(i) B × A (ii) A × A × A
2. Let P = {x, y, z} and Q = {3, 4}. Find the number ofrelations from P to Q.
3. Let f be the exponential function and g be thelogarithmic function. Find (fg)(1).
4. Let f(x) = x2 and g(x) = (3x + 2) be two real functions. �en, �nd (f + g)(x).
5. Let g = {(1, 2), (2, 5), (3, 8), (4, 10), (5, 12), (6, 12)}Is g a function? If yes, �nd its domain and range. Ifno, give reason.
SHORT ANSWER TYPE
6. Let f : Z → Z, g : Z → Z be functions de�ned byf = {n, n2) : n ∈ Z} and g = {(n, |n|2) : n ∈ Z}.Show that f = g.
7. �e function F(x) = 95
32x + is the formula to
convert x° C to Fahrenheit units. Find(i) F(0) (ii) F(–10)(iii) the value of x when F(x) = 212.
8. If f(x) = log ,11
+−
xx
prove that f xx
f x21
22+
= ( ).
9. Let R be a relation on the set of natural numbersN de�ned by xRY ⇔ x + 2y = 41; ∀ x, y ∈ N. Findthe domain and range of R.
10. If P = {a, b} and Q = {x, y, z}, then show thatP × Q ≠ Q × P.
LONG ANSWER TYPE - I
11. Given f (x) = 11( )− x
, g(x) = f {f (x)} and
h(x) = f [f {f (x)}]. �en �nd the value of f (x) ⋅ g(x) ⋅ h(x).
12. Find the domain of the function
yx
x=−
+ +11
210log ( )
13. If A ⊆ B and C ⊆ D, prove that A × C ⊆ B × D.
14. Let f be a real valued function de�ned by
f x e ee e
x x
x x( ) ,| |
| |= −+
−then �nd the range of f.
15. Consider the following :(i) f : R → R : f (x) = logex
(ii) g : R → R : g(x) = x
(iii) h : A → R : h(x) =−
142x
, where A = R – {–2, 2}
Which of them are functions? Also �nd their range, if they are functions.
LONG ANSWER TYPE - II
16. Find the domain and range of the real valued
function f (x) given by f (x) = 44
−−
xx
.
17. If f : R → R is de�ned by f (x) = x3 + 1 and g : R → Ris de�ned by g(x) = x + 1, then �nd(i) f + g (ii) f – g (iii) f ⋅ g
(iv) fg
(v) af (a ∈ R)
18. Find the domain and range of the function
f (x) =−
12 3sin x
19. Let A = {x ∈ N : x2 – 5x + 6 = 0}, B = {x ∈ Z : 0 ≤ x < 2}and C = {x ∈ N : x < 3}, then verify that:(i) A × (B ∪ C) = (A × B) ∪ (A × C)(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
20. Find the domain of the function
f(x) = (log ( ))227 6x x x+ − −
SOLUTIONS
1. (i) B × A = {3, 4, 5} × {1, 2}= {(3, 1), (3, 2), (4, 1), (4, 2), (5, 1), (5, 2)}(ii) A × A × A = {1, 2} × {1, 2} × {1, 2}= {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}2. Given, P = {x, y, z} and Q = {3, 4}\ n(P) = 3 and n(Q) = 2\ n(P × Q) = 3⋅2 = 6Total number of relations from P to Q = number of subsets of P × Q = 26 = 64.3. We have, f : R → R given by f(x) = ex and g : R+ → Rgiven by g(x) = logexDomain (f ) ∩ Domain (g) = R ∩ R+ = R+.\ fg : R+ → R is given by (fg)(x) = f(x) g(x) = ex·logex.Now, (fg)(1) = f(1)g(1) = e1 × loge1 = e × 0 = 0.4. We have (f + g)(x) = f(x) + g(x) = x2 + (3x + 2)5. Yes, dom ( g) = {1, 2, 3, 4, 5, 6},
range ( g) = {2, 5, 8, 10, 12}
| JUNE ‘17 15
6. Given, domain of f = Z and domain of g = ZHence, domain f = domain g = Z …(1)Also, f(n) = n2, for all n ∈ Z and g(n) = |n|2 = n2 for all n ∈ ZHence, f(n) = g(n) for all n ∈ Z …(2)From (1) and (2), we have f = g.
7. F(x) = 95
32x + (given)
(i) F( )0 9 05
32 32= × +
= ⇒ F(0) = 32
(ii) F( ) ( )− = × − +{ } =10 9 105
32 14 ⇒ F(–10) = 14
(iii) F(x) = 212 ⇔ 95
32 212x + =
⇔ 9x = (5 × 180) ⇔ x = 100
8. Given, f x xx
( ) log= +−
11
\ f xx
xxxx
x xx
21
1 21
1 21
1 212
2
2
2
2+
=+
+
−+
= + ++
log log−−
2x
= +−
= +−
=log log ( ).11
2 11
22x
xxx
f x
9. We have, y = 412− ∈x N
Clearly x = 1, 3, 5, 7, …, 39\ Domain R = {x : (x, y) ∈ R; x + 2y = 41}
= {1, 3, 5, 7, …, 39} = set of odd natural numbers less than 40.Now, y can be only those natural numbers for which x ∈ N i.e., x = 41 – 2y ∈ N.Clearly, y = 1, 2, 3, …, 20.\ Range of R = {y : x + 2y = 41} = {1, 2, 3, …, 20}
= set of natural numbers less than 21.10. We have, P = {a, b} and Q = {x, y, z}Now, P × Q = {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z)}Q × P = {(x, a), (x, b), (y, a), (y, b), (z, a), (z, b)}\ P × Q ≠ Q × P.11. Given, g(x) = f {f (x)}
= fx
x
xx
11
1
1 11
1−
=−
−
= − −
and h(x) = f [f {f (x)}] = f {g(x)} = − −
f x
x1
=+ −
=1
1 1 xx
x
\ f(x) ⋅ g(x) ⋅ h(x) = 11
1 1−
− −
⋅ = −x
xx
x
12. For y to be de�ned(i) log10 (1 – x) must be de�ned ⇒ 1 – x > 0 ⇒ x < 1(ii) log10 (1 – x) ≠ 0 ⇒ 1 – x ≠ 100 ⇒ 1 – x ≠ 1 ⇒ x ≠ 0(iii) x + 2 ≥ 0 ⇒ x ≥ –2From (i), (ii) and (iii), we get –2 ≤ x < 1 and x ≠ 0\ –2 ≤ x < 0 or 0 < x < 1Hence domain = [–2, 0) ∪ (0, 1).13. Let (a, b) be an arbitrary element of A × C. �en,(a, b) ∈ A × C⇒ a ∈ A and b ∈ C⇒ a ∈ B and b ∈ D [ A ⊆ B and C ⊆ D]⇒ (a, b) ∈ B × D�us, (a, b) ∈ A × C⇒ (a, b) ∈ B × D for all (a, b) ∈ (A × C).\ A × C ⊆ B × D
14. Let y = f x e ee e
x x
x x( )| |
| |= −+
−
If x ≥ 0 then y ee
x
x= −2
21
2⇒ e
yx2 1
1 21=
−≥ ( x ≥ 0)
⇒ 11 2
1 0−
− ≥y
⇒y
y1 20
−≥ or
yy2 1
0−
≤
\ 0 12
≤ <y ⇒ y ∈
0 12
,
15. f and g are not functions as they are not de�ned fornegative values of x. But h is a function.
\ For range of h, Let y = h(x) = 1
42x −
⇒ x2 – 4 = 1y
⇒ xy
2 4 1= + ⇒ xyy
=+4 1
Hence, range of h = −∞ −
∪ ∞, ( , )14
0
ANSWER KEYMPP-2 CLASS XII
1. (c) 2. (b) 3. (b) 4. (d) 5. (c)6. (c) 7. (a,b) 8. (b,c) 9. (b,c) 10. (b)11. (a,b) 12. (a) 13. (a,b,d) 14. (d) 15. (b)16. (d) 17. (2) 18. (3) 19. (2) 20. (7)
| JUNE ‘1716
16. We have, f x xx
( ) .= −−
44
Domain of f : We observe that f (x) is de�ned for all x except at x = 4. At x = 4, f (x) takes the indeterminate
form 00
. �erefore, Domain (f ) = R – {4}.
Range of f : For any x ∈ Domain (f ) i.e. for any x ≠ 4, we have
f x xx
xx
( ) ( ) .= −−
= − −−
= −44
44
1
\ Range (f ) = {–1}.17. (i) f + g : R → R is de�ned by(f + g) (x) = f (x) + g(x) = x3 + 1 + x + 1 = x3 + x + 2(ii) f – g : R → R is de�ned by(f – g) (x) = f (x) – g(x) = x3 + 1 – x – 1 = x3 – x(iii) f ⋅ g : R → R is de�ned by(fg) (x) = f (x) g(x) = (x3 + 1) (x + 1) = x4 + x3 + x + 1
(iv) fg
: R – {– 1} → R is de�ned by
fg
xf xg x
xx
x x xx
x x
= = ++
= + − ++
= − +( )( )( )
( )( )3 221
11 1
11
(v) af : R → R is de�ned by(af ) (x) = a f (x) = a (x3 + 1) = a x3 + a
18. We have, f (x) =−
12 3sin x
Domain of f : We know that –1 ≤ sin 3x ≤ 1 for all x ∈ R⇒ –1 ≤ – sin 3x ≤ 1 for all x ∈ R⇒ 1 ≤ 2 – sin 3x ≤ 3 for all x ∈ R⇒ 2 – sin 3x ≠ 0 for any x ∈ R
⇒ f (x) =−
12 3sin x
is de�ned for all x ∈ R
Hence, domain ( f) = R.Range of f : 1 ≤ 2 – sin 3x ≤ 3 for all x ∈ R
⇒ 13
12 3
1≤−
≤sin x
for all x ∈ R
⇒ 13
1≤ ≤f x( ) for all x ∈ R
⇒ f (x) ∈ [1/3, 1]Hence, range ( f ) = [1/3, 1]
19. We haveA = {x ∈ N : x2 – 5x + 6 = 0} = {2, 3};B = {x ∈ Z : 0 ≤ x < 2} = {0, 1} and
C = {x ∈ N : x < 3} = {1, 2}\ A = {2, 3}, B = {0, 1} and C = {1, 2}(i) (B ∪ C) = {0, 1, 2}\ A × (B ∪ C) = {2, 3} × {0, 1, 2}= {(2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)}(A × B) = {2, 3} × {0, 1} = {(2, 0), (2, 1), (3, 0), (3, 1)}(A × C) = {2, 3} × {1, 2} = {(2, 1), (2, 2), (3, 1), (3, 2)}\ (A × B) ∪ (A × C) = {(2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)}Hence, A × (B ∪ C) = (A × B) ∪ (A × C)(ii) (B ∩ C) = {0, 1} = {1}\ A × (B ∩ C) = {2, 3} × {1} = {(2, 1), (3, 1)}And,(A × B) ∩ (A × C) = {(2, 1), (3, 1)}Hence, A × (B ∩ C) = (A × B) ∩ (A × C)
20. We have, f(x) = (log ( ))227 6x x x+ − −
= + − −(log ( )) ( )( )2 1 6x x x
For f (x) to be de�ned.(i) (log2(x)) ≥ 0 ⇒ x ≥ 20 ⇒ x ≥ 1(ii) (1 – x) (x – 6) ≥ 0 ⇒ 1 ≤ x ≤ 6�erefore domain of f = [1, 6].
| JUNE ‘1718
Only One Option Correct Type
1. 1
8 81
8 8
8
+ +
− +
i
i
sin cos
sin cos
π π
π πequals
(a) 28 (b) 0(c) – 1 (d) 1
2. If a, b, c are in G.P., then the equationsax2 + 2bx + c = 0, dx2 + 2ex + f = 0 have a common
root if da
eb
fc
, , are in
(a) A.P. (b) G.P.(c) H.P. (d) none of these
3. For positive integers n1 and n2 the value of theexpression ( )1 1+ i n + ( )1 3 1+ i n + ( )1 5 2+ i n + ( )1 7 2+ i n
where i = −1 is a real number i�(a) n1 = n2 (b) n2 = n2 – 1(c) n1 = n2 + 1 (d) ∀ n1 and n2
4. If x, y, z are distinct positive reals such thatlog log logxy z
yz x
zx y−
=−
=−
, then value of xx yy zz is
(a) 1 (b) 0(c) – 1 (d) none of these
5. If the roots of the equation bx2 + cx + a = 0be imaginary, then for all real values of x, theexpression 3b2x2 + 6bcx + 2c2 is
(a) less than 4ab (b) greater than –4ab(c) less than –4ab (d) greater than 4ab
6. Solve for x :log2x + 3 (6x2 + 23x + 21) + log3x + 7 (4x2 + 12x + 9) = 4(a) –4 (b) – 2
(c) − 14
(d) All of these
One or More Than One Option(s) Correct Type
7. ABCD is a rhombus, its diagonals AC and BDintersect at the point R where BD = 2AC. Its pointsD and R represent the complex numbers 1 + iand 2 – i respectively, then the complex numberrepresented by A is(a) (3, – 1/2) or (1, – 1/2)(b) (3, – 1/2) or (1, – 3/2)(c) (– 1/2, – 3/2) or (– 3/2, – 1/2)(d) None of these
8. (1 + i)5+ (1 – i)5 =(a) – 8 (b) 8
(c) −2 54
7 2/ cos π (d) −2 54
7 2/ cos π
9. If a and are non-real cube roots of unity andx = a + b, y = aa + b , z = a + ba, then(a) x + y + z = 1(b) x + y + z = 0(c) x3 + y3 + z3 = 3(a3 + b3)(d) none of these
10. �e equation whose roots are a + and a + ,if a are the roots of the equation ax2 + bx + c = 0and , are roots of the equation a x2 + b x + c = 0,is
Total Marks : 80 Time Taken : 60 Min.
This specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four
marks for correct answer and deduct one mark for wrong answer.
Self check table given at the end will help you to check your readiness.
Class XI
Complex Numbers and Quadratic Equations
| JUNE ‘17 19
(a) aa x2 + bb x + cc = 0 (b) ax2 + (a + a + b )x + bc = 0(c) aa x2 – bb x + cc = 0 (d) none of these
11. If A and B are the points (3, –1) and (2, 1) respectively, then the locus of the points P(z), z = x + yi, x, y ∈R, such that |z – 3 + i| = |z – 2 – i| is(a) a circle containing A and B (b) P is equidistant from A and B(c) right bisector of segment joining A and B(d) none of these
12. �e value of x : |x2 + 2x – 8| + x – 2 = 0 is(a) –5 (b) –2(c) 2 (d) –3
13. If a ∈C be such that |a| = 1, then the equation
11
4+−
iziz
= a has all the roots
(a) real and distinct (b) non-real(c) two real and two non-real (d) none of these
Comprehension Type
Let a be the roots of the equation 6x2 + 6px + p2 = 0, where p is a real number.
14. If both a and are greater than 2, then(a) p < – 4 (b) p < −2 6
(c) p < − −6 2 6 (d) none of these
15. �e equation whose roots are (a + )2 and (a – )2
is(a) 3x2 + 4p2x + p4 = 0 (b) 3x2 – 4p2x + p4 = 0(c) 3x2 – 4p2x – p4 = 0 (d) none of these
Matrix Match Type
16. Match the following :
Column I Column IIP. If a and b are posit ive
numbers and log a b+2
= +12
(log log )a b ab
then
is equal to
1. 3 3
Q. Let A(2 + 0i), B(–1 + 3i ) and C (–1– 3i ) b e t he vertices of ABC. Then, 6 sin A is equal to
2.
R. If one root of the equation (x – 1)(7 – x) = is three times the other, then =
3.
S. Conjugate of the complex number
− −72
3 32
i is
4. –2 + 3
P Q R S(a) 4 2 3 1(b) 3 4 1 2(c) 1 4 3 2(d) 2 1 3 4
Integer Answer Type
17. �e number of values of x in the interval [0, 3 ] satisfying the equation 2sin2x + 5sinx –3 = 0 is
18. �e least integral value of k for which(k– 2)x2 + 8x + k + 4 > 0 for all x ∈R, is
19. If a, are non-real cube roots of unity then (1 + a) (1 + ) (1 + a2) (1 + 2) (1 + a4) (1 + 4)....upto 2n factors is equal to
20. If a be non real cube root of unity, then α equals
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Marks scored in percentage ……
| JUNE ‘1720
CBSE
CLASS XII Series 2
Inverse Trigonometric Functions
X Functions Domain Range
sin–1x [–1, 1]−
π π2 2
,
cos–1x [–1, 1] [0, ]
tan–1x (– , )−
π π2 2
,
IMPORTANT FORMULAE
PROPERTIES OF INVERSE TRIGONOMETRIC FUNCTIONS
1. sin–1(sin x) = x ∀ x ∈ −
π π2 2
, 2. sin(sin–1x) = x, ∀ x ∈[–1, 1]
cos–1(cos x) = x, ∀ x ∈ [0, ] cos(cos–1x) = x, ∀ x ∈ [–1, 1]
tan–1(tan x) = x, ∀ x ∈ −
π π2 2
, tan(tan–1x) = x, ∀ x ∈R
cot–1(cot x) = x, ∀ x ∈ (0, ) cot(cot–1x) = x, ∀ x ∈ R
sec–1(sec x) = x, ∀ x ∈ [0, ] – π2{ } sec(sec–1x) = x, ∀ x ∈ R –(–1, 1)
cosec–1(cosec x) = x, ∀ x ∈ −
−π π2 2
0, { } cosec(cosec–1x) = x, ∀ x ∈ R – (– 1, 1)
3. sin–1(–x) = – sin–1x, ∀ x ∈[–1, 1] cot–1(–x) = – cot–1x, ∀ x ∈R
cos–1(–x) = – cos–1x, ∀ x ∈[–1, 1] sec–1(–x) = – sec–1x, ∀ x ∈ (– , –1] ∪ [1, )
tan–1(–x) = –tan–1x, ∀ x ∈R cosec–1(–x) = –cosec–1x, ∀ x ∈ (– , –1] ∪ [1, )
Functions Domain Rangecot–1x (– , ) (0, )
cosec–1x (– , –1] ∪ [1, )−
−
{ }π π2 2
0,
sec–1x (– , –1] ∪ [1, ) [ , ]02
ππ
− { }
| JUNE ‘17 21
4. sin–1(1/x) = cosec–1x, ∀ x ∈ (– , –1] ∪ [1, ) 5. sin–1x + cos–1x = /2, ∀ x ∈[–1, 1]
cos–1(1/x) = sec–1x, ∀ x ∈ (– , –1] ∪ [1, ) tan–1x + cot–1x = /2, ∀ x ∈R
tan ( / ) cot ,cot ,
−−
−= >− + <
1
1
11 00
x x xx x
forforπ sec–1x + cosec–1x = /2, ∀ x ∈ (– , –1] ∪ [1, )
6.
tan–1x + tan–1y =
+−
> > <
++
−
−
−
tan , , ,
tan ,
1
1
10 0 1
1
x yxy
x y xy
x yxy
x
if
ifπ >> > >
− ++
−
< < >
−
0 0 1
10 0 11
,
tan , ,
y xy
x yxy
x y xy
and
if andπ
tan–1x – tan–1y =
−+
> < > −
+−
+
−
−
tan , , ,
tan ,
1
1
10 0 1
1
x yxy
x y xy
x yxy
if
ifπ xx y xy
x yxy
x y xy
> < < −
− +−
+
< > < −
−
0 0 1
10 0 11
,
tan , ,
and
if andπ
7.
sin sin
sin , ,
− −
−
+ =
− + −{ } − ≤ ≤ + ≤
<1 1
1 2 2 2 21 1 1 1 1
x y
x y y x x y x y
xy
if and
if 00 11 1 0 1 1
2 2
1 2 2 2 2and
or
if andx y
x y y x x y x y+ >
− − + −{ } < ≤ + >
− −
−π
π
sin , ,
siin , ,− − + −{ } − ≤ < + ≥
1 2 2 2 21 1 1 0 1x y y x x y x yif and
sin sin
sin , ,
− −
−
− =
− − −{ } − ≤ ≤ + ≤
>1 1
1 2 2 2 21 1 1 1 1
x y
x y y x x y x y
xy
if and
if 00 11 1 0 1 1 0
2 2
1 2 2 2 2
andif and
orx y
x y y x x y x y+ >
− − − −{ } < ≤ − ≤ ≤ + >−π sin , , 11
1 1 1 0 0 1 11 2 2 2 2− − − − −{ } − ≤ < < ≤ + >
−π sin , ,x y y x x y x yif and
8.
cos coscos , ,
cos
− −
−
−+ =
− − −{ } − ≤ ≤ + ≥
−
1 1
1 2 21 1 1 1 0
2x y
xy x y x y x yif and
π 11 2 21 1 1 1 0xy x y x y x y− − −{ } − ≤ ≤ + ≤
, ,if and
cos coscos , ,
cos
− −
−
−− =
+ − −{ } − ≤ ≤ ≤
− +
1 1
1 2 2
1
1 1 1 1x y
xy x y x y x y
xy
if and
11 1 1 0 0 12 2− −{ } − ≤ ≤ < ≤ ≥
x y y x x y, ,if and
| JUNE ‘1722
9.
2
2 1 12
12
2 1 12
11
1 2
1 2sin
sin ( ) ,
sin ( ) ,−
−
−=
− − ≤ ≤
− − ≤ ≤
−
x
x x x
x x x
if
ifπ
π −− − − ≤ ≤ −
−sin ( ) ,1 22 1 1 12
x x xif
3
3 4 12
12
3 4 12
11
1 3
1 3sin
sin ( ) ,
sin ( ) ,−
−
−=
− − ≤ ≤
− − < ≤
−
x
x x x
x x x
if
ifπ
π −− − − ≤ < −
−sin ( ) ,1 33 4 1 12
x x xif
10.
22 1 0 1
2 2 1 1 01
1 2
1 2cos
cos ( ),
cos ( ),−
−
−=
− ≤ ≤
− − − ≤ ≤
x
x x
x x
if
ifπ
3
4 3 12
1
2 4 3 12
12
2
1
1 3
1 3cos
cos ( ) ,
cos ( ) ,−
−
−=
− ≤ ≤
− − − ≤ ≤x
x x x
x x x
if
ifπ
ππ + − − ≤ ≤ −
−cos ( ) ,1 34 3 1 12
x x xif
11.
2
21
1 1
21
1
12
12tan
tan ,
tan ,−
−
−=
−
− < <
+−
>x
xx
x
xx
x
if
ifπ 11
21
112− +
−
< −
−π tan ,xx
xif
3
31 3
13
13
31 3
1
13
2
13
2tan
tan ,
tan−
−
−=
−−
− < <
+ −−
x
x xx
x
x xx
if
π
>
− + −−
< −
−
,
tan ,
if
if
x
x xx
x
13
31 3
13
13
2π
12.
2
21
1 1
21
1
12
12tan
sin ,
sin ,−
−
−=
+
− ≤ ≤
−+
>x
xx
x
xx
x
if
ifπ 11
21
112− −
+
< −
−π sin ,xx
xif
2
11
0
11
1
12
2
12
2
tancos ,
cos ,
−
−
−
=
−+
≤ < ∞
− −+
x
xx
x
xx
if
if −− ∞ < ≤
x 0
13. sin cos tan− − −= −( ) =
−
1 1 2 12
11
x x x
x= −
=−
=
− − −cot sec12
12
11 1
11x
x x xcosec
cos sin tan− − −= −( ) = −
1 1 2 1 21 1x x x
x=
−
=
=−
− − −cot sec12
1 121
1 1
1x
x x xcosec
tan sin cos cot sec− − − −=+
=+
=
=1 12
12
1
1
1
11x
x
x x x−− −+( ) = +
1 2 1 21 1x x
xcosec
14.
If x1, x2, ... , xn ∈ R, then tan tan . . . tan tan
. . .− − − −+ + + =− + − +
− + −1
11
21 1 1 3 5 7
2 4 61x x x
S S S SS S Sn ++
. . .
where Sk = Sum of the products of x1, x2, ... , xn taken k at a time.
| JUNE ‘17 23
WORK IT OUTVERY SHORT ANSWER TYPE
1. Find the value of cot π4
2 31−
−cot .
2. If tan− =1 43
θ , �nd the value of cos .
3. Find the principal value of sin sin−
1 23π .
4. For the principal values, evaluate the following:
cot ( ) tan ( ) sec− − −− + +
1 1 13 1 23
5. Evaluate : cos cos( )− − °( )1 680
SHORT ANSWER TYPE
6. Evaluate : (i) sin (cot–1 x) (ii) cos (tan–1 x)
7. Prove that sin sin cos .− − −+ =1 1 1817
35
3685
8. Solve for x : cos(tan–1 x) = sin sec .−
1 1312
9. Solve : sin cos− −= +1 16
x xπ
10. If in a ABC, A = tan–12 and B = tan–13, then
show that C is equal to π4
.
LONG ANSWER TYPE-I
11. Write each of the following in the simplest form:
(i) tan–1(sec x + tan x) (ii) sin tan2 11
1− −+
xx
12. Show that 2 11
11
1 12
2tan sin− −+−
+ −
+
=x
xxx
π
13. Evaluate the following :
(i) sin (2 sin–1 0.8) (ii) tan tan2 15 4
1−
−
π
14. Find the value of
tan tan cos cos sec sec− − −
+
+
1 1 156
136
95
π π π
15. If tan ,− + − −
+ + −
=1
2 2
2 2
1 1
1 1
x x
x xα then prove that
x2 = sin 2 a.
LONG ANSWER TYPE-II
16. If a, b, c > 0 such that a + b + c = abc, �nd the value of tan–1 a + tan–1 b + tan–1 c.
17. If cos cos ,− −+ =1 1xa
yb
α prove that
xa
xyab
yb
2
2
2
222− + =cos sin .α α
18. If a1, a2, a3, ..., an is an arithmetic progression with common di�erence d, then show that
tan tan tan− −+
++
1
1 2
1
2 31 1da a
da a
++
+ ++
−
−tan ...1
3 4 11 1da a
da an n
is equal to a a
a an
n
−+
1
11.
19. Solve :
(i) cos sin cos− −+
=1 11
30x
(ii) tan tan tan ( )− − −+−
+ −
= + −1 1 111
1 7xx
xx
π
20. Prove that
(i) sin cos tan− − −+ + =1 1 11213
45
6316
π
(ii) cos tan tan− − −+ =1 1 145
35
2711
SOLUTIONS
1. cot cot cot cotπ π4
2 34
3 12 3
1 12
−
= − −
×
− −
= −
=
⋅ +
−=
+
−=−cot cot
cot
cot
ππ
π443
443
1
43 4
43
1
43
171
2. Given, tan− =1 43
θ , where θ π π∈ −
2 2
,
\ tan .θ = 43
We know that cos > 0, when θ π π∈ −
2 2
, .
\ cossec tan
θθ θ
= =+
=+
=1 1
1
1
1 169
352
| JUNE ‘1724
3. Since, sin sin sin23 3 3π π π π= −
=
\ sin sin sin sin .− −
=
=1 12
3 3 3π π π
Hence, the principal value of sin–1 sin 23π
is π
3.
4. cot ( ) tan ( ) sec− − −− + +
1 1 13 1 23
=
+
+
− − −cot cot tan tan sec sec1 1 156 4 6π π π
= + + =56 4 6
54
π π π π
5. cos cos( ) cos (cos )− −− °{ } = °1 1680 680
= cos cos cos cos− −
= −
1 1349
4 29
π π π
= cos cos−
= = °1 2
929
40π π
6. (i) We have, sin (cot–1 x) = sin cot .−
11x
Now, cot sin− −=+
1 1
2
1
1x
x
Hence, sin (cot–1 x) = sin sin−
+
=+
12 2
1
1
1
1x x
(ii) cos (tan–1 x) = cos tan−
11x
Now, tan cos− −=+
1 1
2
1
1x
x
Hence, cos (tan–1 x) = cos cos−
+
=+
12 2
1
1
1
1x x
7. L.H.S. = sin sin− −+1 1817
35
= cos cos− −+1 11517
45
= cos− ×
− −
⋅ −
1
2 21517
45
1 1517
1 45
= cos cos− −− ⋅
= − ×{ }1 11217
64289
925
1217
817
35
= cos cos− −−{ } =
=1 112
172485
3685
R.H.S.
Hence, sin sin cos .− − −+ =1 1 1817
35
3685
8. Let tan–1 x = f. �en, tanφ = x1
\ cos cosφ φ=+
⇒ =+
−1
1
1
121
2x x
So, cos(tan–1x) = cos cos−
+
=
+
12 2
1
1
1
1x x
Also, let sec .− =1 1312
θ �en, sec θ = 1312
\ sin sinθ θ= ⇒ = −513
513
1
So, sin sec sin sin− −
=
=1 113
125
135
13
�us, 1
1
5132+
=x
⇒ 11
251692( )+
=x
⇒ 1 + x2 = 16925
16925
12⇒ = −x ⇒ x = ± 14425
Hence, x = ± 125
.
9. We have, sin cos− −= +1 16
x xπ
⇒ sin cos− −− =1 16
x x π
⇒ sin sin− −− −
=1 1
2 6x xπ π
⇒ 2 23
1sin− =x π ⇒ sin− =13
x π ⇒ x = 32
10. We have, A = tan–12, B = tan–13 We know that, A + B + C = ⇒ tan–1 2 + tan–1 3 + C =
⇒ π π+ +− ×
+ ∠ =−tan 1 2 3
1 2 3C
⇒ + tan–1(–1) + C = ⇒ π π π− + ∠ =4
C
⇒ 34π π+ ∠ =C ⇒ ∠ =C π
4
11. (i) sec tan sincos
cos
sinx x x
x
x
x+ = + =
− +
+
1 12
2
π
π
=+
+
+
= +
24 2
24 2 4 2
4 2
2sin
sin costan
π
π ππ
x
x xx
| JUNE ‘17 25
| JUNE ‘1726
\ tan–1 tan π π4 2 4 2
+
= +x x
(ii) Put x = cos
\ sin tan sin tan coscos
2 11
2 11
1 1− −−+
= −
+
xx
θθ
=
−sin tan sin /
cos /2 2 2
2 21
2
2θθ
=
= = −−sin tan tan sin22
11 2θ θ x
12. Given expression is 2 11
11
1 12
2tan sin− −+−
+ −
+
xx
xx
Putting x = tan , we get
2 11
11
1 12
2tan tantan
sin tantan
− −+−
+ −
+
θθ
θθ
= 2 4
14
21 1tantan tan
tan tansin (cos )− −
+
− ⋅
+
π θ
π θθ
= 24 2
21 1tan tan sin sin− −+
+ −
π θ π θ
= 24 2
22
22
2π θ π θ π θ π θ π+
+ − = + + − =
13. (i) sin (2 sin–10.8) = sin sin . .− − ( )
1 22 0 8 1 0 8
= sin[sin–1(2 × 0.8 × 0.6)] = sin[sin–1(0.96)] = 0.96
(ii) tan tan tan tan2 15 4
2 15
1 15
1 12
− −−
=
−
π
−
π4
= tan tan tan tan tan− − −−
= −
1 1 1512 4
512
1π
=−
+ ×
= −
− −tan tan tan tan1 1
512
1
1 512
1
717
= − 7
17
14. tan tan cos cos sec sec− − −
+
+
1 1 156
136
95
π π π
⇒ tan tan cos cos
sec se
− −
−
−
+ +
+
1 1
1
62
6π π π π
cc 25
π π−
⇒ tan tan cos cos
sec sec
− −
−
−
+
+
1 1
1
6 6
5
π π
π
= − + + =π π π π6 6 5 5
.
15. We have, tan− + − −
+ + −
=1
2 2
2 2
1 1
1 1
x x
x xα
⇒ 1 1
1 1
2 2
2 2
+ − −
+ + −=x x
x xtan α
⇒ 1 1 1 1
1 1 1 1
2 2 2 2
2 2 2 2
+ − −( ) + + + −( )+ − −( ) − + + −( )
x x x x
x x x x =
+−
tantan
αα
11
⇒ 2 1
2 1
11
1
1
11
2
2
2
2
+
− −=
+−
⇒ −
+=
−+
x
x
x
x
tantan
tantan
αα
αα
⇒ 11
2
2−+
=−+
xx
cos sincos sin
α αα α
⇒ 11
1 21 2
22
22−
+=
−+
⇒ =xx
xsinsin
sinαα
α
16. It is given that a + b + c = abc
\ abcc
ac
bc
= + +1
⇒ ab ac
bc
= + +
1 ⇒ ab a b
c− = +1
⇒ ab – 1 > 0 a b c a bc
, , > ∴ + >
0 0 ⇒ ab > 1 Now, tan–1 a + tan–1 b + tan–1 c
= π + +−
+− −tan tan1 1
1a b
abc [ ab > 1]
= π + −−
+− −tan tan1 1
1abc c
abc
= π +− −
−
+− −tan
( )tan1 11
1c ab
abc
= + tan–1 (–c) + tan–1 c = – tan–1 c + tan–1 c =
| JUNE ‘17 27
17. We have cos cos− −+ =1 1xa
yb
α
⇒ cos− ⋅ − −
−
=12
2
2
21 1xa
yb
xa
y
bα
⇒ xyab
xa
y
b
x y
a b− − − + =1
2
2
2
2
2 2
2 2 cos α
⇒ xyab
xa
y
b
x y
a b− = − − +cos α 1
2
2
2
2
2 2
2 2
Squaring both sides, we have
x y
a b
xyab
xa
y
b
x y
a b
2 2
2 22
2
2
2
2
2 2
2 22
1− + = − − +cos cosα α
⇒ xa
xyab
y
b
2
2
2
222
1− + = −cos cosα α
\ xa
xyab
y
b
2
2
2
222
− + =cos sinα α
18. Given, a1, a2, a3, a4, ... , an is an arithmetic progression, then d = a2 – a1 = a3 – a2 = ... = an –an – 1
\ tan tan tan− −+
++
1
1 2
1
2 31 1da a
da a
++
+ ++
− −
−tan ... tan1
3 4
1
11 1da a
da an n
= tan tan tan− −−+
+−
+
1 2 1
1 2
1 3 2
2 31 1a a
a aa a
a a
+−
+
+ +−
+
− − −
−tan ... tan1 4 3
3 4
1 1
11 1a a
a aa a
a an n
n n
= tan [tan–1 a2 – tan–1 a1 + tan–1 a3 – tan–1 a2 + ... + tan–1an– tan–1 an – 1] = tan[tan–1 an – tan–1 a1]
= tan tan .− −+
=
−+
1 1
1
1
11 1a a
a aa a
a an
n
n
n
19. (i) Here, cos sin cos cos− −+
= = ±
1 113
02
x π
\ sin cos− −+ = ±1 113 2
x π
i.e., cos sin− −= ± −1 12
13
x π
\ x = ± −
−cos sinπ2
13
1
x = −
−cos sinπ2
13
1 x = − −
−cos sinπ2
13
1
=
=−sin sin 1 1
313
= +
−cos sinπ2
13
1
= −
= −−sin sin 1 1
313
\ x = ± 13
(ii) We have,
tan tan tan ( )− − −+
−
+ −
= + −1 1 111
1 7xx
xx
π
⇒
+−
+ −
− +−
−
= + −− −tan tan ( )1 1
11
1
1 11
1 7
xx
xx
xx
xx
π
\ ( ) ( )( ) ( )( )
tan tan ( )x x xx x x x
+ + −− − + −
= + − −1 11 1 1
72
1π
⇒ x2 + x + x2 – 2x + 1 = –7(x2 – x – x2 + 1) ⇒ 2x2 – 8x + 8 = 0 i.e., x2 – 4x + 4 = 0 ⇒ (x – 2)2 = 0 \ x = 2
20. (i) L.H.S. = sin cos tan− − −+ +1 1 11213
45
6316
= tan tan tan− − −+ +1 1 1125
34
6316
= ++
− ×
+− −π tan tan1 1
125
34
1 125
34
6316
= π + −
+− −tan tan1 163
166316
= π π− + =− −tan tan1 16316
6316
(ii) Let cos .− =1 45
θ �en cos θ = 45
.
\ tan coscos
tanθ θθ
θ= − = ⇒ = −1 34
34
21
Consequently, cos tan− −=1 145
34
\ cos tan tan tan− − − −+ = +1 1 1 145
35
34
35
= tan tan− −+
− ⋅
=1 1
34
35
1 34
35
2711
| JUNE ‘1728
Only One Option Correct Type
1. �e number of positive integral solutions of
tan cos sin− − −++
=1 12
1
1310
x y
y is
(a) 0 (b) 1 (c) 2 (d) > 2
2. If cos ,− −+
<1
2
211 3
xx
p then x belongs to the interval
(a) −
13
13
, (b) −
13
13
,
(c) 0 13
,
(d) none of these
3. cos cos− −
1 1715
p is equal to
(a) 1715
p (b) 1315
p
(c) 315p
(d) −1715
p
4. �e domain of the function
f x
xx( ) cos [log( )]=
−
+ −− −1 12
43 is
(a) [–6, 6] (b) [–5, 2) ∪ (2, 3)(c) (2, 3) (d) [–6, 2) ∪ (2, 3)
5. If q are f are the roots of the equation 8x2 + 22x + 5 = 0, then
(a) both sin–1 q and sin–1 f are real(b) both sec–1 q and sec–1 f are real(c) both tan–1 q and tan–1 f are real(d) none of these
6. If sin–1 x + sin–1 y + sin–1z = 32p , then the value of
25 (x + y + z) −+ +
2163 3 3( )x y z
must be
(a) 1 (b) 2(c) 3 (d) none of these
One or More Than One Option(s) Correct Type
7. If cos–1x + (sin–1y)2 = pp2
4and
(cos )(sin )− −1 1 2x y
= p2
16, then
(a) 0 4 1≤ ≤ +pp
(b) p = 2 is the integral value of p(c) p = 0, 1, 2 (integral values)(d) none of these
8. For 0 < f < p2
, if x = cos2
0
n
nf
=
∞
∑ , y = sin2
0
n
nf
=
∞
∑ ,
z = cos sin2 2
0
n n
nf f
=
∞
∑ , then
(a) xyz = xz + y (b) xyz = xy + z(c) xyz = x + y + z (d) xyz = yz + x
9. For the equation 2x = tan(2 tan–1a) + 2 tan(tan–1a + tan–1a3), which of the following is invalid?
(a) a2x + 2a = x (b) a2 + 2ax + 1 = 0(c) a ≠ 0 (d) a ≠ –1, 1
10. tan tan− −−+
+ −+
1 1
1
1 2 1
1 2 1a x ya y x
a aa a + −
+
−tan 1 3 2
2 3 1a aa a
+ + −+
− −
−...... tan 1 1
1 1a aa a
n n
n n+ −tan 1 1
anis equal to
Total Marks : 80 Time Taken : 60 Min.
This specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four
marks for correct answer and deduct one mark for wrong answer.
Self check table given at the end will help you to check your readiness.
Class XII
Inverse Trigonometric Functions
| JUNE ‘17 29
| JUNE ‘1730
(a) tan–1 xy (b) tan–1 xy
(c) tan–1 yx
(d) none of these
11. �e value of q for which
q q q
q= −
+
− −tan ( tan ) sin sin
cos,1 2 12 1
23 2
5 4 2 is/are:
(a) np + tan–1(–2) (b) np, np p+4
(c) np + cot–1(–2) (d) none of these12. If x, y and z are in A.P. and tan–1 x, tan–1 y and tan–1 z
are also in A.P. then (a) x = y = z (b) 2x = 3y = 6z(c) 6x = 3y = 2z (d) 6x = 4y = 3z
13. In DABC, ∠ =C p2
and
sin sin sin− − −=
+
1 1 1x axc
bxc
where a, b and c are the sides of triangle, then the values of x is/are(a) 0 (b) 1 (c) 2 (d) –1
Comprehension Type
tan tan tan− −
−=
− −−
=
−+
= −( )∑ ∑1 1
11
1 11
11x x
x xx xr r
r rr
n
r rr
n
= tan tan ,− −− ∀ ∈1 10x x n Nn
On the basis of above information, answer the following questions:
14. �e value of cosec–1 5 + cosec–1 65 +
cosec–1 ( )325 + .... to ∞ is
(a) p (b) 34p (c) p
2 (d) p
415. �e sum to in�nite terms of the series
tan tan− −
− +
+− +
12 4
12 4
21 1 1
41 2 2
+− +
+−tan ...12 46
1 3 3is
(a) p4
(b) p2
(c) 34p (d) none of these
Matrix Match Type16. Match the following :
Column I Column II
P. If 2 tan–1(2x + 1) = cos–1 (–x), then x is
1. − 12
Q. If 2 cos–1x = sin–1 2 1 2x x−( ),then x is
2. 32
R. If tan tan− −−−
+ ++
1 112
12
xx
xx
= p4
, then isx
3. –1
4. 0
P Q R(a) 2 1 3(b) 3 2 1(c) 3 1 2(d) 4 2 1
Integer Answer Type
17. �e number of solutions for the equation
2 1 32
1 2 1 2sin ( ) cos ( )− −− + + − =x x x x p is
18. If tan q + tan p q3
+
+ tan − +
p q3 = a tan 3q,
then a is equal to
19. If sin–1x + sin–1y + sin–1z = 32p and f(1) = 2,
f(p + q) = f(p). f(q), ∀ p, q ∈ R, then
x f(1) + y f(2) + z f(3) −+ +
+ +( )
( ) ( ) ( )
x y zx y zf f f1 2 3 is equal to
20. If λ = tan 2 15 4
1tan ,− −
p then the value of –17 λ2 must be
Keys are published in this issue. Search now!
Check your score! If your score is> 90% EXCELLENT WORK ! You are well prepared to take the challenge of final exam.
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< 60% NOT SATISFACTORY! Revise thoroughly and strengthen your concepts.
No. of questions attempted ……
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Marks scored in percentage ……
| JUNE ‘17 31
CATEGORY-I (Q. 1 to Q. 50)Only one answer is correct. Correct answer will fetch full marks 1. Incorrect answer or any combination of more than one answer will fetch –¼ marks. No answer will fetch 0 marks.
1. �e number of all numbers having 5 digits, with distinct digits is(a) 99999 (b) 9 × 9P4 (c) 10P5 (d) 9P4
2. �e greatest integer which divides (p + 1)(p + 2) (p + 3).....(p + q) for all p∈N and �xed q ∈N is(a) p! (b) q! (c) p (d) q
3. Let (1 + x + x2)9 = a0 + a1x + a2x2 + ..... + a18x18. �en(a) a0 + a2 + ..... + a18 = a1 + a3 + ... + a17(b) a0 + a2 + ..... + a18 is even(c) a0 + a2 + ..... + a18 is divisible by 9(d) a0 + a2 + ..... + a18 is divisible by 3 but not by 9
4. �e linear system of equations 8 3 5 05 8 3 03 5 8 0
x y zx y zx y z
− − =− + =+ − =
has(a) only zero solution(b) only �nite number of non-zero solutions(c) no non-zero solution(d) in�nitely many non-zero solutions
5. Let P be the set of all non-singular matrices of order 3 over R and Q be the set of all orthogonal matrices of order 3 over R. �en(a) P is proper subset of Q(b) Q is proper subset of P(c) Neither P is proper subset of Q nor Q is proper
subset of P(d) P ∩ Q = f, the void set
6. Let A = x x
x Bx
x+
+
= +
2 33 2
05 2, . �en all
solutions of the equation det (AB) = 0 is(a) 1, –1, 0, 2 (b) 1, 4, 0, –2(c) 1, –1, 4, 3 (d) –1, 4, 0, 3
7. �e value of det A, where
A = 1 0
11 1
coscos cos
cos
θθ θ
θ−
− −
lies
(a) in the closed interval [1, 2](b) in the closed interval [0, 1](c) in the open interval (0, 1)(d) in the open interval (1, 2)
8. Let f : R → R be such that f is injective and f(x)f(y) = f(x + y), ∀x, y ∈R. If f (x), f (y), f (z) are in G.P., then x, y, z are in(a) A.P. always(b) G.P. always(c) A.P. depending on the value of x, y, z(d) G.P. depending on the value of x, y, z
9. On the set R of real numbers we de�ne xPy if and only if xy ≥ 0. �en the relation P is (a) re�exive but not symmetric(b) symmetric but not re�exive(c) transitive but not re�exive(d) re�exive and symmetric but not transitive
10. On R, the relation be de�ned by 'x y holds if and only if x – y is zero or irrational'. �en(a) is re�exive and transitive but not symmetric.(b) is re�exive and symmetric but not transitive.(c) is symmetric and transitive but not re�exive.(d) is equivalence relation
By : Anil Kumar Gupta (akg Classes), Asansol (W.B.) Mob : 09832230099
| JUNE ‘1732
11. Mean of n observations x1, x2, ....., xn is x . If an observation xq is replaced by xq then the new mean is
(a) x – xq + xq (b) ( )n x x
nq− + ′1
(c) ( )n x x
nq− − ′1
(d) nx x x
nq q− + ′
12. �e probability that a non leap year selected at random will have 53 Sundays is(a) 0 (b) 1/7 (c) 2/7 (d) 3/7
13. �e equation sin x(sin x + cos x) = k has real solutions, where k is a real number. �en
(a) 01 2
2≤ ≤
+k (b) 2 3 2 3− ≤ ≤ +k
(c) 0 2 3≤ ≤ −k (d) 1 22
1 22
− ≤ ≤ +k
14. �e possible values of x, which satisfy the trigonometric equation
tan tan− −−−
+++
=1 112
12 4
xx
xx
π are
(a) ± 12
(b) ± 2 (c) ± 12
(d) ± 2
15. Transforming to parallel axes through a point (p, q), the equation 2x2 + 3xy + 4y2 + x + 18y + 25 = 0 becomes 2x2 + 3xy + 4y2 = 1. �en(a) p = –2, q = 3 (b) p = 2, q = –3(c) p = 3, q = –4 (d) p = –4, q = 3
16. Let A(2, –3) and B(–2, 1) be two angular points of ABC. If the centroid of the triangle moves on the
line 2x + 3y = 1, then the locus of the angular point C is given by(a) 2x + 3y = 9 (b) 2x – 3y = 9(c) 3x + 2y = 5 (d) 3x – 2y = 3
17. �e point P(3, 6) is �rst re�ected on the line y = x and then the image point Q is again re�ected on the line y = –x to get the image point Q . �en the circumcentre of the PQQ is(a) (6, 3) (b) (6, –3) (c) (3, –6) (d) (0, 0)
18. Let d1 and d2 be the lengths of the perpendiculars drawn from any point of the line 7x – 9y + 10 = 0 upon the lines 3x + 4y = 5 and 12x + 5y = 7 respectively. �en(a) d1 > d2 (b) d1 = d2 (c) d1 < d2 (d) d1 = 2d2
19. �e common chord of the circles x2 + y2 – 4x – 4y = 0 and 2x2 + 2y2 = 32 subtends at the origin an angle equal to
(a) π3
(b) π4
(c) π6
(d) π2
20. �e locus of the mid-points of the chords of the circle x2 + y2 + 2x – 2y – 2 = 0 which make an angle of 90° at the centre is(a) x2 + y2 – 2x – 2y = 0(b) x2 + y2 – 2x + 2y = 0(c) x2 + y2 + 2x – 2y = 0(d) x2 + y2 + 2x – 2y – 1 = 0
21. Let P be the foot of the perpendicular from focus
S of hyperbola xa
yb
2
2
2
2 1− = on the line bx – ay = 0
and let C be the centre of the hyperbola. �en the area of the rectangle whose sides are equal to that of SP and CP is(a) 2ab (b) ab
(c) ( )a b2 2
2+ (d)
ab
22. B is an extremity of the minor axis of an ellipse whose foci are S and S . If SBS is a right angle, then the eccentricity of the ellipse is
(a) 12
(b) 12
(c) 23
(d) 13
23. �e axis of the parabola x2 + 2xy + y2 – 5x + 5y – 5 = 0 is(a) x + y = 0 (b) x + y – 1 = 0
(c) x y− + =1 0 (d) x y− = 12
24. �e line segment joining the foci of the hyperbola x2 – y2 + 1 = 0 is one of the diameters of a circle. �e equation of the circle is(a) x2 + y2 = 4 (b) x y2 2 2+ =(c) x2 + y2 = 2 (d) x y2 2 2 2+ =
25. �e equation of the plane through (1, 2, –3) and (2, –2, 1) and parallel to X-axis is(a) y – z + 1 = 0 (b) y – z – 1 = 0(c) y + z – 1 = 0 (d) y + z + 1 = 0
26. �ree lines are drawn from the origin O with direction cosines proportional to (1, –1, 1), (2, –3, 0) and (1, 0, 3). �e three lines are(a) not coplanar (b) coplanar
| JUNE ‘17 33
(c) perpendicular to each other(d) coincident
27. Consider the non-constant di�erentiable function f of one variable which obeys the relation f xf y
( )( )
= f(x – y). If f (0) = p and f (5) = q, then f (–5) is
(a) pq
2 (b) q
p (c) p
q (d) q
28. If f(x) = log5 log3 x, then f (e) is equal to(a) e loge 5 (b) e loge3
(c) 15e elog
(d) 13e elog
29. Let F(x) = ex, G(x) = e–x and H(x) = G(F(x)), where
x is a real variable. �en dHdx
at x = 0 is
(a) 1 (b) –1 (c) − 1e
(d) –e
30. If f ″(0) = k, k ≠ 0, then the value of
lim ( ) ( ) ( )x
f x f x f xx→
− +0 2
2 3 2 4 is
(a) k (b) 2k (c) 3k (d) 4k
31. If y = emsin–1x, then (1 – x2) d ydx
x dydx
ky2
2 − − = 0, where k is equal to(a) m2 (b) 2 (c) –1 (d) –m2
32. �e chord of the curve y = x2 + 2ax + b, joining the points where x = a and x = , is parallel to the tangent to the curve at abscissa x =
(a) a b+2
(b) 23
a b+ (c) 23
α β+ (d) α β+2
33. Let f (x) = x13 + x11 + x9 + x7 + x5 + x3 + x + 19. �en f (x) = 0 has(a) 13 real roots(b) only one positive and only two negative real
roots(c) not more than one real root(d) has two positive and one negative real root
34. Let f(x) = x
xx
x
p
q(sin ),
0
02
0
if
, if
< <
=
π, (p, q ∈ R). �en
Lagrange's mean value theorem is applicable to f (x) in closed interval [0, x],(a) for all p, q (b) only when p > q(c) only when p < q (d) for no value of p, q
35. lim(sin ) tan
x
xx→0
2
(a) is 2 (b) is 1(c) is 0 (d) does not exist
36. cos(log )x dx∫ = F(x) + c, where c is an arbitrary constant. Here F(x) = (a) x[cos(log x) + sin(log x)](b) x[cos(log x) – sin(log x)]
(c) x2
[cos(log x) + sin(log x)]
(d) x2
[cos(log x) – sin(log x)]
37. xx x
dx2
4 21
3 1−
+ +(x > 0) is
(a) tan− +
+1 1x
xc (b) tan− −
+1 1x
xc
(c) loge
xx
xx
c+ −
+ ++
1 1
1 1 (d) loge
xx
xx
c− −
− ++
1 1
1 1
38. Let I = sin
.x
xdx
1 810
19
+∫ �en,
(a) I < 10–9 (b) I < 10–7
(c) I < 10–5 (d) I > 10–7
39. Let I1 = [ ] { } ,x dx I x dxn n
and 20 0
=∫ ∫ where [x] and
{x} are integral and fractional parts of x and n∈N – {1}. �en I1/I2 is equal to
(a) 11n −
(b) 1n
(c) n (d) n – 1
40. �e value of lim .....n
nn
nn n→∞ +
++
+ +
2 2 2 21 2
12
is
(a) nπ4
(b) π4
(c) π4n
(d) π2n
41. �e value of the integral e dxx2
0
1
∫(a) is less than 1(b) is greater than 1(c) is less than or equal to 1(d) lies in the closed interval [1, e]
| JUNE ‘1734
42. e dxx x− =∫ [ ]
0
100
(a) e100 1100
− (b) ee
100 11−
−
(c) 100(e – 1) (d) e −1100
43. Solution of (x + y)2 dydx
= a2 ('a' being a constant) is
(a) ( ) tan ,x ya
y ca
c+ = + is an arbitrary constant
(b) xy = a tan cx, c is an arbitrary constant
(c) xa
yc
c= tan , is an arbitrary constant
(d) xy = tan (x + c), c is an arbitrary constant
44. �e integrating factor of the �rst order di�erential equation
x2(x2 – 1) dydx
+ x(x2 + 1)y = x2 – 1 is
(a) ex (b) xx
− 1 (c) xx
+ 1 (d) 12x
45. In a G.P. series consisting of positive terms, each term is equal to the sum of next two terms. �en the common ratio of this G.P. series is
(a) 5 (b) 5 12− (c) 5
2 (d) 5 1
2+
46. If (log5 x)(logx 3x)(log3x y) = logx x3, then y equals(a) 125 (b) 25 (c) 5/3 (d) 243
47. �e expression ( )( )
11 2
+− −
ii
n
n equals
(a) –in + 1 (b) in + 1 (c) –2in + 1 (d) 1
48. Let z = x + iy, where x and y are real. �e points
(x, y) in the X-Y plane for which z iz i
+−
purely imaginary lie on(a) a straight line (b) an ellipse(c) a hyperbola (d) a circle
49. If p, q are odd integers, then the roots of the equation 2px2 + (2p + q)x + q = 0 are(a) rational (b) irrational(c) non-real (d) equal
50. Out of 7 consonants and 4 vowels, words are formed each having 3 consonants and 2 vowels. �e number of such words that can be formed is(a) 210 (b) 25200 (c) 2520 (d) 302400
CATEGORY-II (Q. 51 to Q. 65)Only one answer is correct. Correct answer will fetch full marks 2. Incorrect answer or any combination of more than one answer will fetch –½ marks. No answer will fetch 0 marks.
51. Let A = 1 1 10 1 10 0 1
�en for positive integer n, An is
(a) 100 0
2
2n n
n nn
(b)
1 12
0 10 0 1
n n n
n
+
(c) 100 0
2
2
2
n nn n
n
(d)
1 2 1
0 12
0 0 12
2
n nn n
n
−+
+
52. Let a, b, c be such that b(a + c) ≠ 0.
If a a ab b b
c c c
a b ca b c
a bn n
+ −− + −
− ++
+ + −− − +
− −+ +
1 11 11 1
1 1 11 1 1
1 12 1( ) ( ) (−−1)nc= 0
then the value of n is(a) any integer (b) zero(c) any even integer (d) any odd integer
53. On set A = {1, 2, 3}, relations R and S are given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)} S = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)}. �en
(a) R ∪ S is an equivalence relation(b) R ∪ S is reflexive and transitive but not
symmetric(c) R ∪ S is reflexive and symmetric but not
transitive(d) R ∪ S is symmetric and transitive but not
re�exive54. If one of the diameters of the curve x2 + y2 – 4x – 6y
+ 9 = 0 is a chord of a circle with centre (1, 1), the radius of the circle is(a) 3 (b) 2 (c) 2 (d) 1
55. Let A(–1, 0) and B(2, 0) be two points. A point M moves in the plane in such a way that
MBA = 2 MAB. �en the point M moves along(a) a straight line (b) a parabola(c) an ellipse (d) a hyperbola
56. If f(x) = | |−∫ t dtx
,1
then for any x ≥ 0, f(x) is equal to
(a) 12
(1 – x2) (b) 1 – x2
(c) 12
(1 + x2) (d) 1 + x2
| JUNE ‘17 35
57. Let for all x > 0, f(x) = lim ,n
nn x→∞
−( )1
1 then
(a) f(x) + fx1
=1 (b) f(xy) = f(x) + f(y)
(c) f(xy) = xf(y) + yf(x) (d) f(xy) = xf(x) + yf(x)
58. Let I = ( cos )1 20
100
−∫ xπ
dx, then
(a) I = 0 (b) I = 200 2(c) I = π 2 (d) I = 100
59. �e area of the �gure bounded by the parabolas x = –2y2 and x = 1 – 3y2 is
(a) 43
square units (b) 23
square units
(c) 37
square units (d) 67
square units
60. Tangents are drawn to the ellipse x y2 2
9 51+ =
at the ends of both latusrectum. �e area of the quadrilateral so formed is
(a) 27 sq. units (b) 132
sq. units
(c) 154
sq. units (d) 45 sq. units
61. �e value of K in order that f(x) = sin x – cos x – Kx + 5 decreases for all positive real values of x is given by(a) K < 1 (b) K ≥ 1 (c) K > 2 (d) K < 2
62. For any vector x, the value of ( ) ( ) ( )� � � � � �x i x j x k× + × + ×2 2 2 is equal to(a) | |x 2 (b) 2| |x 2 (c) 3| |x 2 (d) 4| |x 2
63. If the sum of two unit vectors is a unit vector, then the magnitude of their di�erence is(a) 2 units (b) 2 units(c) 3 units (d) 5 units
64. Let a and be the roots of x2 + x + 1 = 0. If n be positive integer, then an + n is
(a) 2 23
cos nπ (b) 2 23
sin nπ
(c) 23
cos nπ (d) 23
sin nπ
65. For real x, the greatest value of x xx x
2
22 4
2 4 9+ ++ +
is
(a) 1 (b) –1 (c) 12
(d) 14
CATEGORY-III (Q. 66 to Q. 75)One or more answer(s) is (are) correct. Correct answer(s) will fetch full marks 2. Any combination containing one or more incorrect answer will fetch 0 marks. Also no answer will fetch 0 marks. If all correct answers are not marked and also no incorrect answer is marked then score = 2 × number of correct answers marked ÷ actual number of correct answers.
66. If a, b ∈ {1, 2, 3} and the equation ax2 + bx + 1 = 0 has real roots, then(a) a > b(b) a ≤ b(c) number of possible ordered pairs (a, b) is 3(d) a < b
67. If the tangent to y2 = 4ax at the point (at2, 2at) where |t| > 1 is a normal to x2 – y2 = a2 at the point (a sec , a tan ), then(a) t = –cosec (b) t = –sec (c) t = 2 tan (d) t = 2 cot
68. �e focus of the conic x2 – 6x + 4y + 1 = 0 is(a) (2, 3) (b) (3, 2) (c) (3, 1) (d) (1, 4)
69. Let f : R → R be twice continuously. Let f(0) = f(1) = f (0) = 0. �en(a) f (x) ≠ 0 for all x(b) f (c) ≠ 0 for some c ∈R(c) f (x) ≠ 0 if x ≠ 0(d) f (x) > 0 for all x
70. If f(x) = xn, n being a non-negative integer, then the values of n for which f (a + ) = f (a) + f ( ) for all a, > 0 is(a) 1 (b) 2 (c) 0 (d) 5
71. Let f be a non-constant continuous function for all x ≥ 0. Let f satisfy the relation f(x) f(a – x) = 1 for
some a ∈ R+. �en I = dxf x
a
10
+∫ ( ) is equal to
(a) a (b) a4
(c) a2
(d) f(a)
72. If the line ax + by + c = 0, ab ≠ 0, is a tangent to the curve xy = 1 – 2x, then(a) a > 0, b < 0 (b) a > 0, b > 0(c) a < 0, b > 0 (d) a < 0, b < 0
73. Two particles move in the same straight line starting at the same moment from the same point in the same direction. �e �rst moves with constant velocity u and the second starts from rest with constant acceleration f. �en
| JUNE ‘1736
(a) they will be at the greatest distance at the end
of time uf2
from the start
(b) they will be at the greatest distance at the end
of time uf
from the start
(c) their greatest distance is uf
2
2
(d) their greatest distance is uf
2
74. �e complex number z satisfying the equation |z – i| = |z + 1| = 1 is(a) 0 (b) 1 + i (c) –1 + i (d) 1 – i
75. On R, the set of real numbers, a relation is de�ned as 'a b if and only if 1 + ab > 0. �en(a) is an equivalence relation(b) is re�exive and transitive but not symmetric(c) is re�exive and symmetric but not transitive(d) is only symmetric
SOLUTIONS
1. (b) : 1st digit must be other than 0 (zero) i.e. 9 ways.\ Other 4 digits can be �lled in 9P4 ways\ Total no. of numbers having 5 (distinct) digits = 9 × 9P42. (b) : (p + 1)(p + 2)(p + 3).....(p + q) is a product of q consecutive positive integers\ It must be always divisible by q!3. (b) : (1 + x + x2)9 = a0 + a1x + a2x2 + ..... + a18x18
Putting x = 1 and –1, we get39 = a0 + a1 + a2 + ..... + a18 ...(i)1 = a0 – a1 + a2 – ..... + a18 ...(ii)Adding (i) & (ii), we get3 1
2
9 + = a0 + a2 + a4 + ..... + a18
⇒ a0 + a2 + a4 + ..... + a18 = 9842, which is even but not divisible by 3 or 9.
4. (d) : 8 3 55 8 33 5 8
0 3 50 8 30 5 8
− −−
−=
− −−
−
[C1 → C1 + C2 + C3] = 0\ Given system of equation has in�nitely many non-zero solutions.5. (b) : Every orthogonal matrix is non-singular but every non-singular matrix may or may not be orthogonal.\ Q is proper subset of P.
6. (b) : AB = x x
xx
x+
+
+
2 33 2
05 2
= x x x x
x x x
2 2
217 3 6
8 10 4 4+ ++ + +
det(AB) = x xxx x( )+++ +217 3
8 10 2 = x(x + 2)(x2 – 5x + 4) = x(x + 2)(x – 1)(x – 4)\ det(AB) = 0 ⇒ x = 1, 4, 0, –2
7. (a) : |A| =1 0
11 1
coscos cos
cos
θθ θ
θ−
− − = (1 + cos2 )
– cos (0) + 0
= 1 + cos2 ∈ [1, 2].8. (c) : f(x)⋅f(y) = f(x + y) ...(i)\ f(y)f(y) = f(y + y) ⇒ {f(y)}2 = f(2y)⇒ f(x)f(z) = f(2y) [ f(x), f(y), f(z) are in G.P.]⇒ f(x + z) = f(2y) [from (i)]⇒ x, y, z are in A.P. but will depend on the value of x, y, z.9. (d) : x2 ≥ 0 \ x⋅x ≥ 0 ⇒ xPx ⇒ Re�exive xy ≥ 0 ⇒ yx ≥ 0 ⇒ Symmetric (–5)(0) ≥ 0 & (0)(7) ≥ 0i.e., (–5, 0) ∈ P & (0, 7) ∈ PBut, (–5)(7) < 0 ⇒ (–5, 7) P\ P is not transitive.10. (b) : Here (x, y) ∈ if x – y is zero or irrational. x – x = 0 for all x ∈ R⇒ is re�exiveIf x – y is zero or irrational then y – x is also zero or irrational.⇒ is symmetric.Let (x, y) ∈ & (y, z) ∈ \ x – y = 0 or irrational & y – z = 0 or irrationalBut, their sum x – z may or may not be 0 or irrationale.g., 2 – 3 is irrational & 3 – 5 both are irrational but their sum 2 – 5 = – 3 is neither zero nor irrational⇒ is not transitive.11. (d) : Mean of n observations x1, x2, ....., xn is x .\ Sum of n observations = nxIf xq is replaced by xq then sum = nx x xq q− + ′
\ New mean = nx x x
nq q− + ′
12. (b) : A non-leap year has 52 weeks & 1 extra day
\ Prob. of 53 sundays = 17
| JUNE ‘17 37
13. (d) : We have, k = sin2 x + sinx cos x = 12
[(1 – cos 2x) + sin 2x]
= 12
12
2 12
2 12
+ ⋅ − ⋅
sin cosx x
= 12
12
2 45+ − °sin ( )x
–1 ≤ sin(2x – 45°) ≤ 1
\ 12
12
12
12
− ≤ ≤ +k
⇒ 1 22
1 22
− ≤ ≤ +k
14. (a) : tan tan tan− − −++
= − −
−
1 1 112
1 12
xx
xx
⇒ xx
xxxx
++
=− −
−
+ −−
12
1 12
1 12
⇒ xx x
++
= −−
12
12 3
⇒ 2x2 = 1 ⇒ x = ± 12
15. (b) : Putting x = x + p and y = y + q in given equation, it becomes2(x + p)2 + 3(x + p)(y + q) + 4(y + q)2 + (x + p) + 18(y + q) + 25 = 0On comparing with 2x2 + 3xy + 4y2 = 1, we get 4p + 3q + 1 = 0 ...(i) 3p + 8q + 18 = 0 ...(ii)and 2p2 + 3pq + 4q2 + p + 18q + 25 = – 1 ...(iii)On comparing (i) & (ii), p = 2, q = –3 by which (iii) is also satis�ed.
16. (a) : Centroid 2 23
3 13
− + − + +
h k,
i.e., h k3
23
, −
lies on 2x + 3y = 1
\ 23
3 23
1h k
+ −
=
⇒ 2h + 3k – 6 = 3⇒ 2x + 3y = 9 is the reqd. locus.17. (d) : P(3, 6) has re�ection on y = x as Q(6, 3)Again re�ection of Q(6, 3) on y = –x will be Q (–3, –6) Slope of PQ × slope of QQ = (–1)(1) = –1\ PQQ is a right angled with PQQ = 90°\ Circumcentre will be mid pt. of hypotenuse PQ i.e. (0, 0)18. (b) : Bisectors of 3x + 4y = 5 & 12x + 5y = 7
are 3 4 5
3 4
12 5 7
12 52 2 2 2
x y x y+ −
+= ± + −
+
A(2, –3)
B (–2, 1)
C (h, k)
⇒ 13(3x + 4y – 5) = ±5(12x + 5y – 7)⇒ 21x – 27y + 30 = 0 & 99x + 77y = 100⇒ 7x – 9y + 10 = 0 & 99x + 77y = 100 7x – 9y + 10 = 0 is one of the bisector.\ Perp. from any pt. on it will be equal to both given lines \ d1 = d2.19. (d) : Equation of common chord i.e., S1 – S2 = 0 will be⇒ (x2 + y2 – 16) – (x2 + y2 – 4x – 4y) = 0⇒ x + y = 4, which subtends 90° at (0, 0)20. (c) : Let M(h, k) be the mid point of chord of (x + 1)2 + (y – 1)2 = (2)2 ...(i)subtending 90° at centre C(–1, 1)\ 2CM2 = CB2
⇒ (h + 1)2 + (k – 1)2 = (2)
C(–1, 1)
BA M45°
45°2
(h, k)
⇒ x2 + y2 + 2x – 2y = 0 is the reqd. locus.
21. (b) : CS = ae, SP = b ae
b a
abe
b a
⋅ −
+=
+
02 2 2 2
\ CP = CS SP2 2−
= a e b a ea b
2 22 2 2
2 2−+
= a e
a b
2
2 2+\ Area of rectangle with sides SP and CP = SP⋅CP
= ab a ea b
ab⋅+
=2 2
2 2 [ b2 = a2(e2 – 1)]
22. (b) : CB = b, CS = ae SBS = 90°\ CB = CS⇒ b = ae ⇒ b2 = a2e2
⇒ a2(1 – e2) = a2e2
⇒ 2e2 = 1 \ e = 12
23. (a) : Parabola is (x + y)2 = 5(x – y + 1) whose axis is x + y = 0.24. (c) : Foci of hyperbola y2 – x2 = 1 are (0, ±be) i.e., (0, ± 2)\ End points of a diameter of reqd. circle are ( , ) ( , )0 2 0 2and − .\ Eqn. of reqd. circle is (x – 0)(x – 0) + ( )( )y y− +2 2 = 0or x2 + y2 = 2
25. (d) : Any plane through (1, 2, –3) is given by a(x – 1) + b(y – 2) + c(z + 3) = 0 ...(i)
P
aeC S(ae, 0)
bx –
ay =
0(0,0)
S
45°
B
C S
x2
a2 + y2
b2 = 1
| JUNE ‘1738
If (i) is || to x-axis then a = 0If (i) passes through (2, –2, 1) then b(–2 – 2) + c(1 + 3) = 0 ⇒ b = c\ (i) becomes b(y – 2) + b(z + 3) = 0⇒ y + z + 1 = 0
26. (b) : Here, 1 1 12 3 01 0 3
−− = 1(–9 – 0) + 1(6 – 0) + 1(0 + 3)
= –9 + 6 + 3 = 0
\ 3 lines with d.r.s (1, –1, 1), (2, –3, 0) & (1, 0, 3) are coplanar.27. (a) : Given, f(x) is non-constant & di�erentiable
s.t.f xf y
( )( )
= f(x – y) ...(i)
Let f(x) = amx satisfying (i). Now, f (x) = mamx
Given f (0) = p ⇒ ma0 = p ⇒ m = p ...(ii)Also, f (5) = q ⇒ ma5m = q
⇒ a5m = qm
qp
= ...(iii)
\ f (–5) = mam(–5) = p⋅a–5m = p pq
⋅ [using (iii)]
= pq
2
28. (c) : f(x) = log5log3x = loge(log3x)log5e = log5e loge(log3x)
f (x) = loglog
(log )53
31e
xd
dxx⋅
= ⋅ ⋅loglog
(log log )53
31e
xd
dxx ee
= loglog
log log log53
3 51 1 1e
xe
xe e
xx⋅ ⋅ ⋅ = ⋅
\ f (e) = log loglog5
1 15
e ee ee
e⋅ =
29. (c) : H(x) = G(F(x)) = G(ex) = e–ex
\ dHdx
e ddx
e e ee x e xx x= ⋅ − = ⋅ −− −( ) ( )
\ dHdx
e e eex
e
= − = − = −=
− −
0
0 10 1( )
30. (c) : limx
f x f x f xx→
− +0 2
2 3 2 4( ) ( ) ( ) 0
0form
= limx
f x f x f xx→
′ − ′ + ′0
2 6 2 4 42
( ) ( ) ( )
00
form
= limx
f x f x f x→
″ − ″ + ″0
2 12 2 16 42
( ) ( ) ( )
= f (0) – 6f (0) + 8f (0) = 3f (0) = 3k
31. (a) : We have, y em x=−sin 1
⇒ dydx
e mx
my
xm x= ⋅ ⋅
−=
−
−sin 1 1
1 12 2
⇒ ( )1 22
2 2−
=x dy
dxm y
⇒ ( )1 2 222
2
2− ⋅ ⋅ − ⋅
x dy
dxd ydx
x dydx
= m y dydx
2 2⋅ ⋅
⇒ ( )1 22
22−
− ⋅ =x d ydx
x dydx
m y ⇒ k = m2
32. (d) : Using mean value theorem, f (c) = f b f ab a
( ) ( )−−
⇒ 2c + 2a = f f( ) ( )β αβ α
−−
= ( ) ( )β β α αβ α
2 22 2+ + − + +−
a b a b = + a + 2a
\ c = α β+2
33. (c) : f(x) = x13 + x11 + x9 + x7 + x5 + x3 + x + 19 f (x) = 13x12 + 11x10 + 9x8 + 7x6 + 5x4 + 3x2 + 1 > 0 ∀ x ∈ R\ f(x) is a strictly increasing function. f(– ) = – , f( ) = , f(0) = 19\ f(x) = 0, will have only one real root34. (b) : LMVT is applicable to f(x) in [0, x]\ f(x) must be continuous in [0, x]
\ limx
f x f→ +
=0
0( ) ( ) ⇒ limx
p
qx
x→ +=
00
(sin )
⇒ limh
p
qh
h→
++
=0
00
0( ){sin( )}
⇒ limh
p q
qh
hh
→
−
=0
0sin
⇒ limh
p qh→
−=
0 10 \ L.H.L exists only if p > q.
35. (b) : Let A = limx
xx→0
2(sin ) tan
log A = limx
x x→0
2 tan log(sin )
= 20
limx
xx→
log(sin )cot
∞∞
form
= 2
1
0 2limcosecxx
x
x→
⋅
−sin
cos= −
→limx
x0
2sin = 0 ⇒ A = 1
36. (c) : Let I = cos(log )x dx∫
= cos(log ) ( ) sin(log ) ( )x x xx
x dx⋅ − − ⋅ ⋅∫ 1
| JUNE ‘17 39
= x x x dxcos(log ) sin(log )+ ∫ = x cos(log x) + sin(log x)(x) – cos(log ) ( )x
xx dx⋅∫ 1
⇒ 2I = x[cos(log x) + sin(log x)]
⇒ I = x2
[cos(log x) + sin(log x)] + c
37. (a) : x
xdx
x xx
22
2 22
1 1
1 3
−
+ +
∫ = 1 1
1 1
2
2
−
+
+
∫ xdx
xx
= dzz
z xx2 11
+= +∫ , where
= tan–1 z + c = tan− +
+1 1x
xc
38. (b) : I = sin xx
dx1 8
10
19
+∫
\ I = sin sinxx
dx xx
dx1 18
10
19
810
19
+≤
+∫ ∫
⇒ I ≤ | sin |xx
dx dxx1 18
10
19
810
19
+≤
+∫ ∫ | | ≤ sin x 1
⇒ I ≤ dxx
x8
7
10
19
10
19
7=
−
−
∫
= 17
10 19 107
107 77
7( )− −−
−− < <
39. (d) : I2 = { } { } { }x dx x dx n x dxn n
0 0
1
0
1
∫ ∫ ∫= =⋅
[ {x} is periodic with period = 1]
= n x dx n
0
1
2∫ =
I1 + I2 = ([ ] { })x x dx x dx x nn n n
+ = =
=∫ ∫
0 0
2
0
2
2 2
\ I1 = n n2
2 2− \
II
n nn
n1
2
21= − = −
40. (b) : limn
nn
nn
nn n→∞ +
++
+ ++
2 2 2 2 2 21 2
.....
= limn n
n nnn
→∞+
+
+
+ +
+
1 1
1 1
1
1 2
1
12 2 2.....
= limn r
n
n rn
→∞ = +
∑1 1
12
1= dx
xx
1 420
11
01
+= =∫ −[tan ] π
41. (d) : 0 ≤ x ≤ 1 \ 1 ≤ ex2 ≤ e
⇒ dx e dx e dxx
0
1
0
1
0
12
∫ ∫ ∫≤ ≤ ⋅
⇒ 12
0
1
≤ ≤∫ e dx ex
42. (c) : e dx e dx e dxx x x x−∫ ∫ ∫= =[ ]( )
{ }( )
{ }
0
100 1
0
100 1
0
1
100
[ {x} is periodic with period 1]
= 1000
1
e dxx∫ = 100(e – 1)
43. (a) : Putting x + y = z ⇒ 1+ =dydx
dzdx
Given eq. becomes z dzdx
a2 21⋅ −
=
⇒ dzdx
az
= +12
2
⇒ zz a
dz dx2
2 2+= ∫∫ ⇒
( )z a az a
dz x2 2 2
2 2+ −
+=∫
⇒ z aa
za
− ⋅ −2 11 tan = x – c
⇒ x + y – a tan–1 x ya
x c+
= −
⇒ y c
ax y
a+
=+
−tan 1 ⇒ x ya
y ca
+ = +
tan ,
c is an arbitrary constant.
44. (b) : We have, dydx
x xx x
y xx x
+ +−
⋅ = −−
( )( ) ( )
2
2 2
2
2 211
11
⇒ dydx
xx x
yx
+ +−
⋅ =2
2 211
1( )
\ I.F. = e e e xx
xx x
dxx
xx
dxx
xe
2
2
211
1 1
1 11
+−
+
− −
∫ ∫
= = ⋅ = −( )log
45. (b) : Let G.P. be a, ar, ar2, .....A.T.Q., a = ar + ar2 ⇒ r2 + r – 1 = 0
⇒ r = − ± + = − ±1 1 42
1 52
Terms are positive \ r < 0 \ r = − +1 52
46. (a) : Using property of logarithm, log5y = 3logxx = 3\ y = 53 = 125
| JUNE ‘1740
47. (c) : 1+−
−i
ii
n
11 2( ) = ( ) ( )1
11 2
2
22+
−
+ −ii
i in
= 1 1 22
2 2 1− +{ } − = − +i i in
n( )
48. (d) : Let z iz i
ki+−
= (k ∈R)
By componendo & dividendo, we have 22
11
zi
kiki
= +−
⇒ z kiki
i z k
ki= +
−⇒ = +
+ −=1
11
11
2 2
2 2.
( ). | |
⇒ x2 + y2 = 1 which represents a circle49. (a) : 2px2 + 2px + qx + q = 0⇒ 2px(x + 1) + q(x + 1) = 0
\ x = –1, −qp2
, which are rational as p & q are odd
integers.50. (b) : 3 out of 7 consonants can be chosen in 7C3 ways and 12 out of 4 vowels can be chosen in 4C2 ways\ Total no. of words that can be formed = 7C3 × 4C2 × 5 = 25200
51. (b) : A2 = 1 1 10 1 10 0 1
1 1 10 1 10 0 1
1 2 30 1 20 0 1
=
Only (b) is satis�ed by putting n = 2
52. (d) : D1 = a a ab b b
c c c
a b ca b ca b c
+ −− + −
− +=
−+ + −− − +
1 11 11 1
1 1 11 1 1
[Interchanging rows & columns]
= ( )−+ + −− − +
−=
+ + −− − +
−1
1 1 11 1 1
1 1 11 1 12
a b ca b c
a b c
a b ca b c
a b c
D2 = a b ca b c
a b cn n n
+ + −− − +
− − −+ +
1 1 11 1 1
1 1 12 1( ) ( ) ( )
= ( )−+ + −− − +
−1
1 1 11 1 1n
a b ca b c
a b c\ D1 + D2 = 0 is possible only when n is any odd integer.53. (c) : R ∪ S = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)}A = {1, 2, 3}(i) (1, 1), (2, 2), (3, 3) ∈ R ∪ S ⇒ Re�exive(ii) (a, b) ∈ R ∪ S
⇒ (b, a) ∈ R ∪ S ∀ a, b {1, 2, 3} ⇒ Symmetric(iii) (2, 1) & (1, 3)∈ R ∪ S but (2, 3) R ∪ S⇒ Not transitive54. (a) : Let radius of circle with centre (1, 1) be aIts eqn. is (x – 1)2 + (y – 1)2 = a2 ...(i)Given circle is x2 + y2 – 4x – 6y + 9 = 0 ...(ii)Eqn. of common chord (S1 – S2 = 0) is 2x + 4y = a2 + 7 ...(iii)If (iii) be a diameter of (ii) then centre (2, 3) will lie on (iii)⇒ 4 + 12 = a2 + 7 ⇒ a2 = 9⇒ Radius = 3 units
55. (d) : tan = kh +1
& tan 2 = kh2 −
⇒ kh
kh
kh
22
1
21
11
2 2−=
−= +
−+
tantan
θθ
= 2 11 2 2
k hh k
( )( )
++ −
⇒ (h + 1)2 – k2
= 2(2 – h)(h + 1)⇒ h2 + 2h + 1 – k2
= 4h + 4 –2h2 – 2h
⇒ 3h2 – k2 = 3 ⇒ h k2 2
1 31− =
\ M moves on a hyperbola
56. (c) : f(x) = | | = | | + | |− −∫ ∫∫t dt t dt t dtx x
1 01
0
= | | + | | = − +∫ ∫ ∫∫−−
x dx x dx x dx x dxx x
0 1
0
01
0
( )
= −
+
= − − + −
= +
−
x x x xx2
1
0 2
0
2 2
2 212
0 12
0 12
( )
57. (b) : f(x) = limn
nk
kn x x
k→∞ →−( ) = −
1
01 1lim , where n = 1
k = loge x
Now, f x fx
x xe e e( ) log log / log+
= + = =1 1 1 0
\ f(xy) = log log loge e exy x y= + = f(x) + f(y)
58. (b) : I = 2 2
0
100
sin x dxπ
∫
hO B(2, 0)
M(h, k)
A(–1, 0)
k
| JUNE ‘17 41
= 2 2 1000
100
0
|sin | |sin |x dx x dx= ⋅∫ ∫π π
[ Period of |sin x|is ]
= 100 2 100 20
0sin cosx dx xπ
π∫ = −[ ]
= 100 2 1 1 200 2( )+ =
59. (a) :
O(–2, 0)
y1
y2
(1, 0)
x = –2 x = 0 x = 1Parabolas are y2 = − x
2 ...(i) & y2 = − −1
31( )x ...(ii)
On solving, − x2
= − −( )x 13
⇒ –2x + 2 = – 3x ⇒ x = –2
\ Reqd. area = 2 22
1
12
0
y dx y dx− −∫ ∫−
[y1, y2 are values of y from (i) & (ii) resp.]
= 2 13 2
2
1
2
0− − −
− −
∫ ∫x dx x dx
= 23
132
22 3
2
3 2
2
1 3 2( ) ( )/ /−
−
− −
−
−
x x−2
0
= − −[ ]+ −[ ]43 3
0 3 43 2
0 23 2 3 2/ /
= 4 83
43
− = square units
60. (a) : We have,
x y2 2
9 51+ = ...(i)
e = 1 59
23
− =
\ L ae ba
, ,2
2 53
≡
Eqn. of tangent to (i) at L 2 53
,
is
x y9
25
53
1⋅ + ⋅ = ⇒ x y9 2 3
1/
+ =
S AC
LB
L
\ CA CB= =92
3,
\ Reqd. area of quad. = 4 × Area of CAB
= 4 12
92
3× ⋅ ⋅ = 27 sq. units.
61. (c) : f(x) = sin x – cos x – Kx + 5
⇒ f (x) = cos x + sin x – K = 24
sin x K+
−π
24
2sin x +
≤π
\ f(x) will be decreasing for all +ve real x if f (x) < 0
24
0sin x K+
− <π
⇒ K > 24
sin x +
π ⇒ K > 2
62. (b) : Let � � � �x ai b j ck= + +\ � � � �x i bk c j× = − + ( ) | |� � � �x i x i b c× = × = +2 2 2 2
Similarly, ( ) & ( )� � � �x j c a x k a b× = + × = +2 2 2 2 2 2
\ ( ) ( ) ( ) ( ) | |� � � � � � �x i x j x k a b c x× + × + × = + + =2 2 2 2 2 2 22 263. (c) : Let a b c, , be three unit vectors such that c a b= +\ c a b2 2= +( ) ⇒ 1 = 1 + 1 + 2a b⋅⇒ 2 a b⋅ = −1Now, | |a b a b− = + − ⋅ =2 1 1 2 3 ⇒ | |a b− = 3 units.64. (a) : Roots of x2 + x + 1 = 0 are & 2
Let a = = − + = +1 32
23
23
i icos sinπ π
and = 2 = − − = −
+ −
1 32
23
23
i icos sinπ π
= cos sin23
23
π π− i \ α β πn n n+ = 2 23
cos
65. (c) : Let y x xx x
= + ++ +
2
22 4
2 4 9⇒ (2y – 1)x2 + 2(2y – 1)x + (9y – 4) = 0For real x, D ≥ 0 ⇒ 4(2y – 1)2 – 4(2y – 1)(9y – 4) ≥ 0⇒ 4y2 – 4y + 1 – 18y2 + 17y – 4 ≥ 0 ⇒ 14y2 – 13y + 3 ≤ 0
⇒ (7y – 3) (2y – 1) ≤ 0 ⇒ 37
12
≤ ≤y
⇒ Greatest value of y i.e. x xx x
2
22 4
2 4 9+ ++ +
is 12
66. (c, d) : a, b ∈ {1, 2, 3} & ax2 + bx + 1 = 0 has real roots⇒ D ≥ 0 i.e., b a2 4≥ ...(i)
| JUNE ‘1742
\ Possible ordered pair (a, b) are (1, 2), (1, 3) & (2, 3) only67. (a, c) : Equation of tangent to y2 = 4ax at (at2, 2at) is y ⋅ 2at = 2a (x + at2)⇒ ty = x + at2 ...(i)
For x2–y2 = a2, dydx
xy
=
⇒ At (a sec , a tan ), dydx
aa
= =sectan sin
θθ θ
1
\ Eqn. of normal to x2 – y2 = a2 at (a sec , a tan ) isy – a tan = – sin (x – a sec )⇒ y = –x sin + 2a tan ...(ii) (i) & (ii) are identical
\ t ata
t1
12 2
2 2=
−= =
sin tan tanθ θ θ⇒ t t= − =cosec θ θ& tan268. (c) : Conic may be written as (x – 3)2 = –4(y – 2)X2 = 4AY ...(i)Here, X = x – 3, Y = y – 2 & 4A = –4 ...(ii)For focus, X = 0, Y = A ⇒ x – 3 = 0, y – 2 = –1⇒ x = 3, y = 1 \ Focus is (3, 1)69. (b)70. (b,c) : f(x) = xn ...(i) f (a + ) = f (a) + f ( ) ...(ii)When n = 2, f(x) = x2 , f (x) = 2x\ f (a + ) = 2(a + ) = 2a + 2 =f (a) + f ( )⇒ (ii) is satis�edWhen n = 0, f(x) = 1 ⇒ f (x) = 0 ⇒ (ii) is satis�ed.(ii) is not satis�ed if n = 1 or 5
71. (c) : I = dxf x
a
10 +∫ ( ) ...(i)
= dxf a x
a
10 + −∫ ( ) = f x f a x
f x f a x f a xdx
a ( ) ( )( ) ( ) ( )
−− + −∫0
I = f x
f xdx
a ( )( ) +∫ 10
...(ii)
Adding (i) & (ii), 2I = dx aa
=∫0 ⇒ I = a
272. (b,d) : ax + by + c = 0 ...(i)
Its slope = −ab
(ab ≠ 0)
Now, xy = 1 –2x ⇒ y + 2 = 1x
...(ii)
Di�erentiating (ii), we get, dydx
= − 12x
\ dydx
< 0
If (i) be a tangent to (ii) then − = − <ab x
1 02
⇒ ba
x= >2 0
\ a & b must be of same sign\ a > 0, b > 0 or a < 0, b < 0
73. (b,c) : u ff= 0
0Dist. between them at any instant is given by
x = u t t ft⋅ − ⋅ +{ }0 12
2
dxdt
u f t u ft d xdt
f= − ⋅ ⋅ = − ⇒ = − <12
2 02
2
For max. distance between them, dxdt
= 0 ⇒ t = uf
Max. distance = u uf
f uf
uf
uf
uf
⋅ − ⋅ = − =12 2 2
2
2
2 2 2
74. (a, c) : |z – i| = |z + 1| = 1⇒ x2 + (y – 1)2 = (x + 1)2 + y2 = 1 ...(i)⇒ –2y + 1 = 2x + 1 ⇒ y = –x ...(ii)When y = –x, we have x2 + 2x + 1 + x2 = 1⇒ 2x(x + 1) = 0 ⇒ x = 0, – 1when x = 0 ⇒ y = 0when x = –1 ⇒ y = 1\ z = 0, –1 + i75. (c) : (i) 1 + a⋅a = 1 + a2 > 0 ⇒ Re�exive(ii) If 1 + ab > 0 then 1 + ba > 0 ⇒ Symmetric
(iii) 1 1 12
32
0 1 12
1 12
1 12
0 12
1
+
= > ⇒
∈
+ −( ) = > ⇒ −
∈
,
,
ρ
ρ
But, 1 + (1)(–1) = 0 /> 0⇒ (1, –1) \ is not transitive.
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MANIPUR at
| JUNE ‘17 43
1. 1 1 1
1p q rp q r +
is equal to
(a) q – p (b) q + p (c) q (d) p (e) 0
2. Let A = 5 01 0
and B =
0 11 0−
.
If 4A + 5B – C = O, then C is
(a) 5 251 0−
(b)
20 51 0−
(c) 5 10 25
−
(d)
5 251 5−
(e) 0 55 25
3. If U =
−
12
12
12
12
, then U–1 is
(a) UT (b) U (c) I (d) 0 (e) U2
4. If A =−
−
0 1 01 0 00 0 1
, then A–1 is
(a) AT (b) A2 (c) A (d) I (e) O
5. If x y x yx z x z+ −+ +
=
2
0 01 1
, then the values of x, y
and z are respectively
(a) 0, 0, 1 (b) 1, 1, 0 (c) –1, 0, 0 (d) 0, 0, 0(e) 1, 1, 1
6. 78
10
50
231
510
+
is equal to
(a) 1627
(b)
2716
(c) 1516
(d) 1615
(e) 1616
7. If 1 2 41 3 51 4 a
is singular, then the value of a is
(a) a = –6 (b) a = 5(c) a = –5 (d) a = 6 (e) a = 0
8. If 1 2 30 4 50 0 1
111
−
=
xyz
, then (x, y, z) is equal to
(a) (1, 6, 6) (b) (1, –6, 1)(c) (1, 1, 6) (d) (6, –1, 1)(e) (–1, 6, 1)
9. If A =
1 50 2
, then
(a) A2 – 2A + 2I = O (b) A2 – 3A + 2I = O(c) A2 – 5A + 2I = O (d) 2A2 – A + I = O (e) A2 + 3A + 2I = O
10. If 2 1 1
0 0x y x yp q p q
+ +− +
=
, then (x, y, p, q) equals
(a) 0, 1, 0, 0 (b) 0, –1, 0, 0(c) 1, 0, 0, 0 (d) 0, 1, 0, 1(e) 1, 0, 1, 0
SOLVED PAPER 2017
Kerala PET
| JUNE ‘1744
11. �e value of 4 2 3 4 2 3+ − − is
(a) 1 (b) 2 (c) 4 (d) 3(e) 5
12. �e value of 82/3 –161/4 –91/2 is(a) –1 (b) –2 (c) –3 (d) –4 (e) –5
13. Let x = 2 be a root of y = 4x2 –14x+q = 0. �en y is equal to(a) (x – 2) (4x – 6) (b) (x – 2) (4x + 6)(c) (x – 2) (–4x – 6) (d) (x – 2) (–4x + 6)(e) (x – 2) (4x + 3)
14. If x1 and x2 are the roots of 3x2 – 2x – 6 = 0, then
x21 + x2
2 is equal to
(a) 509
(b) 409
(c) 309
(d) 209
(e) 109
15. Let x1 and x2 be the roots of the equation x2 + px –3 = 0. If x2
1 + x22 =10, then the value of p is equal to
(a) –4 or 4 (b) –3 or 3 (c) –2 or 2 (d) –1 or 1 (e) 0
16. If the product of roots of the equation mx2 + 6x + (2m – 1) = 0 is –1, then the value of m is
(a) 13
(b) 1 (c) 3 (d) –1
(e) –3
17. If f xx x x x x x x
( ) ,=+ +
−+ +
++
14 4
44 4
422 4 3 2 3 2
then f 12
is equal to
(a) 1 (b) 2 (c) –1 (d) 3 (e) 4
18. If x and y are the roots of the equation x2 + bx + 1 = 0,
then the value of 1 1x b y b+
++
is
(a) 1b
(b) b (c) 12b
(d) 2b
(e) 1
19. �e equations x5 + ax + 1 = 0 and x6 + ax2 + 1 = 0 have a common root. �en a is equal to(a) –4 (b) –2 (c) –3 (d) –1 (e) 0
20. �e roots of ax2 + x + 1 = 0, where a ≠ 0, are in the ratio 1 : 1. �en a is equal to
(a) 14
(b) 12
(c) 34
(d) 1
(e) 0
21. If z2 + z + 1 = 0 where z is a complex number, then
the value of zz
zz
zz
+
+ +
+ +
1 1 122
2
23
3
2
equals (a) 4 (b) 5 (c) 6 (d) 7 (e) 8
22. Let ∆ = − −
1 1 1
1 1
1
2 2
4
w w
w w
, where w ≠ 1 is a complex
number such that w3 = 1. �en equals(a) 3w + w2 (b) 3w2
(c) 3(w – w2) (d) –3w2
(e) 3w2 +1
23. If 3 9 12 9 1
10 9
i ii
i
−− = x + iy, then
(a) x = 1, y = 1 (b) x = 0, y = 1(c) x = 1, y = 0 (d) x = 0, y = 0 (e) x = –1, y = 0
24. If z = cos π π3 3
−
i sin , then z2 – z +1 is equal to
(a) 0 (b) 1 (c) –1 (d) π2
(e)
25. 1
12 12
112 12
+
+
+
−
cos sin
cos sin
π π
π π
i
i
72
is equal to
(a) 0 (b) – (c) 1 (d) 12
(e) − 12
26. If Ak kk k
k=
400 0
and det (A) = 256, then k equals
(a) 4 (b) 5 (c) 6 (d) 7 (e) 8
| JUNE ‘17 45
27. If A =
1 01 1
, then An + nI is equal to
(a) I (b) nA (c) I + nA (d) I – nA (e) nA – I
28. If |z| = 5 and w zz
= −+
55
, then Re(w) is equal to
(a) 0 (b) 125
(c) 25 (d) 1 (e) –1
29. If A =
1 11 1
, then A2017 is equal to
(a) 22015A (b) 22016A (c) 22014A (d) 22017A (e) 22020A
30. If a = ei , then 11
+−
aa
is equal to
(a) cot θ2
(b) tan
(c) icot θ2
(d) i tan θ2
(e) 2 tan 31. �ree numbers x, y and z are in arithmetic
progression. If x + y + z = – 3 and xyz = 8, then x2 + y2 + z2 is equal to(a) 9 (b) 10 (c) 21 (d) 20 (e) 1
32. �e 30th term of the arithmetic progression 10, 7, 4, .... is (a) –90 (b) –87 (c) –77 (d) –67 (e) –57
33. �e arithmetic mean of two numbers x and y is 3 and geometric mean is 1. �en x2 + y2 is equal to(a) 30 (b) 31 (c) 32 (d) 33 (e) 34
34. �e solution of 32x–1 = 811–x is
(a) 23
(b) 16
(c) 76
(d) 56
(e) 13
35. �e sixth term in the sequence is 3, 1, 13
, ... is
(a) 127
(b) 19
(c) 181
(d) 117
(e) 17
36. �ree numbers are in arithmetic progression. �eir sum is 21 and the product of the �rst number and the third number is 45. �en the product of these three numbers is(a) 315 (b) 90 (c) 180 (d) 270 (e) 450
37. If a+1, 2a + 1, 4a – 1 are in arithmetic progression, then the value of a is (a) 1 (b) 2 (c) 3 (d) 4 (e) 5
38. Two numbers x and y have arithmetic mean 9 and geometric mean 4. �en x and y are the roots of(a) x2 –18x –16 = 0 (b) x2 –18x +16 = 0(c) x2 +18x –16 = 0 (d) x2 +18x +16 = 0(e) x2 –17x +16 = 0
39. �ree unbiased coins are tossed. �e probability of getting at least 2 tails is
(a) 34
(b) 14
(c) 12
(d) 13
(e) 23
40. A single letter is selected from the word TRICKS. �e probability that it is either T or R is
(a) 136
(b) 14
(c) 12
(d) 23
(e) 13
41. From 4 red balls, 2 white balls and 4 black balls, four balls are selected. �e probability of getting 2 red balls is
(a) 721
(b) 821
(c) 921
(d) 1021
(e) 1121
42. In a class, 60% of the students know lesson I, 40% know lesson II and 20% know lesson I and lesson II. A student is selected at random. �e probability that the student does not know lesson I and lesson II is
(a) 0 (b) 45
(c) 35
(d) 15
(e) 25
43. Two distinct numbers x and y are chosen from 1, 2, 3, 4, 5. �e probability that the arithmetic mean of x and y is an integer is
| JUNE ‘1746
(a) 0 (b) 15
(c) 35
(d) 25
(e) 45
44. �e number of 3 3 matrices with entries –1 or +1 is(a) 24 (b) 25 (c) 26 (d) 27
(e) 29
45. Let S be the set of all 2 × 2 symmetric matrices whose entries are either zero or one. A matrix X is chosen from S. �e probability that the determinant of X is not zero is
(a) 13
(b) 12
(c) 34
(d) 14
(e) 29
46. �e number of words that can be formed by using all the letters of the word PROBLEM only once is(a) 5! (b) 6! (c) 7! (d) 8!
(e) 9!
47. �e number of diagonals in a hexagon is(a) 8 (b) 9 (c) 10 (d) 11
(e) 12
48. �e sum of odd integers from 1 to 2001 is(a) 10012 (b) 10002 (c) 10022 (d) 10032
(e) 9992
49. Two balls are selected from two black and two red balls. �e probability that the two balls will have no black ball is
(a) 17
(b) 15
(c) 14
(d) 13
(e) 16
50. If z = i9 + i19, then z is equal to(a) 0 + 0i (b) 1 + 0i (c) 0 + i (d) 1 + 2i(e) 1 + 3i
51. �e mean for the data 6, 7, 10, 12, 13, 4, 8, 12 is(a) 9 (b) 8 (c) 7 (d) 6
(e) 5
52. �e set of all real numbers satisfying the inequality x – 2 < 1 is(a) (3, ) (b) [3, ) (c) [–3, ) (d) (– , –3)(e) (– , 3)
53. If xx
−−
>33
0, then
(a) x ∈(–3, ) (b) x ∈(3, )(c) x ∈(2, ) (d) x ∈(1, )(e) x ∈(–1, )
54. �e mode of the data 8, 11, 9, 8, 11, 9, 7, 8, 7, 3, 2, 8 is(a) 11 (b) 9 (c) 8 (d) 3
(e) 7
55. If the mean of six numbers is 41, then the sum of these numbers is(a) 246 (b) 236 (c) 226 (d) 216
(e) 206
56. If f t dt x e xxx ( ) ( ),= + >∫ 20 0 then f(1) is equal to
(a) 1 + e (b) 2 + e (c) 3 + e (d) e
(e) 0
57. xx
dx+ =∫1
1 2/
(a) –x3/2 + x1/2 + c (b) x1/2
(c) 23
23 2 1 2x x c/ /+ + (d) x3/2 + x1/2 + c
(e) x3/2
58. In a �ight 50 people speak Hindi, 20 speak English and 10 speak both English and Hindi. �e number of people who speak at least one of the two languages is (a) 40 (b) 50 (c) 20 (d) 80
(e) 60
59. If f x xx
( ) ,= +−
11
then the value of f(f(x)) is equal to
(a) x (b) 0 (c) –x (d) 1
(e) 2
60. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
(a) 34
(b) 14
(c) 12
(d) 23
(e) 116
61. limx
x xx→
+ − −0
2 2is equal to
(a) 12
(b) 2
(c) 0 (d) Does not exist
(e) 12 2
| JUNE ‘17 47
62. dxe ex x+ +−∫
2 is equal to
(a) 11e
cx ++ (b)
−+
+11e
cx
(c) 1
1++−e
cx (d) 1
1ecx− −
+
(e) 1
1ecx −
+
63. tan tanπ θ π θ4 2 4 2
+
+ −
is equal to
(a) sec (b) 2 sec (c) sec θ2
(d) sin
(e) cos
64. dxx x21
0
2+ +−∫ is equal to
(a) π4
(b) π2
(c) (d) 0
(e) –
65. sinsin cos
xx x+∫02
πdx is equal to
(a) 0 (b) – (c) 32π (d) π
2
(e) π4
66. If (x, y) is equidistant from (a + b, b – a) and (a – b, a + b), then(a) x + y = 0 (b) bx – ay = 0(c) ax – by = 0 (d) bx + ay = 0(e) ax + by = 0
67. If the points (1, 0), (0, 1) and (x, 8) are collinear, then the value of x is equal to (a) 5 (b) –6 (c) 6 (d) 7
(e) –7
68. �e minimum value of the function max{x, x2} is equal to
(a) 0 (b) 1 (c) 2 (d) 12
(e) 32
69. Let f(x + y) = f(x) f (y) for all x and y. If f(0) = 1, f(3) = 3 and f (0) = 11, then f (3) is equal to(a) 11 (b) 22 (c) 33 (d) 44
(e) 55
70. If f(9) = f (9) = 0, then lim ( )x
f xx→
−−9
33
is equal to
(a) 0 (b) f(0) (c) f (3) (d) f(9)
(e) 1
71. �e value of cos cosπ π4 4
+
+ −
x x is
(a) 2 2sin x (b) 2 sin x (c) 2 2cos x (d) 3 cos x(e) 2 cos x
72. Area of the triangle with vertices (–2, 2), (1, 5) and (6, –1) is
(a) 15 (b) 35
(c) 292
(d) 332
(e) 352
73. �e equation of the line passing through(–3, 5) and perpendicular to the line through the points (1, 0) and (–4, 1) is(a) 5x + y + 10 = 0 (b) 5x – y + 20 = 0(c) 5x – y – 10 = 0 (d) 5x + y + 20 = 0(e) 5y – x – 10 = 0
74. �e coe�cient of x5 in the expansion of (1 + x2)5 (1 + x)4 is(a) 30 (b) 60 (c) 40 (d) 10
(e) 4575. �e coe�cient of x4 in the expansion of (1 – 2x)5 is
equal to(a) 40 (b) 320 (c) –320 (d) –32
(e) 8076. �e equation 5x2 + y2 + y = 8 represents
(a) an ellipse (b) a parabola(c) a hyperbola (d) a circle (e) a straight line
77. �e centre of the ellipse 4x2 + y2 –8x + 4y – 8 = 0 is(a) (0, 2) (b) (2, –1) (c) (2, 1) (d) (1, 2)(e) (1, –2)
78. �e area bounded by the curves y = –x2 + 3 and y = 0 is(a) 3 1+ (b) 3 (c) 4 3 (d) 5 3
(e) 6 379. �e order of the di�erential equation
d ydx
d ydx
dydx
3
3
2 2
2
2 5
+
+
= 0 is
| JUNE ‘1748
(a) 3 (b) 4 (c) 1 (d) 5
(e) 6
80. If f(x) = 2 42
xx
+ , then f (2) is equal to
(a) 0 (b) –1 (c) 1 (d) 2
(e) –2
81. �e area of the circle x2 – 2x + y2 – 10y + k = 0 is 25 . �e value of k is equal to(a) –1 (b) 1 (c) 0 (d) 2
(e) 3
82. xx x
dx+ −∫ 40332016
2017 is equal to
(a) 14
(b) 32
(c) 20172
(d) 12
(e) 508
83. �e solution of dydx
y+ tan x = sec x, y(0) = 0 is
(a) y sec x = tan x (b) y tan x = sec x (c) tan x = y tan x (d) x sec x = tan y
(e) y cot x = sec x
84. If the vectors 2 2 6 2 6 2 3i j k i j k i j k+ + + + − +, ,λ are coplanar, then the value of is(a) –10 (b) 1 (c) 0 (d) 10
(e) 2
85. �e distance between (2, 1, 0) and 2x + y +2z + 5 = 0 is
(a) 10 (b) 103
(c) 109
(d) 5
(e) 1
86. �e equation of the hyperbola with vertices (0, ± 15) and foci (0, ± 20) is
(a) x y2 2
175 2251− = (b) x y2 2
625 1251− =
(c) y x2 2
225 1251− = (d) y x2 2
65 651− =
(e) y x2 2
225 1751− =
87. �e value of 15 6 3 6 15 211 4 6 6 36 4 216 1296
3 3+ + ⋅ ⋅ ⋅+ + + +( ) ( ) ( )
is equal to
(a) 297
(b) 719
(c) 617
(d) 2119
(e) 277
88. �e equation of the plane that passes through the points (1, 0, 2), (–1, 1, 2), (5, 0, 3) is(a) x + 2y – 4z + 7 = 0 (b) x + 2y – 3z + 7 = 0(c) x – 2y + 4z + 7 = 0 (d) 2y – 4z – 7 + x = 0(e) x + 2y + 3z + 7 = 0
89. �e vertex of the parabola y2 – 4y – x + 3 = 0 is(a) (–1, 3) (b) (–1, 2) (c) (2, –1) (d) (3, –1)(e) (1, 2)
90. If a b c, , are vectors such that a b c+ + = 0 and
a b c= = =7 5 3, , , then the angle between
c band is
(a) π3
(b) π6
(c) π4
(d)
(e) 0
91. Let f x x ax a x( ) ,= − + +2 9 12 13 2 2 where a > 0. �e minimum of f is attained at a point q and the maximum is attained at a point p. If p3 = q, then a is equal to(a) 1 (b) 3 (c) 2 (d) 2
(e) 12
92. For all real numbers x and y, it is known that the real valued function f satis�es f(x) + f (y) = f(x + y).
If f(1) = 7, then f rr ( )=∑ 1100
is equal to(a) 7 51 102 (b) 6 50 102(c) 7 50 102 (d) 6 25 102(e) 7 50 101
93. �e eccentricity of the ellipse
( )x y− + +
=12
34
116
2 2 is
(a) 12
(b) 1
2 2 (c) 1
2 (d) 1
4
(e) 14 2
94. max{ , }x x dx31
1−∫ is equal to
(a) 34
(b) 14
(c) 12
(d) 1
(e) 0
95. If x y∈ ∈02
02
, , ,π π
and sin x + cos y = 2, then
the value of x + y is equal to
| JUNE ‘17 49
(a) 2 (b) (c) π4
(d) π2
(e) 0
96. Let a, a + r and a + 2r be positive real numbers such that their product is 64. �en the minimum value of a + 2r is equal to(a) 4 (b) (c) 2 (d)
12
(e) 1
97. �e sum S = + + + +19
13 7
15 5
17 3
19! ! ! ! ! ! ! !
is equal to
(a) 28
10
! (b)
210
9
! (c) 2
10
7
! (d)
210
6
!
(e) 28
5
!
98. If f x
x x x
x xx
( ) ,=
2 3
21 2 30 2 6
then f (x) is equal to
(a) x3 + 6x2 (b) 6x3 (c) 3x (d) 6x2 (e) 0
99. xx
dx2
3 21+∫
( ) is equal to
(a) tan–1 (x2) + c (b) 23
1 3tan ( )− +x c
(c) 13
1 3tan ( )− +x c (d) 12
1 2tan ( )− +x c
(e) tan–1 (x3) + c
100. Let fn(x) be the nth derivative of f(x). �e least value of n so that fn = fn+1 , where f(x) = x2 + ex is
(a) 4 (b) 5 (c) 2 (d) 3 (e) 6
101. sin 765° is equal to
(a) 1 (b) 0 (c) 32
(d) 12
(e) 12
102. �e distance of the point (3, –5) from the line 3x – 4y – 26 = 0 is
(a) 37
(b) 25
(c) 75
(d) 35
(e) 1
103. �e di�erence between the maximum and minimum value of the function
f x t t dtx( ) ( )= + +∫ 20 1 on [2, 3] is
(a) 396
(b) 496
(c) 596
(d) 696
(e) 796
104. If a and b are the non zero distinct roots of x2 + ax + b = 0, then the minimum value of x2 + ax + b is
(a) 23
(b) 94
(c) −94
(d) −23
(e) 1105. If the straight line y = 4x + c touches the ellipse
x y2
24
1+ = then c is equal to
(a) 0 (b) ± 65 (c) ± 62 (d) ± 2 (e) ± 13
106. �e equations x – y = 2, 2x – 3y = – and 3x – 2y = –1 are consistent for
(a) = – 4 (b) = 1, 4(c) = 1, – 4 (d) = –1, 4(e) = –1
107. �e set {(x, y):|x|+|y|=1} in the xy plane represents(a) a square (b) a circle(c) an ellipse (d) a rectangle which is not a square(e) a rhombus which is not a square
108. The value of cos tan−
1 34
is
(a) 45
(b) 35
(c) 34
(d) 25
(e) 0109. Let A(6, –1), B(1, 3) and C(x, 8) be three points
such that AB = BC. �e values of x are(a) 3, 5 (b) –3, 5 (c) 3, –5 (d) 4, 5(e) –3, –5
110. In an experiment with 15 observations on x, the
following results were available x2 2830∑ = and
x∑ = 170 . One observation that was 20, was found
to be wrong and was replaced by the correct value 30. �en the corrected variance is
(a) 9.3 (b) 8.3 (c) 188.6 (d) 177.3 (e) 78
| JUNE ‘1750
111. �e angle between the pair of lines x y z− = − = +
−2
21
53
3and x y z+
−= − = −2
14
85
4 is
(a) cos−
1 219 38
(b) cos−
1 239 38
(c) cos−
1 249 38
(d) cos−
1 259 38
(e) cos−
1 269 38
112. Let a be a unit vector. If ( ) ( ) ,x a x a− ⋅ + = 12 then the magnitude of x is
(a) 8 (b) 9 (c) 10 (d) 13 (e) 12
113. �e area of the triangular region whose sides are y = 2x + 1, y = 3x + 1 and x = 4 is
(a) 5 (b) 6 (c) 7 (d) 8 (e) 9
114. If nCr–1 = 36, nCr = 84 and nCr+1 = 126, then the value of r is
(a) 9 (b) 3 (c) 4 (d) 5 (e) 6
115. Let f(x + y) = f(x) f(y) and f(x) = 1 + sin(3x)g(x), where g is di�erentiable. �en f (x) is equal to
(a) 3f(x) (b) g(0) (c) f(x)g(0) (d) 3g(x) (e) 3f(x)g(0)
116. �e roots of the equation x
xx
−−
−=
1 1 11 1 11 1 1
0 are
(a) 1, 2 (b) –1, 2 (c) –1, –2 (d) 1, –2 (e) 1, 1
117. If the 7th and 8th term of the binomial expansion
(2a –3b)n are equal, then 2 32 3a ba b
+−
is equal to
(a) 131
−+
nn
(b) n
n+−
113
(c) 613
−−
nn
(d) n
n−−
113
(e) 2 113
nn
−−
118. Standard deviation of �rst n odd natural numbers is
(a) n (b) ( )( )n n+ +2 13
(c) n2 13− (d) n
(e) 2n
119. Let S = {1, 2, 3, ..., 10}. �e number of subsets of S containing only odd numbers is
(a) 15 (b) 31 (c) 63 (d) 7 (e) 5
120. �e area of the parallelogram with vertices (0, 0), (7, 2) (5, 9) and (12, 11) is
(a) 50 (b) 54 (c) 51 (d) 52 (e) 53
SOLUTIONS
1. (a) : We have,
1 1 1
1p q rp q r +
= 1 1 1 1 1 0
01
p q rp q r
p qp q
+
= 0 + 1 1 0
01
p qp q
( R2 ~ R3 in 1st determinant)
= 1 1 0
01
p qp q
= 1(q – p)
2. (b) : Given, A = 5 01 0
and B =
0 11 0−
Now, 4A + 5B – C = O
⇒ C = 4A + 5B = 45 01 0
+ 5
0 11 0−
= 20 04 0
0 55 0
+
−
=
20 51 0−
3. (a) : Given, U = 1 2 1 2
1 2 1 2
/ /
/ /
−
Since, U–1 = 1
| |U adj(U)
| JUNE ‘17 51
Hence, adj (U) = 1 2 1 2
1 2 1 2
12
12
12
12
/ /
/ /
−
=−
T
|U| = 12
12
12
12
⋅
− − ⋅
= 12
+ 12
= 1
\ U–1 = 1 2 1 2
1 2 1 2
/ /
/ /−
= UT
4. (a) : We have, A = 0 1 01 0 00 0 1
−
−
Hence, adj(A) = 0 1 01 0 0
0 0 1
0 1 01 0 00 0 1
−
=−
T
Now, |A| = 0 + (1)(–1) = –1
\ A–1 = 0 1 01 0 0
0 0 1−
−
= AT
5. (a) : We have, x y x yx z x z+ −+ +
2
= 0 01 1
On comparing, we get x + y = 0 ...(i), x – y = 0 ...(ii)2x + z = 1 ...(iii), x + z = 1 ...(iv)On solving (iii) and (iv), we get x = 0, z = 1.Hence, from (i), we get y = 0\ x = 0, y = 0, z = 1.
6. (b) : We have, 7 1 58 0 0
231
+ 510
= 14 3 516 0 0
50
+ ++ +
+
= 2216
50
+
= 2716
7. (d) : Let A = 1 2 41 3 51 4 a
Since A is singular matrix \ |A| = 0
Hence, 1 2 41 3 51 4 a
= 0
⇒ 1(3a – 20) – 2(a – 5) + 4(4 – 3) = 0 ⇒ a = 6
8. (d) : We have, 1 2 30 4 50 0 1
−
xyz
= 111
⇒ x y z
y zz
+ −+
2 34 5 =
111
On comparing, we getx + 2y – 3z = 1 ...(i) 4y + 5z = 1 ...(ii) and z = 1 ...(iii)On solving (i), (ii) and (iii), we get z = 1, y = –1 and x = 6
9. (b) : We have, A = 1 50 2
A2 = A ⋅ A = 1 50 2
1 50 2
= 1 150 4
3A = 31 50 2
= 3 150 6
2I = 21 00 1
= 2 00 2
\ A2 – 3A + 2I = 1 150 4
– 3 150 6
+ 2 00 2
= 0 00 0
= O
10. (a) : We have, 2x y x yp q p q
+ +− +
= 1 10 0
On comparing, we get 2x + y = 1 ...(i), x + y = 1 ...(ii) p – q = 0 ...(iii), p + q = 0 ...(iv)On solving (i) and (ii), we get x = 0 and y = 1And, on solving (iii) and (iv), we get p = 0 and q = 0\ (x, y, p, q) = (0, 1, 0, 0)
11. (b) : Let | | | |4 2 3 4 2 3+ − − = xOn squaring both sides, we get
4 2 3 4 2 32
2+ − −( ) = x
⇒ (4 + 2 3) + (4 – 2 3) – 2( )4 2 3+ ( )4 2 3− = x2
⇒ 8 – 2 16 12− = x2 ⇒ 8 – 4 = x2
⇒ x2 = 4 ⇒ x = 2
Hence, | | | |4 2 3 4 2 3+ − − = ±2
| JUNE ‘1752
12. (a) : We have, 82/3 – 161/4 – 91/2
= ((2)3)2/3 – ((2)4)1/4 – ((3)2)1/2
= 4 – 2 – 3 = 4 – 5 = –113. (a) : Given, x = 2 is a root of y = 4x2 – 14x + q = 0 then, x = 2 satisfy the given equation i.e., y(2) = 0⇒ 4(2)2 – 14(2) + q = 0⇒ 16 – 28 + q = 0 ⇒ q = 12\ y = 4x2 – 14x + 12 = 0⇒ y = 4x2 – 6x – 8x + 12⇒ y = 2x(2x – 3) – 4(2x – 3)⇒ y = (2x – 4) (2x – 3) ⇒ y = (x – 2)(4x – 6)14. (b) : We have, 3x2 – 2x – 6 = 0\ Roots of given equation (x1, x2)
= − − ± +( )2 4 726
= 2 766
± = 2 2 196
±
= 1 193
±
Hence, x1 = 1 19
3+
and x2 = 1 193
−
Now, x12 + x2
2 = 1 193
1 193
2 2+
+ −
= 19
(1 + 19 + 2 19 + 1 + 19 – 2 19) = 409
15. (c) : Given, x1 and x2 are roots of x2 + px – 3 = 0.Also, x1
2 + x22 = 10
\ Sum of roots = (x1 + x2) = − p1
= –p
And, product of roots = x1x2 = −31
= – 3 \ (x1 + x2)2 = p2
⇒ x12 + x2
2 + 2x1x2 = p2 ⇒ 10 + 2(–3) = p2
⇒ p2 = 4 ⇒ p = 2Hence, the value of p is 2 or –2.16. (a) : Given, mx2 + 6x + (2m – 1) = 0 and product of roots is – 1.
⇒ ( )2 1mm
− = –1 ⇒ 2m – 1 = – m ⇒ 3m = 1 ⇒ m = 13
17. (e) : We have,
f(x) = 14 42x x+ +
– 44 44 3 2x x x+ +
+ 423 2x x+
= 1
2 2( )x + –
422 2x x( )+
+ 4
22x x( )+
\ f(1/2) = 1
12
22
+
– 4
12
12
22 2
+
+ 4
12
12
22
+
= 1
52
2
– 4
14
52
2
+ 4
14
52
= 425
– 6425
+ 325
= − +60 16025
= 10025
= 4
18. (b) : We have, x2 + bx + 1 = 0 and x, y are its roots.\ Sum of roots = (x + y) = – b ...(i)And product of roots = (xy) = 1 ...(ii)
Now, 1x b+
+ 1y b+
= y b x bx b y b
+ + ++ +( )( )
= ( )
( )
x y b
xy b x y b
+ ++ + +
22 =
( )− +− +
b bb b
21 2 2
(Using (i) and (ii)) = b19. (b) : Let y be the common root of x5 + ax + 1 = 0 and x6 + ax2 + 1 = 0�en, y5 + ay + 1 = 0 and y6 + ay2 + 1 = 0⇒ y5 + ay + 1 = y6 + ay2 + 1⇒ y5 – y6 + ay – ay2 = 0⇒ y5(1 – y) + ay(1 – y) = 0⇒ (y5 + ay) (1 – y) = 0 ⇒ y = 1Hence, the common root is 1.i.e., 1 + a + 1 = 0 ⇒ a = –220. (a) : Let x1 and x2 be the roots of ax2 + x + 1 = 0 then,
x1 + x2 = −1a
...(i) and x1x2 = 1a
...(ii)
Also, x1 : x2 = 1 : 1 ⇒ x1 = x2 ...(iii)
Using (iii) in (i), we get 2x1 = −1a
⇒ x1 = −12a
Using (iii) in (ii), we get x12 = 1
a \ 1
4 2a = 1
a
⇒ 4a = 1 ⇒ a = 14
21. (c) : We have, z2 + z + 1 = 0⇒ z = or 2
where , 2 are complex cube roots of unity.
Now, zz
+
1 2 + z
z2
2
21+
+ zz
33
21+
= ωω
+
1 2 + ω
ω2
2
21+
+ ω
ω3
3
21+
| JUNE ‘17 53
= ωω
2 21+
+ ωω
4
2
21+
+ (1 + 1)2
[using 3 = 1]
= −
ωω
2
+ −
ω
ω
2
2
2
+ 4 [using 1 + + 2 = 0]
= 1 + 1 + 4 = 6
22. (b) : We have, =
1 1 1
1 1
1
2 2
4
− − w w
w w
⇒ =
1 1 1
1 11
2 2− − w ww w
=
1 1 1
11
2w ww w
( 1 + w + w2 = 0)Applying C1 → C1 – C2 and C2 → C2 – C3 , we get
=
0 0 1
11 0
2 2− −−
w w w ww w
= 1(0 – (1 – w) (w – w2)) (on expanding along R1) = –(w – w2 – w2 + w3) = –(–1 – w2 – w2 – w2 + 1) = –(–3w2) = 3w2
23. (d) : We have, 3 9 12 9 1
10 9
i ii
i
−− = (x + iy)
⇒ 3i(9i2 + 9) + 9i(2i + 10) + (18 – 90i) = x + iy⇒ 3i(– 9 + 9) + 18i2 + 90i + 18 – 90i = x + iy⇒ – 18 + 18 = x + iy⇒ 0 + 0i = x + iyOn comparing, we get x = 0 and y = 0
24. (a) : We have, z = cos π3
– i sin π3
= 12
– 32
i =
1 32
− i
then, z2 – z + 1 = 1 3
2
2−
i –
1 32
−
i + 1
= 14
(1 – 3 – 2 3 i) + 12
+ 32
i
= − − + +12
32
12
32
i i = 0
25. (c) : Let z = cos sinπ π12 12
+
i
\ 1z
= cos sinπ π12 12
−
i
Now, from given expression, we have
1
12 12
112 12
+
+
+
−
cos sin
cos sin
π π
π π
i
i
72
= 11 1
72++
−
zz
= ( )( )1
1
72++
z zz
= (z)72 = cos sinπ π12 12
72
+
i
= cos sin7212
7212
π π
+
i
(Using De-Moivre's theorem) = cos 6 + i sin 6 = 1
26. (e) : We have, A = 400 0
k kk k
kNow, expanding along C1, we get |A| = 4(k2)But det(A) = 256 (Given)\ On comparing, we get 4k2 = 256 ⇒ k2 = 64 ⇒ k = ±8Hence, |k| = 8
27. (c) : We have, A = 1 01 1
\ A2 = 1 01 1
1 01 1
= 1 02 1
A3 = 1 02 1
1 01 1
= 1 03 1
\ An = 1 0
1n
Hence, An + nI = 1 0
1n
+ n1 00 1
= 1 0
1n
+ n
n0
0
= 1 0
1+
+
nn n
= 1 00 1
+ nn n
0
= 1 00 1
+ n1 01 1
= I + nA28. (a) : Let z = x + iy
\ |z| = 5 ⇒ x y2 2+ = 5 ⇒ x2 + y2 = 25
| JUNE ‘1754
Now, w = zz
−+
55
= x iyx iy
+ −+ +
55
= ( )( )x iyx iy
− ++ +
55
On rationalizing the denominator, we get
(( ) )(( ) )
( ) ( )
x iy x iy
x y
− + + −+ +
5 5
5 2 2 = x y yi
x y
2 2
2 225 10
5
− + ++ +( )
= 10
5 2 2yi
x y( )+ + [Using x2 + y2 = 25]
= 0 + 10
5 2 2yi
x y( )+ + \ R(w) = 0
29. (b) : We have, A = 1 11 1
\ A2 = A ⋅ A = 1 11 1
1 11 1
= 2 22 2
= 21 11 1
= 2A
A3 = A2 ⋅ A = 2 22 2
1 11 1
= 4 44 4
= 41 11 1
= 22 ⋅ A
Similarly An = 2n – 1 ⋅ A\ A2017 = 22017 – 1 ⋅ A = 22016 ⋅ A30. (c) : We have, a = eiq = cos + i sin (polar form)
\ 11
+−
aa
= 1
1+ +
− +cos sin
(cos sin )θ θθ θ
ii
= 2
22
2 2
22
22 2
2
2
cos sin cos
sin sin cos
θ θ θ
θ θ θ
+
−
i
i
= 2
2 2 2
22 2 2
cos cos sin
sin sin cos
θ θ θ
θ θ θ
+
−
i
i
= cotcos sin sin cos
sin cos
θθ θ θ θ
θ θ22 2 2 2
2 2
+
× +
−
i i
i
+
sin cosθ θ
2 2i
[Rationalizing the denominator]
= cotcos sin cos sin cos sin
sin cos
θθ θ θ θ θ θ
θ θ22 2 2 2 2 2
2 2
2 2
2 2
+ + −
+
i i
= i cot θ2
31. (c) : We have, x + y + z = – 3 ...(i)On squaring both sides, we get (x + y + z)2 = (–3)2
⇒ x2 + y2 + z2 + 2xy + 2yz + 2zx = 9 ...(ii)Also, x, y, z are in A.P. ⇒ 2y = x + z⇒ 2y = – 3 – y ⇒ y = – 1 ...(iii)So, xyz = 8 ⇒ xz = – 8 ...(iv)Now, using (iii) and (iv) in (ii), we get x2 + y2 + z2 – 2x – 2z – 16 = 9⇒ x2 + y2 + z2 – 2(x + z) – 16 = 9⇒ x2 + y2 + z2 + 4 – 16 = 9⇒ x2 + y2 + z2 = 9 + 12 = 21Hence, x2 + y2 + z2 = 2132. (c) : We have the series of an A.P. as 10, 7, 4....\ First term (a) = 10Common di�erence (d) = 7 – 10 = – 3\ a30 = a + (30 – 1)d = 10 + 29(–3) = 10 – 87 = – 7733. (e) : Given, arithmetic mean of x and y is 3
i.e., x y+2
= 3 ⇒ x + y = 6 ...(i)
and geometric mean of x and y is 1i.e., xy = 1 ⇒ xy = 1 ...(ii)Squaring (i) on both sides, we get (x + y)2 = (6)2 ⇒ x2 + y2 + 2xy = 36⇒ x2 + y2 + 2 = 36 (Using (ii))⇒ x2 + y2 = 3434. (d) : We have, 32x – 1 = 811 – x
⇒ (3)2x – 1 = ((3)4)1 – x ⇒ (3)2x – 1 = (3)4 – 4x
\ On comparing, we get, 2x – 1 = 4 – 4x
⇒ 6x = 5 or x = 56
35. (c) : Given series 3, 1, 13
, .... forms a G.P.
where �rst term (a) = 3 and common ratio (r) = 13
\ Sixth term, a6 = ar5
= (3)13
5
=
13
4
= 1
8136. (a) : Let three numbers in A.P. be a, b and c�en, according to question a + b + c = 21 ...(i) and ac = 45 ...(ii) a, b and c are in A.P. \ 2b = c + a ...(iii)
| JUNE ‘17 55
Substituting (iii) in (i), we get 3b = 21 ⇒ b = 7Hence, product of these three numbers = abc = 7(45) = 315 (Using (ii))37. (b) : Given, a + 1, 2a + 1, 4a – 1 are in A.P.\ 2(2a + 1) = 4a – 1 + a + 1⇒ 4a + 2 = 5a ⇒ a = 238. (b) : Given, arithmetic mean of x and y is 9
i.e., x y+2
= 9 ⇒ x + y = 18
Geometric mean of x and y is 4 i.e., xy = 4 ⇒ xy = 16Now, sum of roots (x + y) = 18Product of roots (xy) = 16\ Required quadratic equation is x2 – 18x + 16 = 039. (c) : Total number of outcomes = 8 i.e.,{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} Number of favourable outcomes = 4 i.e., {TTH, THT, HTT, TTT} = 4
\ P(getting atleast 2 tails) = 48
= 12
40. (e) : Number of letters in TRICKS = 6Number of favourable outcomes = {T, R} = 2
\ P(either T or R) = 26
= 13
41. (c) : Total number of outcomes = 10C42 red balls can be selected from 4 red balls in 4C2 waysAnd, remaining 2 balls can be selected from 2 white balls and 4 black balls in 6C2 ways.
\ Required probability = 4
26
210
4
C C
C
× = 3
7 or 9
21
42. (d) : Let A and B be two events of knowing lesson I and lesson II respectively.\ According to question,
P(A) = 60100
; P(B) = 40100
P(A ∩ B) = 20100
\ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 60100
+ 40100
– 20100
= 80100
Hence, required probability = P(AC ∩ BC) = P(A ∪ B)C = 1 – P(A ∪ B)
= 1 – 80100
= 20100
= 15
43. (d) : Two distinct numbers can be chosen from 1, 2, 3, 4, 5 in 5C2 ways.
Number of outcomes having arithmetic mean an integer i.e., {(1, 3), (1, 5) (2, 4), (3, 5)} = 4
\ Required probability = 4 4
1052C
= = 25
44. (e) : �ere are 9 elements in 3 × 3 matrices and each element can be �lled in two ways either – 1 or 1.\ Total possible matrices = 29
45. (b) : Total possible set of 2 × 2 symmetric matrices of entries either zero or one = 8Possbile set of matrices having determinant not zero = 4
\ Required probability = 48
= 12
46. (c) : Number of words that can be formed by using all 7 letters of word PROBLEM only once is 7!.47. (b) : Number of diagonals in n-sided polygon
= n n( )− 32
\ Number of diagonals in hexagon = 6 6 32
( )− = 9
48. (a) : We know sum of n odd numbers = n2
Number of odd terms from 1 to 2001 = 1001\ Sum of 1001 odd terms = (1001)2
49. (e) : Total number of ways of selecting two balls = 4C2Number of ways of selecting 2 balls in which no ball is black = 2C2
\ Required probability = 2
24
2
C
C =
16
50. (a) : We have, z = i9 + i19 = (i2)4 ⋅ i + (i2)9 ⋅ i = i + (–i) = 0 = 0 + 0i 51. (a) : Given data is 6, 7, 10, 12, 13, 4, 8, 12
\ Mean = 6 7 10 12 13 4 8 128
+ + + + + + + = 728
= 9
52. (e) : Given, x – 2 < 1 ⇒ x < 1 + 2 = 3Hence, set of all real numbers satisfying the inequality x – 2 < 1 is (– , 3)
53. (b) : Given, | |xx
−−
33
> 0
⇒ |x – 3| > 0 ⇒ x – 3 > 0 ⇒ x > 3\ x ∈ (3, )54. (c) : From the given data 8 has highest frequency.\ Mode of the given data is 8.55. (a) : Let the six numbers be x1, x2, x3, x4, x5, x6Given, mean of six numbers = 41
\ Mean = x x x x x x1 2 3 4 5 6
6+ + + + +
= 41
⇒ x1 + x2 + x3 + x4 + x5 + x6 = 246
| JUNE ‘1756
56. (b) : Given, f t dtx
( )0∫ = x2 + ex (x > 0)
⇒ f(x) = 2x + ex
\ f(1) = 2(1) + e = 2 + e
57. (c) : xx
dx+∫
11 2/ = x dx
xdx∫ ∫+ 1
= 23
x3/2 + 2x1/2 + c
58. (e) : Let H and E be the two events of people speaking Hindi and English respectively.\ n(H) = 50, n(E) = 20 n(H ∩ E) = 10\ n(H ∪ E) = n(H) + n(E) – n(H ∩ E) = 50 + 20 – 10 = 60i.e., number of people who speak atleast one of two languages is 60.
59. (a) : Given, f(x) = xx
+−
11
\ f(f(x)) = f xx
+−
11 =
xxxx
+−
+
+−
−
11
1
11
1
= ( )( )x xx x
+ + −+ − +
1 11 1
= 22x = x
60. (a) : Total number of outcomes = 6 × 6 = 36Favourable number of outcomes = 27
\ Required probability = 2736
= 34
61. (a) : We have, limx
x xx→
+ − −0
2 2
On rationalizing the numerator, we get
lim ( ) ( )( )
lim( )x x
x xx x x
xx x x→ →
+ − −+ + −
=+ + −0 0
2 22 2
22 2
=+ + −
=→
limx x x0
22 2
22 2
= 12
62. (b) : We have, dx
e ex x+ +−∫ 2 =
e dxe e
x
x x2 1 2+ +∫
= e dx
edtt
x
x( )+=∫ ∫1 2 2
[Put (ex + 1) = t ⇒ exdx = dt]
= −1t
+ c = −
+1
1ex + c
63. (b) : We have, tan π θ4 2
+
+ tanπ θ4 2
−
= tan tan
tan tan
π θ
π θ4 2
14 2
+
− +
tan tan
tan tan
π θ
π θ4 2
14 2
−
+
= 1
2
12
+
−
tan
tan
θ
θ + 1
2
12
−
+
tan
tan
θ
θ
= 1
21
2
12
2 2
2
+
+ −
−
tan tan
tan
θ θ
θ =
2 12
12
2
2
+
−
tan
tan
θ
θ
= 2 1cosθ
= 2 sec
64. (None of the options is correct) :
We have, dxx x2
1
0
2+ +−∫
= dx
x x21
0
2 14
14
+ + + −−∫ =
dx
x +
+−∫
12
74
21
0
= dx
x +
+
−∫
12
72
2 21
0 =
17 2
1 27 2
1
1
0
tan //
−
−
+x
= 27
1 27 2
1 27 2
1 1tan //
tan //
− −
− −
= 27
17
17
1 1tan tan− −+
= 47
17
1tan−
65. (e) : Let I = sinsin cos
xx x
dx+∫
0
2π
...(i)
⇒ I = sin
sin cos
π
π π
π2
2 20
2 −
−
+ −
∫x dx
x x
⇒ I = cos
cos sin
x dx
x x+∫0
2π
...(ii)
Adding (i) and (ii), we get
| JUNE ‘17 57
2I = sin cossin cos
x xx x
dx++
∫0
2π
⇒ I = 12
10
2
⋅∫ dxπ
= 12 2
π
= π4
66. (b) : Given, (x, y) is equidistant from (a + b, b – a) and (a – b, a + b)\ Using distance formula, we have
( ( )) ( ( ))x a b y b a− + + − −2 2
= ( ( )) ( ( ))x a b y a b− − + − +2 2
On squaring both sides, we get (x – (a + b))2 + (y – (b – a))2 = (x – (a – b))2 + (y – (a + b))2
⇒ x2 + a2 + b2 + 2ab – 2ax – 2bx + y2 + b2 + a2
– 2ab – 2by + 2ay = x2 + a2 + b2 – 2ab + 2bx – 2ax + y2 + a2 + b2
+ 2ab – 2by – 2ay⇒ – 4bx + 4ay = 0 ⇒ ay – bx = 0 or bx – ay = 067. (e) : Given, (1, 0), (0, 1) and (x, 8) are collinear\ Area of formed by these points is zero
i.e., 12
|1(1 – 8) + 0(8 – 0) + x(0 – 1)| = 0
⇒ 12
|– 7 – x| = 0 ⇒ x = – 7
68. (a) : Graph of max{x, x2} is shown below
Y
1
y = x2
y = x2
y = x
0 X
Hence, min value of max{x, x2} is 0.69. (c) : Given, f(x + y) = f(x) f(y)
f (3) = lim( ) ( )
h
f h fh→
+ −0
3 3
= lim( ) ( ) ( )
h
f f h fh→
−0
3 3 = ff h
hh( ) lim
( )3
10→
−
= ff h f
hh( ) lim
( ) ( )3
00→
− [ f(0) = 1]
= ff h f
hh( ) lim
( ) ( )3
0 00→
+ − = f(3) f (0)
= 3 × 11 = 33 \ f (3) = 33
70. (a) : We have, f f x fxx
′ = −−→
( ) lim ( ) ( )9 999
⇒ =−
×−−→
09
339
lim ( ) ( )( )x
f xx
f xf x
⇒ =−
− +× + −
−
→0
33 3
9 939
lim( )
( )( )( )
( )x
f xx x
f xf x
⇒ =−
−
×
( ) −
− +( )
→ →0
33
3
3 39 9
2 2lim
( )( )
lim( ) ( )
( ) (x x
f xx
f x
f x x
−
− +
→
lim( ( ) ) ( )x f x x9
93 3
⇒ =+
−
×→
03
30lim
( )x
f xx
lim
( )( ( ) )( )x
f xx f→
++
−− +
9
33
99 3 9 3
⇒ =+
−
× +×
→
03
336
93 69
lim( )
x
f xx
⇒+
−
=→
lim( )
x
f xx9
33
0
71. (e) : cos π4
+
x + cos π4
−
x
= cos π4
cos x – sin π4
sin x + cos π4
cos x
+ sin π4
sin x
= 2 cos π4
cos x = 22
cos x = 2 cos x
72. (d) : Area of triangle
= 12
|(–2)(5 + 1) + 1(– 1 – 2) + 6(2 – 5)|
= 12
|(– 2)(6) – 3 – 18| = 12
|–12 – 3 – 18|
= 12
| – 33| = 332
sq. units
| JUNE ‘1758
73. (b) : Equation of line passing through (1, 0) and (–4, 1) is
y −−
01 0
= x −− −
14 1
⇒ y = x −−
15
or x + 5y – 1 = 0Now, equation of line perpendicular to x + 5y – = 0 is 5x – y + = 0 is for some constant Also, 5x – y + = 0 passes through (– 3, 5)\ 5(– 3) – 5 + = 0 ⇒ = 20Hence, 5x – y + 20 = 0 is the required equation of line74. (b) : Given expansion is (1 + x2)5 (1 + x)4
= [5C0(x2)0 + 5C1(x2)1 + 5C2(x2)2 + 5C3(x2)3 + 5C4(x2)4 +5C5(x2)5] [4C0x0 + 4C1x + 4C2x2 + 4C3x3 + 4C4x4]
= [5C0 + 5C1x2 + 5C2x4 + 5C3x6 + 5C4x8 + 5C5x10]
[4C0 + 4C1x + 4C2x2 + 4C3x3 + 4C4x4]\ Coe�cient of x5 = 5C1 ⋅ 4C3 + 5C2 ⋅ 4C1
= 20 + 40 = 60
75. (e) : Given expansion is (1 – 2x)5 = 5C0 – 5C1(2x) + 5C2(2x)2 – 5C3(2x)3
+ 5C4(2x)4 – 5C5(2x)5
\ Coe�cient of x4 = 5C4 . (2)4 = 8076. (a) : We have, 5x2 + y2 + y = 8
⇒ 5x2 + y2 + y – 8 + 14
– 14
= 0
⇒ ( 5 x)2 + (y + 1/2)2 – 334
= 0
⇒ ( 5 x)2 + (y + 1/2)2 = 332
2
which represents an ellipse.77. (e) : We have, 4x2 + y2 – 8x + 4y – 8 = 0⇒ 4(x2 – 2x + 1 – 1) + (y2 + 4y + 4 – 4) – 8 = 0⇒ 4(x – 1)2 – 4 + (y + 2)2 – 4 – 8 = 0
⇒ ( )/
x −11 4
2 + (y + 2)2 = 16
⇒ ( )x −14
2 + ( )y + 2
16
2 = 1
\ Centre (1, –2).
78. (c) :
y
(0, 3)
( 3 , 0)O
y = –x2 + 3
x
y
x( − 3 , 0)
Required area = 2 32
0
3( )− +∫ x dx
= 23
33
0
3− +
x x = 2 3 33
3 3− +
= 2(2 3 ) = 4 3
79. (a) : Order of di�erential equation
d y
dx
3
3
2
+ d y
dx
2
2
2
+
dydx
5
= 0 is 3.
80. (a) : We have, f(x) = 2x + 42x
\ f (x) = 2 ⋅ 12 x
+ 42
⋅ −
12 3 2x /
= 12x
– 23 2( ) /x
⇒ f (2) = 12
– 12
= 0
81. (b) : Let r be the radius of given circle.Given, area of circle x2 – 2x + y2 – 10y + k = 0 is 25i.e., r2 = 25 ⇒ r2 = 25 ...(i)Also, radius from the given equation is
r = ( ) ( )1 52 2+ − k ⇒ r2 = 26 – k
⇒ 25 = 26 – k [Using (i)]⇒ k = 1
82. (d) : Let I = x
x xdx
+ −∫ 40332016
2017 ...(i)
Also, I = 403340332016
2017 −− +∫
xx x
...(ii)
Adding (i) and (ii), we get
| JUNE ‘17 59
2I = x x
x xdx dx
+ −( )+ −( ) = ⋅∫ ∫
4033
40331
2016
2017
2016
2017
⇒ I = 12
[2017 – 2016] = 12
83. (a) : We have, dydx
+ y tan x = sec x, y(0) = 0
�is is a linear di�erential equation
\ I.F. = e x dxtan∫ = elog|sec x| = sec x
\ Solution is given by y ⋅ sec x = sec secx x dx⋅∫⇒ y ⋅ sec x = sec2 x dx∫ ⇒ y sec x = tan x + cNow, we have y(0) = 0
⇒ (0)·sec 0 = tan (0) + c ⇒ c = 0
\ Particular solution is, y sec x = tan x
84. (e) : Since the vectors 2 2 6i j k+ + , 2 6i j k+ +λ ,
2 3i j k− + are coplanar.
\ 2 2 62 62 3 1
λ−
= 0
⇒ 2( + 18) – 2(2 – 12) + 6(– 6 – 2 ) = 0⇒ – 10 + 20 = 0 ⇒ = 285. (b) : Distance of the point (2, 1, 0) from the plane 2x + y + 2z + 5 = 0 is given by
( ) ( ) ( )2 2 1 1 0 2 5
2 1 22 2 2
⋅ + ⋅ + ⋅ +
+ + =
4 1 53
+ + = 10
3
86. (e) : Coordinates of vertices of hyperbola (0, 15) = (0, b)and foci (0, 20) = (0, be)\ b = 15 and be = 20
⇒ e = 2015
= 43
Now, e2 = 1 + ab
2
2 ⇒ 169
= 1 + a2
225
⇒ a2 = 79
× 225 = 175
\ Equation of hyperbola y
bxa
2
2
2
2− = 1
i.e., y2
225 – x2
175 = 1
87. (e) : Given, 15 6 3 6 15 211 4 6 6 36 4 216 1296
3 3+ + ⋅ ⋅ ⋅+ + + +( ) ( ) ( )
= ++
( )( )15 61 6
3
4 =
( )( )217
3
4 = 277
88. (a) : �e equation of the plane passing through the points (1, 0, 2), (–1, 1, 2) and (5, 0, 3) is
x y z− − −− − − −
− − −
1 0 21 1 1 0 2 2
5 1 0 0 3 2 = 0 ⇒
x y z− −−
1 22 1 0
4 0 1 = 0
⇒ (x – 1) (1) – y(–2) + (z – 2)(–4)⇒ x – 1 + 2y – 4z + 8 = 0⇒ x + 2y – 4z + 7 = 089. (b) : Given equation is y2 – 4y – x + 3 = 0⇒ (y – 2)2 – x – 1 = 0 ⇒ (y – 2)2 = x + 1Now, shi�ing the origin to the point (–1, 2) without rotating the axes and denoting the new coordinates w.r.t. these axes by X and Y, we get y – 2 = Y and x + 1 = XNow, substituting (X = 0, Y = 0) y = 2, x = –1Hence, vertex w.r.t to old axes (–1, 2).
90. (a) : Given, a b c+ + = 0 ⇒ a = − +( )b c ⇒ | |a 2 = | ( ) |− +b c 2
⇒ | |a 2 = | |b 2 + | |c 2 + 2 | | | |b c cos ( is angle between b and c )
⇒ 49 = 25 + 9 + 30 cos
⇒ cos = 1530
= 12
⇒ = π3
91. (d) : Let f(x) = 2x3 – 9ax2 + 12a2x + 1\ f (x) = 6x2 – 18ax + 12a2
Now, f (x) = 0 ⇒ 6x2 – 18ax + 12a2 = 0⇒ (x – 2a) (x – a) = 0⇒ x = a or 2aAlso, f (x) = 12x – 18a\ f (a) = 12a – 18a = – 6a < 0 (a > 0)and f (2a) = 24a – 18a = 6a > 0Hence, p = a and q = 2aNow, p3 = q ⇒ a3 = 2a ⇒ a(a2 – 2) = 0
⇒ a = ± 292. (e) : We have, f(x) + f(y) = f(x + y) Put x = y = 1, we get f(1 + 1) = f(1) + f(1) ⇒ f(2) = 7 + 7 = 14 f(2 + 1) = f(2) + f(1) ⇒ f(3) = 14 + 7 = 21
| JUNE ‘1760
Continuing in the same way, we get f (4) = 28, f (5) = 35 and so on.
\ f rr
( )=∑
1
100 = f (1) + f (2) + f (3) +...+ f (100)
= 7 + 14 + 21 +.....+ 700Since, the series forms an A.P.
\ f rr
( )=∑
1
100= 100
2[7 + 700] = 50 × 707 = 50 × 7 × 101
93. (a) : Given, ( )x y− + +
12
34
2 2 = 1
16
or ( )/
x −11 8
2 +
( / )/
y + 3 41 16
2 = 1
\ Eccentricity (e) = 12
2− ba
= 1 1 161 8
− ( / )( / )
= 1 12
− = 12
94. (b) : We have, max{ , }x x dx3
1
1
−∫
= x dx3
1
0
−∫ + x dx
0
1
∫ = x4
1
0
4
−
+ x2
0
1
2
= 14
95. (d) : We have, sin x + cos y = 2It is possible only if sin x = 1 and cos y = 1
⇒ x = π2
and y = 0 \ x + y = π2
+ 0 = π2
96. (a) : Given numbers a, a + r, a + 2r are in A.P. Also their product = 64. �is is possible only when three numbers are equal to 4.i.e., a = a + r = a + 2r = 4\ Minimum value of a + 2r is 4.
97. (b) : We have, S = 19
13 7
15 5
17 3
19! ! ! ! ! ! ! !
+ + + +
= 19
2 2 93
9 8 7 65! ! !
+ ×8× + × × ×
= 19
2 120 2 72 20 3024120!
( )( ) ( )× + +
=
1 614410 12
××!
= 210
9
!
98. (d) : Given, f(x) =
x x x
x xx
2 3
21 2 30 2 6
= x(12x2 – 6x2) – x2(6x) + x3(2 – 0)= 6x3 – 6x3 + 2x3 = 2x3 \ f (x) = 6x2
99. (c) : Let I = xx
dx2
3 21+∫ ( )Put x3 = t ⇒ 3x2dx = dt
\ I = 13 1 2
dtt+∫ = 1
3[tan–1 t] + c = 1
3 tan–1(x3) + c
100. (d) : We have, f(x) = x2 + ex
\ f (x) = 2x + ex
f (x) = 2 + ex
f (x) = ex
Hence, the least value of n so that fn = fn + 1 is 3.101. (e) : sin 765° = sin(2 × 360° + 45°)
= sin 45° = 12
102. (d) : Distance of the point (3, –5) from the line 3x – 4y – 26 = 0 is
3 3 4 5 26
3 42 2
⋅ + − − −
+
( )( ) = 9 20 26
5+ − = 3
5
103. (c) : Given, f(x) = ( )t t dtx
2
01+ +∫ = t t t
x3 2
03 2
+ +
\ f(x) = x x x
3 2
3 2+ +
\ f(x) = x x x
3 2
3 2+ +
on [2, 3] gives
f(2) = 83
+ 42
+ 2 = 203
(minimum)
and f(3) = 273
+ 92
+ 3 = 332
(maximum)
\ Di�erence between the maximum and minimum
value is 332
– 203
= 596
104. (c) : Let, f(x) = x2 + ax + b = 0Since a and b are roots of f(x) \ a + b = –a and ab = b ⇒ a = 1\ 1 + b = – 1 ⇒ b = –2So, f(x) = x2 + x – 2Also, f (x) = 2x + 1. For maximum/minimum f (x) = 0
⇒ x = −12
Now, f (x) = 2 > 0
| JUNE ‘17 61
\ x = −12
is the minimum point.
\ Minimum value = f −
=1
2 −
+ −
12
12
2 – 2
= 14
– 12
– 2 = −94
105. (b) : We have, y = 4x + c and x2
4 + y2 = 1
y = 4x + c touches the given ellipse
\ x2
4 + (4x + c)2 = 1
⇒ x2 + 4(16x2 + c2 + 8xc) = 4⇒ x2 + 64x2 + 4c2 + 32xc – 4 = 0⇒ 65x2 + 32xc + 4c2 – 4 = 0Now, discriminant D = 0 ⇒ (32c)2 – 4(65)(4c2 – 4) = 0⇒ 1024c2 – 1040c2 + 1040 = 0⇒ 16c2 = 1040 ⇒ c2 = 65⇒ c = ± 65106. (d) : Given equations x – y = 2, 2x – 3y = – and 3x – 2y = – 1 are consistent
\ λ
λ− −−−
1 22 33 2 1
= 0
⇒ (–3 + 2 ) + 1(2 – 3 ) – 2(–4 + 9) = 0⇒ –3 + 2 2 + 2 – 3 + 8 – 18 = 0⇒ 2 – 3 – 4 = 0 ⇒ ( – 4)( + 1) = 0\ = –1, 4107. (a) : �e set {(x, y) : |x| + |y| = 1} in xy plane represents
x y x yx y x y
x y x yx y x y
+ = > >− = > <
− + = < >− − = < <
1 0 01 0 0
1 0 01 0 0
; ,; ,
; ,; ,
i.e.,
(0, 1)
(1, 0)
(0, –1)
(–1, 0)Ox
y
y
x
x – y = 1
x + y = 1–x + y
= 1
–x – y = 1
Hence, the given set represents a square.
108. (a) : Let tan–1 34
= ⇒ tan = 3
4
\ cos = 45
Now, cos tan−
1 34
= cos = 45
109. (b) : AB = BC\ By using distance formula, we have
( ) ( )1 6 3 12 2− + + = ( ) ( )x − + −1 8 32 2
On squaring both sides, we get (–5)2 + (4)2 = (x – 1)2 + (5)2
⇒ x2 + 1 – 2x = 16 ⇒ x2 – 2x – 15 = 0⇒ (x + 3)(x – 5) = 0 ⇒ x = –3, 5110. (e) : We have, n = 15, Incorrect x2 = 2830,Incorrect (x) = 170\ Correct (x) = (Incorrect x – Incorrect value)
+ Correct value = (170 – 20) + 30 = 180
\ Correct Mean = Correct Σx15
= 18015
= 12
Similarly, Correct x2 = Incorrect x2 – (Incorrect value)2
+ (Correct value)2
= 2830 – (20)2 + (30)2 = 2830 + 500 = 3330
\ Correct variance = Correct ( )Σxn
2 – (Correct mean)2
= 333015
– (12)2 = 222 – 144 = 78
111. (e) : Let be the angle between the lines
x − 22
= y −15
= z +−
33
and x +−
21
= y − 48
= z − 54
\ cos = | ( ) ( ) ( )( ) |
( ) ( ) ( ) ( ) ( ) ( )
2 1 5 8 3 4
2 5 3 1 8 42 2 2 2 2 2
− + + −
+ + − − + +
= | |− + −2 40 12
38 81 =
269 38
i.e., = cos–1 269 38
112. (d) : We have, ( ) ( )x a x a− ⋅ + = 12⇒ | | | |x a2 2− = 12Since, a be a unit vector \ | |a = 1⇒ | |x 2 – 1 = 12 ⇒ | |x 2 = 13 ⇒ | |x = 13113. (d) : Sides of triangular region are y = 2x + 1; y = 3x + 1 and x = 4.
| JUNE ‘1762
(4, 13)
(0, 1)
(–1/2, 0)
O
A
B(4, 9)
x = 4y = 2x + 1y = 3x
+ 1
X
Y
X
C
\ Area of shaded region i.e., ABC where A = (0, 1), B = (4, 9), C = (4, 13)
= 12
|0(9 – 13) + 4(13 – 1) + 4(1 – 9)| = 12
|4(12) – 32|
= 12
× 16 = 8 sq. units
114. (b) : Given, nCr – 1 = 36, nCr = 84, nCr + 1 = 126
Since, n
rn
r
C
Cn r
r−= − +
1
1
\ 8436
= n rr
− +1 ⇒ n rr
− +1 = 73
⇒ 7r = 3n – 3r + 3 ⇒ 10r = 3n + 3 ...(i)
Also, n
rn
r
C
C +1 = r
n r+−
1 ⇒ 84126
= rn r
+−
1 ⇒ rn r
+−
1 = 23
⇒ 3r + 3 = 2n – 2r ⇒ 5r = 2n – 3 ...(ii)Solving (i) & (ii), we get r = 3
115. (e) : f (x) = lim( ) ( )
h
f x h f xh→
+ −0
= lim( ) ( ) ( )
h
f x f h f xh→
−0
[ f(x + y) = f(x)f(y)]
= f xf h
hh( ) lim
( )→
−0
1 = f xh g hhh
( ) limsin ( )
→
+ −0
1 3 1
= f x g h hhh
( ) lim ( ) sin→0
3
= f x g hhh
( ) ( ) lim sin0 330→
× 3 = 3 f(x) g(0)
116. (b) : Given, x
xx
−−
−
1 1 11 1 11 1 1
= 0
⇒ (x – 1)[(x – 1)2 –1] – 1[x – 1 – 1] + 1[1 – x + 1] = 0⇒ (x – 1)[x2 + 1 – 2x – 1] – x + 2 + 2 – x = 0⇒ (x – 1) x(x – 2) – 2x + 4 = 0⇒ x(x – 1)(x – 2) – 2(x – 2) = 0⇒ (x – 2)[x2 – x – 2] = 0 ⇒ (x – 2)(x – 2)(x + 1) = 0⇒ x = 2, –1
117. (None of the options is correct) : Given, T7 = T8⇒ nC6(2a)n – 6 (–3b)6 = nC7(2a)n – 7 (–3b)7
⇒ n
n!
( )! !− 6 6(2a)n – 6 (–3b)6 =
nn
!( )! !− 7 7
(2a)n – 7 (–3b)7
⇒ 1
6( )n −(2a) = ( )−3
7b ⇒ 2
3ab
= − −( )n 67
Applying componendo & dividendo, we get
2 32 3
6 76 7
131
131
a ba b
nn
nn
nn
+−
= − +− −
= −− −
= −+
118. (c) : Mean of �rst n odd natural numbers
x nn
nn
n= + + + + − = =1 3 5 2 1 2.... ( )
Sum of square of �rst n odd natural numbers i.e.,
12 + 32 + 52 + ... + (2n – 1)2 = n n n( )( )2 1 2 13
+ −
\ Standard deviation = n n n
nn( )( )2 1 2 1
32+ − −
= n2 1
3−
119. (b) : Given, S = {1, 2, 3, ..., 10}. \ Set containing odd numbers of S = {1, 3, 5, 7, 9}\ Number of subsets of S containing only odd numbers = (2)5 – 1 = 32 – 1 = 31120. (e) : Let the vertices of parallelogram be A(0, 0), B(7, 2), C(5, 9) and D(12, 11)
C
B
D
A
(12, 11) (5, 9)
(0, 0) (7, 2)Area of || gm ABCD = area of ABC + area of ADC
Now, area of ABC = 12
|0(2 – 9) + 7(9 – 0) + 5(0 – 2)|
= 12
|63 – 10| = 532
\ Area of ADC = 12
|0(11 – 9) + 12(9 – 0) + 5(0 – 11)|
= 12
|108 – 55| = 532
\ Area of || gm ABCD = 532
532
+ = 53 sq. units
| JUNE ‘17 63
1. �e distance of the point (–2, 4, –5) from the line x y z+ = − = +3
34
58
6 is
(a) 3710
(b) 3710
(c) 3710
(d) 3710
2. If A is a square matrix of order 3 × 3, then |KA| is equal to(a) K|A| (b) K2|A| (c) 3K|A| (d) K3|A|
3. Equation of line passing through the point (1, 2) and perpendicular to the line y = 3x – 1 is(a) x – 3y = 0 (b) x + 3y = 0(c) x + 3y – 7 = 0 (d) x + 3y + 7 = 0
4. General solution of di�erential equation dydx
y y+ = ≠1 1( ) is
(a) log 11 −
= +y
x C (b) log|1 – y| = x + C
(c) log|1 + y| = x + C (d) log 11 −
= − +y
x C
5. �e value of C in mean value theorem for the function f(x) = x2 in [2, 4] is
(a) 2 (b) 4 (c) 72
(d) 3
6. �e value of lim coscosθ
θθ→
−−0
1 41 6 is
(a) 94
(b) 49
(c) 93
(d) 34
7. If y x xx x
dydx
= +−
−tan sin coscos sin
, then1 is equal to
(a) 0 (b) 12
(c) π4 (d) 1
8. If 11
1+−
=i
i
m, then the least positive integral
value of m is(a) 4 (b) 1 (c) 2 (d) 3
9. | |x dx+−∫ 25
5 is equal to
(a) 28 (b) 29 (c) 27 (d) 30
10. cos coscos cos
2 2xx
dx−−∫
θθ
is equal to
(a) 2(sinx + x cos ) + C (b) 2(sinx – x cos ) + C (c) 2(sinx + 2x cos ) + C (d) 2(sinx – 2x cos ) + C
11. �e area of the region bounded by the curve y = x2 and the line y = 16 is
(a) 2563
sq . units (b) 128
3sq . units
(c) 323
sq . units (d) 643
sq . units
12. If A and B are �nite sets and A B, then(a) n(A ∪ B) = n(B) (b) n(A ∩ B) = n(B)(c) n(A ∩ B) = f (d) n(A ∪ B) = n(A)
13. If a matrix A is both symmetric and skew symmetric, then(a) A is diagonal matrix (b) A is a zero matrix(c) A is scalar matrix (d) A is square matrix
14. If 3
13 24 1
xx
= then x is equal to
(a) 8 (b) 4 (c) ± 2 2 (d) 2
Karnataka CETSOLVED PAPER 2017
| JUNE ‘1764
15. �e integrating factor of the di�erential equation
x dydx
y x⋅ + =2 2 is (x ≠ 0)
(a) elog x (b) log |x|(c) x (d) x2
16. �e perpendicular distance of the point P(6, 7, 8) from XY-plane is(a) 7 (b) 6 (c) 8 (d) 5
17. �e shaded region in the �gure is the solution set of the inequations
(a) 5x + 4y ≤ 20, x ≤ 6, y ≤ 3, x ≥ 0, y ≥ 0 (b) 5x + 4y ≥ 20, x ≤ 6, y ≥ 3, x ≥ 0, y ≥ 0 (c) 5x + 4y ≥ 20, x ≤ 6, y ≤ 3, x ≥ 0, y ≥ 0 (d) 5x + 4y ≥ 20, x ≥ 6, y ≤ 3, x ≥ 0, y ≥ 0
18. If an LPP admits optimal solution at two consecutive vertices of a feasible region, then(a) the required optimal solution is at the midpoint
of the line joining two points. (b) the optimal solution occurs at every point on
the line joining these two points.(c) the LPP under consideration is not solvable.(d) the LPP under consideration must be
reconstructed.
19. 3 + 5 + 7 + .... to n terms is(a) n2 (b) n(n – 2)(c) n(n + 2) (d) (n + 1)2
20. If 21 30
01 2
5 61 8x
y
+
=
, then the value of
x and y are(a) x = 3, y = 3 (b) x = –3, y = 3 (c) x = 3, y = –3 (d) x = –3, y = –3
21. �e derivative of cos–1(2x2 – 1) w.r.t cos–1 x is
(a) 2 (b) 2x
(c) 1 – x2 (d) −
−
1
2 1 2x
22. A box has 100 pens of which 10 are defective. �e probability that out of a sample of 5 pens drawn one by one with replacement and atmost one is defective is
(a) 9
10 (b) 1
29
10
4
(c) 910
12
910
5 4
+
(d) 12
910
5
23. If y = log (log x), then d ydx
2
2 is equal to
(a) ( log )
log1
2+ x
x x (b)
− +( log )
( log )
12x
x x
(c) ( log )
( log )
12
+ x
x x (d)
− +( log )
log
12
x
x x
24. ( )
( )
x e
xdx
x+
+∫3
4 2 is equal to
(a) e
xC
x
( )++
4 (b)
ex
Cx
( )++
4 2
(c) e
xC
x
( )++
3 (d)
14 2( )x
C+
+
25. 12 2 2 2
0
2
a x b xdx
sin cos
/
+∫π
is equal to
(a) πa
b4 (b)
πba4
(c) π2ab
(d) πa
b2
26. Let f : R → R be de�ned by f(x) = x4, then(a) f is one-one but not onto (b) f is neither one-one nor onto(c) f is one-one and onto (d) f may be one-one and onto
27. �e point on the curve y2 = x where the tangent
makes an angle π4
with X-axis is
(a) (1, 1) (b) 14 2
, 1
(c) 12 4
, 1
(d) (4, 2)
28. �e total number of terms in the expansion of (x + a)47 – (x – a)47 a�er simpli�cation is
(a) 24 (b) 96 (c) 47 (d) 48
| JUNE ‘17 65
29. �e function f(x) = x2 + 2x – 5 is strictly increasing in the interval(a) [–1, ) (b) (– , –1)(c) (– , –1] (d) (–1, )
30. �e degree of the di�erential equation
12 2
2
2+
=dy
dxd ydx
is
(a) 1 (b) 4 (c) 2 (d) 3
31. Binary operation * on R – {–1} de�ned by
a b ab* =
+ 1 is
(a) * is associative and commutative (b) * is neither associative nor commutative(c) * is commutative but not associative (d) * is associative but not commutative
32. �e plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1 (a) with X-axis. �e value of a is equal to
(a) 3
2 (b) 2
7 (c) 2
3 (d)
37
33. If coe�cient of variation is 60 and standard deviation is 24, then arithmetic mean is
(a) 207
(b) 720
(c) 140
(d) 40
34. �e contrapositive statement of the statement “If x is prime number, then x is odd” is(a) If x is not a prime number, then x is odd. (b) If x is not a prime number, then x is not odd.(c) If x is a prime number, then x is not odd. (d) If x is not odd, then x is not a prime number.
35. �e probability distribution of X is
X 0 1 2 3P(X) 0.3 k 2k 2k
�e value of k is(a) 0.7 (b) 0.3 (c) 1 (d) 0.14
36. x x dx2 2 5+ +∫ is equal to
(a) ( )x x x+ + +1 2 52
− + + + + +2 1 2 52log x x x C
(b) 12
1 2 52( )x x x+ + +
+ + + + + +2 1 2 52log x x x C
(c) ( )x x x+ + +1 2 52
+ + + + + +2 1 2 52log x x x C
(d) ( )x x x+ + +1 2 52
+ + + + + +1
21 2 52log x x x C
37. If nC12 = nC8 then n is equal to(a) 12 (b) 26 (c) 6 (d) 20
38. If yf x g x h x
l m na b c
dydx
=( ) ( ) ( )
, then is equal to
(a)
f x g x h xl m na b c
′ ′ ′( ) ( ) ( )
(b) l m n
f x g x h xa b c′ ′ ′( ) ( ) ( )
(c) f x l ag x m bh x n c
′′′
( )( )( )
(d) l m na b c
f x g x h x′ ′ ′( ) ( ) ( )
39. If tan tan− −+ =1 1 45
x y π , then cot–1x + cot–1y is equal to
(a) π5
(b) 35π
(c) 25π (d)
40. �e range of the function f x x( ) = −9 2 is(a) [0, 3] (b) (0, 3] (c) (0, 3) (d) [0, 3)
41. Two events A and B will be independent if (a) P(A ∩ B ) = (1 – P(A)) (1 – P(B))(b) A and B are mutually exclusive (c) P(A) + P(B) = 1 (d) P(A) = P(B)
42. �e eccentricity of the ellipse x y2 2
36 161+ = is
(a) 2 56
(b) 2 134
(c) 2 54
(d) 2 136
43. If a b& are unit vectors, then angle between a band for 3 a b− to be unit vector is(a) 45º (b) 30º (c) 90° (d) 60°
| JUNE ‘1766
44. If � � � � � � � �a i j k b i j k= + + = + +2 2 3λ and are orthogonal, then value of is
(a) 32 (b) 1 (c) − 5
2 (d) 0
45. �e value of cos245° – sin215° is
(a) 3 12 2
+ (b) 34
(c) 3 12 2
− (d) 3
246. �e range of sec–1x is
(a) [0, ] (b) [ , ]02
π π− { }(c)
−
π π2 2
, (d) −
π π2 2
,
47. If a b c, , are unit vectors such that a b c+ + = 0 , then the value of a b b c c a⋅ + ⋅ + ⋅ is equal to
(a) − 32
(b) 3 (c) 32
(d) 1
48. dxe xsin
/
/
+−∫ 12
2
π
π is equal to
(a) 1 (b) 0 (c) π2
(d) − π2
49. �e rate of change of volume of a sphere with respect to its surface area when the radius is 4 cm is(a) 2 cm3/cm2 (b) 4 cm3/cm2
(c) 8 cm3/cm2 (d) 6 cm3/cm2
50. tan
cot tan
/ 7
7 70
2 xx x
dx+∫
π is equal to
(a) π4
(b) π2
(c) π6
(d) π3
51. If |x –2| ≤ 1, then(a) x ∈ (1, 3) (b) x ∈ (–1, 3)(c) x ∈ [1, 3] (d) x ∈ [–1, 3)
52. [ ].
.x dx
0 2
3 5
∫ is equal to
(a) 3.5 (b) 4.5 (c) 3 (d) 453. �e area of triangle with vertices (K, 0), (4, 0), (0, 2)
is 4 square units, then value of K is(a) 8 (b) 0 or –8 (c) 0 (d) 0 or 8
54. If f x Kx xx
( ) ifif
= ≤>
2 23 2
is continuous at x = 2, then
the value of K is
(a) 43
(b) 34
(c) 3 (d) 4
55. If Ax x
x x=
− −
− −
11 1
1 1π
ππ
ππ
sin ( ) tan
sin cot ( )
Bx x
x x=
−
−
− −
− −
cos ( ) tan
sin tan ( )
1 1
1 1
ππ
ππ
then A – B is equal to
(a) 12
I (b) I (c) O (d) 2I
56. If f(x) = 8x3, g(x) = x1/3, then fog(x) is (a) 83x (b) 8x3
(c) 8x (d) (8x)1/3
57. Let ∆ ∆= =
Ax x
By y
Cz z
A B Cx y zzy zx xy
2
2
21
1
1
1
and then
(a) 1 = – (b) 1 = (c) 1 = 2 (d) 1 ≠
58. If sin , tan , thenx tt
y tt
dydx
=+
=−
21
212 2 is equal
to(a) 1 (b) –1 (c) 2 (d) 0
59. Re�ection of the point (a, , ) in XY plane is (a) (0, 0, ) (b) (a, , – )(c) (–a, – , ) (d) (a, , 0)
60. Area of the region bounded by the curve y = cos x, x = 0 and x = is(a) 2 sq. units (b) 3 sq. units(c) 4 sq. units (d) 1 sq. units
ANSWER KEYMPP-2 CLASS XI1. (c) 2. (a) 3. (d) 4. (a) 5. (b)
6. (c) 7. (b) 8. (a,c) 9 . (b,c) 10. (d)
11. (b,c) 12. (a,c,d) 13. (a) 14. (a) 15. (b)
16. (d) 17. (4) 18. (5) 19. (1) 20. (1)
| JUNE ‘17 67
SOLUTIONS
1. (c) : Given, point is A(–2, 4, –5),
Line (l) is x y z+ = − = + =33
45
86
λ (say)
Co-ordinates of B are (3 – 3, 5 + 4, 6 – 8)\ Direction ratios of AB are (3 – 1, 5 , 6 – 3)
Now, (3 – 1)(3) + (5 )(5) + (6 – 3)(6) = 0
⇒ λ = 310
∴ = − + −AB i j k� ��� � � �1
1032
65
∴ = = + +d AB| |� ��� 1
10094
3625
= + + = =1 225 144
100370100
3710
2. (d) : We have |KA| = Kn|A|, Here n = 3\ |KA| = K3|A|
3. (c) : Equation of required line is y – 2 = – 13
(x – 1)⇒ x + 3y – 7 = 0
4. (a) : Given, dydx
y+ = 1
�is is a linear di�erential equation.
∴ = =∫I.F. e edx x1
\ Solution is given by, ye e dx e Cx x x= ⋅ = +∫ 1 1⇒ ex(y – 1) = C1 ⇒ x + log|y – 1| = log C1 ⇒ –x – log|y – 1| = – logC1
⇒−
= +log 11y
x C [where – logC1 = C]
or log 11 −
= +y
x C
5. (d) : We have, f(x) = x2 in [2,4]\ According to mean value theorem,
We have, ′ = −−
f C f b f ab a
( ) ( ) ( )
[where a = 2 and b = 4]
∴ = −−
= − =2 4 24 2
4 22
122
2 2C f f( ) ( ) ( ) ( )
⇒ 2C = 6 ⇒ C = 3
6. (b) : We have, lim coscosθ
θθ→
−−0
1 41 6
=
→lim sin
sinθ
θθ0
2
223
= =( )( )23
49
2
2
7. (d) : We have, y x xx x
= +−
−tan sin coscos sin
1
= +
−
= +
= +− −tan tan
tantan tan1 11
1 4 4xx
x xπ π
∴ =dydx
1
8. (a) : Given, 11
1+−
=ii
m ⇒ im = i4 ⇒ m = 4
9. (b) : Let I x dx= +−∫ | |25
5
= − + + +
−−
−
∫∫ ( ) ( )x dx x dx2 22
5
5
2
= − +
+ +
−
−
−
x x x x2
5
2 2
2
5
22
22
= + = =9
2492
582
29
10. (a) : Let I xx
dx= −−∫
cos coscos cos
2 2θθ
= − − −−∫
( cos ) ( cos )cos cos
2 1 2 12 2xx
dxθθ
= −−
= +∫ ∫2 2
2 2(cos cos )cos cos
(cos cos )xx
dx x dxθθ
θ
= 2(sinx + x cos ) + C11. (a) : Required area = Area of shaded portion
| JUNE ‘1768
= ∫20
16ydy = ⋅ = =2 2
343
4 2563
3 2016 3y / [ ] sq. units
12. (a) : Given, A B ⇒ A ∩ B = A \ n(A ∪ B) = n(A) + n(B) – n(A ∩ B) = n(A) + n(B) –n(A) = n(B)
13. (b) : Given, A is symmetric ⇒ aij = aji ...(i) i ≠ jA is skew symmetric ⇒ aij = –aji ...(ii) and aii = 0Adding (i) and (ii) we get 2aij = 0 ⇒ aij = 0\ A is a zero matrix.
14. (c) : We have, 3
13 24 1
xx
=
⇒ 3 – x2 = 3 – 8 ⇒ = ⇒ = ±x x2 8 2 2
15. (d) : Given, x dydx
y x dydx
yx
x⋅ + = ⇒ + =2 22
∴ = = =∫
I.F. e e xxdx x
222log
16. (c) : Perpendicular distance of the point P(6,7,8) from XY-plane is 8.
17. (c) : Clearly, shaded region represents 5x + 4y > 20, x < 6, y < 3, x > 0, y > 0
18. (b) : �e optimal solution occurs at every point on the line joining these two points.
19. (c) : Given series is in A.P. with �rst term (a) = 3, common di�erence (d) = 2
∴ = × + − = +S n n n nn 22 3 1 2 2[ ( ) ] ( )
20. (a) : Given 2 6
1 2 25 61 8
++
=
yx
⇒ 2 + y = 5 and 2x + 2 = 8 ⇒ y = 3, x = 3
21. (a) : Let u = cos–1(2x2 – 1) and v = cos–1 x ⇒ cos v = x\ u = cos–1(2cos2v – 1) = cos–1(cos2v) = 2v
∴ =dudv
2
22. (c) : p q p n= = = − = =10100
110
1 910
5, ,
P X x Cx
x x( )= =
−5
5110
910
\ P(X ≤ 1) = P(X = 0) + P(X = 1)
=
+
50
0 55
1
1 4110
910
110
910
C C
23. (b) : We have, y = log (log x) ⇒ dydx x x
= 1log
⇒ = − ⋅ + ⋅
=− +d y
dx x xx
xx
x
x x
2
2 2 21 1 1
1
( log )log
( log )
( log )
24. (a) : Let I = ( )( )x ex
dxx+
+∫34 2 = + −
+∫( )
( )x e
xdx
x4 14 2
=+
−+
=
++∫
14
14 42x x
e dx ex
Cxx
( )
e f x f x dx e f x Cx x( ( ) ( )) ( )+ ′ = + ∫
25. (c) : Let Ia x b x
dx=+∫1
2 2 2 20
2
sin cos
/π
=
+∫sectan
/ 2
2 2 20
2 xdxa x b
π
Put tan x = t ⇒ sec2 x dx = dt
∴ =+
=
+
∞ ∞
∫ ∫I dta t b
dt
a t ba
2 2 20 2 2
2
20
=
=−∞1
221
0aab
atb ab
tan π
26. (b) : Given f(x) = x4
Now, f(1) = f(–1) but 1 ≠ –1\ f is not one-one Also, co-domain of f is R and range of f is [0, )\ f is not onto
27. (b) : We have, y2 = x ⇒ 2yy = 1
⇒ ′ = = ⇒ =yy
y12 4
12
tan π
When y x y= = =
=12
12
14
22
,
28. (a) : Number of terms in (x + a)47 – (x – a)47
= + =47 12
24
[Number of terms in (a + b)n – (a – b)n is n + 1
2,
when n is odd number]
| JUNE ‘17 69
29. (d) : f is strictly increasing ⇒ f (x) > 0⇒ 2x + 2 > 0 ⇒ x > –1
30. (a) : Highest order derivative is d ydx
2
2 and its power is 1.
Hence, degree of di�erential equation is 1.
31. (b) : We have, a * b = a
b + 1
1 2 13
2 1 22
1* *= = =but
�us 1 * 2 ≠ 2 * 1 \ * is not commutative
Now, (1 * 2) * 3 = 13 * 3 = 1 3
41
12/ =
and ( )1 2 3 1 12
112
1
23* * *= =
+=
\ * is not associative [Infact, * is not a binary operation !!!]
32. (b) : Let f be the angle made by plane 2x – 3y + 6z – 11 = 0 and X-axis i.e., (1,0,0)
∴ = × − × + ×+ + + +
sin | |φ 2 1 3 0 6 04 9 36 1 0 0
⇒ =
=− −φ αsin sin ( )1 127
33. (d) : We have C.V. = σx
× 100
⇒ = × ⇒ =60 24 100 40x
x
34. (d) : �e contrapositive statement of the statement “If x is prime number, then x is odd” is “If x is not odd, then x is not a prime number.”
35. (d) : We know, P X( ) =∑ 1⇒ 0.3 + k + 2k + 2k = 1 ⇒ 5k = 0.7 ⇒ 0.14
36. (b) : Let I x x dx x dx= + + = + +∫ ∫2 2 22 5 1 2( )
= + + + + + + + + +( ) logx x x x x x C12
2 5 2 1 2 52 2
37. (d) : Given, nC12 = nC8 ⇒ 8 + 12 = n ⇒ n = 20 [ nCx = nCy ⇒ x = y or x + y = n]
38. (a, c, d)
39. (a) : Given, tan–1 x + tan–1 y = 45π
⇒ + = − =− −cot cot1 1 45 5
x y π π π
40. (a) : Let y = 9 2− x ⇒ y2 = 9 – x2
⇒ x2 = 9 – y2 ⇒ x = 9 2− y
Clearly, 9 – y2 > 0 ⇒ y2 < 9⇒ –3 < y < 3But y > 0. Hence 0 < y < 3.
41. (a) : If A and B are independent�en A and B are also independent⇒ P(A ∩ B ) = P(A )P(B ) = (1 – P(A))(1–P(B))
42. (a) : e a ba
= − = − =2 2 36 16
62 5
643. (b) : Given, | | | |a b= = 1
Now, | | | | | | ( )3 3 2 32 2 2a b a b a b− = + − ⋅
⇒ 1 = 3(1) + 1 − 2 3 | || | cosa b θ
⇒ 1 = 4 −2 3 1 1( )( )cosθ
⇒ = ⇒ = °cosθ θ32
30
44. (c) : For, a band to be orthogonal a b⋅ = 0⇒ (2)(1) + ( )(2) + (1)(3) = 0
⇒ 5 + 2 = 0 ⇒ = − 52
45. (b) : cos245° – sin215° = cos(45°+15°)cos(45°–15°)
= cos60° cos30° = ⋅ =12
32
34
46. (b) : Range of sec–1 x is [ , ]02
π π−
47. (a) : Here, a b c, , are unit vectors and a b c+ + = 0 ∴ + + =( )a b c 2 0
⇒ + + + ⋅ + ⋅ + ⋅ =| | | | | | ( )a b c a b b c c a2 2 2 2 0
⇒ + + + ⋅ + ⋅ + ⋅ =1 1 1 2 0( )a b b c c a
⇒ ⋅ + ⋅ + ⋅ = −a b b c c a 32
48. (c) : Let I dxe x=
+−∫ sin/
/
12
2
π
π ...(i)
⇒ =+
=+−
− −∫ ∫I dx
ee dx
ex
x
xsin/
/ sin
sin/
/
1 12
2
2
2
π
π
π
π ...(ii)
| JUNE ‘1770
Adding (i) and (ii), we get
2 112
2I e
edx
x
x= ++−
∫sin
sin/
/
π
π
⇒ = = = ⇒ =−−∫2 1
222
2
2I dx x I[ ] /
/
/
/
ππ
π
ππ π
49. (a) : dVdS
dV dtdS dt
r drdt
r drdt
r= = = = =//
/cm4
8 242
22
3 2π
πcm
50 . (a) : tan
cot tan
/ 7
7 70
2
4x
x xdx
+=∫
ππ
f x
f x f xdx( )
( )
/
+ −
=
∫ πππ
240
2
51. (c) : Here, |x –2| ≤ 1 ⇒ –1 ≤ x – 2 ≤ 1 ⇒ 1 ≤ x ≤ 3⇒ x ∈[1,3]
52. (b) : Here, [ ]..
. .x dx dx dx dx dx= + + +∫∫ ∫ ∫ ∫0 1 2 3
0 2
1
0 2
3 5
1
2
2
3
3
3 5
= 0 + [x]12 + [2x]2
3 + [3x]33.5 = 1 + 2(1) + 3(0.5) = 4.5
53. (d) : Given, area of triangle with vertices (K, 0),(4, 0), (0, 2) is 4 sq. units i.e.,
± = ⇒ − + = ±4 12
0 14 0 10 2 1
2 82
4K
K
⇒ K = 0 or 854. (b) : Given, f(x) is continuous
⇒ = =→ →− +lim ( ) lim ( ) ( )
x xf x f x f
2 22
⇒ = ⇒ =→
lim ( )x
Kx K2
2 3 4 3 ⇒ =K 34
55 . (None of the options is correct) :
If matrix B will be, Bx x
x x=
−
−
− −
− −
11 1
1 1π
ππ
ππ
cos tan
sin tan
�en, A B I− = 12
56. (c) : fog(x) = f(g(x)) = = =f x x x( ) ( )/ /1 3 1 3 38 8
57. (b) : We have, ∆ = =
Ax x
By y
Cz z
xyz
A xx
B yy
C zz
2
2
2
1
1
1
1
1
1
= = =A x yzB y xzC z xy
A B Cx y zzy zx xy
∆1
58. (a) : We have, x tt
t=+
=− −sin tan12
121
2
and tan tany tt
t=−
=− −12
121
2
∴ = ⇒ =y x dydx
1.
59. (b) : Re�ection of the point (a, , ) in XY plane is(a, ,– ) .
60. (a) : Given, curves are y = cos x ; x = 0 and x =
\ Required area = 20
2cos
/xdx
π∫
= [ ]2 02sin /x π = 2 sq. units
| JUNE ‘1772
JEE MAIN
1. If a1 = 11, a2 = 75, a3 = 20, a4 = 23 and an = an – 1 – an – 2+ an – 3 – an – 4, n 5, then a31 – a53 + a75 =(a) 53 (b) 58(c) 65 (d) 75
2. If xyz = 1, x + 1z
= 5, y + 1x
= 29, then z + 1y
=
(a) 13
(b) 14
(c) 2 (d) 34
3. In a triangle ABC, D 1 1 1 12
12
22
32r r r r
+ + +
=
(a) 2 tanA (b) 2 cotA(c) 4 tanA (d) 4 cotA
4. A straight line through the point (a, b) meets x-axisat A and y-axis at B. O is the origin. If (a, b) = (4, 1),then the minimum value of OA + OB is(a) 7 (b) 8 (c) 9 (d) 10
5. If x ≠ a, y ≠ b, z ≠ c anda y zx b zx y c
= 0, then
x ax a
y by b
z cz c
+−
+ +−
+ +−
=
(a) –1 (b) 0(c) 1 (d) 2
JEE ADVANCED
6. A bag contains 30 tokens numbered serially from0 to 30. �e number of ways of selecting 3 tokensfrom the bag, so that the sum of the numbers onthem is 30, is divisible by(a) 2 (b) 3 (c) 5 (d) 7
COMPREHENSION
Let f x x x x( ) [sin ] [cos ], [ , ]= + ∈− −1 1 0 1
7. �e number of points at which f(x) is notdi�erentiable is(a) 0 (b) 1 (c) 2 (d) 3
8. �e solution set of the equation f(x) = 0 is an interval of length
(a) 2 14
sin −
p (b) 2 14
sin +
p
(c) 2 14
cos −
p (d) 2 14
cos +
p
INTEGER TYPE
9. Ten boys and two girls are to be seated in a rowsuch that there are atleast 3 boys between thegirls. �e number of ways this can be done isλ · 12!, where λ =
MATRIX MATCH
10. �e curve f (x, y) = 0 lies in the �rst quadrant.�e tangent at a point on it meets the positivex and y axes at A and B and O is the origin
List-I List-IIP. f(x, y) = 4xy – 1 1. AB = 1Q. f(x, y) = x2 + y2 – 1 2. OA + OB = 1
R. f x y x y( , ) = + − 1 3. OA · OB = 1
S. f(x, y) = x2/3 + y2/3 – 1 4.1 1 12 2OA OB
+ =
P Q R S (a) 2 1 3 4(b) 1 2 4 3(c) 4 3 2 1 (d) 3 4 2 1
Maths Musing was started in January 2003 issue of 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.
See Solution Set of Maths Musing 173 on page no. 85
| JUNE ‘17 73
1. Let x, y, z be real numbers such thatcos x + cos y + cos z = 0 andcos 3x + cos 3y + cos 3z = 0, thencos 2x cos 2y cos 2z(a) ≤ 0(b) ≥ 0(c) depends on x, y, z values(d) data insu�cient
2. Eliminate from the system : cos 2 = cos ( + a)and sin 2 = 2 sin ( + a)(a) (cos a)2/3 – (sin a)2/3 = 2/3
(b) (cos a)2/3 + (sin a)2/3 = 2/3
(c) (cos a)1/3 – (sin a)1/3 = 1/3
(d) (cos a)1/3 – (sin a)1/3 = 1/3
3. �e bisector of BAC intersects the circumcircleof ABC at D. If AB2 + AC2 = 2AD2 then angle ofintersection of AD and BC is(a) 30° (b) 45°(c) 60° (d) 90°
4. �e laws of points (x, y) satisfying the equationsx2 + y cos2 a = x sin a cos a andx cos 2a + y sin 2a = 0 lies on (a is a parameter)(a) circle (b) parabola(c) ellipse (d) hyperbola
5. In ABC, ABC = ACB = 40°. If P is a point inthe interior of the triangle such that PBC = 20°and PCB = 30° then(a) BP = BA (b) BP = 2BA
(c) BP = 12 BA (d) BP = 3BA
6. Let ABC be a triangle of area 12
, then minimum value of a2 + cosec A is(a) 2 (b) 3 (c) 4 (d) 5
7. Let a, , , be positive numbers such that for all x,sin ax + sin x = sin x + sin x, then + = (a) a (b) 2a (c) 3a (d) 4a
8. Let S be the set of all triangles ABC for which
5 1 1 1 3 6AP BQ CR AP BQ CR r
+ +
− =min. { , , }
,
where r is inradius and P, Q, R are points of tangency of incircle with sides AB, BC, CA respectively then all the triangles in the set S are(a) scalene (b) isosceles(c) equilateral (d) right angled
9. Let ABC be a triangle such that max. {A, B} = C + 30°
andRr
= +3 1 , R is circumradius, r is inradius
then ABC is(a) scalene (b) isosceles(c) equilateral (d) right angled
10. Let ABCDEFGHIJKL be a regular do-decagon thenABAF
AFAB
+ =
(a) 1 (b) 2 (c) 3 (d) 4
11. Let a, b, c, d ∈ [0, ] such that 2 cos a + 6 cos b + 7 cos c + 9 cos d = 0 and 2 sin a – 6 sin b + 7 sin c – 9 sin d = 0 then
cos ( )cos ( )
a db c
++
=
(a) 73
(b) 37
(c) 35
(d) 53
12. If sin x cos y + sin y cos z + sin z cos x = 32
, then
(a) sin x = cos 2y (b) sin x = sin y(c) sin x = cos y (d) sin x = cos z
20 CHALLENGINGPROBLEMS
For Entrance Exams
| JUNE ‘1774
13. In ABC, = sin A sin B + sin B sin C + sin C sin Aand (1 + sin A)(1 + sin B) (1 + sin C) = 2( + 1), then
ABC is(a) scalene (b) isosceles(c) equilateral (d) right angled
14. Let ABC be a triangle such thatsin2B + sin2C = 1 + 2 sinB sinC cosA, then ABC is(a) scalene (b) isosceles(c) equilateral (d) right angled
15. In ABC, cot cot cotA B C sr2
42
92
67
22 2
2
+ + =
where s → semiperimeter and r → inradius, thenthe ratio of the sides of the triangle is(a) 45 : 40 : 13 (b) 50 : 55 : 45(c) 55 : 65 : 70 (d) 60 : 65 : 80
16. Let n be a positive integer and for real numbers and akl . (k, l = 1, 2, 3, ... n) (k > l), we have
sinsin
cos ( )2
21
2 2nxx
a k l xkll k n
= + −≤ < ≤∑λ
(x ≠ m , m ∈I) then = (a) l (b) k (c) n (d) k – l
17. In ABC, A = 30° and an inscribed circle of �xedradius r is drawn. If ABC has least perimeter then
B =(a) 30° (b) 45° (c) 60° (d) 75°
18. If A1, A2, A3, A4 are the angles of a convex quad.then max. value of
sin sin sin sinA A A A1 2 3 4
2 2 2 2
+
+
+
is
(a) 2 (b) 2 2 (c) 3 2 (d) 4 2
19. If a, b, c, k are constants and a are variablessubject to the relation a tan a + b tan + c tan = kthen minimum value of tan2 a + tan2 + tan2 is
(a) k
a b c
2
2( )+ +(b)
ka b c
2
2 2 2+ +
(c) ( )a b c
k+ + 2
2(d)
a b ck
2 2 2
2
+ +
20. For x ∈ R, the minimum value of|sin x + cos x + tan x + sec x + cosec x + cot x| is(a) 2 2 (b) 2 2 – 1
(c) 2 2 + 1 (d) 2 1− BP
SOLUTIONS
1. (a) : Using cos 3x = 4 cos3x – 3 cos x.We have, cos3 0x =∑using identity a abc a a ab3 23− = −∑ ∑ ∑∑ ( )( )We have, abc = cos x cos y cos z = 0Let cos z = 0. So, cos x = – cos y andcos 2x = cos 2y and cos 2z = –1So, cos 2x cos 2y cos 2z = – cos2 2x ≤ 0.
2. (b) : Expanding the given equations, we have(cos cos a – sin sin a) = (cos2 – sin2 )and sin cos a + cos sin a = sin cos ⇒ cos a = cos3 and sin a = sin3 Hence, (cos a)2/3 + (sin a)2/3 = 2/3.
3. (b) : Let CAB = a, ABC = , BCA = thenusing sine rule, we have
ABsin γ
= AC ADsin sin ( / )β α β
=+2
So, AB2 + AC2 = 2AD2 becomes sin2 + sin2 = 2 sin2 (a/2 + )
Simplifying, cos ( – )[1 + cos ( + )] = 0i.e., cos ( – ) = 0. So, a + 2 = /2
So, AEC = α β π2 4
45+ = = °
4. (b) : Simplifying the �rst equation, we havex sin 2a – y cos 2a = 2x2 + yand the other given equation isx cos 2a + y sin 2a = 0Solving, we have
sin 2a = x x y
x yy x yx y
( )cos
( )22
22
2 2
2
2 2
++
=− +
+and α
Hence, sin2 2a + cos2 2a = 1 gives
( )2 2
2 2
x yx y
++
2
= 1 i.e., 4x2 + 4y – 1 = 0, Parabola.
5. (a) : Let us assume that AB = AC = 1 unit thenBC = 2 cos 40° and BPC = 130°Applying sine rule in BPC, we have
BPsin ( )30°
= BCsin ( )130°
⇒ BP =BC ⋅ °
°=
° ⋅ °°
=sin
coscos sin
cos30
402 40 30
401
i.e., BP = AB.
| JUNE ‘17 75
6. (d) : Given, area = 12
⇒ 12
bc sin A = 12
⇒ cosec A = bc and [bc ≥ 1]Now, a2 + cosec A = a2 + bc = b2 + c2 – 2bc cos A + bc
[cosine rule]
= b c bc A bc2 2 22 1+ − − +sin
= b c b c bc A bc2 2 2 2 22+ − − +( sin )
= b bc c b c2 2 2 22 1+ + − −
≥ 3 2 12 2bc b c− −
Let x = bc (≥ 1) then y = 3 2 12x x− − gives5x2 – 6xy + y2 + 4 = 0. As x ∈ R, D ≥ 0 gives(–6y)2 – 20(y2 + 4) ≥ 0 i.e., y ≥ 5 .
7. (b) : Di�erentiating the given identity three times,we have, a3 cos ax + 3 cos x = 3 cos x + 3 cos xAlso, acos ax + cos x = cos x + cos xIn particular for x = 0, we have
a + = + and a3 + 3 = 3 + 3
i.e., (a + )3 = ( + )3
⇒ a = on simpli�cation.So, (a – )(a – ) = a2 – a( + ) +
= a2 – a(a + ) + a = 0⇒ = a and = a i.e., + = 2a
8. (b) : Let us assume that min. {AP, BQ, CR} = APand let tan (A/2) = x, tan (B/2) = y, tan (C/2) = z.
So, AP = rx
, BQ = ry
and CR = rz
Now, the relation in question becomes 2x + 5y + 5z = 6and in any , we know that xy + yz + zx = 1.Now, eliminating (x) from these two equations, we have 5y2 + 5z2 + 8yz – 6y – 6z + 2 = 0i.e., (3y – 1)2 + (3z – 1)2 = 4(y – z)2
or, 5a2 + 5 2 + 8a = 0 [where 3y – 1 = a,3z – 1 = ]
⇒ a = 0 = for real solutions.
i.e., y = z = 13
and so, x = 43
i.e., isosceles triangle.9. (d) : Let max. {A, B} = A then A.T.Q, A – C = 30°
and using the identity
r = 42 2 2
R A B Csin . sin . sin
We have, r r A B C= +
⋅
⋅
4 3 1
2 2 2sin sin sin
i.e., 3 14− = sin cos cosB A C A C
2 2 2−
−+
i.e.,3 14−
= sin cos cosB B2
302
1802
°
−° −
i.e., sin sin .2
2 26 2
43 14
0B B−
+
+
−
=
Solving, sin (B/2) = 6 24
22
−or
i.e., B2
= 15° or 45°.
But B = 90° is not possible. Hence, B = 30° and A = 90°, C = 60°, i.e., [Right angled ]
10. (d) : Let R be the circumradius then
AB = 212
R sin π
and AF = 2 512
R sin π
So, the required quantity is 22 5
2 52
RR
RR
sinsin
sinsin
θθ
θθ
+
where θ π=
12
= sin sin
sin sin
2 2 55
θ θθ θ+
= 1 2 1 10
4 6− + −
−cos coscos cos
θ θθ θ
= 4 (on simpli�cation)11. (a) : Rearranging the two equations, we have
2sin a – 9 sin d = 6 sin b – 7 sin cand 2 cos a + 9 cos d = – 6 cos b – 7 cos cSquaring and adding both the equations, we have85 + 36 cos (a + d) = 85 + 84 cos (b + c)
i.e.,cos ( )cos ( )
a db c
++
= =8436
73
12. (c) : �e given equation can be rewritten as,(sin x – cos y)2 + (sin y – cos z)2 + (sin z – cos x)2 = 0i.e., sin x = cos y
13. (d) : Equating the two values and simplifying,we have(1 – sin A)(1 – sin B) (1 – sin C) = 0i.e., sin A = 1 or sin B = 1 or sin C = 1i.e., ABC is right angled triangle.
| JUNE ‘1776
14. (d) : Using cosine rule, we have a2 + 2bc cos A = b2 + c2
And using sine rule, here, we have sin2 B + sin2 C = sin2 A + 2 sin B sin C cos ANow comparing this with equation given in question, we have, sin2 A = 1 i.e., A = 90°
15. (a) : In any triangle, we have
cot cot cot cot .cot .cotA B C A B C sr2 2 2 2 2 2
+ + = =
So, the given relation in question becomes,
( ) cot cot cot6 3 22
22
32
2 22 2
+ +
+
+
22A B C
= 62
62
62
2
cot cot cotA B C+ +
i.e., Equality in Cauchy-Schwarz inequality.
So, cot ( / ) cot ( / ) cot ( / )A B C2
62 2
33 2
2= =
i.e., cot ( / ) , cot ( / ) ( / )A B C2 7 2 74
2 79
= = =and cot
⇒ sin , sin , sinA B C= = =725
5665
126130
i.e., Sides of the triangle are 26, 80 and 90.
16. (c) : Using the two identities, we have
S1 = cos ( )sin cos ( )
sin2
1
1mx
nx n xxm
n=
⋅ +
=∑
and S2 = sin ( )sin sin ( )
sin2
1
1mx
nx n xxm
n=
⋅ +
=∑
S S12
22+ =
sinsin
nxx
2
But S S12
22+ = (cos 2x + cos 4x + ... + cos nx)2
+ (sin 2x + sin 4x + ... + sin nx)2
= n k x l x kx lx+ +∑2 2 2 2 2(cos cos sin sin )1 ≤ l < k ≤ n
i.e., S S12
22+ = n k l x
l k n+ −
≤ < ≤∑ cos ( )2
1
i.e., = n
17. (d) : In any triangle, we have cot ( / )A sr
2 =∑
So, perimeter 2s = 2 2r A. cot ( / )∑
Since A, r are �xed, s is min. when cot ( / )A 2∑ is min.i.e., cot (B/2) + cot (C/2) is min.
i.e.,cos ( / )
sin ( / ) sin ( / )A
B C2
2 2 is min.
or, sin (B/2) . sin (C/2) is max.
i.e.,12 2 2
cos sinB C A−
−
is max.
i.e., cosB C−
2
= 1 for max. i.e., B = C.
\ ABC is an isosceles with B = C = 75°.18. (b) : A1 + A2 + A3 + A4 = 2 ,
So, A A A A1 2 3 4
2 2 2 2+ + + =
Now, f (x) = sin x is a concave function.
So, sin sin sin sinA A A A1 2 3 4
2 2 2 2
+
+
+
is max.
When A A A A1 2 3 4
2 2 2 2= = = = each = π
4[Result of Jensen's inequality]
So, required max. value is 4 ⋅ sin ( /4) = 2 2 .19. (b) : Use the identity
( tan tan )b cγ β−∑ 2
= (a2 + b2 + c2)(tan2 a + tan2 + tan2 )– (a tan a + b tan + c tan )2
We have, RHS ≥ 0⇒ ( ) ( tan )a k2 2 2 0∑ ∑⋅ − ≥α
i.e., tanmin
22
2α∑( ) = kaΣ
20. (b) : Let E = |sin x + cos x + tan x + sec x + cosec x+ cot x|
Putting a = sin x, b = cos x, c = a + bWe have on simpli�cation, the given expression
E = cc
cc
+−
= − +−
+21
1 21
1
where c ∈ −[ , ]2 2
Using AM ≥ GM on (c – 1) and 21c −
, we have
Emin = | |− +2 2 1 i.e., 2 2 1−
| JUNE ‘17 77
1. �e sum of all the non-real roots of(x2 + x – 2) (x2 + x – 3) = 12 is(a) 1 (b) –1(c) 6 (d) –2
2. Statement-1 : �e equation sinx + xcosx = 0 has atleast one root in the interval (0, ).Statement-2 : Between any two roots of f(x) = 0, there exists atleast one root of f (x) = 0.(a) Statement-1 is true, Statement-2 is true; Statement-2
is a correct explanation for Statement-1.(b) Statement-1 is true, Statement-2 is true; Statement-2
is not correct explanation for Statement-1.(c) Statement-1 is true, Statement-2 is false.(d) Statement-1 is false, Statement-2 is true.
3. If LCM of p, q is r2t4s2, where r, s, t are primenumbers and p, q are positive integers . �en the number of ordered pairs (p, q) is(a) 252 (b) 254(c) 225 (d) 224
4. Let f(x) = max{x, x3} x ∈ R the set of points wheref(x) is not di�erentiable is(a) {–1, 1} (b) {–1, 0}(c) {0, 1} (d) {–1, 0, 1}
5. Statement-1: �e sum of the �rst 30 terms of thesequence 1, 2, 4, 7, 11, 16, 22, 29, 37, 46 ... is 4520.Statement-2: �e successive di�erences of the terms of the sequence form an A.P.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.(d) Statement-1 is false, Statement-2 is true.
6. tan−
=
=∑ 12
1
12rr
n
(a) tan–1n (b) tan−+
11
nn
(c) tan−+
12
nn
(d) tan− ++
1 12
nn
7. �e statement p → (q → p) is equivalent to(a) p → (p → q) (b) p → (p q)(c) p → (p q) (d) p → (p q)
8. �e remainder le� out when 82n – (62)2n + 1 isdivisible by 9 (where n ∈ N)(a) 0 (b) 2 (c) 7 (d) 8
9. Consider all functions that can be de�ned from the set A = {1, 2, 3} to the set B = {1, 2, 3, 4, 5}. A function f(x) is selected at random from these functions. �e probability that, selected function satis�es f(i) ≤ f(j) for i < j, is equal to
(a) 625
(b) 1225
(c) 25
(d) none of these
Math Archives, as the title itself suggests, is a collection of various challenging problems related to the topics of IIT-JEE Syllabus. This section is basically aimed at providing an extra insight and knowledge to the candidates preparing for IIT-JEE. In every issue of MT, challenging problems are offered with detailed solution. The readers’ comments and suggestions regarding the problems and solutions offered are always welcome.
BESTPROBLEMS10
B y : b . : 0 9 3 3 4 8 7 0 0 2 1
10. [cot ]x dx0
π
∫ =
(where [⋅] denotes the greatest integer function)
(a) π2
(b) 1 (c) –1 (d) − π2
SOLUTIONS
1. (b) : Put x2 + x = y then, we have y2 – 5y – 6 = 0⇒ (y – 6) (y + 1) = 0⇒ x2 + x – 6 = 0 or x2 + x + 1= 0⇒ x = –3, 2 or x = , 2
Sum of non-real roots = + 2 = –12. (c) : Take f(x) = xsinx, which is continuous in[0, ] and di�erentiable in (0, ). Also f(0) = f( ) = 0By Rolle’s theorem, there exists at least one root of f (x) = 0 ⇒ xcosx + sinx = 03. (c) : No. of ordered pairs
= (2(2) + 1)(2(4) + 1)(2(2) + 1)= 5 × 9 × 5 = 225
4. (d) :–1 0 1
5. (d) : a2 – a1 = 1, a3 – a2 = 2, a4 – a3 = 3, ...
an n n n
n = +−( ) = − +1
12
22
2
\ Sum = ⋅ ⋅ − ⋅ + ⋅
=12
30 31 616
30 312
2 30 4525
6. (b) : 12
24
2 1 2 1
1 4 12 2 2r rr r
r= =
+( ) − −( )+ −( )
Sum = (tan–1(2r + 1) – tan–1 (2r – 1))
= tan–1(2n + 1) – tan–1(1) =+
−tan 11
nn
7. (b) : p → (q → p) ~ p (q → p) ~ p (~ q p) p → (p q)
8. (b) : (1 + 63)n – (63 – 1)2n + 1
Remainder is 2.9. (d) : Total Function = 53
10. (d) : Let = ⇒ = −∫∫[cot ] [cot( )]x dx x dxπππ
00
∴ + = − + − =− ∉
∈
∫� � ∵( ) [ ] [ ]1
10
0dx x x
x zx z
π ifif
∴ = − π2
| JUNE ‘1780
1. a1 , ... , ak, ak + 1, ..., an are positive numbers (k < n).Suppose that the values of ak + 1, ...., an are �xed. How should one choose the values of a1,...., an in
order to minimize aa
i
ji j i j, , ≠∑ ?
2. Let m be a positive integer. De�ne the sequencea0 , a1 , a2, .... by a0 = 0, a1 = m and an + 1 = m2an – an – 1 for n = 1, 2, 3, .... . Prove that an ordered pair (a, b) of non-negative integers, with a ≤ b, gives a solution to the equation
a bab
m2 2
21
++
=
if and only if (a, b) is of the form (an , an + 1) for some n ≥ 0.
3. In a ABC, C = 2 B. P is a point in theinterior of ABC satisfying that AP = AC andPB = PC. Show that AP trisects A.
4. Determine all the possible values of the sum of thedigits of the perfect squares.
5. ABCD is a convex quadrilateral and O is theintersection of its diagonals. Let L, M, N be themid-points of DB, BC, CA respectively. Supposethat AL, OM, DN are concurrent. Show thateither AD BC or [ABCD] = 2[OBC].
SOLUTIONS
1. To minimize the given rational function, choose
aa a
a a
iik n
k n
=+ +
+ +
= ⋅ =+
+
1
1
1 2
1 21 1
1 2...
...( ) , , ,
/
/A H ....,k
where A is the arithmetic mean and H is the harmonic mean of ak + 1 , ..., an. To prove this, we will be forgiven if we change notation : let xi = ai , i = 1, 2, ..., k and br = ak + r , r = 1, ..., m with k + m = n and denote the given rational function F(x1, ..., xk). �en we have F(x1, ..., xk) = X + Y + B, where
Xxx
xx
i
j
j
ii j k= +
∑
≤ < ≤1,
Yxb
bxi k
i
r
r
ir m= ∑ +
∑
≤ ≤ ≤ ≤1 1,
Bbb
bb
r
s
s
rr s m= +
∑
≤ < ≤1.
Note that B is �xed and Y can be improved to
Yb
x bxrr m
i ir m ii k
= ∑
+ ∑
∑
≤ ≤ ≤ ≤≤ ≤
1 11 11
= +
∑ m x m
xiii HA
where A is the arithmetic mean and H is the harmonic mean of the br .
Now we recall that the simple function α βxx
+(with a, , x all positive) assumes its minimum
when α βxx
= ; that is x = β α/ . Thus each
of the terms in Y (and so Y itself ) assumes its minimum when we choose, for i = 1, 2, ..., k,
x mmi = =A
HAH
( / ), as asserted.
But there is more. It is also known that each term in X, (and so X itself) assumes its minimum when xi = xj, with 1 ≤ i < j ≤ k. �us choosing all xi = AH minimizes both X and Y and, since B
| JUNE ‘17 81
is �xed, minimizes F(x1, ..., xk) as claimed.2. Let us �rst prove by induction that
a aa a
mn n
n n
21
2
1
21
+⋅ +
=+
+ for all n ≥ 0.
Proof : Base case (n = 0) : a aa a
m m02
12
0 1
22
100 1
+⋅ +
= ++
= .
Now, let us assume that it is true for n = k, k ≥ 0.
�en, a aa a
mk k
k k
21
2
1
21
+⋅ +
=+
+
a a m a a mk k k k2
12 2
12+ = ⋅ ⋅ ++ +
a m a m a a ak k k k k+ + ++ − ⋅ ⋅ +12 4
12 2
122
= m2 + m4a2k + 1 – m2 · ak . ak + 1
a2k + 1 + (m2ak + 1 – ak)2 = m2 + m2ak + 1(m2ak + 1 – ak)
a a m m a ak k k k+ + + ++ = + ⋅ ⋅12
22 2 2
1 2
Therefore, a aa a
mk k
k k
+ +
+ +
+⋅ +
=12
22
1 2
21
, proving the
induction. Hence (an, an+1) is a solution to a bab
m2 2
21
++
= for all n ≥ 0.
Now, consider the equation a bab
2 2
1++
= m2 and suppose
(a, b) = (x, y) is a solution with 0 ≤ x ≤ y. �en
x yxy
m2 2
21
++
= ...(1)
If x = 0 then it is easily seen that y = m, so (x, y) = (a0, a1). Since we are given x ≥ 0, suppose now that x > 0.Let us show that y ≤ m2x.Proof by contradiction : Assume that y > m2x. �en y = m2x + k where k ≥ 1.Substituting into (1) we get
x m x kx m x k
m2 2 2
22
1+ +
+ +=( )
( )( )
x2 + m4x2 + 2m2xk + k2 = m4x2 + m2kx + m2
(x2 + k2) + m2(kx – 1) = 0.No w, m 2( k x – 1 ) ≥ 0 s i n c e k x ≥ 1 a n d x2 + k2 ≥ x2 + 1 ≥ 1 so (x2 + k2) + m2(kx – 1) ≠ 0. �us we have a contradiction, so y ≤ m2x if x > 0.Now substitute y = m2x – x1, where 0 ≤ x1 < m2x, into (1).
We havex m x xx m x x
m2 2
12
21
2
1+ −
− +=
( )( )
x2 + m4x2 – 2m2x · x1 + x12 = m4x2 –m2x · x1 + m2
x2 + x12 = m2(x · x1 + 1)
x xx x
m2
12
1
21
+⋅ +
= ....(2)
If x1 = 0, then x2 = m2. Hence x = m and(x1, x) = (0, m) = (a0, a1). But y = m2x – x1 = a2, so (x, y) = (a1, a2). �us suppose x1 > 0.Let us now show that x1 < x.Proof by contradiction: Assume x1 ≥ x.Then m2x – y ≥ x, since y = m2x – x1, and
x yxy
x y x2 2
1++
− ≥ , since (x, y) is a solution to
a bab
m2 2
21
++
= .
So x3 + xy2 ≥ x2y + xy2 + x + y. Hence x3 ≥ x2y + x + y, which is a contradiction since y ≥ x > 0.With the same proof that y ≤ m2x, we have x ≤ m2x1. So the substitution x = m2x1 – x2 with x2 ≥ 0 is valid.Subst itut ing x = m2x1 – x2 into (2) g ives
x xx x
m12
22
1 2
21
+⋅ +
= .
If x2 ≠ 0, then we continue with the substitution
x m xi x ii= −
+ +1
22(*) until we get
x x
x xmj j
j j
21
2
1
21
+
⋅ +=+
+
and xj + 1 = 0. (�e sequence xi is decreasing, non-negative and integer.)So, if xj + 1 = 0, then x2
j = m2 so xj = m and(xj + 1, xj) = (0, m) = (a0, a1).�en (xj, xj –1) = (a1, a2) since xj –1 = m2xj – xj + 1 (from (*)).Continuing, we have (x1, x) = (an – 1, an) for some n. �en (x, y) = (an, an + 1).
Hence a bab
m2 2
21
++
= has solutions (a, b) if and
only if (a, b) = (an, an + 1) for some n.
| JUNE ‘1782
3. Let PAC and BAP be 2a and respectively.�en, since C = 2 B, we deduce fromA + B + C = 180° that2a + + 3B = 180°. ...(1)�e angles at the base of theisosceles triangle PAC are each 90° – a. Also BPC isisosceles, having base anglesC – (90° – a) = 2B + a – 90°, and so BPA = 180° – ( PBA + BAP)= 180° – [B – (2B + a – 90°) + 180° – 2a – 3B]= 4B + 3a – 90°As usual, let a, b and c denote the lengths of thesides BC, AC and AB. By the Law of Cosines,applied to BPA, where PA = b and PB = PC= 2b sin a,c2 = b2 + (2b sin a)2 – 2⋅b⋅2b sin a⋅cos (4B + 3a
– 90°),so thatc2 = b2 [1 + 4 sin2 a – 4 sina sin(4B + 3a)] ...(2)We now use the fact that C = 2 B is equivalent to the condition c2 = b(b + a).
Since a = 2⋅ PC ⋅ cos(2B + a – 90°) = 4b sin a sin (2B + a), we have
c2 = b2 [1 + 4 sin a sin(2B + a)] ...(3)�erefore, from (2) and (3), we getb2 [1 + 4 sin2 a – 4 sin a sin (4B + 3a)]
= b2 [1 + 4 sin a sin (2B + a)],which simpli�es to sin a – sin (4B + 3a) = sin (2B + a).Since sin a – sin (4B + 3a) = –2 cos(2B + 2a) sin(2B + a), this equation may be rewritten as sin(2B + a). [1 + 2 cos (2B + 2a)] = 0Since, from (1), 2B + a < 180°, we must have 1 + 2 cos(2B + 2a) = 0, giving cos(2B + 2a) = – 1/2; that is,2B + 2a = 120° ...(4)Since, again from (1), 2B + 2a < 180°Finally, we may eliminate B between (1) and (4) to obtain a = . �e result follows.
4. �e squares can only be 0, 1, 4 or 7 mod 9.�us the sum of the digits of a perfect square cannotbe 2, 3, 5, 6 or 8 mod 9, since the number itselfwould then be 2, 3, 5, 6 or 8 mod 9.We shall show that the sum of the digits of aperfect square can take every value of the form0, 1, 4 or 7 mod 9.(10m – 1)2 = 102m – 2 ⋅ 10m + 1
= ≥− −
9 9 9 8 0 01 11 1
... ... , .m m
m��� �
A
B C
Pa a
�e sum of the digits is 9m, giving all the values greater than or equal to 9 congruent to 0 mod 9(10m – 2)2 = 102m – 4 ⋅ 10m + 4
= ≥− −
9 9 9 6 0 0 4 11 1
... ... , .m m
m��� �
�e sum of the digits is 9m + 1, which gives all values greater than or equal to 10 congruent to 1 mod 9.(10m – 3)2 = 102m – 6 ⋅ 10m + 9
= ≥− −
9 9 9 4 0 0 9 11 1
... ... , .m m
m��� ���
�e sum of the digits is 9m + 4, which takes every value greater than or equal to 13 which is congru-ent to 4 mod 9(10m – 5)2 = 102m – 10m+1 + 25
=− −
9 9 0 0 0 2 51 1
... ... .m m������
�e sum of the digits is 9(m – 1) + 7 = 9m – 2, from which we get every value greater than or equal to 7 congruent to 7 mod 9.We have taken care of all the integers apart from 0, 1, 4, which are the sums of the digits of 02, 12 and 22 respectively.
5. Let O be the origin of a coordinate system whereA, B, C, D are represented by (a, 0), (0, b), (c, 0),(0, d) with a, b positive and c, d negative. �us Lis the point
02 2 2 2
0, ( ) , , , ( ) ,b d M c b N a c+( ) ( ) +( )is is and
AL : (b + d)x + 2ay – a(b + d) = 0OM : bx – cy = 0DN : 2dx + (a + c) y – d(a + c) = 0.�ese lines are concurrent if and only if
b cb d a a b d
d a c d a c
−+ − +
+ − +=
02
20( )
( ).
�is equation reduces (a�er some manipulation) to (ab – cd) [(a – c) (b – d) + 2bc] = 0.Consequently, either(a) ab = cd, in which case AD BC, or
(b) 12
2 12
( )( )sin sina c b d bc− − = −( )α α
(where a = AOB), in which case [ABCD] = 2 [OBC].
| JUNE ‘1784
1. For r = 0, 1, ........., 10, let Ar, Br, Cr denote respectively, the coe�cients of xr in the expansion of (1 + x)10,
(1 + x)20, (1 + x)30. �en r =∑
1
10 Ar (B10Br – C10 Ar) =
(Ram Krishan, West Bengal)Ans. (1 + x)10 = A0 + A1x + A2x2 + ........... + A10x10 (x + 1)20 = B0x20 + B1x19 + B2x18 + ............+ B20. Considering the coe�cient of x20 in the product, we getA0B0 + A1B1 + A2B2 + .......... + A10 B10 = coe�cient of x20 in the expansion of (1 + x)10(x + 1)20 = (1 + x)30 which is C20
\ A B C Cr rr
= ==∑ 20 10
0
10
But n n n n
nn
n0 1 222 2 2 2
+
+
+ +
=
........
\ A02 + A1
2 + A22 + .............. + A Bn
210
2010
=
=
A Brr
210
0
10=
=∑ . Now, A B B C Ar r r
r( )10 10
1
10−
=∑
= −==∑∑B A B C Ar r r
rr10 10
2
1
10
1
10
= B10 (C10 – 1) – C10 (B10 – 1) = C10 – B10.
2. If
4 4 1
4 4 1
4 4 1
112
2
2
2
a a
b b
c c
fff
−
( )( )( )
=
3 3
3 3
3 3
2
2
2
a a
b b
c c
+
+
+
, f (x) is a
quadratic function and its maximum value occurs at a point V. A is a point of intersection of y = f (x) with x-axis and point B is such that the chord AB subtends a right angle at V. Find the area enclosed by y = f (x) and the chord AB. (Suresh Prasad, Jharkhand)
Ans. �e given equation implies, (4f (–1) – 3) x2 + (4f (1) – 3) x + f (2) = 0 is satis�ed by 3 roots a, b, c
⇒ It is an identity \ f (–1) = 34 , f (1) =
34 , f (2) = 0
If f (a) = a x2 + x + , then = 0, a = − 14 , = 1
⇒ f (x) = − + = −x x2 2
41 4
4�e maximum point, V = (0, 1) ; A(– 2, 0). Taking B = (2t, 1 – t2)
AVB = π2 ⇒ 1
2 2. −
t = – 1 ⇒ t = 4 \ B = (8, – 15)
Equation of AB is 3x + 2y + 6 = 0
�e area required is 44
3 22
2
2
8 − + +
−
∫ x x( )dx = 41.67
3. Let ABC be an equilateral triangle inscribed in thecircle x2 + y2 = a2. Suppose perpendiculars from A,
B, C to the major axis meet the ellipse xa
y
b
2
2
2
21+ =
(a > b) respectively at P, Q, R so that P, Q, R lie on the same side of the major axis as A, B, C respectively.Prove that the normals to the ellipse at P, Q, R areconcurrent. (Priyanshu Sharma, Bihar)
Ans. A, B, C are the vertices of an equilateral triangle.A (a cos , a sin ),
B a acos , sin ,θ π θ π+
+
23
23
C a acos , sinθ π θ π+
+
43
43
Hence, P (a cos , b sin ),
Q a bcos , sin ,θ π θ π+
+
23
23
R a bcos , sinθ π θ π+
+
43
43
�e normals to the ellipse at P, Q, R are respectively,
L1 ax sin – by cos – a b2 2
2− sin 2 = 0;
L2 ax sin θ π+
23
– by cos θ π+
23
– a b2 2
2−
sin2 θ π+
23
= 0; L3 ax sin θ π+
43
– by cos θ π+
43
– a b2 2
22 4
3− +sin ( )θ π = 0
Since, sin + sin θ π+
23
+ sin θ π+
43
= 0
cos + cos θ π+
23
+ cos θ π+
43
= 0
sin 2 + sin 2 θ π+
23
+ sin 2 θ π+
43 = 0
Hence L1, L2, L3 are concurrent.
Do you have a question that you just can’t get answered?Use the vast expertise of our MTG team to get to the bottom of the question. From the serious to the silly, the controversial to the trivial, the team will tackle the questions, easy and tough.The best questions and their solutions will be printed in this column each month.
Y UASKWE ANSWER
| JUNE ‘17 85
SOLUTION SET-173
1. (c) : Anil Balu ProbabilityR, R R, R 1/33B, B B, B 1/33R, B R, B 10/33
\ Required probability = + = =233
1033
1233
411
2. (b) : Let (h, k) be the mid point of the chord ofcircle x2 + y2 = a2, then the equation of the chord be S1 = T
⇒ xh + yk – a2 = h2 + k2 – a2
⇒ h2 + k2 = xh + yk⇒ h2 + k2 = ah + bk (As it passes through(a, b))⇒ x2 + y2 = ax + by (By changing the locus of
h, k → x, y)
3. (b) : Let I = dx
x xI x
xdx
cos sinsec
tan
/ /
6 60
2 6
60
2
1+⇒ =
+∫ ∫π π
⇒ I = ( tan )
tan
/ 11
2 3
60
2 ++∫
xx
dxπ
⇒ I =1 3 1
1
6 2 2
60
2 + + ++∫
tan tan ( tan )tan
/ x x xx
dxπ
I = 13
1
2 2
3 20
2
++∫
tan sec(tan )
/ x x dxx
π
Put tan3 x = t ⇒ dt = 3tan2x sec2x dx
⇒ I = dx dtt
++∫ ∫
∞
0
2
20 1
π/
= π π π π2 2 2
1
0+ = + =−
∞
tan ( )t
4. (d) : Consider Cr =24r
(1 – x)24 = C0 – C1x + C2x2 – .....+ C16x16 + .....(1 – x)–1 = 1 + x + x2 + .....+ x16 + .....Considering the coe�cient of x16 in the product of these two,C0 – C1 + C2 – .....+ C16 = coe�. of x16 in (1 – x)23
=
=
2316
237
5. (a) : Let the A.P.s. be a, a + a, a + 2a,... andb, b + , b + 2 ,...⇒ ab = a1b1 = 120, (a + a)(b + ) = a2b2 = 143,(a + 2a)(b + 2 ) = a3b3 = 154⇒ ab = 120, a + ba = 29, a = –6\ a8b8 = (a + 7a)(b + 7 ) = ab + 7(a + ba) + 49a = 120 + 7 × 29 – 49 × 6 = 29
6. (b, c) : f (x) = loge [ ]x x3 6 1+ +
\ f (–x)= loge [ ]− + +x x3 6 1\ f (x) + f (–x) = loge1 = 0
f (–x) = – f (x)Hence f(x) is an odd function. \ (b) is correct.
Again, f (x) = 1
13 6
2 13 6
25
6x xx x
x+ ++
+
= 1
13 1
13 6
26 3
6x xx x x
x+ +
+ +
+
=+
>3
10
2
6
x
x\ f (x) is an increasing function, so (c) is correct.
7. (b) : DBC = a ⇒ BD = 20, EBD = aBy sine rule for EBD, we getsin sin3
20 8α α=
A
B
3� 2�
�
E 8 D 20 C
⇒ − =3 4 52
2sin α
⇒ = =sin , cos2 18
2 34
α α .
8. (d) : AE + 28 = AB cot a and AE + 8 = AB cot 2aOn subtracting, we get 20 = AB (cot a – cot 2a)
\ AB = 20 sin 2a = 20 1 916
5 7− =
\ AE AB+ = = =8 2 5 7 37
15cot α ⋅ ⇒ AE = 7.
9. (8) : a + b + c = 0 ⇒ a, b, c are the roots ofx3 + qx + r = 0 ...(i)
ab = q, abc = –r, a + b + c = 0⇒ a3 = 3abc \ abc = 1, r = –1, a2 = –2qFrom (i), we get a5 + q a3 + r a2 = 0 ⇒ 10 + 3q + 2q = 0 ⇒ q = –2From (i), x3 – 2x – 1 = 0.\ a3 – 2 a – = 0 ⇒ a4 = 2 a2 + a
a4 = 2(–2q) + 0 = (–4) –2) = 8. 10. (b) : P. 232 = – 1 mod 53, 2322 = – 1 mod 53, 2323 = – 23 mod 53 = 30 mod 53Q. X = x + 2, Y = y + 1, Z = z, U = u – 1⇒ X + Y + Z + U = 5
\ Number of solutions = 83
56
=
R. Coe�cient of x2y in (1 + x + 2y)5 is 52 2
2 60!
! !.
⋅=
S. r rrr=
∑ − = + + + =1
10 11 1 2 10 55( ) ... .
