Page 1
CBSE XII | Mathematics
Sample Paper – 6 Solution
Mathematics
Class XII
Sample Paper – 6 Solution
SECTION – A
1. A matrix is of order m × n, then it has mn elements.
So, mn = 12
Which means we have to find values of m and n, such that it will satisfy the
above condition
So all the possible cases are
(1, 12), (2, 6), (3, 4), (4, 3), (6, 2), (12, 1).
2.
Let f(x) = sin(2x
2
)
? ? ? ? ? ? ? ?
? ? ? ?
? ?
22
2
2
d d d
f x sin 2x 2x
dx dx dx
cos 2x 4x
4xcos 2x
?
?
?
3.
Sol:
Order: 3
Degree: 1
4.
The Cartesian form of a line is given by
1 1 1
x x y y z z
a b c
? ? ?
??
Substituting point (2, -1, 4) and d. r. s. 1, 1, -2
We get the equation as
x 2 y 1 z 4
1 1 2
? ? ?
??
?
Page 2
CBSE XII | Mathematics
Sample Paper – 6 Solution
Mathematics
Class XII
Sample Paper – 6 Solution
SECTION – A
1. A matrix is of order m × n, then it has mn elements.
So, mn = 12
Which means we have to find values of m and n, such that it will satisfy the
above condition
So all the possible cases are
(1, 12), (2, 6), (3, 4), (4, 3), (6, 2), (12, 1).
2.
Let f(x) = sin(2x
2
)
? ? ? ? ? ? ? ?
? ? ? ?
? ?
22
2
2
d d d
f x sin 2x 2x
dx dx dx
cos 2x 4x
4xcos 2x
?
?
?
3.
Sol:
Order: 3
Degree: 1
4.
The Cartesian form of a line is given by
1 1 1
x x y y z z
a b c
? ? ?
??
Substituting point (2, -1, 4) and d. r. s. 1, 1, -2
We get the equation as
x 2 y 1 z 4
1 1 2
? ? ?
??
?
CBSE XII | Mathematics
Sample Paper – 6 Solution
OR
The equation can be written as
1
y
x 2 z 5
2
3 1 1
?
??
??
?
D.R.S. of this line is proportional to 3, 1,-1.
Also the line passes through (1,-1, 0)
x 1 y 1 z 0
3 1 1
? ? ?
??
?
SECTION – B
5.
(i) For all
ab
a,b N,a * b
2
Now,
b a a b
b * a a * b
22
Thus, the binary operation * is commutative.
(ii) Let a,b,c N
bc
a
b c 2a b c
2
a * b * c a *
2 2 4
ab
c
a b a b 2c
2
a * b * c * c
2 2 4
a * b * c a * b * c
Thus, the binary operation * is not associative.
Page 3
CBSE XII | Mathematics
Sample Paper – 6 Solution
Mathematics
Class XII
Sample Paper – 6 Solution
SECTION – A
1. A matrix is of order m × n, then it has mn elements.
So, mn = 12
Which means we have to find values of m and n, such that it will satisfy the
above condition
So all the possible cases are
(1, 12), (2, 6), (3, 4), (4, 3), (6, 2), (12, 1).
2.
Let f(x) = sin(2x
2
)
? ? ? ? ? ? ? ?
? ? ? ?
? ?
22
2
2
d d d
f x sin 2x 2x
dx dx dx
cos 2x 4x
4xcos 2x
?
?
?
3.
Sol:
Order: 3
Degree: 1
4.
The Cartesian form of a line is given by
1 1 1
x x y y z z
a b c
? ? ?
??
Substituting point (2, -1, 4) and d. r. s. 1, 1, -2
We get the equation as
x 2 y 1 z 4
1 1 2
? ? ?
??
?
CBSE XII | Mathematics
Sample Paper – 6 Solution
OR
The equation can be written as
1
y
x 2 z 5
2
3 1 1
?
??
??
?
D.R.S. of this line is proportional to 3, 1,-1.
Also the line passes through (1,-1, 0)
x 1 y 1 z 0
3 1 1
? ? ?
??
?
SECTION – B
5.
(i) For all
ab
a,b N,a * b
2
Now,
b a a b
b * a a * b
22
Thus, the binary operation * is commutative.
(ii) Let a,b,c N
bc
a
b c 2a b c
2
a * b * c a *
2 2 4
ab
c
a b a b 2c
2
a * b * c * c
2 2 4
a * b * c a * b * c
Thus, the binary operation * is not associative.
CBSE XII | Mathematics
Sample Paper – 6 Solution
6.
The corresponding values of two equal matrices are equal
a + b = 6 and ab = 8
Therefore,
8
b
a
?
Substituting in first condition we get,
? ? ? ?
2
8
a6
a
a 6a 8 0
a 4 a 2 0
a 4 or a 2 and
b 2 or b 4 respectively.
??
? ? ?
? ? ?
??
??
7.
x
sin4x 4
Let I e dx
1 cos4x
?
? ??
?
??
?
??
x
x2
2
sin2(2x) 4
e dx
1 cos2(2x)
2sin2xcos2x 4
e dx [Using,sin2x 2sin x.cosxand2sin x 1 cos(2x)
2sin (2x)
?
?
?? ?
?
??
?
??
??
?
? ? ? ?
??
??
??
? ?
xx
2 2 2
x2
2(sin(2x)cos(2x) 4 sin(2x)cos(2x) 2
e dx e dx
2sin 2x sin 2x sin 2x
e cot(2x) 2cosec 2x dx
??
?
? ? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
??
Now, let f(x) = cot(2x) then f’(x) = -2cosec
2
2x
I ? ?
x
e f(x) f '(x) dx
?
??
So, I = e
x
f(x) + C = e
x
cot 2x + C , where C is a constant
Therefore,
? ?
xx
sin4x 4
e dx e cot 2x + C
1 cos4x
?
? ??
?
??
?
??
Page 4
CBSE XII | Mathematics
Sample Paper – 6 Solution
Mathematics
Class XII
Sample Paper – 6 Solution
SECTION – A
1. A matrix is of order m × n, then it has mn elements.
So, mn = 12
Which means we have to find values of m and n, such that it will satisfy the
above condition
So all the possible cases are
(1, 12), (2, 6), (3, 4), (4, 3), (6, 2), (12, 1).
2.
Let f(x) = sin(2x
2
)
? ? ? ? ? ? ? ?
? ? ? ?
? ?
22
2
2
d d d
f x sin 2x 2x
dx dx dx
cos 2x 4x
4xcos 2x
?
?
?
3.
Sol:
Order: 3
Degree: 1
4.
The Cartesian form of a line is given by
1 1 1
x x y y z z
a b c
? ? ?
??
Substituting point (2, -1, 4) and d. r. s. 1, 1, -2
We get the equation as
x 2 y 1 z 4
1 1 2
? ? ?
??
?
CBSE XII | Mathematics
Sample Paper – 6 Solution
OR
The equation can be written as
1
y
x 2 z 5
2
3 1 1
?
??
??
?
D.R.S. of this line is proportional to 3, 1,-1.
Also the line passes through (1,-1, 0)
x 1 y 1 z 0
3 1 1
? ? ?
??
?
SECTION – B
5.
(i) For all
ab
a,b N,a * b
2
Now,
b a a b
b * a a * b
22
Thus, the binary operation * is commutative.
(ii) Let a,b,c N
bc
a
b c 2a b c
2
a * b * c a *
2 2 4
ab
c
a b a b 2c
2
a * b * c * c
2 2 4
a * b * c a * b * c
Thus, the binary operation * is not associative.
CBSE XII | Mathematics
Sample Paper – 6 Solution
6.
The corresponding values of two equal matrices are equal
a + b = 6 and ab = 8
Therefore,
8
b
a
?
Substituting in first condition we get,
? ? ? ?
2
8
a6
a
a 6a 8 0
a 4 a 2 0
a 4 or a 2 and
b 2 or b 4 respectively.
??
? ? ?
? ? ?
??
??
7.
x
sin4x 4
Let I e dx
1 cos4x
?
? ??
?
??
?
??
x
x2
2
sin2(2x) 4
e dx
1 cos2(2x)
2sin2xcos2x 4
e dx [Using,sin2x 2sin x.cosxand2sin x 1 cos(2x)
2sin (2x)
?
?
?? ?
?
??
?
??
??
?
? ? ? ?
??
??
??
? ?
xx
2 2 2
x2
2(sin(2x)cos(2x) 4 sin(2x)cos(2x) 2
e dx e dx
2sin 2x sin 2x sin 2x
e cot(2x) 2cosec 2x dx
??
?
? ? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
??
Now, let f(x) = cot(2x) then f’(x) = -2cosec
2
2x
I ? ?
x
e f(x) f '(x) dx
?
??
So, I = e
x
f(x) + C = e
x
cot 2x + C , where C is a constant
Therefore,
? ?
xx
sin4x 4
e dx e cot 2x + C
1 cos4x
?
? ??
?
??
?
??
CBSE XII | Mathematics
Sample Paper – 6 Solution
OR
? ?
2
1x
dx
x 1 2x
?
?
?
Here
? ?
2
1x
x 1 2x
?
?
is an improper rational fraction.
Reducing it to proper rational fraction gives
? ? ? ?
2
1 x 1 1 2 x
x 1 2x 2 2 x 1 2x
??
??
??
??
??
??
??
…………(1)
Now, let
? ?
2 x A B
x 1 2x x (1 2x)
?
??
??
? ? ? ?
? ?
2 x A(1 2x) Bx
2 x A x(2A B)
x 1 2x x 1 2x
Equating the coefficientsweget,A 2 and B 3
2 x 2 3
So,
x 1 2x x (1 2x)
? ? ?
? ? ? ? ? ? ?
??
??
?
??
??
Substituting in equation (1), we get
? ?
2
1 x 1 1 2 3
x 1 2x 2 2 x (1 2x)
?? ?
? ? ?
??
??
??
? ?
2
1 x 1 1 2 3
i.e dx dx
x 1 2x 2 2 x (1 2x)
??
?? ?? ?
? ? ?
?? ??
??
?? ??
dx dx 3 dx x 3 1
log x log 1 2x C
2 x 2 (1 2x) 2 2 ( 2)
x3
log x log 1 2x C
24
? ? ?
? ? ? ? ? ? ? ? ?
??
? ? ? ? ?
Page 5
CBSE XII | Mathematics
Sample Paper – 6 Solution
Mathematics
Class XII
Sample Paper – 6 Solution
SECTION – A
1. A matrix is of order m × n, then it has mn elements.
So, mn = 12
Which means we have to find values of m and n, such that it will satisfy the
above condition
So all the possible cases are
(1, 12), (2, 6), (3, 4), (4, 3), (6, 2), (12, 1).
2.
Let f(x) = sin(2x
2
)
? ? ? ? ? ? ? ?
? ? ? ?
? ?
22
2
2
d d d
f x sin 2x 2x
dx dx dx
cos 2x 4x
4xcos 2x
?
?
?
3.
Sol:
Order: 3
Degree: 1
4.
The Cartesian form of a line is given by
1 1 1
x x y y z z
a b c
? ? ?
??
Substituting point (2, -1, 4) and d. r. s. 1, 1, -2
We get the equation as
x 2 y 1 z 4
1 1 2
? ? ?
??
?
CBSE XII | Mathematics
Sample Paper – 6 Solution
OR
The equation can be written as
1
y
x 2 z 5
2
3 1 1
?
??
??
?
D.R.S. of this line is proportional to 3, 1,-1.
Also the line passes through (1,-1, 0)
x 1 y 1 z 0
3 1 1
? ? ?
??
?
SECTION – B
5.
(i) For all
ab
a,b N,a * b
2
Now,
b a a b
b * a a * b
22
Thus, the binary operation * is commutative.
(ii) Let a,b,c N
bc
a
b c 2a b c
2
a * b * c a *
2 2 4
ab
c
a b a b 2c
2
a * b * c * c
2 2 4
a * b * c a * b * c
Thus, the binary operation * is not associative.
CBSE XII | Mathematics
Sample Paper – 6 Solution
6.
The corresponding values of two equal matrices are equal
a + b = 6 and ab = 8
Therefore,
8
b
a
?
Substituting in first condition we get,
? ? ? ?
2
8
a6
a
a 6a 8 0
a 4 a 2 0
a 4 or a 2 and
b 2 or b 4 respectively.
??
? ? ?
? ? ?
??
??
7.
x
sin4x 4
Let I e dx
1 cos4x
?
? ??
?
??
?
??
x
x2
2
sin2(2x) 4
e dx
1 cos2(2x)
2sin2xcos2x 4
e dx [Using,sin2x 2sin x.cosxand2sin x 1 cos(2x)
2sin (2x)
?
?
?? ?
?
??
?
??
??
?
? ? ? ?
??
??
??
? ?
xx
2 2 2
x2
2(sin(2x)cos(2x) 4 sin(2x)cos(2x) 2
e dx e dx
2sin 2x sin 2x sin 2x
e cot(2x) 2cosec 2x dx
??
?
? ? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
??
Now, let f(x) = cot(2x) then f’(x) = -2cosec
2
2x
I ? ?
x
e f(x) f '(x) dx
?
??
So, I = e
x
f(x) + C = e
x
cot 2x + C , where C is a constant
Therefore,
? ?
xx
sin4x 4
e dx e cot 2x + C
1 cos4x
?
? ??
?
??
?
??
CBSE XII | Mathematics
Sample Paper – 6 Solution
OR
? ?
2
1x
dx
x 1 2x
?
?
?
Here
? ?
2
1x
x 1 2x
?
?
is an improper rational fraction.
Reducing it to proper rational fraction gives
? ? ? ?
2
1 x 1 1 2 x
x 1 2x 2 2 x 1 2x
??
??
??
??
??
??
??
…………(1)
Now, let
? ?
2 x A B
x 1 2x x (1 2x)
?
??
??
? ? ? ?
? ?
2 x A(1 2x) Bx
2 x A x(2A B)
x 1 2x x 1 2x
Equating the coefficientsweget,A 2 and B 3
2 x 2 3
So,
x 1 2x x (1 2x)
? ? ?
? ? ? ? ? ? ?
??
??
?
??
??
Substituting in equation (1), we get
? ?
2
1 x 1 1 2 3
x 1 2x 2 2 x (1 2x)
?? ?
? ? ?
??
??
??
? ?
2
1 x 1 1 2 3
i.e dx dx
x 1 2x 2 2 x (1 2x)
??
?? ?? ?
? ? ?
?? ??
??
?? ??
dx dx 3 dx x 3 1
log x log 1 2x C
2 x 2 (1 2x) 2 2 ( 2)
x3
log x log 1 2x C
24
? ? ?
? ? ? ? ? ? ? ? ?
??
? ? ? ? ?
CBSE XII | Mathematics
Sample Paper – 6 Solution
8.
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
22
2
2x
I dx
x 1 x 3
Let x z
2xdx dz
dz
I
z 1 z 3
By partialfraction,
1 A B
z 1 z 3 z 1 z 3
1 A z 3 B z 1
?
?
?
??
?
??
??
??
??
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
22
22
2
2
Putting z 3, we obtain :
1 2B
1
B
2
1
A
2
1
1
1 2
2
z 1 z 3 z 1 z 3
dz 1 dz dz
z 1 z 3 2 z 1 z 3
11
log z 1 log z 3 C
22
2xdx 1 1
log x 1 log x 3 C
22
x 1 x 3
1 x 1
log C
2
x3
? ? ?
?
??
??
??
??
??
?
??
??
? ? ?
? ? ? ?
? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
??
?
??
?
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