maits 2015 unit test-1 (main) sol. 11-10-14 … · maits 2015_unit test-1 (main)_sol._11-10-14...

MAITS 2015_Unit Test-1 (Main)_Sol._11-10-14 (MATHEMATICS) [ 3 ]

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MATHEMATICS

SOLUTION31. (C)

Circumcentre of PQR is (1, 2). P

Q

R

(1, 2)

x+3y–7=0

3x–2y+1=0Straight line through it of slope ‘m’ is y – 2 = m(x – 1)

intersect axes at 2A 1 ,0 and B(0,2 m)m

. Now area of OAB is

= 12

21m

(2 – m) = 1 44 m 4.2 m

32. (B)m = 3 and n = 4 x2 – 4x + 3 < 0 (x – 1)(x – 3) < 0 x (1, 3)

x = 2 is only integer solution.33. (A)

If n(A) = n then number of relations those are symmetric as well as reflexive on it are =2n n22

Here2n n22

= 1024 n = 5

Number of reflexive relations on A = 220

Number of symmetric relations on A = 215

Number of relations on A those are either symmetric or reflexive = 220 + 215 – 210.Number of relations on A those are neither reflexive nor symmetric = 225 – 220 – 215 + 210.

34. (C)

tan(3x – 2) is a periodic function with period 3

. The function f(x) = {x} is periodic with period

1. The function in (D) can be written as3 3 3 3cos x sin x sin x cos xf(x) 1 1

sinx cos x sinx cos x sinx cos x

2 2(sin x cos x)(sin x cos x sinx cos x)1sin x cos x

1 11 1 sin2x sin2x2 2

which is periodic with period . The function x + cosx is non-periodic as x is non-periodic.


[ 4 ] (MATHEMATICS) MAITS 2015_Unit Test-1 (Main)_Sol._11-10-14

35. (C)We have,

f(a) = 33164a

a = 3

31(4a)

a =

31 1 14a 3.4a. 4aa a a

= (3)3 – 12·3 = 27 – 36 = –9. [Since, a, b are roots of 4x + 1x

= 3.

14a 3a

]

Similarly, f(b) = –9. f(a) = f(b) = –936. (C)

Let perpendicular bisector of AB is 3x + 4y – 20 = 0and perpendicular bisector of AC is 8x + 6y – 65 = 0.Image of A w.r.t. 3x + 4y – 20 = 0 is Band image of A w.r.t. 8x + 6y – 65 = 0 is C.

For B,x 10 y 10 30 40 202

3 4 25

B = (–2, –6)

For C,x 10 y 10 80 60 652

8 6 100

C = (–2, 1) B (–2, –6)

C(–2, 1)

A (10, 10)

Area of ABC = 12

(10 + 2)(1 + 6) = 42.

37. (C)

3 3 1tan , tan4 2(3) 2

y

O

3A

D3/2

x

B4

C

Slope of

OB tan ( )2

cot( )

3 11 14 23 1 24 2



38. (D)By observation, it is clear that 3 vertices of triangle lie on circle x2 + y2 = 25.

(3,4)

(5 cos , 5 sin )

(5 sin , – 5 cos )

centroid

5 sin 5cos 3 5 sin 5cos 4G ,3 3

Circumcentre O (0, 0) Orthocentre, H (h, k)

5sin 5cos 33 2(o) h3

or h 5 sin 5cos 3 ...(1)

5sin 5cos 43 2(o) k3 or k 5 sin 5cos 4 ...(2)

By eq. (1) + eq.(2) h + k = 10 sin + 7By eq. (1) – eq.(2) h – k +1 = 10 cos

(x + y – 7)2 + (x – y + 1)2 = 100 is the required locus.39. (B)

Image of A (2, – 1) w.r.t.3x – 2y + 5 = 0, A is given by

x 2 y 1 6 2 52. 23 2 13

3x – 2y + 5 = 0

B(–3, – 2)

(2, –1)A C

A

A ( 4,3)

Equation of BC is

y – 3 =

3 2 (x 4)4 3

5x + y + 17 = 040. (A)

For point of interaction,

3x + 4mx + 4 = 9 x = 5

3 4m 3 4m 1, 5

1 1m , 1, , 22 2number of integral values of m is 2



41. (D)

a 2 bc b c

2 2

b c a 0 b c a 0

a x b y c 0 passes through fixed point (– 1, 1)

42. (B)

2 2

1

a 2a tan tanP

sec

2(a tan )sec

2

3(b tan )P

sec

2

2

ab (a b) tan tanP

sec (a tan )(b tan )

sec

22 1 3P PP

43. (A)

3( 1) 4(4) 12 12( 1) 5(4) 7 0

equation of bisector containing (– 1, 4) in its region is

3x 4y 12 12x 5y 7

5 13

21x + 27y – 121 = 0

44. (A)Equating slopes of P ’A and P ’Q are equal

1 3 12 7

(0, )

(–2, 1)

A

P P

Q (5, 3)

(2, 1)

11 11A 0,7 7



45. (B)

Image of A say A w.r.t x – 2y + 1 = 0 lies on BC

A(1, 2)

B(2, 1)

x – 2y + 1 = 0

C

Here,

2x 1 y 2 (1 4 1) 42

1 2 51 2

9 2A ,5 5

Equation of BC joining 9 2A ,5 5

and B (2, 1) is

21 35y 1 (x 2) x 29 125

3x – y – 5 = 0 a b 3 1 2

46. (B)

y f(x) 3 sinx cos x 2 2sin x 26 ...(1)

since f(x) is one-one and onto. f is invertible.

form (1),

y 2sin x6 2

or, 1 y 2x sin

2 6

or,

1 1 x 2f (x) sin2 6

47. (B)

1sin (3 x)f(x)log(1x) 2)

Let g(x) = sin–1 (3 – x) 1 3 x 1

The domain of g(x) is D1 is [2, 4]Let h(x) = log (|x| – 2) i.e., |x| – 2 > 0 or |x| > 2



i.e., x < – 2 or x > 2 Domain D2 is ( , 2) (2, )

We know that domain of 1 2f(x) is defined x D D – x : g(x) 0g(x)

Therefore, the domain of f(x) is (2,4] {3} (2,3) (3, 4]

48. (C)

Clearly, form the graph

2

3

1 1,0 x64 8

1f(x) x , x 18

x ,x 1

y y=x3

y=x2

y=1/64x

0 11/8 1/4

49. (C)f(7) + f(– 7) = – 10

or, f(7) = – 17or, f(7) + 17 cos x = – 17 + 17 cos xwhich has the range [–34, 0]

50. (D)

Since Co-domain =

0,2 for f to be onto range is

0,2

This is possible only when 2x x a 0, x R.

Thus, 21 4a 0 1for a ,4

f(x) is well defined but for f(x) to be onto a = 14

.

51. (B)

2 4x F(x) F(1 x) 2x x

Replacing x by 1 – x, we get(1 – x)2 F(1 – x) + F(x) = 2(1 – x) – (1 – x)4

Eliminating F(1 – x) form (1) and (2), we get F(x) = 1 – x2.52. (B)

2 2 21 1 1x x 1 1 x

2 2 2

Thus from domain point of view,

If 2 21 1x 0, 1 then x 1, 0

2 2



i.e, 1 1 1 1sin (1) cos (0) and sin (0) cos ( 1)

Hence range is { }

53. (D)

f(x) 1 f(1 x) 1 0

So, g(x) g(1 x) 0

Replacing 1x by x ,2

we get

1 1g x g x 02 2

So, it is symmetric about point

1,02

54. (B)

42 y

216 yf

y

4y

216f 1y

42 y4f 1

y

222 y4f 1

y

= 52

55. (A)

{y} [y] |x|h(x) = log f(x).g(x) log e {y} [y] e sgn x (where |x|y e sgn(x) )

x

|x|

x

e , x 0e sqn x 0 , x 0

e , x 0

x

x

e , x 0h( x) 0 , x 0

e , x 0

h(x) h( x) 0 for all x

56. (D)Slope of bisector = k – 1

middle point k +1 7,2 2

=

Equation of bisector is

7 (k 1)y k 1 x2 2



Put x = 0 and y = –4

k 4

57. (D)

R is reflexive xRx as x = x, ( = 1)

R is symmetric 1xRy yRx as x y then y x

(If is rational then 1

is also

rational)

R is transitive xRy & yRz xRz

as 1 2x y and y z then 1 2x ( )z [ 1 2& rational 1 2 rational]

R is equivalence relation.

S is reflexive as m mRn n

[n 0 & mn mn]

S is symmetric as m p p mR R [q, n 0 & mq pn]n q q n

S is transitive as m p p r m rR & R Rn q q s n s

mq pnn,q, s 0, ms nr

sp qr

S is equivalence relation.58. (C)59. (D)

V x R, x – x = 0 and 0 I T is reflexive relation.

If x y I y x I

T is symmetric relation

If 1x y I and 2y z I

Then 1 2x z (x y) (y z) I I I

T is also transitive.Hence T is an equivalence relation.

Clearly x x 1 (x,x) S

S is not reflexive.



60. (C)

Given 3f(x) x 5x 1

Now 2f '(x) 3x 5 0, x R

f(x) is strictly increasing function

f is one-oneclearly, f(x) is a continuous function and also increasing on R,f(x) is odd degree polynomial therefore range of f(x) is R f(x) is onto.

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