WBJEE Answer Keys by Aakash Institute, Kolkata Centre

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WBJEE - 08 Answer Keys by, Kolkata Centre MATHEMATICS Q.No. 0 A B C D 0 C D A B 0 B D A C 04 C B A B 05 C A C C 06 A C D C 07 B A C C 08 B *C,D C A 09 C D D B 0 D A C D B A B C C D A B B A A C 4 C C B A 5 C B B A 6 C C D C 7 A C D A 8 B A B A 9 D B A A 0 C B C C B C A D C D *C,D C A B D C 4 A C A D 5 C B A C 6 A C D B 7 A C A A 8 A C C A 9 C A B B 0 D B C B C D C D C C A D D B B B 4 C C B A 5 B A C C 6 A A D A 7 A C B *C,D 8 B A C D 9 B A B A 40 D A C A 4 D C C D 4 B D C A 4 A C A C 44 C C B B 45 A D D C 46 *C,D C C C 47 D B B A 48 A A C B 49 A A A B 50 D B A C 5 A C C A 5 B B B C 5 A C C D 54 C A B A 55 D B C B 56 A A A D 57 B C B A 58 D D A A 59 A A C C 60 A B D B 6 C D A C 6 B A B B 6 C A D C 64 B C A A 65 C B A B 66 A,B,C,D A,C A,C B 67 C,D C A,B A,D 68 B A,B,C,D B B,C 69 A,D C,D A,C A,C 70 B,C B C A,B 7 A,C A,D A,B,C,D B 7 A,B B,C C,D A,C 7 B A,C B C 74 A,C A,B A,D A,B,C,D 75 C B B,C C,D * More than one option correct instead of single option

Code - CATEGORY - I (Q to Q50) Carry mark each and only one option is correct. In case of incorrect answer or any combination of more than one answer, ¼ marks will be deducted.. The approximate value of sin º is > 0.5 (B) > 0.6 < 0.5 < 0.4 sin 0º < sinº < sin 7º 0.5 < sin º < 0.6 d. Let f (s) = e x f (x) f n (x), f (x) = e,...f n(x) e for all n.the for any fixed n, f n (x) is dx f n (x) (B) f n (x) f n (x) f n (x) f n (x)... f (x) f n (x)... f (x)e x ANSWERS & HINT for WBJEE - 08 SUB : MATHEMATICS d f n (x) dx d d df (x) dx n f n (x) e f n(x) = d d f (x) e f (x) f n (x) n n df n(x) dx d f (x)f (x)f (x)... f (x) f (x) dx = n n n = n n n f (x)f (x)f (x)...f (x). The domain of definition of f(x) = x is x (, ) (, ) (B) [, ] (, ) (, ) (, ) (, ) [, ] (, ) Here (a, b) {x : a < x < b} & [a, b] {x : a x b} x 0 x 0 x as x > x x x, (, ) [, ] ()

4. Let f : [a, b] be differentiable on [a, b] & k. Let f(a) = 0 = f(b) Also let J(x) = f(x) + k f (x). Then J(x) > 0 for all x [a, b] (B) J(x) < 0 for all x [a, b] J(x) = 0 has at least one root in (a, b) J(x) = 0 through (a, b) Let g(x) = kx f(x) which is continuous in [a, b] and differentiable in (a, b) g(a) = 0 = g(b) ) g(c) = 0 for same c (a, b) ( by Rolle s theorem) kf(c) + kcf(c) = 0 j(x) = 0 for same x = c (a, b) 5. f( h) f() Let f(x) = x 0 7x 8 + 5x 6 x + x 7. Then lim h0 h h does not exsit (B) is 50 f( h) f() f() f( h) h lim lim h0 h 0 h h h h h is 5 is 5 = f () 6. Let f : [a, b] be such f is differentiable in (a, b), f is continuous at x = a & x = b and moreover f(a) = 0 = f(b). Then there exists at least one point c in (a, b) such that f(c) = f(c) (B) f(x) = f(x) does not hold at any poit in (a, b) at every point of (a, b), f(x) > f(x) at every point of (a, b), f(x) < f(x) Let, g(x) = e x f(x) which is continuous in [a, b] and differentiable in (a, b) Now, g(a) = g(b) = 0 k (a, b) so that k k g(k) = 0 e f (k) e f(k) 0 f(k) = f(k) 7. Let f : be a twice continuously differentiable function such that f(0) = f() = f(0) = 0. Then f(0) = 0 (B) f(c) = 0 for some c if c 0, then f(c) 0 f (x) > 0 for all x 0 f(0) = f() = 0 f(k) = 0 for same k (0, ) by Rolle s Theorem. Again, f(o) = f(k) = 0 f(c) = 0 for same c (0, k) by Rolle s Theorem. ()

8. If sinx. xcos x sinx sinx e dx e.f(x) c cos x, where c is constant of integration, then f(x) = sec x x (B) x sec x tan x x x tan x e sin x xcos x cos x sinx sin x e xcos x sec x tan x sin x sin x sinx e xcos x sec x tan x = sin x d[e (x sec x)] e cos x(x sec x) e ( sec x tanx) 9. If f(x) sin x cos x dx = logf(x) c, (b a ) where c is the constant of integration, then f(x) = (b a )sinx (B) ab sin x (b a )cosx abcos x d logf(x) c dx (b a ) f (x) f(x)sin xcos x f(x) (b a ) = (by question) dy y sinx = dx b a (Take f(x) = y) dy y (b a )cosx (b a ) sin x dx y 0. If M = / /4 cos x sin x cos x dx, N dx x, then the value of M N is (x ) 0 0 (B) 4 4 4 Ans : N 4 0 sinxdx x let x t, dt =dx, N 0 dt sint t 4 N 0 sint dt t sint cost dt t t 0 0 N M M N 4 4 ()

04 tan x. The value of the integral I dx is x I/04 log04 4 (B) log04 log 04 log04 I 04 /04 tan x dx x /04 tan / t t dt (put x / t) t 04 04 /04 cot t dt t 04 dx I / I log 04 log04 x 4 /04 / sinx. Let I dx. Then x / 4 I (B) 4 I 0 I 8 6 I / sinx I dx x / 4 4 sin I sin 4 4 4 I 8 6. The value of I 5 / / e e tan sinx tan sin x tan cosx e dx, is (B) e / I 5 tan 0 e e sinx tan cos x tan sin x e dx 0 tan sin x e 5 dx tan sin x tan cosx I I e e 4 4 n 4. The value of lim sec sec... sec n n 4n 4n 4n is log e (B) 4 e Required limit = n r lt sec n r n 4n = x sec dx 4 0 = 4 x tan 4 0 = 4 (4)

5. The differential equation representing the family of curves y = d x d where d is a parameter, is of order (B) degree degree degree 4 y dx d dy y d dx Degree = 6. Let y(x) be a solution of dy x xy 4x dx = 0 and y(0) =. Then y() is equal to (B) 6 dy x 4x y dx x x x dx, x IF e = ln x e = +x 4x x y y x 4 = 6 = 6 7. The law of motion of a body moving along a straight line is x vt, x being its distance from a fixed point on the line at time t and v is its velocity there. Then acceleration f varies directly with x (B) acceleration f varies inversely with x acceleration f is constant acceleration f varies directly with t dx dv x vt v t dt dt v v t f v x f f t t 8. Number of common tangents of y = x and y = x + 4x 4 is (B) 4 y = x P (, ) is a point on this parabola. y = (x ) y = x -----() is a tangent x = x + 4x 4 x +x( ) + (4 ) = 0 Discriminant = 0 4( ) 4(4 ) = 0 4 4 4 0 = 0 = 0, = (5)

9. Given that n number of A.Ms are inserted between two sets of numbers a, b and a, b where a, b. Suppose further that the m th means between these sets of numbers are same, then the ratio a : b equals n m + : m (B) n m + : n n : n m + m : n m + Ans : b = (n+)th term = a+(n+)d b a d n b a mth mean = a m n Similarly b = (n+)th term = a+(n+)d b a d n mth mean = a+m b a n b a b a a m a m n n a m b n m 0. If x + log 0 (+ x ) = x log 0 5 + log 0 6 then the value of x is x (B) x log x log 5 log 6 0 0 0 log 0 log log 5 log 6 x x x 0 0 0 0 x (+ x ) = 6 y(+y) = 6 y = x = x = r r. If Zr sin icos then 0 r0 Z r (B) 0 i i r r Zr icos isin r i i e 0 0 r r0 r0 r i Z i e = i 0 = 0 (6)

. If z and z be two non zero complex numbers such that z z z, then the origin and the points represented by z z and z lie on a straight line (B) form a right angled triangle form an equilateral triangle from an isosceles triangle z + z + z = z z + z z + z z is the condition for z z z to be the vertices of an equilateral triangle. Putting z = 0 z + z = z z. If b b = (c +c ) and b, b, c, c are all real numbers, then at least one of the equations x +b x + c = 0 and x +b x+c = 0 has real roots (B) purely imaginary roots roots of the form a+ib (a, b, ab 0) rational roots D = b 4c D = b 4c D +D = b + b 4(c +c ) = b + b b b = (b b ) 0 At least one of D, D non-negative. 4. The number of selection of n objects from n objects of which n are identical and the rest are different is n (B) n n n + Ways of selections are n identical and no different = way n identical and one from different elements = n c. 0 identical rest from different = n c n... n n n n c 0 c c... c n = n 5. If ( r n), then n C r +. n C r+ + n C r+ is equal to. n C r+ (B) n+ C r+ n+ C r+ n+ C r n C r + n C r+ + n C r+ = n C r + n C r+ + n C r+ + n C r+ = n+ C r+ + n+ C r+ = n+ C r+ 6. The number (0) 00 is divisible by 0 4 (B) 0 6 0 8 0 (0) 00 = ( + 00) 00 = [ + 00 C 00 + 00 C 00 +...+ 00 C 00 (00) 00 ] = 0 4 ( + 00 C + 00 C 0 +...+ 00 C 00 (00) 98 ) = 0 4 ( + an integer multiple of 0) (7)

7. If n is even positive integer, then the condition that the greatest term in the expansion of (+x) n may also have the greatest coefficient is n n x n n (B) n n x n n n n x n n n x n n n n x(n ) n x n n x n n 8. If 7 0 4 = A, then 5 7 5 5 is A (B) A A + I A A + I A + 5A 4I 7 0 P 4 then adj P = 5 7 5 5 A = P and A = Adj (P) 9. If a r = (cos r + i sin r) /9, then the value of a a a a a a 4 5 6 a7 a8 a is 9 (B) 0 Let i /9 e then given determinant = 4 5 6 7 8 9 = =0 0. If S r = r x n n 6r y n n 4r nr z n n, then the value of n Sr is independent of r x only (B) y only n only x, y, z and n Ans : n n x n n n Sr n n y n n = 0 r n n z n n (8)

. If the following three linear equations have a non-trivial solution, then x + 4ay + az =0 x + by + bz = 0 x + cy + cz = 0 a, b, c are in A.P. (B) a, b, c are in G.P a, b, c are in H.P. a + b + c = 0 4a a b b 0 To have non-trivial solution, c c c a b. On, a relation is defined by xy if and only if x y is zero or irrational. Then is equivalence relation (B) is reflexive but neither symmetric nor transitive is reflexive and symmetric but not transitive is symmetric and transitive but not reflexive On the set xy x y = 0 or x y Q c x x 0 xx (Reflexive) if x y = 0 y x = 0 or x y Q c y x Q c (Symmetric) Take x = ; y = ; z x y = c Q and y z = Q c Here xry and yrz but x is not related to z Not transitive. On the set of real numbers, the relation is defined by xy, (x, y) If x y < then is reflexive but neither symmetric nor transitive (B) If x y < then is reflexive and symmetric but not transitive If x y then is reflexive and transitive but not symmetric If x > y then is transitive but neither reflexive nor symmetric Ans : for option D, x > x is not true hence not reflexive Take x =, y =, clearly x > y but y > x doe not hold hence not symmetric Now, Let x > y and y > z x, y >0. Rewriting, x > y and y > z x > z hence transitive 4. If f : be defined by f(x) = e x and g : be defined by g(x) = x. The mapping g f : be defined by (g f ) (x) = g[f(x)] x, Then g f is bijective but f is not injective (B) g f is injective and g is injective g f is injective but g is not bijective g f is surjective and g is surjective g is neither injective nor surjective g f(x) = e x, x g f is injective 5. In order to get a head at least once probability 0.9, the minimum number of time a unbiased coin needs to be tossed is (B) 4 5 6 Let x = no. of heads appear in n tossed X Bin n, Now, P ( x ) = P (x = 0) = 0.9 n n n 4 0 minimum number of tosses = 4 (9)

6. A student appears for tests I, II and III. The student is successful if he passes in tests I, II or I, III. The probabilities of the student passing in tests, I, II and III are respectively p, q and. If the probability of the student to be successful is. Then p ( + q) = (B) q (+p) = pq = Let x - He passes in test-i x - He passes in test-ii x - He passes in best - III x - He is successful x x x x x x x x x x px px.p x.p x p x.p x.p x p x.p x.p x p q pq. p q pq p pq p(+q)= 7. If sin 6+ sin 4 + sin = 0, then general value of is n n n,n (B),n,n (n is integer) 4 4 6 4 sin 6+ sin 4 + sin = 0 sin 4. cos + sin 4 = 0 sin 4 = 0 or cos + = 0 n,n 4 6 4 = n cos = = n, n n 4 n n 8. If 0 A 4 tana tan cot A tan cot A, then tan is equal to 4 (B) 0 9. Without changing the direction of the axes, the origin is transferred to the point (, ). Then the equation x + y 4x 6y + 9 = 0 changes to x + y + 4 = 0 (B) x + y = 4 x + y 8x y + 48 = 0 x + y = 9 x x +, y y + in given curve x + y = 4 (0)

40. The angle between a pair of tangents drawn from a point P to the circle x + y + 4x 6y + 9 sin + cos = 0 is. The equation of the locus of the point P is x + y + 4x + 6y + 9 = 0 (B) x + y 4x + 6y + 9 = 0 x + y 4x 6y + 9 = 0 x + y + 4x 6y + 9 = 0 Ans : A C (,) P(h,k) sin = CA CP [(h+) + (k ) ]sin = 4sin h + k + 4h 6k + 9 = 0 4. The point Q is the image of the point P(, 5) about the line y = x and R is the image of the point Q about the line y = x. The circumcenter of the PQR is (5, ) (B) ( 5, ) (, 5) (0, 0) Ans : P(,5) O R(, 5) Q(5,) Q = 90 Circumcentre = mid point of P and R i.e. (0, 0) 4. The angular points of a triangle are A (, 7), B(5, ) and C(, 4). The equation of the bisector of the ABC is x = 7y + (B) 7y = x + y = 7x + 7x = y + Equation of AB : y 4x + 7 = 0 Equation of BC : 4y + x 9 = 0 Equation of bisectors 4y + x 9 = ± (y 4x + 7) (using position of points A and C with respect to bisectors) 7y = x + 4. If one of the diameters of the circle, given by the equation x + y + 4x + 6y = 0, is a chord of a circle S, whose centre is (, ), the radius of S is 4 unit (B) 5 unit 5 unit 5 unit A C(, ) r N (, ) B Equation chord AB : x = A ( + r cos90, + r sin 90 ) A(, r ) Putting A(, r ) in the circle r = ± r = 5 CN = 4 CA = 4 ()

44. A chord AB is drawn from the point A (0, ) on the circle x + 4x + (y ) = 0, and is extended to M such that AM = AB. The locus of M is x + y 8x 6y + 9 = 0 (B) x + y + 8x + 6y + 9 = 0 x + y + 8x 6y + 9 = 0 x + y 8x + 6y + 9 = 0 A (0,) B M (x,y) AM = AB x y B,. Putting in the circle x + y + 8x 6y + 9 = 0 x 45. Let the eccentricity of the hyperbola a equals y be reciprocal to that of the ellipse x + 9y = 9, then the ratio a : b b 8 : (B) : 8 9 : : 9 b 9 a 8 b a 8 a 8 b 46. Let A, B be two distinct points on the parabola y = 4x. If the axis of the parabola touches a circle of radius r having AB as diameter, the slope of the line AB is r Ans : (C,D) (B) r r r Slope of AB = t t B (t ) (h,±r) A (t ) x t t ± r t + t = ± r Slope = r, r 47. Let P(at, at), Q, R(ar, ar) be three points on a parabola y = 4ax. If PQ is the focal chord and PK, QR are parallel where the co-ordinates of K is (a, 0), then the value of r is t t (B) t t t t t t Ans : Slope of QR = r t at t Slope of Pk = at a t t r t t r t t ()

x y 48. Let P be a point on the ellipse and the line through P parallel to the y-axis meets the circle x + y = 9 at 9 4 Q, where P, Q are on the same side of the x-axis. If R is a point on PQ such that PR, then the locus of R is RQ x 9y x y (B) 9 49 49 9 P( cos, sin) Equation of line // to y axis is x = cos It meets circle at Q( cos, sin) PR : RQ = : x y 9 49 9x y 49 9 7sin R cos, x 9y 9 49 49. A point P lies on a line through Q(,, ) and is parallel to the line x y z. If P lies on the plane 4 5 x + y 4z + = 0, then segment PQ equals to 4 units (B) units 4 units 5 units Let P ( +, 4, 5 + ) It lies on x + y 4z + = 0 ( + ) + (4 ) 4(5 + ) + = 0 6 = 6 = P = (,, 8) PQ 4 5 = 4 50. The foot of the perpendicular drawn from the point (, 8, 4) on the line joining the points (0,, 4) and (,, ) is (4, 5, ) (B) ( 4, 5, ) (4, 5, ) (4, 5, ) Ans : Equation of line is x y z 4 9 8 Let foot of be P (, 8, + 4) DR s of line joining (, 8, 4) and P is (, 8 9, ) ( ) + 8 (8 9) ( ) = 0 4 + 64 + 9 = + 5 77 = 54 = P is (4, 5, ) ()

CATEGORY - II (Q5 to Q65) Carry marks each and only one option is correct. In case of incorrect answer or any combination of more than one answer, ½ mark will be deducted. 5. A ladder 0 ft long leans against a vertical wall. The top end slides downwards at the rate of ft per second. The rate at which the lower end moves on a horizontal floor when it is ft from the wall is 8 (B) 6 5 7 4 y x 0 x + y = 0, dx dt y. dy x dt dx 6 8 dt 5. For 0 p and for any positive a, b ; let (p) = (a + b) p, J(p) = a p + b p, then (p) > J (p) (B) (p) J (p) (p) J (p) in P 0, & (p) J (p) in P, (p) J (p) in P, & J (p) (p) in P 0, n let p = n, then (ap + b p ) /p n n = a b = a + b + k, k 0 a p + b p (a+b) p a+b 5. Let i j k, i j k and i j k be three vectors. A vector, in the plane of and, whose projection on is, is given by i j k (B) i j k i j k i j k 54. Let,, be three unit vectors such that.. 0 and the angle between and is 0. Then is (B) ± = sin 0 = = ± z 55. Let z and z be complex numbers such that z z and z z. If Re (z ) > 0 and m(z ) < 0, then z One (B) real and positive real and negative purely imaginary z z is Ans : let z = x + iy, z = x + iy z z x y x y...(i) Now x x y y i. xy xy z z z z x x y y which is purely imaginary (by (i)) (4)

56. From a collection of 0 consecutive natural numbers, four are selected such that they are not consecutive. The number of such selections is 84 7 (B) 85 7 84 6 85 6 cos sin 4 4 57. The least positive integer n such that sin cos 4 4 n is an identity matrix of order is 4 (B) 8 6 0 A 4 0 8 0 =, A, A correct option is (B) 0 0 0 58. Let be a relation defined on, the set of natural numbers, as = (x,y) : x y 4 Then is an equivalence relation (B) is only reflexive relation is only symmetric relation is not transitive Ans : xy, yz x + y = 4 and y + z = 4 which do not imply x + z = 4 59. If the polynomial f (x) = a b x x a x x b x x b a, then the constant term of f(x) is. b + b (B) +. b + b +. b b. b b f(0) = b b b =.b + b 60. A line cuts the x-axis at A (5, 0) and the y-axis at B (0, ). A variable line PQ is drawn perpendicular to AB cutting the x-axis at P and the y-axis at Q. If AQ and BP meet at R, then the locus of R is x + y 5x + y = 0 (B) x + y + 5x + y = 0 x + y + 5x y = 0 x + y 5x y = 0 Y Q R P (0, ) B A(5,0) X ARB = / Locus of R is a circle with A, B as end points of a diameter, which is x + y 5x + y = 0 (5)

6. Let A be the centre of the circle x + y x 4y 0 = 0. Let B (, 7) and D (4, ) be two points on the circle such that tangents at B and D meet at C. The area of the quadrilateral ABCD is 50 sq. units (B) 50 sq. units 75 sq. units 70 sq. units Y B (,7) y = 7 C (6, 7) (,) A O x-4y-0 = 0 X Area of the quadrilateral ABCD =. ( ABC) =..(5).(5) = 75 D (4,-) 6. Let f(x) = sinx, if x A sin x B, if x. Then cos x, if x f is discontinuous for all A and B (B) f is continuous for all A = and B = f is continuous for all A = and B = f is continuous for all real values of A, B At x, LHL RHL A B At x, LHL RHL A B 0 A, B f(x)is continuousx 6. The normals to the curve y = x x +, drawn at the points with the abscissa x = 0, x = and x = 5 are parallel to each other (B) are pair wise perpendicular are concurrent are not concurrent The three normals are x y + = 0 x y + 0 = 0 x + 8y - 4 = 0 (6)

64. The equation x log x = x has no root in (, ) (B) has exactly one root in (, ) x log x ( x) > 0 in [, ] x log x ( x) < 0 in [, ] ut f(x) = x log x + x f (x) = logx + > 0 f() =, f() = log, f().f() <0 Hence Exactly one root in x (,) as f (x) > 0 65. Consider the parabola y = 4x. Let P and Q be points on the parabola where P. (4, 4) & Q (9, 6). Let R be a point on the arc of the parabola between P & Q. Then the area of PQR is largest when PQR = 90 0 (B) R (4, 4) R, 4 R, 4 Q (9,6) R (t, t) Area is Maximum When RS is maximum S P(4, -4) Equation of PQ is RS y = x t t 5 (t t 6) RS t t 5 5 RS is maximum at t Hence R is(, ) 4 (7)

CATEGORY - III (Q66 to Q75) Carry marks each and one or more option(s) is/are correct. If all correct answers are not marked and also no incorrect answer is market then score = x number of correct answers marked actual number of correct answers. If any wrong option is marked or if any combination including a wrong option is marked, the answer will condisered wrong, but there is no negative marking for the same and zero marks will be awarded. 66. Let I = 0 x cosx dx.then x (B) Ans : (A, B, C, D) 67. A particle is in motion along a curve y = x. The rate of change of its ordinate exceeds that of abscissa in < x < (B) x = x < x > Ans : (C, D) dy dx x dt dt dy dt x, x 4, x, ] [, dx dt 68. The area of the region lying above x-axis, and included between the circle x + y = ax & the parabola y = ax, a > 0 is 8a (B) a 4 6a 9 7 a 8 Area a a 4 69. If the equation x cx + d = 0 has roots equal to the fourth powers of the roots of x + ax + b = 0, where a > 4b, then the roots of x 4bx + b c = 0 will be both real (B) both negative both positive one positive and one negative Ans : (A, D) x 4bx + b c =0 let, be roots, = b c = b (a 4 4a b + b ) = a (4b a ) < 0 Hence one positive and one negative (8)

70. On the occasion of Dipawali festival each sutdent of a class sends greeting cards to others. If there are 0 students in the class, the number of cards sends by students is 0 C Ans : (C, B) (B) 0 0 0 P C P 0 C 7. In a third order matrix A, a ij denotes the element in the i-th row and j-th column. If a ij = 0 for i = j = for i > j = for i < j Then the matrix is skew symmetric (B) symmetric not invertible non-singular Ans : (A, C) a ij = a ji Hence Matrix is skew symmetric 7. The area of the triangle formed by the intersection of a line parallel to x-axis and passing through P(h, k), with the lines y = x and x + y = is h. The locus of the point P is x = y (B) x = (y ) x = + y x = ( + y) Ans : (A, B) y = x B ( k, k) P(h, k) A (k, k) (, ) C y = k x + y = Area ( ACB) = h k k h 4 k 4h y x y x and y x (9)

7. A hyperbola, having the transverse axis of length sin is confocal with the ellipse x + 4y =. Its equation is x sin y cos = (B) x cosec y sec = (x + y ) sin = + y x cosec = x + y + sin Focus of Ellipse is (, 0) 74. Let f x For Hyperbola a = sin and a e = e = cosec b a e = cos Equation of Hyperbola is x sin y cos cos x, x 0 then assuming k as an integer, (B) f(x) increases in the interval, k k f(x) decreases in the interval, k k f(x) decreases in the interval, k k f(x) increases in the interval, k k Ans : (A, C) f x sin x x f x 0 f x 0 x, k k x, k k (0)

75. Consider the function y log a x x, a > 0, a. The inverse of the function does not exist (B) is x log y y a is x = sinh (y ln a) is x cosh yln a y a x x y a x x y y a a x x sinhylna ()