Page 1
CBSE XII | Mathematics
Sample Paper 4 Solution
CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 4
SECTION – A
1. Given that
OR
|3 A | = k |A|
|3 A | = 3
3
|A|
|3 A | = 27|A|
k = 27
2. We have,
3x 7
10
24
12x 14 10
12x 24
x2
?
? ? ?
??
??
2
2
x2
2
f(x) kx x 2
3 x 2
lim kx 3
k(2) 3
4k 3
3
k
4
?
??
??
?
?
?
?
Page 2
CBSE XII | Mathematics
Sample Paper 4 Solution
CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 4
SECTION – A
1. Given that
OR
|3 A | = k |A|
|3 A | = 3
3
|A|
|3 A | = 27|A|
k = 27
2. We have,
3x 7
10
24
12x 14 10
12x 24
x2
?
? ? ?
??
??
2
2
x2
2
f(x) kx x 2
3 x 2
lim kx 3
k(2) 3
4k 3
3
k
4
?
??
??
?
?
?
?
CBSE XII | Mathematics
Sample Paper 4 Solution
3. We have,
?
?
?
?
?
??
?
?
?
?
??
? ? ?
2
3
3
2
2
3
x
I dx
1x
Put 1 x t
3x dx dt
dt
x dx
3
dt
3
I
t
1 dt
I
3t
1
I log t c
3
1
I log 1 x c
3
4. 2x + y – z = 5
Dividing both sides by 5,
2x y z
1
5 5 5
? ? ?
? ? ?
?
x y z
1
5
55
2
It is known that the equation of a plane in intercept form is
x y z
a b c
?? = 1, where a,
b and c are the intercepts cut off by the plane at x, y, and z-axes respectively.
Thus, the intercept cut off by the given plane on the x-axis is
5
2
.
Page 3
CBSE XII | Mathematics
Sample Paper 4 Solution
CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 4
SECTION – A
1. Given that
OR
|3 A | = k |A|
|3 A | = 3
3
|A|
|3 A | = 27|A|
k = 27
2. We have,
3x 7
10
24
12x 14 10
12x 24
x2
?
? ? ?
??
??
2
2
x2
2
f(x) kx x 2
3 x 2
lim kx 3
k(2) 3
4k 3
3
k
4
?
??
??
?
?
?
?
CBSE XII | Mathematics
Sample Paper 4 Solution
3. We have,
?
?
?
?
?
??
?
?
?
?
??
? ? ?
2
3
3
2
2
3
x
I dx
1x
Put 1 x t
3x dx dt
dt
x dx
3
dt
3
I
t
1 dt
I
3t
1
I log t c
3
1
I log 1 x c
3
4. 2x + y – z = 5
Dividing both sides by 5,
2x y z
1
5 5 5
? ? ?
? ? ?
?
x y z
1
5
55
2
It is known that the equation of a plane in intercept form is
x y z
a b c
?? = 1, where a,
b and c are the intercepts cut off by the plane at x, y, and z-axes respectively.
Thus, the intercept cut off by the given plane on the x-axis is
5
2
.
CBSE XII | Mathematics
Sample Paper 4 Solution
Section-B
5. Let A = IA
1 1 2 3
2 2 1 3 3 1
23
2 3 3 1 0 0
2 2 3 0 1 0 A
3 2 2 0 0 1
R R R R
1 1 4 1 1 1
2 2 3 0 1 0 A
3 2 2 0 0 1
Applying R R 2R ,R R 3R
1 1 4 1 1 1
0 0 5 2 1 2 A
0 5 10 3 3 4
RR
1 1 4
0 5 10
0 0 5
? ? ? ? ?
? ? ? ?
?
? ? ? ?
? ? ? ? ?
? ? ? ?
???
? ? ? ? ?
? ? ? ?
?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ?
?
??
??
? ? ?
??
?? ?
??
1 1 1
3 3 4 A
2 1 2
? ??
??
??
??
?? ??
??
2 2 3 3
1 1 3 2 2 3
11
R R , R R
55
1 1 1
1 1 4
3 3 4
0 1 2 A
5 5 5
0 0 1
2 1 2
5 5 5
R R 4R ,R R 2R ,
3 1 3
5 5 5
1 1 0
11
0 1 0 0 A
55
0 0 1
2 1 2
5 5 5
??
??
??
?? ?
??
??
?
??
??
?
??
??
??
??
??
?
??
??
? ? ? ?
??
?
??
????
?
????
?
????
????
??
?
??
??
??
Page 4
CBSE XII | Mathematics
Sample Paper 4 Solution
CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 4
SECTION – A
1. Given that
OR
|3 A | = k |A|
|3 A | = 3
3
|A|
|3 A | = 27|A|
k = 27
2. We have,
3x 7
10
24
12x 14 10
12x 24
x2
?
? ? ?
??
??
2
2
x2
2
f(x) kx x 2
3 x 2
lim kx 3
k(2) 3
4k 3
3
k
4
?
??
??
?
?
?
?
CBSE XII | Mathematics
Sample Paper 4 Solution
3. We have,
?
?
?
?
?
??
?
?
?
?
??
? ? ?
2
3
3
2
2
3
x
I dx
1x
Put 1 x t
3x dx dt
dt
x dx
3
dt
3
I
t
1 dt
I
3t
1
I log t c
3
1
I log 1 x c
3
4. 2x + y – z = 5
Dividing both sides by 5,
2x y z
1
5 5 5
? ? ?
? ? ?
?
x y z
1
5
55
2
It is known that the equation of a plane in intercept form is
x y z
a b c
?? = 1, where a,
b and c are the intercepts cut off by the plane at x, y, and z-axes respectively.
Thus, the intercept cut off by the given plane on the x-axis is
5
2
.
CBSE XII | Mathematics
Sample Paper 4 Solution
Section-B
5. Let A = IA
1 1 2 3
2 2 1 3 3 1
23
2 3 3 1 0 0
2 2 3 0 1 0 A
3 2 2 0 0 1
R R R R
1 1 4 1 1 1
2 2 3 0 1 0 A
3 2 2 0 0 1
Applying R R 2R ,R R 3R
1 1 4 1 1 1
0 0 5 2 1 2 A
0 5 10 3 3 4
RR
1 1 4
0 5 10
0 0 5
? ? ? ? ?
? ? ? ?
?
? ? ? ?
? ? ? ? ?
? ? ? ?
???
? ? ? ? ?
? ? ? ?
?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ?
?
??
??
? ? ?
??
?? ?
??
1 1 1
3 3 4 A
2 1 2
? ??
??
??
??
?? ??
??
2 2 3 3
1 1 3 2 2 3
11
R R , R R
55
1 1 1
1 1 4
3 3 4
0 1 2 A
5 5 5
0 0 1
2 1 2
5 5 5
R R 4R ,R R 2R ,
3 1 3
5 5 5
1 1 0
11
0 1 0 0 A
55
0 0 1
2 1 2
5 5 5
??
??
??
?? ?
??
??
?
??
??
?
??
??
??
??
??
?
??
??
? ? ? ?
??
?
??
????
?
????
?
????
????
??
?
??
??
??
CBSE XII | Mathematics
Sample Paper 4 Solution
1 1 2
1
R R R
23
0
55
1 0 0
11
0 1 0 0 A
55
0 0 1
2 1 2
5 5 5
23
0
55
11
Hence A 0
55
2 1 2
5 5 5
?
??
???
??
????
?
????
?
????
????
??
?
??
??
??
???
??
??
?
??
?
??
??
?
??
??
??
6. Let P(x, y) be any point on the given curve x
2
+ y
2
– 2x – 3 = 0.
Tangent to the curve at the point (x, y) is given by
dy
dx
.
Differentiating the equation of the curve w .r. t. x we get
? ? ?
??
??
dy
2x 2y 2 0
dx
dy 2 2x 1 x
dx 2y y
Let P(x1, y1) be the point on the given curve at which the tangents are parallel to the
x-axis.
? ?
11
x ,y
dy
dx
?
?
?
?
= 0
?
1
1
1x
y
?
= 0
? 1 – x1 = 0
? x1 = 1
To get the value of y1 just substitute x1 = 1 in the equation x
2
+ y
2
– 2x – 3 = 0, we get
? ?
2
2
1
2
1
2
1
1
1 y 2 1 3 0
y 4 0
y4
y2
? ? ? ? ?
? ? ?
??
? ? ?
So the points on the given curve at which the tangents are parallel to the x-axis are
(1, 2) and (1, -2).
Page 5
CBSE XII | Mathematics
Sample Paper 4 Solution
CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 4
SECTION – A
1. Given that
OR
|3 A | = k |A|
|3 A | = 3
3
|A|
|3 A | = 27|A|
k = 27
2. We have,
3x 7
10
24
12x 14 10
12x 24
x2
?
? ? ?
??
??
2
2
x2
2
f(x) kx x 2
3 x 2
lim kx 3
k(2) 3
4k 3
3
k
4
?
??
??
?
?
?
?
CBSE XII | Mathematics
Sample Paper 4 Solution
3. We have,
?
?
?
?
?
??
?
?
?
?
??
? ? ?
2
3
3
2
2
3
x
I dx
1x
Put 1 x t
3x dx dt
dt
x dx
3
dt
3
I
t
1 dt
I
3t
1
I log t c
3
1
I log 1 x c
3
4. 2x + y – z = 5
Dividing both sides by 5,
2x y z
1
5 5 5
? ? ?
? ? ?
?
x y z
1
5
55
2
It is known that the equation of a plane in intercept form is
x y z
a b c
?? = 1, where a,
b and c are the intercepts cut off by the plane at x, y, and z-axes respectively.
Thus, the intercept cut off by the given plane on the x-axis is
5
2
.
CBSE XII | Mathematics
Sample Paper 4 Solution
Section-B
5. Let A = IA
1 1 2 3
2 2 1 3 3 1
23
2 3 3 1 0 0
2 2 3 0 1 0 A
3 2 2 0 0 1
R R R R
1 1 4 1 1 1
2 2 3 0 1 0 A
3 2 2 0 0 1
Applying R R 2R ,R R 3R
1 1 4 1 1 1
0 0 5 2 1 2 A
0 5 10 3 3 4
RR
1 1 4
0 5 10
0 0 5
? ? ? ? ?
? ? ? ?
?
? ? ? ?
? ? ? ? ?
? ? ? ?
???
? ? ? ? ?
? ? ? ?
?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ?
?
??
??
? ? ?
??
?? ?
??
1 1 1
3 3 4 A
2 1 2
? ??
??
??
??
?? ??
??
2 2 3 3
1 1 3 2 2 3
11
R R , R R
55
1 1 1
1 1 4
3 3 4
0 1 2 A
5 5 5
0 0 1
2 1 2
5 5 5
R R 4R ,R R 2R ,
3 1 3
5 5 5
1 1 0
11
0 1 0 0 A
55
0 0 1
2 1 2
5 5 5
??
??
??
?? ?
??
??
?
??
??
?
??
??
??
??
??
?
??
??
? ? ? ?
??
?
??
????
?
????
?
????
????
??
?
??
??
??
CBSE XII | Mathematics
Sample Paper 4 Solution
1 1 2
1
R R R
23
0
55
1 0 0
11
0 1 0 0 A
55
0 0 1
2 1 2
5 5 5
23
0
55
11
Hence A 0
55
2 1 2
5 5 5
?
??
???
??
????
?
????
?
????
????
??
?
??
??
??
???
??
??
?
??
?
??
??
?
??
??
??
6. Let P(x, y) be any point on the given curve x
2
+ y
2
– 2x – 3 = 0.
Tangent to the curve at the point (x, y) is given by
dy
dx
.
Differentiating the equation of the curve w .r. t. x we get
? ? ?
??
??
dy
2x 2y 2 0
dx
dy 2 2x 1 x
dx 2y y
Let P(x1, y1) be the point on the given curve at which the tangents are parallel to the
x-axis.
? ?
11
x ,y
dy
dx
?
?
?
?
= 0
?
1
1
1x
y
?
= 0
? 1 – x1 = 0
? x1 = 1
To get the value of y1 just substitute x1 = 1 in the equation x
2
+ y
2
– 2x – 3 = 0, we get
? ?
2
2
1
2
1
2
1
1
1 y 2 1 3 0
y 4 0
y4
y2
? ? ? ? ?
? ? ?
??
? ? ?
So the points on the given curve at which the tangents are parallel to the x-axis are
(1, 2) and (1, -2).
CBSE XII | Mathematics
Sample Paper 4 Solution
OR
Consider the given equation,
3
2
2
1
y x 2x 4
Differentiating the above function with respect to x, we have,
dy
3x 2
dx
m 3x 2
? ? ?
??
? ? ?
2
12
Given that the tangents to the given curve are
perpendicular to the line x+14y+3=0
1
Slope of this line, m
14
Since the given line and the tangents to
the given curve are perpendicular, we have,
m m 1
?
?
? ? ?
? ?
2
2
2
2
3
3
1
3x 2 1
14
3x 2 14
3x 12
x4
x2
If x = 2, y = x 2x 4
y 2 2 2 4
y8
? ??
? ? ? ?
??
??
? ? ?
??
??
? ? ?
??
? ? ? ? ?
??
? ? ? ?
3
3
If x = 2, y = x 2x 4
y 2 2 2 4
y 16
? ? ?
? ? ? ? ? ? ?
? ? ?
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