Bitsat 01 Math Sol

8/10/2019 Bitsat 01 Math Sol

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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


2/16

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)


3/16

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



4/16

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


5/16

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)]



6/16

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



7/16

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



8/16

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


9/16

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


10/16

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



11/16

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.



12/16

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).



13/16

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



14/16


15/16

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.



16/16

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


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