Solution - Career Point

85.[C] Letters of word 'STATISTICS' are1A, 2I, 1C, 3S, 3T total = 10Letters of word 'ASSISTANT' are2A, 1I, 1N, 3S, 2T total =9common letters are A, I, S & T1 2 2probability of choosing A is = × = 10 9 90probability of choosing I = 102 × 91 = 902probability of choosing S = 103 × 93 = 909probability of choosing T = 103 × 92 = 9062 2 9 6 19total = + + + = 90 90 90 90 9086.[D] A = { – 2, –1, 0, 1, 2} ⇒ n(A) = 5B = {0, 1, 2, 3} ⇒ n(B) = 4C = {1, 2} ⇒ n(C) = 2D = {(1, 7),(2, 6),(3, 5),(4, 4),(5, 3)(6, 2),(7, 1),}87.[B]⇒ n(D) = 7(A ∪ B ∪ C) = {– 2, – 1, 0, 1, 2, 3}⇒ n(A ∪ B ∪ C ) = 6n(D) = 7n(B ∪ C) = 4this is of the formdyf '(y) + f(y). f(x) = Q(x)dxput tan y = z∴ sec 2 dy dzy = dx dxdz∴ + 2xz = x3dx∴ e∫ pdx= ∫ 2xdxe =∴ sol n is z= 21= 21∫xe 22xe. x∫e t . tdt2=∫2x dx2xexe 2. x3dx= 21 e t (t – 1) + c2x 1 2x∴ tan y e = e (x 2 – 1) + c288.[D] The given curves are y 2 = 8x ...(1)and xy = – 1 ...(2)any tangent to (1) is2y = mx + m...(3)⎧Q tan gent to y2= 4ax⎪ a⎨is y = mx +⎪ m⎪∴4a= 8 ⇒ a = 2⎩we shall find m so that it touches (2)∴ from (2) & (3)⎛ 2 ⎞x⎜mx+ ⎟ = −1⇒ m 2 x 2 + 2x + m = 0 ...(4)⎝ m ⎠(3) touches (2) if quadratic (4) has equal roots∴ D = 0⇒ 4 – 4m 3 = 0 ⇒ m 3 = 1 ⇒ m = 1∴ required common tangent is y = x + 289.[B] h(x) = (f(x)) 2 + (g(x)) 2∴ h'(x) = 2f(x)f '(x) + 2g(x) g'(x)= 2f(x)g(x) + 2g(x)g'(x) {Q f '(x) = g(x)}Also f '(x) = g(x)∴ f"(x) = g'(x) ⇒ – f(x) = g'(x)∴ h'(x) = 2f(x) g(x) – 2g(x)f(x)h'(x) = 0∴ h(x) = constant for all xbut h(5) = 11 ∴ h(x) = 1 for all x∴ h(10) = 1190.[A] for m 1 : Let I & N are assumed single letter inIN order Now total no. of letter are IN T E G ER = 6 letterthese 6 letter can be arranged in row6= × 2 = 720 ways2∴ Letter of words INTEGER can be arranged in7a row = = 2520 ways2Now no. of ways in which IN are not together ism 1 = 2520 – 720 = 1800Now for m 2 : I - - - - - R → rest five can arrange5= = 60 ways22x⇒ tan y = c e − 1+ (x 2 – 1) 2∴m 1 =60m21800 = 30XtraEdge for IIT-JEE 109APRIL 2010