Sample Solution Paper 5 - Math, Class 12

# Sample Solution Paper 5 - Math, Class 12 | Mathematics (Maths) Class 12 - JEE PDF Download

``` Page 1

CBSE XII | Mathematics
Sample Paper  5 Solution

CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 5

SECTION – A

1. Matrix A is a matrix of order 2.
Identity matrix of second order is
10
01
??
??
??

For A to be an identity matrix,
1 0 cos sin
0 1 sin cos
cos 1 andsin 0
o
cos cos0 and sin sin0
0
Thus, for 0 , A is an identity matrix
? ? ? ? ? ? ?
?
? ? ? ?
??
? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ?

2. We have,

? ?
? ? ? ?
2
x 25
f(x) x 5
x5
k x 5
2
x 25
lim f 5
x5 x5
x 5 x 5
lim k
x5 x5
lim x 5 k
x5
5k
?
??
?
??
?
?
? ?
??
?
? ?
??
?
?

Page 2

CBSE XII | Mathematics
Sample Paper  5 Solution

CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 5

SECTION – A

1. Matrix A is a matrix of order 2.
Identity matrix of second order is
10
01
??
??
??

For A to be an identity matrix,
1 0 cos sin
0 1 sin cos
cos 1 andsin 0
o
cos cos0 and sin sin0
0
Thus, for 0 , A is an identity matrix
? ? ? ? ? ? ?
?
? ? ? ?
??
? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ?

2. We have,

? ?
? ? ? ?
2
x 25
f(x) x 5
x5
k x 5
2
x 25
lim f 5
x5 x5
x 5 x 5
lim k
x5 x5
lim x 5 k
x5
5k
?
??
?
??
?
?
? ?
??
?
? ?
??
?
?

CBSE XII | Mathematics
Sample Paper  5 Solution

OR

? Since f(x) is continuous at x 3.
?
?
?
?
?
??
??
??
?
? ? ? ?
??
?
??
??
?
? ? ? ? ? ?
? ? ?
??
x3
2
x3
x3
x3
x3
lim f(x) f(3)
(x 3) 36
lim k
x3
(x 3 6)(x 3 6)
lim k
x3
(x 9)(x 3)
lim k
x3
lim (x 9) k ( x 3, x 3 0)
3 9 k
k 12

3. We have,
1
I= dx
2
9 25x
11
dx
52
3
2
x
5
1 5 x
1
sin c
3
53
5
1 5x
1
sin c
33
?
?
?
?
??
?
??
??
??
??
?
? ? ?
??
??
??
??
?
??
??
??

4. Equation of line through (-2, 1, 3) and parallel to
x 3 y 4 z 8
3 5 6
? ? ?
?? is

x 2 y 1 z 3
3 5 6
? ? ?
??

Page 3

CBSE XII | Mathematics
Sample Paper  5 Solution

CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 5

SECTION – A

1. Matrix A is a matrix of order 2.
Identity matrix of second order is
10
01
??
??
??

For A to be an identity matrix,
1 0 cos sin
0 1 sin cos
cos 1 andsin 0
o
cos cos0 and sin sin0
0
Thus, for 0 , A is an identity matrix
? ? ? ? ? ? ?
?
? ? ? ?
??
? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ?

2. We have,

? ?
? ? ? ?
2
x 25
f(x) x 5
x5
k x 5
2
x 25
lim f 5
x5 x5
x 5 x 5
lim k
x5 x5
lim x 5 k
x5
5k
?
??
?
??
?
?
? ?
??
?
? ?
??
?
?

CBSE XII | Mathematics
Sample Paper  5 Solution

OR

? Since f(x) is continuous at x 3.
?
?
?
?
?
??
??
??
?
? ? ? ?
??
?
??
??
?
? ? ? ? ? ?
? ? ?
??
x3
2
x3
x3
x3
x3
lim f(x) f(3)
(x 3) 36
lim k
x3
(x 3 6)(x 3 6)
lim k
x3
(x 9)(x 3)
lim k
x3
lim (x 9) k ( x 3, x 3 0)
3 9 k
k 12

3. We have,
1
I= dx
2
9 25x
11
dx
52
3
2
x
5
1 5 x
1
sin c
3
53
5
1 5x
1
sin c
33
?
?
?
?
??
?
??
??
??
??
?
? ? ?
??
??
??
??
?
??
??
??

4. Equation of line through (-2, 1, 3) and parallel to
x 3 y 4 z 8
3 5 6
? ? ?
?? is

x 2 y 1 z 3
3 5 6
? ? ?
??

CBSE XII | Mathematics
Sample Paper  5 Solution

Section-B

5. We shall prove by principle of mathematical induction
Here, let
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??

? ?
cosn sinn
n
P n : A ,
sinn cosn
?? ??
?
??
? ? ?
??

? ?
cos sin
1
So, A
sin cos
P 1 is true.
?? ??
?
??
? ? ?
??
?

Assuming result to be true for n = k i.e. P(k) to be true
? ?
cosk sink
k
P k : A
sink cosk
?? ??
?
??
? ? ?
??

We have to prove P (k +1) is true,
? ?
k 1 1 k
P k 1 : A A A
cos sin cosk sink
k1
A
sin cos sink cosk
?
??
? ? ? ? ? ? ? ?
?
??
? ? ? ?
? ? ? ? ? ?
? ? ? ?

? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
cos cosk sin sink cos sink sin cosk
sin cosk cos sink sin sink cos cosk
cos k sin k
sin k cos k
cos k 1 sin k 1
sin k 1 cos k 1
? ? ? ? ? ? ? ? ? ? ??
?
??
? ? ? ? ? ? ? ? ? ? ? ?
??
?? ? ? ? ? ? ?
?
??
? ? ? ? ? ? ?
??
?? ? ? ? ?
?
??
? ? ? ? ?
??

? P (k + 1) is true.

Thus by principle of mathematical induction
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??
for all nN ?

Page 4

CBSE XII | Mathematics
Sample Paper  5 Solution

CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 5

SECTION – A

1. Matrix A is a matrix of order 2.
Identity matrix of second order is
10
01
??
??
??

For A to be an identity matrix,
1 0 cos sin
0 1 sin cos
cos 1 andsin 0
o
cos cos0 and sin sin0
0
Thus, for 0 , A is an identity matrix
? ? ? ? ? ? ?
?
? ? ? ?
??
? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ?

2. We have,

? ?
? ? ? ?
2
x 25
f(x) x 5
x5
k x 5
2
x 25
lim f 5
x5 x5
x 5 x 5
lim k
x5 x5
lim x 5 k
x5
5k
?
??
?
??
?
?
? ?
??
?
? ?
??
?
?

CBSE XII | Mathematics
Sample Paper  5 Solution

OR

? Since f(x) is continuous at x 3.
?
?
?
?
?
??
??
??
?
? ? ? ?
??
?
??
??
?
? ? ? ? ? ?
? ? ?
??
x3
2
x3
x3
x3
x3
lim f(x) f(3)
(x 3) 36
lim k
x3
(x 3 6)(x 3 6)
lim k
x3
(x 9)(x 3)
lim k
x3
lim (x 9) k ( x 3, x 3 0)
3 9 k
k 12

3. We have,
1
I= dx
2
9 25x
11
dx
52
3
2
x
5
1 5 x
1
sin c
3
53
5
1 5x
1
sin c
33
?
?
?
?
??
?
??
??
??
??
?
? ? ?
??
??
??
??
?
??
??
??

4. Equation of line through (-2, 1, 3) and parallel to
x 3 y 4 z 8
3 5 6
? ? ?
?? is

x 2 y 1 z 3
3 5 6
? ? ?
??

CBSE XII | Mathematics
Sample Paper  5 Solution

Section-B

5. We shall prove by principle of mathematical induction
Here, let
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??

? ?
cosn sinn
n
P n : A ,
sinn cosn
?? ??
?
??
? ? ?
??

? ?
cos sin
1
So, A
sin cos
P 1 is true.
?? ??
?
??
? ? ?
??
?

Assuming result to be true for n = k i.e. P(k) to be true
? ?
cosk sink
k
P k : A
sink cosk
?? ??
?
??
? ? ?
??

We have to prove P (k +1) is true,
? ?
k 1 1 k
P k 1 : A A A
cos sin cosk sink
k1
A
sin cos sink cosk
?
??
? ? ? ? ? ? ? ?
?
??
? ? ? ?
? ? ? ? ? ?
? ? ? ?

? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
cos cosk sin sink cos sink sin cosk
sin cosk cos sink sin sink cos cosk
cos k sin k
sin k cos k
cos k 1 sin k 1
sin k 1 cos k 1
? ? ? ? ? ? ? ? ? ? ??
?
??
? ? ? ? ? ? ? ? ? ? ? ?
??
?? ? ? ? ? ? ?
?
??
? ? ? ? ? ? ?
??
?? ? ? ? ?
?
??
? ? ? ? ?
??

? P (k + 1) is true.

Thus by principle of mathematical induction
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??
for all nN ?

CBSE XII | Mathematics
Sample Paper  5 Solution

OR

? Since A is skew symmetric matrix.
??
? ? ?
? ? ?
? ? ?
? ? ? ? ?
??
??
T
T
T3
T
T
Therefore,A A
AA
A ( 1) A
AA
A A ......(Since A A )
2 A 0
A0

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

? ?
dy
dx
x ,y
11
?
?
?
?
= 0
?
1x
1
y
1
?
= 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 y 2 1 3 0
1
2
y 4 0
1
2
y4
1
y2
1
? ? ? ? ?
? ? ?
??
? ? ?

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  5 Solution

CBSE Board
Class XII Mathematics
Board Paper Solution
Sample Paper – 5

SECTION – A

1. Matrix A is a matrix of order 2.
Identity matrix of second order is
10
01
??
??
??

For A to be an identity matrix,
1 0 cos sin
0 1 sin cos
cos 1 andsin 0
o
cos cos0 and sin sin0
0
Thus, for 0 , A is an identity matrix
? ? ? ? ? ? ?
?
? ? ? ?
??
? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ?

2. We have,

? ?
? ? ? ?
2
x 25
f(x) x 5
x5
k x 5
2
x 25
lim f 5
x5 x5
x 5 x 5
lim k
x5 x5
lim x 5 k
x5
5k
?
??
?
??
?
?
? ?
??
?
? ?
??
?
?

CBSE XII | Mathematics
Sample Paper  5 Solution

OR

? Since f(x) is continuous at x 3.
?
?
?
?
?
??
??
??
?
? ? ? ?
??
?
??
??
?
? ? ? ? ? ?
? ? ?
??
x3
2
x3
x3
x3
x3
lim f(x) f(3)
(x 3) 36
lim k
x3
(x 3 6)(x 3 6)
lim k
x3
(x 9)(x 3)
lim k
x3
lim (x 9) k ( x 3, x 3 0)
3 9 k
k 12

3. We have,
1
I= dx
2
9 25x
11
dx
52
3
2
x
5
1 5 x
1
sin c
3
53
5
1 5x
1
sin c
33
?
?
?
?
??
?
??
??
??
??
?
? ? ?
??
??
??
??
?
??
??
??

4. Equation of line through (-2, 1, 3) and parallel to
x 3 y 4 z 8
3 5 6
? ? ?
?? is

x 2 y 1 z 3
3 5 6
? ? ?
??

CBSE XII | Mathematics
Sample Paper  5 Solution

Section-B

5. We shall prove by principle of mathematical induction
Here, let
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??

? ?
cosn sinn
n
P n : A ,
sinn cosn
?? ??
?
??
? ? ?
??

? ?
cos sin
1
So, A
sin cos
P 1 is true.
?? ??
?
??
? ? ?
??
?

Assuming result to be true for n = k i.e. P(k) to be true
? ?
cosk sink
k
P k : A
sink cosk
?? ??
?
??
? ? ?
??

We have to prove P (k +1) is true,
? ?
k 1 1 k
P k 1 : A A A
cos sin cosk sink
k1
A
sin cos sink cosk
?
??
? ? ? ? ? ? ? ?
?
??
? ? ? ?
? ? ? ? ? ?
? ? ? ?

? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
cos cosk sin sink cos sink sin cosk
sin cosk cos sink sin sink cos cosk
cos k sin k
sin k cos k
cos k 1 sin k 1
sin k 1 cos k 1
? ? ? ? ? ? ? ? ? ? ??
?
??
? ? ? ? ? ? ? ? ? ? ? ?
??
?? ? ? ? ? ? ?
?
??
? ? ? ? ? ? ?
??
?? ? ? ? ?
?
??
? ? ? ? ?
??

? P (k + 1) is true.

Thus by principle of mathematical induction
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??
for all nN ?

CBSE XII | Mathematics
Sample Paper  5 Solution

OR

? Since A is skew symmetric matrix.
??
? ? ?
? ? ?
? ? ?
? ? ? ? ?
??
??
T
T
T3
T
T
Therefore,A A
AA
A ( 1) A
AA
A A ......(Since A A )
2 A 0
A0

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

? ?
dy
dx
x ,y
11
?
?
?
?
= 0
?
1x
1
y
1
?
= 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 y 2 1 3 0
1
2
y 4 0
1
2
y4
1
y2
1
? ? ? ? ?
? ? ?
??
? ? ?

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  5 Solution

7. Here,

33
x asin t , y bcos t ??
Differentiating (1) wrt t

dx
2
3asin t cost
dt
?? and

dy
2
3bcos t sint
dt
? ? ?

dy
2
dy 3bcos t sint b
dt
cot t
2
dx
dx a
3asin t cost
dt
??
? ? ? ?
?

? Slope of the tangent at t
2
?
?

dy b
cot 0
dx a 2
2
? ?
? ? ?
?
?
?

Hence, equation of tangent is given by

3
y bcos 0  or  y 0
2
?
? ? ?

8. The volume of a sphere(V) with radius (r) is given by,

4
3
Vr
3
??
?Rate of change of volume (V) w.r.t. (t) is given by,

?
??
??
??
??
??
?
? ? ?
??
??
?
?
dV dV dr
.
dt dr dt
d 4 dr
3
r.
dr 3 dt
dr
2
4 r .
dt
dV
3
It is given that 900 cm / s
dt
dr 900 225
22
dt
4 r r
Therefore, when radius = 15 cm
dr 225 1
2
dt
(15)
Hence,the rate at which the radius of the balloon
increases whe
?
1
n the radius is 15 cm is cm / s.

```

## Mathematics (Maths) Class 12

204 videos|288 docs|139 tests

## FAQs on Sample Solution Paper 5 - Math, Class 12 - Mathematics (Maths) Class 12 - JEE

 1. What are the different types of integrals in calculus?
Ans. In calculus, there are mainly two types of integrals: indefinite integrals and definite integrals. Indefinite integrals give the antiderivative of a function, while definite integrals calculate the area under a curve between two specific points.
 2. How do you find the derivative of a function?
Ans. To find the derivative of a function, you can use differentiation rules such as the power rule, product rule, quotient rule, and chain rule. These rules allow you to find the rate of change of a function at a given point.
 3. What is the relationship between integrals and derivatives?
Ans. The relationship between integrals and derivatives is given by the Fundamental Theorem of Calculus. It states that if a function f(x) is continuous on an interval [a, b], and F(x) is its antiderivative, then the definite integral of f(x) from a to b is equal to F(b) - F(a). In other words, the integral of a function gives you its antiderivative.
 4. How do you solve a limit in calculus?
Ans. To solve a limit in calculus, you can use various techniques such as direct substitution, factoring, rationalization, L'Hôpital's rule, and special limit theorems. These methods help determine the value that a function approaches as it gets arbitrarily close to a certain point.
 5. What is the concept of continuity in calculus?
Ans. Continuity in calculus refers to the smoothness and unbroken nature of a function. A function is continuous at a point if it is defined at that point, and the limit of the function as x approaches that point exists and is equal to the value of the function at that point. Continuity is an important concept in calculus as it allows for the application of many fundamental theorems and techniques.

## Mathematics (Maths) Class 12

204 videos|288 docs|139 tests

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