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MAITS 2015_Unit Test-1 (Main)_Sol._11-10-14 (MATHEMATICS) [ 3 ]
Mentors Eduserv: Plot No.-136/137, Parus Lok Complex, Boring Road Crossing,Patna-1, Ph. No. : 0612-3223680 / 81, 7781005550 / 51
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.
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[ 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
MAITS 2015_Unit Test-1 (Main)_Sol._11-10-14 (MATHEMATICS) [ 5 ]
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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
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[ 6 ] (MATHEMATICS) MAITS 2015_Unit Test-1 (Main)_Sol._11-10-14
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
MAITS 2015_Unit Test-1 (Main)_Sol._11-10-14 (MATHEMATICS) [ 7 ]
Mentors Eduserv: Plot No.-136/137, Parus Lok Complex, Boring Road Crossing,Patna-1, Ph. No. : 0612-3223680 / 81, 7781005550 / 51
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
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[ 8 ] (MATHEMATICS) MAITS 2015_Unit Test-1 (Main)_Sol._11-10-14
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
MAITS 2015_Unit Test-1 (Main)_Sol._11-10-14 (MATHEMATICS) [ 9 ]
Mentors Eduserv: Plot No.-136/137, Parus Lok Complex, Boring Road Crossing,Patna-1, Ph. No. : 0612-3223680 / 81, 7781005550 / 51
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
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[ 10 ] (MATHEMATICS) MAITS 2015_Unit Test-1 (Main)_Sol._11-10-14
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.
MAITS 2015_Unit Test-1 (Main)_Sol._11-10-14 (MATHEMATICS) [ 11 ]
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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.