Solution - Career Point
Solution - Career Point Solution - Career Point
77.[C]4∫−1f (x) dx = 42⇒∫f (x)−12⇒∫f (x)−14dx +∫24dx = 4 –∫2f (x) dx = 4f (x) dx ...........(i)4Q∫(3 – f(x) ) dx = 7 (given)24 4⇒∫3 dx –∫f (x) dx = 72 24⇒ 3[x] 4 2 = 7 +∫f (x) dx24⇒ 3 × 2 = 7 +∫2.....(ii)f (x)4dx ⇒∫2put this value from (ii) in (i), we get2∫−1f (x) dx = 4 – (–1) = 5−1⇒∫f (x) dx = – 52f (x) dx = – 178.[A] lines – 2x – y + 6 = 0 & 4x – 2y + 7 = 0(make c 1 & c 2 +ve)now a 1 a 2 + b 1 b 2 = – 8 + 2 = – 6 < 0bisector by +ve is acute and contains origin eq n−2 x − y + 6 ( 4x − 2y + 7)given by= +52 5– 4x – 2y + 12 = 4x – 2y + 78x = 5∫79.[C] [ f g(x) ]21−1f 'g(x). g'(x) dxput f g(x) = t ⇒ f 'g(x) g'(x) dx = dt∴ required integral2[ log f g(x) ]1log (f g(2)) – log (f g(1)) = 0{Q g(1) = g(2)}2x + sin x80.[C] f(x) =∫sec 2 x dx21+x⎛22=∫ ⎟ ⎞⎜x + 1−cos xsec 2 x dx2⎝ 1+x ⎠281.[B]⎛2=∫ ⎟ ⎞⎜cos x1−sec 2 x dx2⎝ 1+x ⎠⎛ 2 1 ⎞=∫⎜secx − ⎟ dx2⎝ 1+x ⎠= tan x – tan –1 x + c∴ f(0) = 0 ⇒ tan0 –tan –1 0 + c = 0⇒ c = 0∴ f(x) = tan x – tan –1 x∴ f(1) = tan1 – tan –1 1= tan1 – π/4we know x –[x] = {x}∴ Domain of {x} is R & {x} ∈ [0, 1)but in f(x), {x} is in denominator & it shouldnot be equal to zero∴ {x} 0 ⇒ x Iand domain of sec –1 x is R –(–1, 1)∴ domain of f(x) isR – (–1, 1) – I82.[C] Q sin –1 x is defined for |x| ≤ 1 andsec –1 x is defined for |x| ≥ 1 thereforeboth defined for |x| = 1 ⇒ x = {1,–1}∴ D f = {–1, 1}further f(–1) = sin –1 (–1) + sec –1 (–1)= – π/2 + π = π/2and f(1) = sin –1 (1) + sec –1 (1) = π/2 + 0 = π/2Hence R f = {π/2}83.[B]Let A(z 1 ), B(z 2 ) and C(z 3 ) be the vertices of thetriangle then|z 1 | = |z 2 | = |z 3 |⇒ |OA| = |OB| = |OC|O being the origin⇒ O is circumcentre of the triangle, Also, thetriangle is equilateral, therefore circumcentrecoincide with the centroid⇒ origin is centroidz1 + z2+ z3⇒= 03⇒ z 1 + z 2 + z 3 = 084.[B] Total students n = 100Average marks x = 72Total boys n 1 = 70average marks of boys x 1 = 75Total girls n 2 = 30n1 x1+ n1x2now x =n∴ x 2 =x 2 =nx− n1x1n 2100 × 72 − 70×75= 6530XtraEdge for IIT-JEE 108APRIL 2010
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
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77.[C]4∫−1f (x) dx = 42⇒∫f (x)−12⇒∫f (x)−14dx +∫24dx = 4 –∫2f (x) dx = 4f (x) dx ...........(i)4Q∫(3 – f(x) ) dx = 7 (given)24 4⇒∫3 dx –∫f (x) dx = 72 24⇒ 3[x] 4 2 = 7 +∫f (x) dx24⇒ 3 × 2 = 7 +∫2.....(ii)f (x)4dx ⇒∫2put this value from (ii) in (i), we get2∫−1f (x) dx = 4 – (–1) = 5−1⇒∫f (x) dx = – 52f (x) dx = – 178.[A] lines – 2x – y + 6 = 0 & 4x – 2y + 7 = 0(make c 1 & c 2 +ve)now a 1 a 2 + b 1 b 2 = – 8 + 2 = – 6 < 0bisector by +ve is acute and contains origin eq n−2 x − y + 6 ( 4x − 2y + 7)given by= +52 5– 4x – 2y + 12 = 4x – 2y + 78x = 5∫79.[C] [ f g(x) ]21−1f 'g(x). g'(x) dxput f g(x) = t ⇒ f 'g(x) g'(x) dx = dt∴ required integral2[ log f g(x) ]1log (f g(2)) – log (f g(1)) = 0{Q g(1) = g(2)}2x + sin x80.[C] f(x) =∫sec 2 x dx21+x⎛22=∫ ⎟ ⎞⎜x + 1−cos xsec 2 x dx2⎝ 1+x ⎠281.[B]⎛2=∫ ⎟ ⎞⎜cos x1−sec 2 x dx2⎝ 1+x ⎠⎛ 2 1 ⎞=∫⎜secx − ⎟ dx2⎝ 1+x ⎠= tan x – tan –1 x + c∴ f(0) = 0 ⇒ tan0 –tan –1 0 + c = 0⇒ c = 0∴ f(x) = tan x – tan –1 x∴ f(1) = tan1 – tan –1 1= tan1 – π/4we know x –[x] = {x}∴ Domain of {x} is R & {x} ∈ [0, 1)but in f(x), {x} is in denominator & it shouldnot be equal to zero∴ {x} 0 ⇒ x Iand domain of sec –1 x is R –(–1, 1)∴ domain of f(x) isR – (–1, 1) – I82.[C] Q sin –1 x is defined for |x| ≤ 1 andsec –1 x is defined for |x| ≥ 1 thereforeboth defined for |x| = 1 ⇒ x = {1,–1}∴ D f = {–1, 1}further f(–1) = sin –1 (–1) + sec –1 (–1)= – π/2 + π = π/2and f(1) = sin –1 (1) + sec –1 (1) = π/2 + 0 = π/2Hence R f = {π/2}83.[B]Let A(z 1 ), B(z 2 ) and C(z 3 ) be the vertices of thetriangle then|z 1 | = |z 2 | = |z 3 |⇒ |OA| = |OB| = |OC|O being the origin⇒ O is circumcentre of the triangle, Also, thetriangle is equilateral, therefore circumcentrecoincide with the centroid⇒ origin is centroidz1 + z2+ z3⇒= 03⇒ z 1 + z 2 + z 3 = 084.[B] Total students n = 100Average marks x = 72Total boys n 1 = 70average marks of boys x 1 = 75Total girls n 2 = 30n1 x1+ n1x2now x =n∴ x 2 =x 2 =nx− n1x1n 2100 × 72 − 70×75= 6530XtraEdge for IIT-JEE 108APRIL 2010
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