QUESTION BOOKLET (MHT-CET - 2018)Subjects : Paper I : Mathematics

Instructions to Candidates1. This question booklet contains 150 Objective Type Questions in the subjects of Physics (50) , Chemistry (50) and

Mathematics (50)2. The question paper and OMR (Optical Mark Reader) Answer Sheet is issued separately at the start of the examination.3. Choice and sequence for attempting questions will be as per the convenience of the candidate.4. Candidate should carefully read the instructions printed on the Question Booklet and Answer Sheet and make the correct

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Day and Date : Thursday, 10th May, 2018 Duration : 3 hoursTotal Marks : 200

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2 Kalrashukla Classes

MATHEMATICS

1. If K

20

dx ,2 18x 24

then the value of K is

(A) 3 (B) 4 (C)13 (D)

14

Sol. (C)K K K

2 2 20 0 0

dx 1 dx dx2 18x 24 2 1 9x 24 1 (3x) 12

K1 10

1 1tan (3x) tan 3K or 3K 1 K3 12 4 3

2. The cartesian coordinates of the point on the parabola y2 = 16x, whose parameter is 12 , are

(A) (2, 4) (B) (4, 1) (C) (1, 4) (D) (1, 4)Sol. (D)

Given y2 = –16x = –4(4)xIn usual notations a = 4, if t is the parameter t = 1/2, then its cartesian coordinates are

(–at2, 2at) = (–4t2, 8t) or 1 14 , 8 ( 1, 4)2 2

3. 2

1 dxsin x cos x

(A) sec x + log |sec x + tan x| + c (B) sec x . tan x + c(C) sec x + log |sec x tan x| + c (D) sec x + log |cosec x cot x| + c

Sol. (D)

2

1I dxsin x cos x x

= 2 2 2 2

2 2 2

sin x cos x sin x cos xdx dxsin x cos x sin x cos x sin x cos x

= sec x tan x dx cosec x dx sec x log | cosec x cot x | c

4. If 3 3

10 3 3

x ylog 2x y

, then

(A)xy (B) –

yx (C) –

xy (D)

yx

MH-CET - 2018 3

Sol. (D)3 3

2 3 3 3 33 3

x y 10 100 x y 100x 100yx y

3 399or y x

101 but

3

3

99 y101 x

or 2 2dy 993y 3xdx 101

2 3 2

2 3 2

dy 99x y x ydx 101y x y x

5. If f : R {2} R is a function defined by f(x) = 2x 4 ,

x 2

then its range is

(A) R (B) R {2} (C) R {4} (D) R {2, 2}Sol. (C)

f : R {2} R2x 4f (x) x 2

x 2

f(x) = x + 2

As – < x < 2 and 2 < x < – < x + 2 < 4 and 4 < x + 2 < f(x) R {4}

6. If f(x) = x2 + for x 0

= 22 x 1 for x < 0

is continuous at x = 0 and 1f 22

then 2 + 2 is

(A) 3 (B)825 (C)

258 (D)

13

Sol. (C)Given f(x) is continous = 2 + or = – 2

Now, 21 1 7 1f 2 2

2 2 4 4

2 2 258

7. If 2

1 2 2 2 22

d y dxy (tan x) , then (x 1) 2x(x 1)dx dy

(A) 4 (B) 2 (C) 1 (D) 0Sol. (B)

Consider y = (tan–1 x)2

1 22 1 2

2 2 2

dy 2(tan x) dy d y dy 2(1 x ) 2(tan x) (1 x ) 2xdx 1 x dx dx dx (1 x )

22 2 2d y dy(1 x ) 2x(1 x ) 2

dx dx

4 Kalrashukla Classes

8. The line 5x + y 1 = 0 coincides with one of the lines given by 5x2 + xy kx 2y + 2 = 0 then the value of k is(A) 11 (B) 31 (C) 11 (D) 31

Sol. (C)Given 5x + y – 1 = 0 coincides with one of the lines 5x2 + xy – kx – 2y + 2 = 0

coefficient of x2 = 5, coefficient of y2 = 0, coefficient of xy = 12 , constant term = 2

The other line can be (x – 2) = 0 Their combined equation is

(5x + y – 1) (x – 2) = 0 5x2 + xy 11x 2y + 2 = 0 comparing it with 5x2 + xy kx 2y + 2 = 0 k = 11

9. If 1 2 3

A 1 1 21 2 4

then (A2 – 5A)A–1 =

(A)

4 2 31 4 2

1 2 1

(B)

4 2 31 4 2

1 2 1

(C)4 1 1

2 4 23 2 1

(D)1 2 1

4 2 31 4 2

Sol. (B)

(A2 – 5A)A–1 = A2 A–1 – 5AA–1 = A – 5I = 4 2 31 4 2

1 2 1

10. The equation of line passing through (3, 1, 2) and perpendicular to the lines ˆ ˆ ˆ ˆ ˆ ˆr (i j k) (2i 2 j k)

and ˆ ˆ ˆ ˆ ˆ ˆr (2i j 3k) (i 2 j 2k) is

(A)x 3 y 1 z 2

2 3 2

(B)x 3 y 1 z 2

3 2 2

(C)x 3 y 1 z 2

2 3 2

(D)x 3 y 1 z 2

2 2 3

Sol. (C)A (3, –1, 2)Given lines are

1ˆ ˆ ˆ ˆ ˆ ˆL r (i j k) (2i 2j k) b p

2ˆ ˆ ˆ ˆ ˆ ˆL r (2i j 3k) (i 2j 2k) c q

ˆ ˆ ˆi j kˆ ˆ ˆp q 2 2 1 2i 3j 2k

1 2 2

The desired line is x 3 y 1 z 2

2 3 2

i.e. x 3 y 1 z 2

2 3 2

MH-CET - 2018 5

11. Letters in the word HULULULU are rearranged. The probability of all three L being together is

(A)3

20 (B)25 (C)

328 (D)

523

Sol. (C)The word is HULULULU1H, 3L’s and 4U’s

They can be re-arranged in 8!

3!4! ways

Now when all 3L’s are together considere them as 1 group.

Thus we have 1 group, 1H and 4U’s to arrange they can be arrnged in 6!4! ways

Desired probability =6!/ 4! 6! 3! 4! 6 3

8!/ (3! 4!) 4! 6 7 8 56 28

12. The sum of the first 10 terms of the series 9 + 99 + 999 + …., is

(A) 109 (9 1)8

(B) 9100 (10 1)9

(C) 109 – 1 (D) 10100 (10 1)9

Sol. (B)Consider the sum S = 9 + 99 + 999 + …….. upto 10 terms

= (10 – 1) + (102 – 1) + (103 – 1) + ... + (1010 – 1) = 10 + 102 + 103 + ... + 1010 – 10

= 10(10 1)10 10

10 1

= 10 1010 1 10 1010 1 109 9

= 9100 (10 1)9

13. If A, B, C are the angles of ABC then cot A. cot B + cot B. cot C + cot C. cot A =(A) 0 (B) 1 (C) 2 (D) 1

Sol. (B)We know that, in ABCtan A + tan B + tan C = tan A tan B tan C

1 1 1 1

tan B tan C tan A tan C tan A tan B

cot B cot C + cot A cot C + cot A cot B = 1

14. If 1

2

dx Asin (Bx) C16 9x

, then A + B =

(A)94 (B)

194 (C)

34 (D)

1312

Sol. (D)

1

2 2

dx 1 3x 1 3 1 3 13sin C A and B A B3 4 3 4 3 4 124 (3x)

6 Kalrashukla Classes

15. x 2 sin 2xe dx1 cos 2x

(A) ex tan x + c (B) ex + tan x + c (C) 2ex tan x + c (D) ex tan 2x + cSol. (A)

x x x 2 x2

2 sin 2x 2 2sin x cos xe dx e dx e [sec x tan x] e tan x c1 cos 2x 2cos x

16. A coin is tossed three times. If X denotes the absolute difference between the number of heads and the number oftails then P(X = 1) =

(A)12 (B)

23 (C)

16 (D)

34

Sol. (D)Given n = 3, X = | number of head – number of tails |, {HHH, HHT, HTH, THH, TTT, TTH, THT, HTT }

P(X = 1) = P[(2H + 1T) + (1H, 2T)] = 3 3 6 38 8 8 4

17. If 2sin cos ,3 6

then tan =

(A) 3 (B)13

(C)13 (D) – 3

Sol. (D)

2sin cos 2sin cos 2cos sin cos cos sin sin3 6 3 3 6 6

3 1 3sin 3 cos cos sin sin cos tan 32 2 2

18. The area of the region bounded by x2 = 4y, y = 1, y = 4 and the y-axis lying in the first quadrant is ______ squareunits.

(A)223 (B)

283 (C) 30 (D)

214

Sol. (B)

We have x2 = 4y x 2 y

44 3/2 4

11 1

y 2 4 28A 2 y dy 2 2 y y (8 1)(3 / 2) 3 3 3

MH-CET - 2018 7

19. If f(x) = 2x

2

e cos x ,x

for x 0 is continuous at x = 0, then value of f(0) is

(A)23 (B)

52 (C) 1 (D)

32

Sol. (D)2x

2 2x 0 x 0

e 1 1 cos x 1 3f (0) lim lim 1x x 2 2

20. The maximum value of 2x + y subject to 3x + 5y 26 and 5x + 3y 30, x 0, y 0 is(A) 12 (B) 11.5 (C) 10 (D) 17.33

Sol. (A)Given z = 2x + y

Let 1x yL 1 (3x 5y 26)26 263 5

x y 1

8.67 5.2

10

8

6

4

2

1 2 3 4 5 6 7 8 9 10

Y

X X

Y L1

L2

B

A

0 C

2x yL 16 10

(5x + 3y = 30)

Solving L1 and L2 we get their point of

intersection as 9 5,2 2

OABCO is the feasible region, where O (0, 0);

A (0, 5.2); 9 5B , , C (6,0)2 2

z0 = 0, zA = 10.4, zB = 11.5, zc = 12 Max z at C (6, 0)

21. If a, b, c are mutually perpendicular vectors having magnitudes 1, 2, 3 respectively, then

[a b c b a c] (A) 0 (B) 6 (C) 12 (D) 18

Sol. (C)Given | | 1, | | 2, | | 3a b c

[ ] a b c b a c

= ( ) ( )a b c b c a c = 2[ ] [ ]a b c b a c = 2( )a b c = 2ˆ ˆ ˆ22 3 | 2 |c c c = 12

8 Kalrashukla Classes

22. If points P(4, 5, x), Q(3, y, 4) and R(5, 8, 0) are collinear, then the value of x + y is(A) 4 (B) 3 (C) 5 (D) 4

Sol. (D)Drs of PQ 1, 5 – y, x – 4, Drs of QR –2, y – 8, 4

1 52 8

yy

=

44

x x = 4 – 2 = 2 y – 8 = –10 + 2y y = 2 x + y = 4

23. I f the slope of one of the lines given by ax2 + 2hxy + by2 = 0 is two times the other then(A) 8h2 = 9ab (B) 8h2 = 9ab2 (C) 8h = 9ab (D) 8h = 9ab2

Sol. (A)Given 2 22 0ax hxy by

2

2 0

y yb h ax x 2 2 0bm hm a

1 122 hm mb

and 212 am

b m1 =

23

hb

2

22(4 )

9

h abb

8h2 = 9ab

24. The equation of the line passing through the point (3, 1) and bisecting the angle between co-ordinate axes is(A) x + y + 2 = 0 (B) x + y + 2 = 0 (C) x y + 4 = 0 (D) 2x + y + 5 = 0

Sol. (A, C)Note : The interpretation of bisecting the angle between coordinate axis is not to be considered passingthrough origin otherwise the Q vanishes. y – 1 = (x + 3) y – 1 = –x – 3 or y –1 = x + 3 x + y + 2 = 0 or x – y + 4 = 0

25. The negation of the statement: “Getting above 95% marks is necessary condition for Hema to get the admissionis good college”.(A) Hema gets above 95% marks but she does not get the admission in good college(B) Hema does not get above 95% marks and she gets admission in good college(C) If Hema does not get above 95% marks then she will not get the admission in good college(D) Hema does not get above 95% marks or she gets the admission in good college

Sol. (B)Given p qi.e. q is necessary for pq getting above 95% p Hema to get the admission in good collegeIts negation is p ~q i.e. ~q pHema does not get above 95% but she gets admission in good college

26. Cos 1Cos2Cos3…… Cos179=

(A) 0 (B) 1 (C) 12 (D) 1

Sol. (A)Cos 1Cos2Cos3…… Cos179= 0 ( cos 90º = 0)

MH-CET - 2018 9

27. If planes x cy bz = 0, cx y + az = 0 and bx + ay z = 0 pass through a straight line then a2 + b2 + c2 =(A) 1 abc (B) abc 1 (C) 1 2abc (D) 2abc 1

Sol. (C)Given x – cy – bz = 0, cx – y + az = 0 and bx + ay – z = 0They pass through a common line of intersection.The vectors (1, –c, –b), (c, –1, a) and (b, a, –1) are perpendicular to the desired line The vectors are coplanar

1

1 01

c bc ab a

(1 – a2) + c(–c – ab) –b (a c + b) = 0

1 – a2 – c2 – abc – abc – b2 = 0 a2 + b2 + c2 = 1 – 2abc

28. The point of intersection of lines represented by x2 y2 + x + 3y 2 = 0 is

(A) (1, 0) (B) (0, 2) (C)1 3,2 2

(D)1 1,2 2

Sol. (C)The lines are x2 – y2 + x + 3y – 2 = 0, In usual notations

a = 1, b = –1, g = 12

f = 32

, c = –2, h = 0

Yn = 2 2,hf bg gh afab h ab b

= 1/ 2 3 / 2,

1 1

=

1 3,2 2

29. A die is rolled. If X denotes the number of positive divisors of the outcome then the range of the random variableX is(A) {1, 2, 3} (B) {1, 2, 3, 4} (C) {1, 2, 3, 4, 5, 6} (D) {1, 3, 5}

Sol. (B) { Number of the divisiors of outcome},

Set of divisiors = (1), (1, 2), (1,3), (1, 2, 4), (1,5), (1, 2,3,6)1 2 2 3 2 4

Range of X = {1, 2, 3, 4}

30. A die is thrown four times. The probability of getting perfect square in at least one throw is

(A)1681 (B)

6581 (C)

2381 (D)

5881

Sol. (B)

n = 4, 2 1 2,6 3 3

p q

P (at least once) = ( 1) 1 ( 0)P X P X

= 0 4

40

1 213 3

C = 16 65181 81

10 Kalrashukla Classes

31.4

2

0

x sec x dx

(A) log 24 (B) log 2

4 (C) 1 log 2 (D)

11 log 22

Sol. (B)/4

2

0

secx xdx

= /4

0tan 1 tan

x x x dx = /4 /4

0 0tan logsecx x x = log 24

32. In ABC, with usual notations, if a, b, c are in A.P. then a 2 2C Acos c cos2 2

(A)a32 (B)

c32 (C)

b32 (D)

3abc2

Sol. (C)Given a, b, c are in A.P. 2b = a + c

2 2cos cos2 2c Aa c =

1 cos 1 cos ( cos cos )2 2 2

C A a c a C c Aa c = 3

2 2

a c b b

33. If x = e(sin cos ), y = e(sin + cos ) then dydx at 4

is

(A) 1 (B) 0 (C)12 (D) 2

Sol. (A)

x = (sin cos ) dxd = (cos sin ) (sin cos )e e = 2 (sin ) e

y = (sin cos )e dyd = (cos sin ) (sin cos ) e e = 2 (cos ) e

dydx = cot

4

1

dydx

34. The number of solutions of sin x + sin 3x + sin 5x = 0 in the interval , 32 2

is

(A) 2 (B) 3 (C) 4 (D) 5Sol. (B)

sin sin 5 sin 3 0x x x 2sin 3 cos 2 sin 3 6x x x sin 3x = 0 OR cos 2x = 12

3

nx for n = 0, n = 1, n = 2, n = 3

2 4, ,3 3

x

MH-CET - 2018 11

OR 2x = 2m 23

, 2 44 0, 1 ,

3 3 3

x m m x

35. If tan1 2x + tan1 3x = 4

, then x =

(A) –1 (B)13 (C)

16 (D)

12

Sol. (C)

1 1tan 2 tan 34

x x

22 31 6

x xx

= 1 5x = 1– 6x2 or 6x2 + 5x – 1 = 0

6x2 + 6x – x – 1 = 0 6x (x + 1) –1 (x + 1) = 0 x = –1 (reject) or 16

x

36. Matrix A =

1 2 31 1 52 4 7

then the value of a31 AA31 + a32 A32 + a33 A33 is

(A) 1 (B) 13 (C) 1 (D) 13Sol. (C)

We know that a31 A31 + a32 A32 + a33 A33 = | A |

| A | =

1 2 31 1 52 4 7

= 2(10 – 3) – 4(5 – 3) + 7 (1 – 2) = –1

37. The contrapositive of the statement: “If the weather is fine then my friends will come and we go for a picnic.”(A) The weather is fine but my friends will not come or we do not go for a picnic(B) If my friends do not come or we do not go for picnic then weather will not be fine(C) If the weather is not fine then my friends will not come or we do not go for a picnic(D) The weather is not fine but my friends will come and we go for a picnic

Sol. (B)Given p (q r)Its contrapositive is ~ ( ) ~q r p

(~ ~ ) ~ q r pp weather is fine, q my friends will come, r we go for a picnic

38. If f(x) = 2

xx 1 is increasing function then the value of x lies in

(A) R (B) (, 1) (C) (1, ) (D) (1, 1)Sol. (D)

f(x) = 2 1x

x f (x) =

2

2 2( 1)(1) (2 )

( 1)x x x

x

=

2

2 21

( 1)x

x

f (x) > 0 1 – x2 > 0

x2 < 1 x (–1, 1)

12 Kalrashukla Classes

39. If X = {4n 3n 1: n N} and Y = {9 (n 1) : n N}, then X Y =(A) X (B) Y (C) (D) {0}

Sol. (A)

X {4n – 3n –1 : n N} 0, 9, 54, 243, 1008, ...

Y {9(n –1): nN} {0, 9, 18, 27, ...}, X Y X Y = X

40. The statement pattern p (~ p q) is(A) a tautology (B) a contradiction(C) equivalent to p q (D) equivalent to p q

Sol. (B)p (~p q) = (p ~p) q = f q f

41. If the line y = 4x 5 touches to the curve y2 = ax3 + b at the point (2, 3) then 7a + 2b =(A) 0 (B) 1 (C) 1 (D) 2

Sol. (A)Given y = 4x – 5 touches the curve y2 = ax3 + b

2 dyydx = 3a x2

dydx =

232ax

y (2,3)

dydx

=

3 46a

= 2a = 4

(2, 3) lies on y2 = 2x3 + b 9 = 16 + b b = 7

4

7a + 2b = 14 – 14 = 0

42. The sides of a rectangle are given by x = a and y = b. The equation of the circle passing through the verticesof the rectangle is(A) x2 + y2 = a2 (B) x2 + y2 = a2 + b2

(C) x2 + y2 = a2 b2 (D) (x a)2 + (y b)2 = a2 + b2

Sol. (B)

y = b

y = –b

x = –a x = a x2 + y2 = (4a2 + 4b2)x2 + y2 = a2 + b2

MH-CET - 2018 13

43. The minimum value of the function f(x) = x log x is

(A)1e

(B) –e (C)1e (D) e

Sol. (A)Given f(x) = xlogx

1( ) log 1 logf x x x xx

, ( ) 0 f x 1xe

( ) f x 11x

= 1 + e > 0 min of f at 1xe

and min. 1 1 1logfe e e

44. If X ~ B (n, p) with n = 10, p = 0.4 the E (X2) = ?(A) 4 (B) 2.4 (C) 3.6 (D) 18.4

Sol. (A)n = 10, p = 0.4, q = 0.6 E(X) = np = 4, Var(X) = npq = 2.4 (E(X2)) – (E(X))2 = 2.4 E(X2) = 2.4 + 16 = 18.4

45. The general solution of differential equation dxdy = cos (x + y) is

(A)x ytan y c

2

(B)

x ytan x c2

(C)

x ycot y c2

(D)

x ycot x c2

Sol. (A)

dxdy = cos (x + y)

Put x + y = t 1 dx dtdy dy 1dt

dy = cos t or

dtdy =

22cos2t

2sec 2

2t dt dy

2 tan 2 22t y c tan

2x y y c

46. If planes ˆ ˆ ˆr (pi j 2k) 3 0 and ˆ ˆ ˆr (2i pj k) 5 0 include angle 3

then the value of p is

(A) 1, 3 (B) 1, 3 (C) 3 (D) 3Sol. (D)

Give angle between (p, –1, 2) and (2, –p, –1) is 3

2 2

2 2 125 5

p p

p p

6p – 4 = p2 + 5 or p2 – 6p + 9 = 0 (p – 3)2 = 0 p = 3

14 Kalrashukla Classes

47. The order of the differential equation of all parabolas, whose latus rectum is 4a and axis parallel to the x-axis, is(A) one (B) four (C) three (D) two

Sol. (D)Its equation is (y – k)2 = 4a (x – h), h and k are two arbitrary constants second order differential equation

48. If lines x 1 y 1 z 12 3 4

and y kx 3 z

2

intersect then the value of k is

(A)92 (B)

12 (C)

52 (D)

72

Sol. (A)The point of intersection is P (2r +1, 3r –1, 4 + 1) (p + 3, 2p + k, p)

2r + 1 – 3 = 4r + 1 2r –2 = 4r + 1 r = 32

p = 2r – 2 = –3 – 2 = –5

P 112, , 52

(–2, –10 + k, – 5) –10 + k = 112

k = 10 – 5.5 = 4.5 = 92

49. If a line makes angles 120and 60with the positive directions of X and Z axes respectively then the angle madeby the line with positive Y-axis is(A) 150 (B) 60 (C) 135 (D) 120

Sol. (C)Given = 120º, = 60º

cos2 = 1 – cos2120º – cos260º = 1 – 14 –

14 =

12 cos2

12

50. L and M are two points with position vectors 2a b and a 2b respectively. The position vector of the pointN which divides the line segment LM in the ratio 2 : 1 externally is(A) 3b (B) 3b (C) 5b (D) 3a 4b

Sol. (C)2 , 2l a b m a b

N is point on LM ratio 2 : 1 is 2m l = 2 4 (2 )a b a b = 5b