May 2011 - Career Point

MATHEMATICSSECTION – ISingle Correct Choice TypeThis section contains 8 multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.41. Let P(6, 3) be a point on the hyperbola2 2x y− = 1. If the normal at the point P2 b2aintersects the x-axis at (9, 0), then the eccentricityof the hyperbola is -Ans.Sol.(A)52(B)32(C) 2 (D) 3[B]Equation of the normal at (6, 3) isa 2 x b 2 y+ = a 2 + b 26 3it passes through (9, 0)9a 2so = a 2 + b 262a⇒ b 2 =2Now b 2 = a 2 (e 2 –1)∴ e 2 1–1 = 2e 2 = 23⇒ e =42. Let (x, y) be any point on the parabola y 2 = 4x. LetP be the point that divides the line segment from(0, 0) to (x, y) in the ratio 1 : 3. Then the locus ofP is(A) x 2 = y(B) y 2 = 2x(C) y 2 = x(D) x 2 = 2yAns. [C]2tSol. h = 42t, k = 4(0, 0)1323• P(h, k)(t 2 , 2t)t 2 = 4h, t = 2kso 4k 2 = 4h∴ k 2 = hhence required locus is y 2 = x43. Let f(x) = x 2 and g(x) = sin x for all x∈R. Then the setof all x satisfying (f o g o g o f) (x) = (g o g o f) (x),where (f o g) (x) = f(g(x)), is(A) ± n π, n ∈{0, 1, 2...}(B) ± n π, n ∈{1, 2...}π(C) + 2nπ,n ∈{....,− 2, −1,0, 1, 2, ....}2(D) 2nπ, n∈{…., –2, –1, 0, 1, 2, ….}Ans. [A]Sol. gof(x) = gf(x) = g(x 2 ) = sin x 2go (gof(x)) = g(sin x 2 ) = sin (sin x 2 )fo(gogof(x)) = f(sin (sin x 2 )) = (sin(sin x 2 )) 2∴ (sin (sin x 2 )) 2 = sin (sin x 2 )sin (sin x 2 ) (sin (sin x 2 ) –1) = 0sin (sin x 2 ) = 0 or sin (sin x 2 ) = 1sin x 2 = nπsin x 2 π= 2nπ + 2At n = 0 At n = 0sin x 2 = 0sin x 2 π= 2x 2 = nπx = ±n πNot possible; n ∈ {0, 1, 2, ….}44. Let f : [–1, 2] → [0, ∞) be a continuous functionsuch that f(x) = f(1 – x) for all x∈[–1, 2]. LetAns.Sol.R21 =∫x f x)−1( dx , and R 2 be the area of the regionbounded by y = f(x), x = – 1, x = 2, and the x-axis.Then(A) R 1 = 2R 2 (B) R 1 = 3R 2(C) 2R 1 = R 2 (D) 3R 1 = R 2[C]2R 1 =∫x f ( x)dx… (i)2−1R 1 =∫( 1−x ) f (1 − x)dx =∫( 1−x ) f ( x)dx ...(ii)−1(i) + (ii)22R 1 =∫−1f ( x)dx = R 22−1∴ 2R 1 = R 2XtraEdge for IIT-JEE 91 MAY 2011