mht-cet 2016 : mathematics - actual test...

MHT-CET 2016 : Mathematics - Actual Test Paper

(Solution at the end)

This is a Simulator Test. Do not attempt this Test Paper until you are ready to take this Test under strict

examination conditions.

IMPORTANT INSTRUCTIONS

1. The test is of 1½ hours duration 2. The Test Booklet consists of 50 questions. The maximum marks are 100. 3. Each question is allotted 2 (two) marks for each correct response. 4. No deduction from the total score will be made if incorrect response or no answer is marked. 5. There is only one correct answer for each question.

6. All calculations / writing work are to be done on the, space provided for this purpose in the Test Booklet itself, marked ‘Space for Rough Work’. 7. Use of Electronic / Manual Calculator is prohibited.

8. Use Black Ball Point Pen only for writing particulars / marking responses on the Answer Sheet.

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(2) CET Score Booster : Mathematics

1. Let X ~ B (n, p). If E(X) = 5, Var(X) = 2.5, then P(X < 1) = ________

(A) 11

1

2

(B) 10

1

2

(C) 6

1

2

(D) 9

1

2

2. Derivative of 1

2

xtan

1 x

with respect to sin1(3x 4x3) is _______

(A) 2

1

1 x (B)

2

3

1 x (C) 3 (D)

1

3

3. The differential equation of the family of circles touching yaxis at the origin is

(A) 2 2 dyx y 2xy 0

dx (B) 2 2 dy

x y 2xy 0dx

(C) 2 2 dyx y 2xy 0

dx (D) 2 2 dy

x y 2xy 0dx

4. If

1 1 0

A 2 1 5

1 2 1

, then a11A21 + a12A22 + a13A23 = __________

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

5. If Rolle‟s theorem for f(x) = ex (sinx cosx) is verified on 5

,4 4

, then the value of c is

(A) 3

(B)

2

(C)

3

4

(D)

6. The joint equation of lines passing through the origin and trisecting the first quadrant is

(A) 2 2x 3xy y 0 (B) 2 2x 3xy y 0 (C) 2 23x 4xy 3y 0 (D) 3x2 y2 = 0

7. If 2 tan1

(cosx) = tan1

(2cosecx), then sinx + cosx =

(A) 2 2 (B) 2 (C) 1

2 (D)

1

2

8. Direction cosines of the line x 2 2y 5

, z 12 3

are

(A) 4 3

, ,05 5 (B)

3 4 1, ,

5 5 5 (C)

3 4, ,0

5 5 (D)

4 2 1, ,

5 5 5

9. 2

1dx

8 2x x

(A) 11 x 1sin c

3 3

(B) 1 x 1

sin c3

(C) 11 x 1

sin c3 3

(D) 1 x 1

sin c3

10. The approximate value of f(x) = x3 + 5x2 7x + 9 at x = 1.1 is

(A) 8.6 (B) 8.5 (C) 8.4 (D) 8.3

11. If r.v. X : waiting time in minutes for bus and p.d.f. of X is given by

1, 0 x 5

f(x) 5

0, otherwise,

then probability of waiting time not more than 4 minutes is =

(A) 0.3 (B) 0.8 (C) 0.2 (D) 0.5

12. In ABC : (a b)2cos2 2 2C C(a b) sin

2 2 =

(A) b2 (B) c2 (C) a2 (D) a2 + b2 + c2

MHT-CET 2016 : Mathematics Paper (3)

13. Derivative of log(sec + tan ) with respect to sec at = 4

is

(A) 0 (B) 1 (C) 1

2 (D) 2

14. The joint equation of bisectors of angles between lines x = 5 and y = 3 is

(A) (x 5)(y 3) = 0 (B) x2 y

2 10x + 6y +16 = 0

(C) xy = 0 (D) xy 5x 3y +15 = 0

15. The point on the curve 6y = x3 + 2 at which ycoordinate is changing 8 times as fast as xcoordinate is

(A) (4, 11) (B) (4, 11) (C) (4, 11) (D) (4, 11)

16. If the function f(x) defined by

1

f(x) x sin ,x

for x 0

= k, for x = 0

is continuous at x = 0, then k =

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

2

17. If y = 1m sin xe and

2

2 2dy(1 x ) Ay

dx

, then A =

(A) m (B) m (C) m2 (D) m2

18. If x

x

x

4e 25dx Ax B log | 2e 5 | c

2e 5

, then

(A) A = 5, B = 3 (B) A = 5, B = 3 (C) A = 5, B = 3 (D) A = 5, B = 3

19. 1 1

1 1

tan 3 sec ( 2)

1cosec ( 2) cos

2

(A) 4

5 (B)

4

5 (C)

3

5 (D) 0

20. For what value of k, the function defined by

0

2

log(1 2x) sinxf(x) for x 0

x

k for x 0

is continuous at x = 0 ?

(A) 2 (B) 1

2 (C)

90

(D)

90

21. If 2 2

10 2 2

x y dylog 2, then

dxx y

_______ .

(A) 99x

101y (B)

99x

101y (C)

99y

101x (D)

99y

101x

22. /2

/2

2 sinxlog dx

2 sinx

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


23.

12 (x tan x)

2

(x 2)adx

x 1

=

(A) 1x tan xloga a c

(B) 1(x tan x)

cloga

(C)

1x tan xac

loga

(D) log a . (x + tan1x) + c

24. The degree and order of the differential equation

7/33 2

2

dy d y1 7

dx dx

respectively are

(A) 3 and 7 (B) 3 and 2 (C) 7 and 3 (D) 2 and 3

25. The acute angle between the line ˆ ˆˆ ˆ ˆ ˆr (i 2 j k) (i j k) and the plane ˆˆ ˆr (2i j k) 5

(A) 1 2cos

3

(B) 1 2sin

3

(C) 1 2tan

3

(D) 1 2sin

3

26. The area of the region bounded by the curve y = 2x x2 and xaxis is

(A) 2

3 sq.units (B)

4

3sq. units (C)

5

3sq.units (D)

8

3sq.units

27. If f(x)

dx log logsinx c,log(sinx)

then f(x) =

(A) cot x (B) tan x (C) sec x (D) cosec x

28. If A and B are foot of perpendicular drawn from point Q (a, b, c) to the planes YZ and ZX respectively,

then the equation of plane through the points A, B and O is

(A) x y z

0a b c (B)

x y z0

a b c (C)

x y z0

a b c (D)

x y z0

a b c

29. If ˆ ˆ ˆˆ ˆ ˆ ˆ ˆa i j 2k, b 2i j k and c 3i k and c ma nb then m + n = _____

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

30. /2 n

n n0

sec xdx

sec x cosec x

(A) 2

(B)

3

(C)

4

(D)

6

31. If the p.d.f. of r.v. X is given as

xi 2 1 0 1 2

P(X = xi) 0.2 0.3 0.15 0.25 0.1

then F(0) =

(A) P(X < 0) (B) P(X > 0) (C) 1 P(X > 0) (D) 1 P(X < 0)

32. The particular solution of the differential equation dx

y(1 logx) xlogx 0dy

when x = e, y = e2 is

(A) y = ex log x (B) ey = x logx (C) xy = e logx (D) y logx = ex

33. M and N are the midpoints of the diagonals AC and BD respectively of quadrilateral ABCD, then

AB D CB CD =

(A) 2MN (B) 2NM (C) 4MN (D) 4NM

34. If sin x is the integrating factor (I.F.) of the linear differential equation dy

Py Qdx

, then P is

(A) log sin x (B) cos x (C) tan x (D) cot x


35. Which of the following equation does not represent a pair of lines?

(A) x2 x = 0 (B) xy x = 0 (C) y2 x + 1 = 0 (D) xy + x + y + 1 =0

36. Probability of guessing correctly atleast 7 out of 10 answers in a “True” or “False” test is = ______

(A) 11

64 (B)

11

32 (C)

11

16 (D)

27

32

37. Principal solutions of the equation sin 2x + cos 2x = 0, where < x < 2 are

(A) ,8 8

(B) ,

8 8

(C)

15,

8 8

(D)

15 19,

8 8

38. If line joining points A and B having position vectors 6a 4b 4c and 4c respectively, and the line

joining the points C and D having position vectors a 2b 3c and a 2b 5c intersect, then their

point of intersection is

(A) B (B) C (C) D (D) A

39. If A = 2 2 0 1

, B ,3 2 1 0

then

11 1B A

____

(A) 2 2

2 3

(B) 2 2

2 3

(C) 2 3

2 2

(D) 1

2 3

40. If p : Every square is a rectangle

q : Every rhombus is a kite,

then truth values of p q and p q are _________ and _______ respectively.

(A) F, F (B) T, F (C) F, T (D) T, T

41. If G(g), H(h) and C(c) are centroid, orthocenter and circumcenter of a triangle and xc yh zg 0 ,

then (x, y, z) = _____

(A) 1, 1, 2 (B) 2, 1, 3 (C) 1, 3, 4 (D) 2, 3, 5

42. Which of the following quantified statements is true?

(A) The square of every real number is positive

(B) There exists a real number whose square is negative

(C) There exists a real number whose square is not positive

(D) Every real number is rational

43. The general solution of the equation tan2x = 1 is

(A) n4

(B) n

4

(C) n

4

(D) 2n

4

44. The shaded part of given figure indicates the feasible region

then the constraints are

(A) x, y 0, x + y 0, x 5, y 3 (B) x, y 0, x y 0, x 5, y 3

(C) x, y 0, x y 0, x 5, y 3 (D) x, y 0, x y 0, x 5, y 3


45. Direction ratios of the line which is perpendicular to the lines with direction ratios 1, 2, 2 and 0, 2, 1 are

(A) 1, 1, 2 (B) 2, 1, 2 (C) 2, 1, 2 (D) 2, 1, 2

46. If Matrix 1 2

A4 3

such that Ax = 1, then x = ________

(A) 1 31

2 15

(B)

4 21

4 15

(C)

3 21

4 15

(D)

1 21

1 45

47. If ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆa i j k, b 2i j k and c i j 4k and a (b c) 10 , then is equal to

(A) 6 (B) 7 (C) 9 (D) 10

48. If r.v. X ~ B1

n 5,p ,3

then P(2 < x < 4) = _______

(A) 80

243 (B)

40

243 (C)

40

343 (D)

80

343

49. The objective function Z = x1 + x2 subject to x1 + x2 10, 2x1 + 3x2 15, x1 6, x1, x2 0 has

maximum value _______ of the feasible region.

(A) at only one point

(B) at only two points

(C) at every point of the segment joining two points

(D) at every point of the line joining two points

50.

Symbolic form of the given switching circuit is equivalent to______ (assume S1 = p and S2 = q)

(A) p ~q (B) p ~ q (C) p q (D) ~(p q)

Solution to MHT-CET 2016 : Mathematics - Actual Test Paper

ANSWER KEY

Q. No. Ans. Q. No. Ans. Q. No. Ans. Q. No. Ans. Q. No. Ans.

1. (B) 2. (D) 3. (B) 4. (B) 5. (D)

6. (C) 7. (B) 8. (A) 9. (D) 10. (A)

11. (B) 12. (B) 13. (B) 14. (B) 15. (A)

16. (A) 17. (C) 18. (B) 19. (B) 20 (C)

21. (A) 22. (D) 23. (C) 24. (B) 25. (B)

26. (B) 27. (A) 28. (A) 29. (C) 30. (C)

31. (C) 32. (A) 33. (C) 34. (D) 35. (C)

36. (A) 37. (C) 38. (A) 39. (A) 40. (D)

41. (B) 42. (A) 43. (C) 44. (B) 45. (B)

46. (C) 47. (A) 48. (B) 49. (C) 50. (D)

S1 2S

S2 1S

L


DETAILED SOLUTIONS

1. (B)

Here, E(x) = np = 5 and Var(x) = npq = 2.5

2.5 1

q5 2

1

p2

and np = 5 5 5

n 101p

2

Now, P(x < 1) = P(x = 0)

=

0 10 0 10 1010

0

1 1 1 1C 1

2 2 2 2

2. (D)

Let u = 1

2

xtan

1 x

and v = sin1(3x 4x3)

Put x = sin = sin1x

u = 1

2

sintan

1 sin

and v = sin1(3sin 4sin3)

= 1 sin

tancos

and v = sin1(sin3)

= 1tan tan and v = sin1(sin3)

u = = sin1x and v = 3 = 3sin1x

2

du 1

dx 1 x

and

2

dv 3

dx 1 x

2

2

1du

du 11 xdxdvdv 33dx

1 x

3. (B)

As required circle touches yaxis at the origin, let centre of the circle be (a, 0) and radius is a.

Equation of circle will be,

(x a)2 + (y 0)2 = a2

x2 2ax + a2 + y2 = a2

x2 + y2 2ax = 0 …(i)

By differentiating above equation w.r.t. x, we get

dy

2x 2y 2a 0dx

2a = dy

2x 2ydx

…(ii)

From (i) and (ii),

x2 + y2

dy2x 2y x 0

dx

x2 + y2 2x2 2xydy

0dx

2 2 dy

x y 2xy 0dx

x2 y2 + 2xydy

dx = 0


4. (B)

We know that, aij stands for element in matrix A in ith row and jth column, and

Aij stands for cofactor of element aij of matrix A.

a11 = 1, a12 = 1 and a13 = 0

2 1

21

1 0A ( 1) 1

2 1

2 2

22

1 0A ( 1) 1

1 1

2 3

23

1 1A ( 1) 1

1 2

a11A21 + a12A22 + a13A23 = 1 (1) + 1(1) + 0 (1) = 0

5. (D)

Given, f(x) = ex (sin x cos x)

x x xf (x) e cosx sin x sin x cosx e e cosx sin x sin x cosx

xf (x) 2e sin x

To verify Rolle‟s Theorem,

f (c) 0 c2e sin c 0 sin c 0 c 2 , 3 , ...... [ as ec 0]

c = 5

as ,4 4

6. (C)

As given, both lines pass through (0, 0), and 1 2,6 3

Equation of first line is :

y 0 = tan x 06

1y x

3 3y x

x 3 y 0 …(i)

Equation of second line is :

y 0 = tan x 03

y 3x

3 x y 0 …(ii)

Hence, joint equation of these lines is

(x 3y) ( 3x y) 0

2 23x xy 3xy 3y 0

2 23x 3y 4xy 0

7. (B)

As given, 2tan1(cos x) = tan1 (2cosec x)

tan1(cos x) + tan1(cos x) = tan1(2cosec x)

tan11

2

2cos xtan (2cosecx)

1 cos x

… 1 1 1 A B

tan A tan B tan1 AB

2

2cos x2cosec x

sin x

cos x 1 1cot x 1

sin x sin x sin x

x4

1 1 2sin x cos x sin cos 2

4 4 2 2 2

2nd line

1st line

1 2

y

x


8. (A)

The equation of line is

x 2 2y 5

, (z 1) i.e.2 3

52 y

x ( 2) 2, z 1

2 3

, i.e.

5y

x ( 2) 2, z 1

32

2

Hence the line is passing through the point 5

2, , 12

and

direction ratios of the line are 3

2, , 02

i.e. 4, 3, 0

direction cosines of the line are

2 2 2 2 2 2

4 3 0 4 3, , i.e. , , 0

5 5(4) (3) 0 (4) (3) 0 (4) (3) 0

9. (D)

Let 2

dxI

8 2x x

2 2 2 2

dx dx dx

9 2x x 1 9 (x 2x 1) (3) (x 1)

1 x 1

I sin c3

1

2 2

dx xsin c

aa x

10. (A)

f(x) = x3 + 5x2 7x + 9

2f (x) 3x 10x 7

Let a = 1 and h = 0.1

f (a) = f(1) = (1)3 + 5(1)2 7(1) + 9 = 15 7 = 8

2f (a) f (1) 3(1) 10(1) 7 13 7 6

f(1.1) = f(1) + (0.1) f (1)

= 8 + (0.1)(6) = 8 + (0.6) = 8.6

11. (B)

Required probability = P(X 4)

=

4 4

0 0

1 1dx dx

5 5 =

4

0

1x

5 =

14 0.8

5

12. (B)

2 2 2 2C C

(a b) cos (a b) sin2 2

= 2 2 2 2 2 2C Ca 2ab b cos (a 2ab b )sin

2 2

= 2 2 2 2 2 2 2 2C C C Ca b cos 2abcos (a b )sin 2absin

2 2 2 2

= 2 2 2 2 2 2C C C Ca b cos sin 2ab cos sin

2 2 2 2

= 2 2a b 1 2ab cosC

= 2 2 2a b 2abcosC c …[By cosine rule]


13. (B)

Let y1 = log(sec + tan )

21dy 1sec tan sec

d sec tan

1dy sec [sec tan ]

d [sec tan

1dysec

d

…(i)

Let y2 = sec

2dysec tan

d

…(ii)

1

1

22

dy

dy d

dydy

d

= sec

cotsec tan

1

24

dycot 1

dy 4

14. (B)

We can say both line passes through point (5, 3) and makes angle 45° and 135° with x axis

Equation of 1st line is,

y 3 = tan 45°(x 5) y 3 = x 5

y x + 2 = 0 …(i)

Equation of 2nd line is

y 3 = tan 135°(x 5) y 3 = 1(x 5)

y + x 8 = 0 …(ii)

Required joint equation of line is

(y x + 2) (y + x 8) = 0

y2 xy + 2y + xy x2 + 2x 8y + 8x 16 = 0

x2 + y2 + 10x 6y 16 = 0

x2 y2 10x + 6y + 16 = 0

15. (A)

As required point is on the curve : 6y = x3 + 2, we will go by options.

We find that only point (4, 11) satisfies the given equation.

This problem can be alternatively solved as follows :

We have 6y = x3 + 2 and dy dx

8dt dt

2dy dx

6 3xdt dt

2 2dx dx

6 8 3x 48 3xdt dt

x2 = 16 x = 4

3x 2y

6

3(4) 2y

6

or

3( 4) 2y

6

= 64 2

6

or

64 2

6

y = 11 or y = 62

6

Hence required points are (4, 11) and 62

4,6

x = 5

y = 3

x

(5, 3)

y

45° 135°

2nd line 1st line


16. (A)

Since f(x) is continuous at x = 0, we write

f(0) = x 0lim f (x)

x 0

1k lim x sin 0 (finite number) 0

x

k = 0

17. (C)

Given, 1msin xy e

1msin x

2

dy 1e m

dx 1 x

1

22 2msin x 2

2

dy 1e (m)

dx 1 x

= 1

2 22msin x 2

2 2

1 y me m

1 x 1 x

2

2 2 2dy1 x m y

dx

Comparing, we write A = m2

18. (B)

Let I =

x

x

4e 25dx

2e 5

=

x x

x

10e 25 6edx

2e 5

x x

x x

5(2e 5) 6 edx dx

2e 5 2e 5

=

x

x

2e5 dx dx

2e 5

I = 5x 3log(2ex 5 ) + c

Comparing we get, A = 5 and B = 3

19. (B)

Let y = 1 1

1 1

tan 3 sec ( 2)

1cos ec ( 2) cos

2

= 1 1

1 1

tan 3 sec (2)

1cos ec ( 2) cos

2

=

1 1

1 1

1tan 3 cos

2

1 1sin cos

22

=

2

3 3 3 37 5

4 3 12 12

= 12 4

3 5 5

20. (C)

Since f(x) is continuous at x = 0, we write

0

2x 0

log(1 2x) sin xlim k

x


0

x 0

log(1 2x) sin xlim

x x

= k

c

x 0

xsin

log(1 2x) 180lim k

2x x

x 0 x 0

sin x2 log(1 2x) 180

lim lim k2x 180

x180

2 k180

k

90

21. (A)

Given,

2 2

10 2 2

x ylog 2

x y

2 22

2 2

x y10 100

x y

…[By definition of logarithm]

x2 y2 = 100x2 + 100y2 101y2 = 99 x2

99x2 + 101 y2 = 0

Differentiating, we get

(2 99)x + (2 101)y dy

0dx

dy

99x 101y 0dx

dy

101y 99xdx

dy 99x

dx 101y

22. (D)

Let I =

/2

/2

2 sin xlog dx

2 sin x

Let f(x) = log2 sin x

2 sin x

log(2 sinx) log(2 + sinx)

f(x) = log2 sin( x) 2 sin x

log2 sin( x) 2 sin x

= log(2 + sinx) log(2 sinx)

= [log(2 sinx) log(2 + sinx)]

= f(x)

Hence f(x) is odd function.

I = 0

23. (C)

Let I =

12 x tan x

2

(x 2) adx

1 x

Put 1x tan xa t

1x tan x

2

1a 1 (log a)dx dt

1 x

1

2x tan x

2

1 x 1a (log a)dx dt

1 x

1

2x tan x

2

2 x dta dx

log a1 x


dt 1 t

I dt clog a log a log a

1x tan xaI c

log a

24. (B)

We have

7/33 2

2

dy d y1 7

dx dx

Raising both sides to power 3, we get

7 33 23

2

dy d y1 (7)

dx dx

Hence order = 2, degree = 3 degree and order are respectively 3 and 2.

25. (B)

We have line ˆ ˆ ˆ ˆ ˆ ˆr (i 2 j k) (i j k) and plane ˆ ˆ ˆr (2i j k) 5

Standard equation of line and plane are r a b and r n p respectively

ˆ ˆ ˆ ˆ ˆ ˆb i j k and n 2i j k

Let be the required angle.

ˆ ˆ ˆ ˆ ˆ ˆi j k 2i j kb n

sinˆ ˆ ˆ ˆ ˆ ˆ| b | | n | i j k 2i j k

2 2 2 2 2 2

(1)(2) (1)( 1) (1)(1) 2 1 1

3 6(1) (1) (1) (2) ( 1) (1)

= 2

18 =

2 2

33 2

1 2sin

3

26. (B)

Point of intersection of y = 2x x2 and x axis i.e. y = 0 is

0 = 2x x2 x(2x) = 0 x = 0 or x = 2

When x = 0, y = 0 and when x = 2, y = 0

Hence point of intersection are (0, 0) and (2, 0)

Hence required area is

22

0

A (2x x )dx

2 22 2 2 32

0 0 0 0

x xA 2 x dx x dx

2 3

=

23

2

0

xx

3

=

84

3

=

4

3 sq. unit

27. (A)

Given f (x)

dx log logsin x clog(sin x)


Differentiating both sides w.r.t. x, we get

f (x) d

log(log(sin x)) clog(sin x) dx

f (x) 1 1

cos x 0log(sin x) log(sin x) sin x

f (x) 1

cot xlog(sin x) log(sin x)

f(x) = cot x

28. (A)

A and B are foot of perpendicular from point (a, b, c) on planes YZ and ZX respectively.

A ≡ (0, b, c) and B ≡ (a, 0, c)

We have to find equation of plane passing through A(0, b, c), B(a, 0, c) and O(0, 0, 0)

ˆ ˆ ˆ â bj ck , b ai ck , o 0

Hence required vector equation of plane is

r (AB AO ) a (AB AO )

ˆ ˆ ˆ ˆ ˆ ÂB b a (ai ck bj ck) ai bj

ˆ ˆ ˆ ÂO a bj ck) bj ck

ˆ ˆ î j k

AB AO a b 0

0 b c

= ˆ ˆ î(bc) j( ac) k( ab) = ˆ ˆ ˆbc i ac j ab k

a (AB AO ) ˆ ˆ ˆ ˆ ˆ(bj ck) (bc i ac j ab k) abc abc 0

Required equation is ˆ ˆ ˆr (bc i ac j ab k) 0

If ˆ ˆ ˆr x i y j z k , then above equation becomes ˆ ˆ ˆ ˆ ˆ ˆ(x i y j z k) (bc i ac j ab k) 0

bc x + ac y ab z = 0

Dividing throughout by abc, we get

x y z

0a b c

29. (C)

From given data, we write

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ3i k m( i j 2k) n( 2i j k)

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ3i k m i mj 2mk 2ni nj nk

ˆ ˆ ˆ ˆ ˆ3i k m 2n) i (m n) j (n 2m)k

Comparing, we get

m+2n = 3 …(i)

m n = 0 …(ii)

n 2m = 1 …(iii)

Solving, we get m = 1 and n = 1 m + n = 2

30. (C)

Let I =

/2 n

n n0

sec xdx

sec x cosec x

…(i)

I =

n/2

0 n n

sec x2

dx

sec x cosec x2 2

…(ii)


=

/2 n

n n0

cosecxdx

cosecx sec x

Adding equation (i) and (ii), we get

2I =

/2 n n

n n0

sec x cosec xdx

sec x cosec x

2I =

/2

0

dx

2I = /2

0x

2I

2

I =

4

31. (C)

F(0) = P(X 0)

= P(X = 2) + P(X = 1) + P(X = 0)

= (0.2) + (0.3) + (0.15) = 0.65

P(X>0) = P(1) + P(2)

= 0.25 + 0.1 = 0.35

F(0) = 1 P(X > 0)

32. (A)

We have dx

y(1 log x) x log x 0dy

dx

y(1 log x) x log xdy

1 log x dy

dxx log x y

Integrating both sides, we get

1 log x dy

dxx log x y

In LHS, put xlogx = t

x

log x dx dtx

(1 log x) dx dt

Given equation becomes

dt dy

t y

log t = log y + log c

log(xlog x) = log y + log c log(xlog x) = log(y . c)

xlog x = y . c …(i)

As x = e, y = e2

e log e = e2 . c e(1) = e2 . c c = 1

e

Putting 1

ce

in eq.(i) we get

y

x log xe

i.e. y = ex log x

33. (C)

AB AD CB CD

= (AM MN NB) (AM MN ND)

(CM MN NB) (CM MN ND)

A B

D

N M

C


= 2AM 4MN 2NB 2ND 2CM

= 4MN 2(AM CM NB ND)

= 4MN 2(AM MC NB DN)

= 4MN ...[AM MC and NB DN , as M & N are mid points of AC & BD resp. given]

34. (D)

It is given that Integrating factor of dy

Py Qdx

is sin x

Pdx

e sin x P dx ln(sin x)

Differentiating both sides w.r.t. x, we get

d

P ln(sin x)dx

P = 1

cos xsin x

P = cotx

35. (C)

We will go by options

Option A :

x2 x = 0 x(x1) = 0 x = 0 and x = 1 are two lines

Option B :

xy x = 0 x(y 1) = 0 x = 0 and y = 1 are two lines

Option D :

xy + x + y + 1 x(y + 1) + 1(y + 1) = 0 x + 1 = 0 and y + 1 = 0 are two lines

Option C :

y2 x + 1 = 0 y2 = x 1 This is equation of parabola

Alternatively this problem can be solved as follows :

When ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of straight lines, then

abc + 2fgh af2 bg2 ch2 = 0

Students may check all options by using this formula.

36. (A)

Let „p‟ be the probability of guessing correctly in a „True‟ or „False‟ test.

1

p2

1 1

q 12 2

There are 10 questions n = 10

From given condition, we write

P(x 7) = P(x = 7) + P(x = 8) + P(x = 9) + P(x = 10)

=

7 3 8 2 9 1 1010 10 10 10

7 8 9 10

1 1 1 1 1 1 1C C C C

2 2 2 2 2 2 2

=

1010 10 10 10

7 8 9 10

1C C C C

2

= 1

120 45 10 11024

176 22 11

1024 128 64

37. (C)

We have, sin2x + cos2x = 0

Multiplying by 1

2 on both sides, we get

1 1

sin 2x cos 2x 02 2

sin 2x cos cos 2x sin 04 4


sin 2x 04

sin 2x sin(0)

4

2x n4

2x = n

1 4n 1n

4 4 4

(4n 1)

x8

11

x8

and

15

8

….[ < x < 2] (Putting n = 3, 4 gives required answer)

38. (A)

Vector equation of line passing through A and B is

r = (6a 4b 4c) [( 4c) (6a 4b 4c)]

= (6a 4b 4c) ( 6a 4b 8c)

xa yb zc = (6 6 )a ( 4 4 )b (4 8 )c

Comparing, we get

x = (6 6), y = (4 + 4), z = (4 8)

Thus, coordinates of any point on line AB are

[(6 6), (4 + 4), (4 8)] …(i)

Similarly vector equation of line passing through C and D is

r = ( a 2b 3c) [(a 2b 5c) ( a 2b 3c) ]

( a 2b 3c) 2a 4b 2c]

xa yb zc = (1 + 2) a ( 2 4 ) b ( 3 2 )c

Comparing, we get

x = (1 + 2), y = ( 2 4 ), z ( 3 2 )

Thus coordinates of any point on line CD are

[(1 + 2), ( 2 4 ), ( 3 2 )] …(ii)

It is given that lines AB and CD intersect.

From equation (i) and (ii), we write

6 6 = 1 + 2 6 + 2 = 7 …(1)

4 + 4 = 2 + 4 4 4 = 2 …(2)

4 8 = 3 2 8 2 = 7 …(3)

Solving equation (1) and (2), we get = 1 and 1

2

These values of and satisfy equation (3)

Hence point of intersection is

[(66), (4 + 4), (4 8)] = (6 6, 4 + 4, 4 8) i.e. (0, 0, 4) which is point B.

Students may get the answer by using value of .

39. (A)

1 1 1 1 1 1 1(B A ) (A ) (B ) AB

1 1 1 2 2 0 1

(B A )3 2 1 0

0 2 2 0 2 2

0 2 3 0 2 3

40. (D)

p : Every square is a rectangle : Truth value of statement p is T

q : Every rhombus is a kite : Truth value of statement q is T

p q ≡ T T ≡ T

p q ≡ T T ≡ T


41. (B)

We know that centroid, orthocenter and circumcentre of a triangle are collinear and distance between

centroid and orthocenter is twice the distance between centroid and circumcentre.

H : Orthocentre

G : Centroid

C : Circumcentre

GH = 2 GC

Thus G divides segment HC in the ratio 2 : 1

2c h

g 3g 2c h i.e. 2c h 3g 02 1

We have xc yh zg 0 …(given)

Comparing, we get x = 2, y = 1, z = 3

42. (A)

By fundamental concepts about real numbers, we find that only option (A) is correct.

43. (C)

tan2x = 1

tan x = 1 tan x = 1 or tan x = 1

tan x = 4

or tan x =

3

4

x = n4

44. (B)

By fundamental concepts of LPP, the inequalities of feasible region are as follows :

Line OC : x = y x y 0

Line AB : x = 5 x 5

Line BC : y = 3 y 3

Feasible region is in 1st quadrant x 0, y 0

45. (B)

Let direction ratios of the required line be a, b, c

1a + 2b + 2c = 0

0a + 2b + 1c = 0

a b c

2 2 1 2 1 2

2 1 0 1 0 2

a b c

2 4 1 0 2 0

a b c

2 1 2

a b c

2 1 2

Hence direction ratios of the required line are 2, 1, 2

46. (C)

Given : Ax = I A1Ax = A1 I Ix = A1

x = A1

We have 1 2

A4 3

| A | = 3 8 = 5

and adj A = 3 2

4 1

1 3 21

A x4 15

3 21x

4 15

H G C

D

F E


47. (A)

1 1 1

a (b c) 2 1 10

1 1 4

1(4 + 1) 1(8 1) + 1(2) = 10

4 + 1 7 2 = 10 3 9 + 1 = 10 3 = 18 = 6

48. (B)

We have, n = 5, p = 1

3 q = 1 p =

1 21

3 3

P(2 < x <4) = P(x = 3)

=

3 25

3

1 2C

3 3

= 5 4 1 4

2 27 9

=

40

243

49. (C)

First we draw lines AB, CD, EF whose equations are x1 + x2 = 10, 2x1 + 3x2 = 15, x1 = 6 respectively

Line Equation Point on X axis Point on Y axis Same/Opposite side of origin

AB 1 2x x 10 A(10, 0) B(0, 10) Same

CD 2x1 + 3x2 = 15 C(7.5, 0) D(0, 5) Same

EF x1 = 6 E(6, 0) Same

Point of intersection of AB and CD is P(3, 7)

Point of intersection of AB and EF is Q(6, 4)

Point of intersection of CD and EF is R(6, 9)

The feasible region is shown shaded in figure.

The vertices of feasible region are O(0, 0), D(0, 5), P(3, 7), Q(6, 4) and E(6, 0).

The values of objective function Z = x1 + x2 at these vertices are as follows.

Z(O) = 0 + 0 = 0

Z(D) = 0 + 5 = 5

Z(P) = 3 + 7 = 10

Z(Q) = 6 +4 = 10

Z(E) = 6 + 0 = 6

Z has maximum value at P and Q. Hence Z has maximum value at every point of the segment joining P

and Q.

50. (D)

We have, S1 ≡ p and S2 ≡ q

The symbolic form of given switching circuit is

(p ~q) (~p q) ≡ ~(p q) …[By fundamental concept]

7 8

C

O E 6 10

A

B

R

F

D

y

10

Q

x

5

P

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