mht-cet – 5 (mathematics) mathematics test-05 mht-cet maths... · 2018. 4. 26. · mht-cet – 5...

MHT-CET – 5 (MATHEMATICS)

CENTERS: MUMBAI /DELHI / AKOLA / LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO /DUBAI # 1

MATHEMATICS 1. (C)

Let P a cos , bsin be any point on the ellipse 2 2

2 2

x y 1a b

Equation of tangent is

x cos ysin 1

a b

It meets the coordinate axes at A and B

a bA ,0 .B 0,cos sin

Mid – point of AB is

a b

2cos 2sin

a bcos sin2 2

Squaring and adding

2 2

2 22 2

a bcos sin 14 4

Locus of , is

2 2

2 2

a b 14x 4y

2 2 2 2 2 2a y b x 4x y 2. (A)

p q p q ~ p (p q) ~ p T T T F T T F T F T F T T T T F F F T T

It is clear that p q ~ p is a tautology. 3. (C)

q p q → p p → (q → p) p q p (p q) T T T T T T T F F T T T F T T T T T F F T T F T

∴Statement p →(q → p) is equivalent to p p q . 4. (B)

Let three numbers be a ,a,arr

Then according to the question



a a ar 3375r

3a 3375 a 15 (i)

a a ar 65r

21 r ra 65

r

2r r 1 65 13

r 15 3

23r 3r 3 13r 23r 10r 3 0 23r 9r r 3 0 3r 1 r 3 0

1r or33

5. (C)

Given, 2 1 3 x 91 3 1 y 43 2 1 z 10

It can be written as AX = B Now, A 2 3 2 1 1 3 3 2 9 7

1

5 7 8 5 7 81adj A 4 7 5 A 4 7 57

7 7 7 7 7 7

From Eq. (i) 5 7 8 9

1x 4 7 5 47

7 7 7 10

1

AX BX A B

x 7

1y 147

z 21

x 1, y 2 and z 3 6. (A) The given equation can be rewritten as

2 11 cos cos4

2 3cos cos 04

24cos 4cos 3 0



4 16 48 4 8 1 3cos ,8 8 2 2

Since, 3cos2

is not possible, so we take

1cos cos 2n2 3 3

…….. (i)

For the given interval, put n = 0 and n 1 in Eq. (i), we get

5,

3 3

7. (B)

Given, 1 1sin x cos x 12 2

cos 14

2n 14

2n4

32n4

8. (D) Here, a 6,2h 1,b 4c

Now, 1 2 1 22h 1 a 6m m ,m mb 4c b 4c

One line of given pair of lines is 3x 4y 0

Slope of line 13 m4

[say]

23 1m4 4c

21 3m4c 4

And 3 1 3 64 4c 4 4c

6 41 3c3

3c 9 c 3 9. (C) (1) c a b c a b …….. (i) a b c a b c ………. (ii) Since a, b, c are in A. P.

a cb a c b 2b a c b

2



Now a c 2a a c a ca b a

2 2 2

……… (iii)

And a c a c 2c a cb c c

2 2 2

………. (iv)

From (iii) & (iv) a b b c b c , c a , a b are in A. P.

(2) If a b c, ,

abc abc abcare in A. P.

Then a, b, c are in A. P.

1 1 1, ,bc ac ab

are in A. P.

(3) a b c, ,

abc abc abcare in A. P.

Then 1 1 1, ,bc ac ab

are in A. P.

So, 1 1 1,ab, bc ac

are not in A. P.

(4) 1 1 1, ,b c c a a b

are in A. P.

If 1 1 1 1c a b c a b c a

b a c bc a a c a b c a

b a c bb c a b

b a c b 2b c a i.e. a, b, c are in A. P., which is true.

1 1 1, ,b c c a a b

are in A. P.

10. (A) Given that, , be the angle between the lines represented by equaton 2 23x 7xy 4y 0

1 1

492 3 4 14tan tan3 4 7

…….. (ii)

And 2 be the angle between the lines represented by equation 2 26x 5xy y 0

2 2

252 6 14tan tan6 1 7

……. (ii)



From Eqs. (i) and (ii), we get 1 2tan tan 1 2 11. (C) Given that, a i j k, a.b 1 And a b j k As, we know a a b a.b a a.a b

i j k j k i j k 3b a.a 1 1 1 3

2i j k i j k 3b 2i j k i j k 3b

3b 3i b i 12. (B) R 3,3 , 6, 2 , 9,1

Range = 3, 2,1 1, 2,3 13. (A)

To determine the angle between a and b, we can use the formula a.bcosa b

It is given that a 3, b 2 and a.b 6 Let be the required angle, then

a.b 6 1cosa b 3 2 2

1 11cos cos cos4 42

14. (C) 2 2 2sin sin sin 2 2 21 cos 1 cos 1 cos

2 2 23 cos cos cos 3 1 2 15. (A or B) Given, 1 1 1a ,b , c p,1, 2 And 2 2 2a , b , c 2, p, 1

1 2 1 2 1 22 2 2 2 2 21 1 1 2 2 2

a a b b c ccosa b c a b c

2 2 2 2 2 2

p 2 1 p 2 11cos3 2 p 1 2 2 p 1



2 2

1 3p 22 p 5 p 5

2

1 3p 22 p 5

2p 6p 1 0

26 6 4 1 6 32p2 1 2

6 4 22

3 2 2 16. (A) The given equation of line is

5yx 2 2y 5 x 2 z 12, z 1 32 3 2 02

So, its vector equation is

5 3r 2i j k 2i 0k2 2

Hence, x = 0 17. (C) R 3, 2 , 5, 2 , 5, 4 Range = {2, 4} 18. (B) Since, the line passing through the points (4, -1, 2) And 3, 2,3 . So, the DR’s the line is 4 3, 1 2, 2 3 i.e. 7, 3, 1 Since, the line is perpendicular to the plane therefore, DR’s of this line is proportional to the normal of the plane. Required equation of plane is 7 x 10 3 y 5 1 z 4 0 7x 3y z 89 0 19. (B)

1x y 2 z 121 2 a

If any line is parallel to the plane, then 1 2 1 2 1 2a a b b c c 0 Here DR’s of a plane 1 1 1a , b , c 2, 1,1 And DR’s of a line 2 2 2a , b , c 1, 2, a



2 1 2 1 a 1 0 a 4 20. (C)

Time taken to cover x km x h25

Time taken to cover y km y h

40

Since, he has only one hour.

x y 125 40

Total cost of petrol 2x 5y Since, he has only Rs. 100 to spend 2x 5y 100 Since, and y are distances, so x and y cannot be negative. x 0, y 0 Hence, inequalities represent the given data are

x y2x 5y 100, 1, x 0, y 025 40

21. (C) When inequality 2x y 8 At 0,0 ,0 0 8 (not true) So shaded region is away from the origin. When inequality x 2y 10 . At 0,0 ,0 0 10 (not true) So shaded region is away for the origin. 22. (D)

We know 1 1sin x cos x2

We have 1 1 bbsin x bcos x2

And 1 1a sin x bcos x c Adding the above 2 equations.

1 1 bbsin x a sin x c2

1 ba b sin x c2

1b c

2sin xa b

Similarly 1a c

2cos xa b



1 1 ab c a b

a sin x bcos xa b

23. (D) Since, f x is continuous at x = 0 LHL RHL f 0 ……. (i)

Now, 2

2 2x 0 x 0 x 0

1 cos 4x 2sin 2xlim f x lim limx x

2

2

x 0

sin 2xLHL lim 8 8 1 82x

x 0 x 0

xRHL lim f x lim16 x 4

x 0

x 16 x 4lim

16 x 4 16 x 4

x 0

x 16 x 4lim

16 x 16

x 0lim 16 x 4 8

From Eq (i), 8 8 a a 8 24. (D)

Given, x ,if x 1

f x cx k, , if1 x 42x ,ifx 4

Since, f x is continuous everywhere.

x 1lim f x f 1

and x 4limf x f 4

x 1lim cx k 1

and x 4lim cx k 2 4

c k 1 and 4c k 8 On solving, we get c 3,k 4 25. (A) Let y tan x On differentiating both sides w.r.t. x, we get

dy 1 d tan xdx dx2 tan x

21 1sec x2 x2 tan x



2sec x

4 x tan x

26. (B)

Give, 1 1sin x sin y2

1 1sin x sin y2

1 1sin x cos y

1 2 1cos 1 x cos y On differentiating both sides w.r.t. x, we get

2

dy 1 x2xdx y2 1 x

27. (C) 1 1 1 1y tan x cot x sec x cos ec x

dy 0

2 2 dx

28. (C) Let u sin x3 and v cos x3 On differentiating both functions w.r.t. x, we get

3 2du cos x .3xdx

and 3 2dv sin x .3xdx

2 3

32 3

du du dx 3x cos x cot xdv dv dx 3x sin x

29. (B) Given, y xy log y ylog x

1 dy 1 dy. y. .log xy dx x dx

2dy 1 y dy ylog xdx y x dx x 1 y log x

30. (B)

Given, sin yx

sin a y

On differentiating both sides w.r.t. y, we get

2

sin a y cos y sin ycos a ydxdy sin a y

2

sin a y ysin a y



2sin a ydydx sin a

31. (A) Draw BP AC , the longer diagonal. In right angled triangle APB, we have

oAP cos30AB

o 3AP ABsin 30 10 5 3cm2

Also, oBP sin 30AB

o 1BP ABcos30 10 5cm

2

In 2 2 2BPC, BP PC BC 2 2 25 PC 6 2 2 2PC 6 5 11 AC AP PC 5 3 11 cm

option (a) is correct 32. (C) They will encounter if 2 210 6t 3 t t 6t 7 0 t 7 At t 7s, speed of first point 1v

1d 10 6t 6cmsdt

At t 7s , speed of second point 2v

2 1d 3 t 2t 2 7 14cmsdt

Resultant velocity 12 1v v 14 6 8cms

33. (B)

2

1 dxdx1 cos ax 2cos ax 2

2 tan ax 21 ax 1sec dx .2 2 2 a 2



1 axtan Ca 2

34. (A)

Let

4

3 24

1 xI dx1 x

33 3

3 2 3 23 2 2

2 2

1 1x x xx xdx dx

1 1x x xx x

Put 22

1 x zx

3

1 dzx dxx 2

1 23 2 4

1 dz xI z C C2 z 1 x

35. (A)

Put 12

dxtan x t dt1 x

12

tan x t 22

1 x xe dx e tan t sec t dt1 x

1t tan xe tan t C xe C

36. (C)

4 3 4 3 3 3 2

x 1 xdx dx 1 1dx dxx x x x x x 1 x x x 1

3 2

1 1 1 dxx x x x 1

3 2

1 1 1 1 dxx x x x 1

2

1 1 log x log x 1 C2x x

2

1 1 xlog C2x x x 1

1A and B 12

37. (B)

2

2

1 1 1dx dx4 99x 4x x x

4



2 2

2

1 1 dx2 9 9 9x x

4 8 8

2 2

1 1 dx2 9 9x

8 8

1 1

9x1 1 8x 98sin C sin C92 2 98

38. (C) Let 17 4f x x cos x f x is an odd function.

1 17 4

1x cos x dx 0

39. (A)

Let a xf x loga x

a xf x loga x

a xloga x

f x is an odd function.

10

10f x dx 0

40. (A)

Given pair of lines 2 2x y 2xy 0

p q h

OR 2 21 1 1x y 2 xy 0P q h

1 2 1 2

22q qhm m m m1 h p

q

2 1 12qm 2m given 3mh

21 2

q 4q q2m 2p 9h p

2

pq 9h 8

2pq : h 9 :8



41. (A) Given xy 2x 2y 4 0 And x y 2 0 From eq. (i) and (ii) we get xy 0 x 0 y 0 Vertices of triangle 2,0 0,0 and 0, 2 In a right angled triangle circumcentre is mid point of hypotenuse. 42. (B)

Intersection points of both the curves are (0, 0) and 2

16 16,m m

. Therefore, required area

216 m

0

216x mx dx3

216 m2

3 2

0

2 mx 24, x3 2 3

3 2 2

2 2

8 16 m 16 23 m 2 m 3

3 3 3

8 64 16 16 1 512 256 2.3 m 2m m 3 2 3

3 128 3m 643 2

m 4 43. (B) Given, 2 2x y 2ax 0 On differentiating w.r.t. x, we get 2x 2yy' 2a 0 a x yy' On putting the value of a in Eq. (i), we get 2 2x y 2x x yy ' 0

2 2y x 2xyy ' 44. (D) Given differential equation can be rewritten as



x x

x x

ydy e dx 1 e1 dy dxy 1 e 1 y 1 e 1

xy log y 1 log e 1 log C [Integrating both sides]

xe 1 y 1

y logC

x ye 1 y 1 Ce 45. (C)

PQ lines x 2y 1 Equation of PQ

1y 5 x 32

2y 10 x 3 x 2y 7 …….. (1) Solving 2x 3y 14 and x 2y 7 0 We get B 1,4

2 2PQ 3 1 5 4 5 46. (B) Equation of family of parabolas with focus at (0, 0) and X – axis as axis is 2y 4a x a ….. (i) On differentiating Eq. (i), we get

12yy 4a 1dyHere, ydx

On putting the value of a in Eq. (i),

2 11

yyy 2yy x2

21 1y 2xy yy

2dy dyy 2x y

dx dx

47. (A) We know that sum of pdf is one 2 2 20 k 2k 2k 3k k 2k 7k k 1 210k 9k 1 0 10k 1 k 1 0



1k k 1

10

P x 3 P x 0 P x 1 P x 2

30 k 2k 3k

10

P x 6 P x 7

2 7 1 177k k100 10 100

P 0 x 3 P x 1 P x 2

3k 2k 3k

10

48. (B) The equation of line through 1 2A L L 0 2x 3y 1 ax by 1 0

It passes through 0 0,0 1 So 2 a x 3 b y 0 Side AD BC

2 a 1 13 b 2

a 2b 8 …… (2) Similarly BE AC a b 0 ……… (3) b 8 and a 8 Hence a, b is 8,8 49. (B)

Given, n = 4 and 16P X 081

Let p be the probability of success and q be the probability of failure in a trial.

Then 16P X 081



4 o 40

16C p q81

4

4 2 2q q3 3

1p3

4

4 4 o 44

1 1P X 4 C p q p3 81

50. (B) The probability of suffering of a disease is 10%.

10 1p

100 10 and

9q10

Total number of patients, n = 6 Required probability

3 3

63

1 9C10 10

6.5.4 1 9 9 93.2.1 1000 1000

55

2 729 1458 1010

mht-cet – 5 (mathematics) mathematics test-05 mht-cet maths... · 2018. 4. 26. · mht-cet – 5...

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