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Answer Sheet (Mock Test-1)(MATHEMATICS)
81. (b) 82. (c) 83. (a) 84. (d) 85. (b)
86. (b) 87. (b) 88. (a) 89. (b) 90. (c)
91. (b) 92. (d) 93. (c) 94. (d) 95. (d)
96. (b) 97. (a) 98. (a) 99. (a) 100. (d)
101. (a) 102. (c) 103. (c) 104. (d) 105. (c)
106. (a) 107. (b) 108. (b) 109. (d) 110. (d)
111. (a) 112. (c) 113. (a) 114. (a) 115. (a)
116. (b) 117. (c) 118. (c) 119. (b) 120. (d)
121. (c) 122. (c) 123. (b) 124. (d) 125. (c)
Hints & Solutions81. The given determinant Mock Test1 Solutions (Mathematics)
=
1 +a11
1
1 +b1
11
1 +c
=(abc)
1
a+11b1c
1
a1b+11c
1
a1b1c+1
[Taking a, b, c from R1, R2and R3respectively]
=(abc)
1a
+1b
+1c
+1
11b1c
1
1b+11c
1
1b1c+1
[Performing R1R1+R2+R3 andtaking (1a+1b+1c+1) common]
=(abc)1a
+1b+1c +1
00
1
01
1
11b
1c+1
[C1(C1C3)and C2(C2C3)]
=(abc)
1a
+1b
+1c
+1. (1)
01
11
[Expanding by 1st row]=(abc)
1a
+1b
+1c
+1. 1
=abc(0 +1)=abc...
1a
+1b
+1c=0 is given
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82.
1 +i1
i
n
=1
1 +i1 i
.1 +i1 +i
n
=1
1 +i+i2+i1 i2
n
=1
1 +2i+i2
1 i2
n
=1
1 +2i11 (1)
=1 [... i2=1]
2i2
n
=1 in=1
n=2 [... i2
=1]
83. We haveA=
32
64
|A|=(3 4)(6 2)=12 12=0
Since |A| =0, then A is singular.
84. 0
11 x1 +x
dx= 0
11 x1 +x
.1 x1 x
dx
= 0
1 (1 x)1 x2
dx
= 01 dx1 x2
x1 x2 dx
=sin1x0
1
0
1 x
1 x2dx
... sin1x= 11 x2
Putting 1 x2=t for x=1 2xdx=dt t=1 1 =0
xdx=dt2
for x=0
t=1 0 =1So, we get
=sin1x0
1+1
2
0
1 dtt
=sin11 sin10 +
12
t12
1
0
.
.
.
t
12
=t1
2 +1
12 +1
=(2 0)+[t ]1
0
=(2 0)+(0 1)=(2 1).
Mock Test1 Solutions (Mathematics)
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85. (x+1)ex
cos
2
(xex
)
dx
Putting xex=t,we get (xex+ex)dx=dt (x+1)exdx=dt
(x+1)ex
cos2(xex)dx= dt
cos2t
=sec2
tdt
=tan t ...
ddx
tan x=sec2x
=tan (xex).
86. LetA=
ah
g
hb
f
gf
c
Now A=
ahg
hbf
gfc
[Calculating
Transpose A of A by interchanging therows and columns of A]Since A=A, therefore matrix A issymmetric.
87. sin1(cos x)
=sin1sin
2
x
=2x
88. Lines are 3x+4y=0 ...(1)
Slope m1=43
and 6x+8y=15 ...(2)
... Slope m2=86=4
3since m1=m2, the given lines are parallel.Let y=0 in first equation, we get
3x=9, x=3So, let point are first is (3, 0) Required distance
=ax+byca2+b2
[Length of perpendicular
from (3, 0)on second line]
=6 (3)+8 (0)15
36 +64
=18 15
10
= 3
10
Mock Test1 Solutions (Mathematics)
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89. (f0f)(x)=f(f(x))
=f(x2
+1) [...f(x)=x
2
+1]=(x2+1)2+1
=x4+2x2+1 +1
=x4+2x2+2.
90. Derivative will exist if
1 x2>0i.e. 1 >x2
i.e. x2
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93. Let I= 1
2xf(x)dx ...(1)
Now f(3 x)=f(x)
I= 1
2(3 x)f(x)dx
I= 1
23f(x)dx
1
2xf(x)dx
I=
1
23f(x)dxI
2I=3 1
2f(x)dx
I=32
1
2f(x)dx.
94. Using Binomial
(a+b)n=a+n(b)+n(n1)b2
2 !+
(1 +4x+x2)12=1 +
12
(4x+x2)+12 (12 1)
2 !
(4x+x2)2+
=1 +2x+x2
2 1
8(16x2
+x4
+8x3
)+
Co-efficient of x2=12
2 =32
95. A=sin2+cos4
=sin2+(1 sin2)2
=sin2+1 +sin42 sin2=1 sin2+sin4
=1 sin2(1 sin2)
=1 sin2cos2=1 (sin cos )2
=1 1
4(2 sin cos )2
=1 14
(sin 2)2=114
sin22 ...(1)
Now 0 sin221
1 sin220
1
4
1
4
sin220
1 14
1 14
sin221
34
1 14
sin221
34
A1 [from (1)]
Mock Test1 Solutions (Mathematics)
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96. Let p be any arbitray element of AB.Then p(AB)
pAandpB
pAandpBc
pABc
ABABc ...(i)Again, let q be any arbitrar element of
ABc
. Then,qABc
qAand qBc
qAand qB qAB
ABc(AB) ...(ii)Hence from (i) and (ii), we have
AB=ABc
97. limn
(4n+5n)1n
= limn (5n
)1n
1 +
4
5
n
1n
= limn
51 +
45
n
1n
=limn
5 (1 +0)1n
=5 10=5
98. We have
3 +1x
=2
3 +1x
=2
So, 3 +1x=+2
1x
=2 3 =1
x=1 ...(1)
Also 3 +1x
=2
1x=2 3 =5
x=15
...(2)
i.e. x=1, 1
5
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99. We have =
x2
31
0y
59
00
z0
00
0
expanding along R1we get
=x
y59
0z0
00
=x.yz0
0
0 +0
[expanding along R1]
=x.y[z0]=xyz
100. |3z1 | =3 |z2 ||3 (x+iy)1 | =3 |(x+iy)2 |
[...z=x+iy]|(3x1)+3iy| =3 |(x2)+iy| (3x1)2+9i2y2=9 [(x2)2+i2y2]
[Squaring both sides]
9x2+1 6x9y2=9 [x2+4 4xy2]9x2+1 6x9y2=9x2+36 36x9y2
1 6x=36 36x 30x=35
x=76
i.e. a line parallel to y-axis.
101. The probability that one test is held
=2 15
45
=8
25
Probability that test is held on both days
=1
51
5= 1
25Thus, probability that the student misses
at least one test = 825
+ 125
= 925
.
102. Let Sidenote the event of getting an ace
in the ithdraw.
Probability of getting aces in both thedraws=P(S1S2)=P(S1)P(S2)
[Multiplication theorem]
= 452
452
= 113 113
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103. 113+123+133+ +203
=(13
+23
+ +203
)(13
+23
+ +103
)=
20 (20 +1)2
2
10 (10 +1)2
2
... 13+23+ +n3=
n(n+1)2
2
.
.
.
n3
=(n)
3
=
n(n+1)2
2
=[10 (21)]2[5 (11)2]=(10 21)2(5 11)2
=44100 3025=41075=an odd integer divisible by 5.
104. In the case of each book we may take 0,1, 2, 3, ... m copies. We may deal with each book in(m+1) ways and therefore with all thebooks in (m+1)nways.But this includes the case where all books
are rejected and no selection is made.So, the number of ways in whichselection can be made
=(m+1)n1
105. Let 2nbe the total number of terms
S2n
=5 [T1+T
3+ +T
2n1]
where T1, T3 T2n1 are terms
occupying odd places
a[1 r2n]
1 r=5 [a+ar2+ +ar2n2]
[Sum of G.P., where a is first term andr is common ratio]
a[1 r2n]
1 r=5a[1 +r2+ +r2n2]
a[1 r2n]
1 r=
5a[1 r2n]1 r2
1 = 51 +r
1 +r=5
r=4
106. Let x be fraction
Let f(x)=xx2
f(x)=1 2xFor maxima and minima, put f(x)=0
1 2x=0 x=1
2f(x)=2
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107. The equation of circle is
x
2
+y2
+2gx+2fy+c=0 ...(1)comparing the equationx2+y2+2x+8y25 =0 with (1), we getg=1,f=4 Centre of first circle C1(1, 4)
also radius =g2+f2c=1 +16 +25 =42
comparing the equationx2+y24x10y+19 =0 with (1), we getg=2,f=5 Centre of second circle C2(2, 5)also radius =4 +25 19 =10now C1C2=(2 +1)2+(5 +4)2
=32+92
=9 +81 =310 ...(2)Also r1+r2=42 +10 r1+r2>40 +10
r1+r2>210 +10r1+r2>310r1+r2>C1C2
C1C2
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109. Mixed doubles includes 2 men and 2women.
Since 2 men are selected of 6 men
number of ways =6C2
Also 2 women are selected of 4 women
number of ways =4C2
But they also can be interchanged in 2
ways.
Total no. of ways =6C24C22
=6 51 2
4 31 2
2
=180
110. Since (a+b)is perpendicular to b
(a
+b). b
=0
a
. b
+b
. b
=0
a
. b
=b
. b
...(1)
Also (2b
+a)is perpendicular to a
a
. (2b
+a)=0
2a
b
+a
. a
=0
2 (b
. b
)+a
. a
=0 [From (1)] a22b2=0
a2=2b2
a=2 b.
111. The number of ways of selecting 6
boys =6 !The number of ways of selecting 6 girls=6 !Since girls and boys sit alternatively. Required number of ways
=2 6 ! 6 !=2 720 720
=1036800
Mock Test1 Solutions (Mathematics)
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112. The position vectors of two given pointsare
a= i j+3kand b=3i+3j+3k^
and the equation of the given plane is
r
. (5i+2j7k^ )+9 =0
or r
. n
+d=0
We have
a
. n +d
=(i j+3k). (5i+2j7k)+9
=5 2 21 +9 0
So, the points a
and b
are on the oppositesides of the plane.
113.dydx
=x+1
x2
Integrating both sides, we get
y=x2
2
1x
+c ...(1)
This equation passes through (3, 9)
9=32
21
3+c
9 =92
13
+c
c=9 92
+13
c=296 Putting value of c in (1), we get
y=x2
2
1x
+296
y=3x36 +29x
6x
6xy=3x36 +29x.
Mock Test1 Solutions (Mathematics)
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114. x=at2,y=2at
dxxt =2at,
dydt =2a
dydx
=dydt
dtdx
=2a 12at
=1t
Slope of tangent =1t
Also slope of normal
= 1slope of tangent
=t equation of normal is
yy1=slope of normal (xx1) y2at=t(xat2) tx+y=2at+at3.
1
15. The equation of the circle touching theco-ordinate axes is
x
2
+y2
2cx2cy+c2
=0i.e. (xc)2+(yc)2=c2 ...(1)This touches the line
x3
+y4
=1 ...(2)
i.e.4x+3y
12=1
i.e. 4x+3y12 =0 ...(3)Perpendicular distance between circle(c, c)and line is
ax+bycx12+x22
=c
i.e.
4c+3c1242+32
=c
|7c12 |5
=c
7c12 =5cor 5c
7c12 =5cor 5c
c=6 or c=1
Hence c=(1, 6).
Mock Test1 Solutions (Mathematics)
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116. Let r=(xi+yj
+zk
)
r(i+2j+k)=ik
(xi+yj
+zk
)(i
+2j
+k)=i
k
2xk
xjyk
+yi
+2j
2zi
=ik
[Doing cross product]
(y2z)i+(zx)j+(2xy)k=ikComparing the co-efficients of i, j, k on
both sides
y2z=1 ...(i)zx=0 ...(ii)
2xy=1 ...(iii)
x=z=t, where t is a scalar y2z=1 y=1 +2t
r
=xi+yj
+zk
=ti+(1 +2t)j
+tk
=j+t(i+2j+k).
117. g(x)=xf(x), wheref(x)=xsin 1x
at x=0g(0)=0f(0)
=0 (0)=0
g(0)=Limx0
g(x)g(0)x
=Limx0
xf(x)0
x
=Limx0
f(x)=0
also g(x)=xf(x)+f(x)Limx0
g(x)=Limx0
[xf(x)+f(x)]
=0 +f(0)=0 +0 =g(0) Lim
x0g(x)=g(0)
Hence g is continuous, and g(x) isdifferentiable.
118. 1 radian =180
degree =1803.14
=57 1744.8
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123. f(x)=sin xcos xkx+b
f(x)=cos x+sin xk
=2
12
cos x+ 12
sin xk
[multiplying and dividing by 2]
=2
sin
4
cos x+cos
4
sin x
k
... sin
4
=cos 4
= 12
=2 sin
4
+x
k
If f(x) decreases for all x f(x) is negative
i.e. kmax of 2 sin
4
+x
i.e. k2 .
124.
1cos ()cos ()
cos ()1cos ()
cos ()cos ()
1
=
cos2+sin2cos cos +sin sin cos cos +sin sin
cos cos +sin sin
cos2+sin2cos cos +sin sin
cos cos +sin sin cos sin +sin sin
cos2+sin2
=
cos cos cos
sin sin sin
000
cos cos cos
sin sin sin
000
=0 0 =0.
Mock Test1 Solutions (Mathematics)
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125. Let E1, E2, E3be the respective events of
solving the problem and E1
___, E2
___, E3
___be the
respective events of not solving theproblem. Then
P(E1)=12
, P(E2)=13
, P(E3)=14
P(E1___
)=1 12
=12
P(E2___
)=1 13=23
P(E3___
)=1 14
=34
P (none solves the problem)
=P[(not E1)and (not E2)and (not E3)]
=P(E1___
E2___
E3___
)
=P(E1___). P(E2___). P(E3___)
[... E1
___, E2
___, E3
___are independent]
=12
23
34
=14
Hence P (the problem will be solved)
=1 P (none solves the problem)=1
14
=34
Mock Test1 Solutions (Mathematics)