May 2011 - Career Point

Ans.Sol.[A,D]r = xiˆ + yj ˆ + zkˆis coplanar with the given vectorsox y z∴ 1 1 2 = 01 2 1So, 3x = y + z ...(1)→∴ r ⊥ iˆ + ˆj+ kˆ→So, r . (ˆ i + ˆj+ kˆ)= 0So, x + y + z = 0 ...(2)On solving (1) & (2)So, x = 0 ∴ y + z = 0 ∴ (A) & (D) Satisfy55. Let f : R → R be a function such thatf(x + y) = f(x) + f(y), ∀x, y ∈ RIf f(x) is differentiable at x = 0, then(A) f(x) is differentiable only in a finite intervalcontaining zero(B) f(x) is continuous ∀ x ∈ R(C) f '(x) is constant ∀ x ∈ R(D) f(x) is differentiable except at finitely manypointsAns. [B,C]Sol. f(x + y) = f(x) + f(y)By Partial differentiation with respect to xf ' (x + y) = f ' (x)f ' (y) = f '(0)f(y) = (f '(0))y + cf(y) = ky +c∴ f(y) = ky as f(0) = 0∴ f(x) = kxAlternatef ( x + h)− f ( x)f '(x) = limh→0hf ( x)+ f ( h)− f ( x)f ( h)= lim= limh→0hh →0h= λ (let)f(x) = λx + c As f(0) = 0 ⇒ c = 0f(x) = λx56. Let M and N be two 3 × 3 non-singularskew-symmetric matrices such that MN = NM.If P T denotes the transpose of P, thenM 2 N 2 (M T N) –1 (MN –1 ) T is equal to(A) M 2 (B) –N 2(C) –M 2(D) MNAns. [C]Sol.Q MN = NMM 2 N 2 = MN MN Q (M T ) –1 = (–M) –1 = –M –1Given, M 2 N 2 (M T N) –1 . (MN –1 ) T= – MN MN N –1 M –1 N –1 MI= –M NN –1 M = – M 2The most suitable answer is (C), althoughgiven information is contradictory as Skewsymmetric matrix of odd order cannot be nonsingular57.x yLet the eccentricity of the hyperbola –2 2a b= 1be reciprocal to that of the ellipse x 2 + 4y 2 = 4. Ifthe hyperbola passes through a focus of theellipse, then2 2x y(A) the equation of the hyperbola is – = 13 2(B) a focus of the hyperbola is (2, 0)(C) the eccentricity of the hyperbola is53(D) the equation of the hyperbola is x 2 – 3y 2 = 3Ans. [B,D]Sol. Let e 1 = eccentricity of hyperbolae 2 = eccentricity of ellipse1∴ e 1 =e 2so eccentricity of ellipse =3 = e22eccentricity of ellipse =2 = e13Now focus of ellipse is (± ae 2 , 0) ≡ (± 3 , 0)Hyperbola passes through it2( 3)So, – 0 = 1 ⇒ a 2 = 32aalso b 2 = a 2 (e 2 1 – 1)b 2 ⎛ 4 ⎞= 3 ⎜ – 1⎟ = 1⎝ 3 ⎠and hyperbola22x − y = 13 1also focus (± ae 1 , 0) ≡ (± 2, 0)22XtraEdge for IIT-JEE 76 MAY 2011