GuruAanklan Foundation / MHT-CET / Examination Mathematics Set - [A .

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / SolutionsGuruAanklan Foundation / MHT-CET / ExaminationMathematicsSet - [A] - SolutionsANSWER )11.(A)12. (B)13.(D)14.(D)15.(B)16.(A)17.(C)18. (C)19.(B)20.(A)21.(A)22.(A)23.(A)24. (C)25.(C)26.(C)27.(C)28.(C)29.(B)30. (C)31.(C)32.(C)33.(B)34.(B)35.(A)36. (D)37.(D)38.(D)39.(C)40.(A)41.(A)42. (B)43.(D)44.(D)45.(D)46.(D)47.(B)48. (D)49.(C)50.(A)1(C)

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / SolutionsMATHEMATICS 1.(A)4 5 (Here major axis of an allipse is along y-axis)PS PS' 2a 202.(A) 1 x 1fe x sin 1x dx e sin x c21 x [ fe x f (x) f 1 (x ) dx e x f (x) c]3.(D)22If f x e x then xe x dx 212x e x dx 21 x21e c f x c22By trial and errorChecking each choice and integrating in case of choice D Let f x e x22 x f x dx xex dx 212xex dx 22f x ex c c224.(D)I sin xdxsin 3x sin xdx3sin x 4sin 3 x 1dx3 4sin 2 x sec 2 xdx3sec 2 x 4 tan 2 x sec 2 xdx3 tan 2 xLet tan x t sec 2 xdx dt 1dt 3 t212 3 tdt22

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / Solutions 5.12 3log3 tan x c3 tan x(C)2 2 x x x x 0 e 2 4 dx 0 2 e 4 e dxx2 x 1 e x e x e x 24 0 2 1 1 e2 e 2 e 24 2 4 5 1 1 e2 4 2 2 46. since xe x dx xe x e x c (C)The distance between the given lines is equal to25 The required line must be to both lines let its equation be 2x y k 0As it passes through ( 5, 4), 10 4 k 0Hence k 14Therefore equation of lime is 2x y 14 07.(B)bbsince f x dx f a b x dxaabI x f x dx given ab I a b x f a b x dxab a b x f x dxabb I a b f x dx xf x dxaab I a b f x dx Ia3

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / Solutions I 8. a b b f2 x dxa(A)x 2 y and y 4xsolving x 0, 4 y 0,16A 4,16 44Required area A y dx ydx between line & parabola0404y x24 4x dx x dx 4x x dx200yA0xx04 x3 2x 2 3 0 32 9.y64 32 sq.units33(D)dy cos 1 adxIntegrating both sidesy cos 1 a x c 2 0 cx 0, c 2y cos 1 a x cy c cos 1 ax y c cos a x 10.y 4x2(A)4

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / SolutionsLet 4x y 1 vDifferentiating w.r.t. x4 dy dvdy2 but 4x y 1 v 2dx dxdx4 v2 dvdxdv dx4 v2Integrating both sides1vtan 1 x c '22v tan 2x 2c ' 2put 2c ' cv tan 2x c 2v 2 tan 2x c 4x y 1 2 tan 2x c replace c by k11.(A)As the length of sides of a triangles in G.P.Let they are 9, 9r and 9r 2 r 1 Now perimeter 9 9r 9r 2 37 9r 2 9r 28 012. 3r 7 3r 4 0 r 74 7or r but r 0, r is rejected 333 Hence the sides of the triangle are 9, 12, 16 The lengths of the other two sides are 12 and 16(B)sincedx kxdtlog x kt cdx kdtx5r 43

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / SolutionsAt t 0, x x 0 , then c log x 0 log x kt Clog x kt log x 0At t 2, x 600, then1600log 600 2k log x 0 k log2x0 600 t log x log log x 02 x0 At t 8, x 75000, thenlog 75000 600 8log log x 02 x0 log 75000 4log 600 4log x 0 log x 0 600 4 3log x 0 log 75000 log600 600 600 60075000 3log x 0 log6 6 6 6 1000003 5 5 log 2 6 6 6 4 1000 log 2 2 2 6 6 6 10 10 10 3log120 x 0 12013.(D)Sum of all probabilities is 1, pi 1 P x 2 P x 1 P x 0 p x 1 p x 2 P x 3 1 0.1 K 0.2 2k 0.3 k 14k 0.6 16

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / Solutions4k 0.4 k 0.1x 2n ni 2 10123P f n pi 0.1 0.1 0.2 0.2 0.3 0.1Expected value E x x x i pi 2 0.1 1 0.1 0 0.2 1 0.2 2 0.3 3 0.1 0.2 0.1 0.2 0.6 0.3 0.3 0.2 0.6 0.3 0.814.(D)xi01234pi0.450.350.150.030.02x i pi00.350.300.090.08 xi pi 0.82Var x x i2 p i 2x i2 pi00.350.600.270.32 x i2 pi 154since expected value E x x i pi 0.822 1.54 0.82 1.54 0.6724 0.867615.(B)Mean n.p variance npqnp x k C k p k q n k np x k 1 Ck 1p k 1q n k 1n C k p k k InCk 1 q n k I n kn nCk pC k 1 qsince n Ck 1 n!n!and n Ck 1k! n k ! n k 1 ! k 1 !7

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / Solutionsn!k! n k !p n!q n k 1 ! k 1 !16. n k 1 ! k 1 ! pk! n k !q n k 1 n k ! k I ! pk k I ! n k !q n k 1 p kq(A)x 2 y2 116 9Equation of tangent to hyperbola having slope m isy mx 16m 2 9It touches the circle Distance of this line from centre of the radius of the circle16m 2 9m2 1 37m 2 18m 327Equation of tangents is y 317. 15x 77(C)Contrapositive of p q is r p q r p q r p q 18.19.(C)p q q p qT T FTT F TTF T FFF F TT(B)Given expression p q q ] p q 8

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / Solutions p T p q p p q p p p q T p q 20. p q(A)Given A 2 0Now, A(I A) n A( n Co I n C1A n C 2 A 2 n ( n A n ) A(I nA) A nA 2 A21.(A)8 Parabola is y 2 k x ORk y 2 4AXWhere 4A k, Y y, X x Its directrix is X A or x k k8 k or x k4k 4Comparing with x 1, we get 1 32 k 24kk 2 4k 32 0 k 8 k 4 0k 4 or k 822.(A) 1 0 0 Id A 2 1 0 3 2 1 0 Au 2 1 0 0 Au 3 0 1 0 0 0 Au 2 Au 3 1 0 1 0 1 1 9

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / Solutions 0 x A u 2 u 3 1 , let u 2 u 3 y 1 z 1 0 0 x 0 A u 2 u 3 2 1 0 y 1 3 2 1 z 1 x 0, 2x y 1, 3x 2y z 1 x 0, y 1, z 1 0 u 2 u 3 1 1 23.(A)cosA sin B2sin Cb 2 c 2 a 2 bk b 2bc2ck 2c b 2 c2 a 2 b2 c a is an isosceles 24.(C)c a cos B b cos A,b c cos A a cos C 25.c b cos A a cos B cos B b c cos A a cos C cos C(C) 1 1 tan tan 1 tan 1 2 3 1 1 1 2 3 tan tan 1 1 6 26. 1 1 tan tan 1 7 7 (C)22 m m Here, radius 5 5 2 2 10

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / Solutions2m 2 2m 119 01 2391 239 m 22 7.2 m 8.2 approximately 7.2 m 8.2 approximately m 7, 6,.,5,6, 7,827.(C)f(x) is continuous at x 0 limf x f 0 x 02e x cos x f 0 limx 0x22e x 1 1 cos x limx 0x22e x 1 1 cos x lim 2 x 0xx2 log e 1 121232(C) 28.x K yT x y K TTaking log on both sidesKlog x Tlog y K T log x y Differentiating w.r.t. x K T dy K T dy 1 x y dx x y dx K T dy K T K y x y dx x y x xT yT Ky yT dyy x y dx xK xT Kx Kyx x y 11

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / Solutions29. xT Ky dy xT Ky ydxx dy y dx x(B)y log x log x log x .i.e. y log x ysquaring y 2 log x yDifferentiating w.r.t. x2ydy 1 dy dx x dx 2y 1 30.dy 1 dx x dy1 dx x 2y 1 1x 2y 1 (C) x 2 y2 Let cos 2 2 a2 x y puty tan x y tan 1 x y 2 1 1 x a cos y 2 1 x 1 tan a2 cos 11 tan 2 cos 1 cos 2 a12

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / Solutions 2 a a2 y a tan 1 x 2ya tanx2Differentiating w.r.t. x, we getx dy ydx 0x2dy ya tandx x2d2y 2 0dx31.(C)The probability of getting a defective bulb from the box isprobability is 0.9 32.5(C)f x 2x 3 15x 2 144x 7 f ' x 6x 2 30x 144 f ' x 6 x 8 x 3 For decreasing function f ' x 0 6 x 3 x 8 0 x 3 x 8 0 x 8 0 and x 3 0 x 8 and x 3 3 x 833.(B)x 160t 16t 2 v dx 160 32tdtat t 1, v 160 – 32 128131. Hence using binomial distribution, the required10

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / Solutionsat t 9, v 160 – 288 –128velocities are equal and opposite.34.(B)Given general second degree equation represents a pair of lines if a h g h b f 0 g f c 3 5 0 5 4 8 0 0 8 k 3 3k 64 5 5k 0 0 0 9k 192 25k 0 16k 192 k 1235.(A)Required lines are x 9 and x –9 joint equation is (x – 9) (x 9) 036.x 2 81 0(D)Suppose x families live in the townA {families have scooter}B {families have car} m( A) 30 x40 x50 xn( B ) and n( A B ) ' 10010010050 x10020 x n( A B ) 10020 x 2000100 x 10000 n( A B ) 37.(D)Given equation kx 2 4xy y 2 0a k, 2h 4, b 114

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / Solutionsm1 m2 2h4a k 4, m1m2 kb 1b 1m1 m2 8 m1 m 2 8 m1 m 2 2 m1 m 2 2 64 4m1m 2 6416 4 k 6416 4k 644k 48 k 1238.(D)1Volume of tetrahedron PQ PR PS 6P i 2 j 3kq 3j 2 j kr 2i j 3ks i 2 j 4kPQ 4i 4 j 2k as PQ q pPR 3i j 0k as PR r pPS 0i 0 j k as PS s p1Volume of tetrahedron PQ PR PS 6 4 4 2 1 3 1 0 6 0 0 1 39. 1 4 1 4 3 2 0 6 8 4 cu.units6 3(C)[a b c].[b q r] [a b c].1 b x c c x a c x b [a b c] 15

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / Solutions 40.3[ a b c] 3[ a bc](A)EAC AB BC a bDCFAD 2BC 2bband AD AC CDA a BCD AD ACCD 2b a b b a41.(A)2 f x sin x sin x cos cos x sin cos x cos x cos sin x sin 33 33 22 sin x cos x 33 sin x cos x cos x sin x 22 2 2 2555 sin 2 x cos 2 x 44442.(B)D.R.S. of line AB are a 2, b – 1, c 8 line AB is parallel to the line whose D.R.S. are 6, 2, 3then, a 2 6, b 1 2, c 8 3a 4,b 3,c 543.(D)since and cos2 cos2 cos2 1 3cos 2 1 cos2 44.13 cos 13(D) Use dist d a 2 a1 b1 b 2 b1 b 2 a1 4i j, a 2 9 j 2k16

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / Solutions a 2 a1 3i 2k b1 i 2 j 3k , b 2 i 4 j 5ki j k b1 b 2 1 2 31 4 5 b1 b 2 2i 2 j 2k and b1 b2 4 4 4 2 3 a 2 a1 b1 b 2 2 using distance 45. a 2 a1 b1 b2 b1 b 2 2 11 2 333(D)A 5,5, , B 1,3, 2 ,C 4, 2, 2 Equation of line BC. r b c b r i 3j 2k 3i j 4k A, B, C all collinear 5i 5j k i 3j 2k 3i j 4k comparing coefficients of i5 1 3 3 6. i comparing coefficient of k. ii 2 4 2 4 2 46. 10(D)If is angle between line x x1 y y1 z z1 and planea 2 x b 2 y c 2z d 2a1b1c1then sin a1a 2 b1b 2 c1c 221a b12 c12 a 22 b 22 c 22But a1, b1,c1 1, 2, 2 a12 b12 c12 1 4 4 317

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / Solutionsa 2 , b 2 ,c 2 2, 1, a 22 b 22 c 22 4 1 5 cos 83 sin 131 1 2 2 1 2 , squaring both sides33 5 14 9 9 5 47.5 4 5 3 5 3(B)Equation of planex y z 1a b c A a,0,0 B 0, b,0 C 0,0,c 1 2 4 centroid of ABC is , , 3 3 3 1 a 0 0 2 0 b 0 4 0 0 c , , 333333 a 1, b 2, c 4 equation of required planex y z 11 2 4 4x 2y z 448.(D)Distance between parallel planesax by cz d1 0 and ax by cz d 2 0d d1 d 2a 2 b2 c2Here d1 3d2 52a, b, c 2, 1, 218

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / Solutions a 2 b 2 c2 4 1 4 352 6 5 132 3 6d d 49.(C)Equation of line AB is x y 20Equation of line CD is 2x 5y 80Feasible region is closed as per figure. Answer is (C)50.(A)Here A {2, 4, 6} and B {2, 3, 5} A B contains3 3 9elementsNo.of relations 2919

GuruAanklan Foundation / MHT-CET / Examination / Set-A / Mathematics / Solutions 4 b a a b I f x dx 2 8. (A) x y and y 4x2 solving x 0, 4 y 0,16 A 4,16 Required area 4 4 0 0 A y dx ydx between line & parabola 4 4 4 2 2 0 0 0 4xdx x dx 4x x dx 0 y 4x A x y x y