NCERT Solutions Class 12 Mathematics Part II . 674 Pages (668-1342). Free Flip-Book by Study Innovations

Class XII Chapter 8 – Application of Integrals Maths

Question 7:
Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12
Answer
The area enclosed between the parabola, 4y = 3x2, and the line, 2y = 3x + 12, is
represented by the shaded area OBAO as

The points of intersection of the given curves are A (–2, 3) and (4, 12).
We draw AC and BD perpendicular to x-axis.
∴ Area OBAO = Area CDBA – (Area ODBO + Area OACO)

Page 35 of 53

Class XII Chapter 8 – Application of Integrals Maths

Question 8: and the line
Find the area of the smaller region bounded by the ellipse

Answer , and the line,
The area of the smaller region bounded by the ellipse,

, is represented by the shaded region BCAB as

Page 36 of 53

Class XII Chapter 8 – Application of Integrals Maths

∴ Area BCAB = Area (OBCAO) – Area (OBAO)

Page 37 of 53

Class XII Chapter 8 – Application of Integrals Maths
and the line
Question 9:

Find the area of the smaller region bounded by the ellipse

Answer , and the line,
The area of the smaller region bounded by the ellipse,

, is represented by the shaded region BCAB as

∴ Area BCAB = Area (OBCAO) – Area (OBAO)
Page 38 of 53

Class XII Chapter 8 – Application of Integrals Maths

Question 10:
Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and x-
axis
Answer
The area of the region enclosed by the parabola, x2 = y, the line, y = x + 2, and x-axis
is represented by the shaded region OABCO as

Page 39 of 53

Class XII Chapter 8 – Application of Integrals Maths

The point of intersection of the parabola, x2 = y, and the line, y = x + 2, is A (–1, 1).
∴ Area OABCO = Area (BCA) + Area COAC

Page 40 of 53

Class XII Chapter 8 – Application of Integrals Maths

Question 11:
Using the method of integration find the area bounded by the curve

[Hint: the required region is bounded by lines x + y = 1, x – y = 1, – x + y = 1 and – x
– y = 11]
Answer

The area bounded by the curve, , is represented by the shaded region ADCB
as

The curve intersects the axes at points A (0, 1), B (1, 0), C (0, –1), and D (–1, 0).
It can be observed that the given curve is symmetrical about x-axis and y-axis.

∴ Area ADCB = 4 × Area OBAO

Page 41 of 53

Class XII Chapter 8 – Application of Integrals Maths

Question 12: , is represented by the

Find the area bounded by curves
Answer

The area bounded by the curves,
shaded region as

It can be observed that the required area is symmetrical about y-axis.

Question 13:
Using the method of integration find the area of the triangle ABC, coordinates of whose
vertices are A (2, 0), B (4, 5) and C (6, 3)

Page 42 of 53

Class XII Chapter 8 – Application of Integrals Maths

Answer
The vertices of ∆ABC are A (2, 0), B (4, 5), and C (6, 3).

Equation of line segment AB is
Equation of line segment BC is
Equation of line segment CA is

Page 43 of 53

Class XII Chapter 8 – Application of Integrals Maths

Area (∆ABC) = Area (ABLA) + Area (BLMCB) – Area (ACMA)

Question 14:
Using the method of integration find the area of the region bounded by lines:
2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0
Answer
The given equations of lines are
2x + y = 4 … (1)
3x – 2y = 6 … (2)
And, x – 3y + 5 = 0 … (3)

Page 44 of 53

Class XII Chapter 8 – Application of Integrals Maths

The area of the region bounded by the lines is the area of ∆ABC. AL and CM are the
perpendiculars on x-axis.
Area (∆ABC) = Area (ALMCA) – Area (ALB) – Area (CMB)

Question 15: , is represented as

Find the area of the region
Answer

The area bounded by the curves,

Page 45 of 53

Class XII Chapter 8 – Application of Integrals Maths

The points of intersection of both the curves are .
The required area is given by OABCO.
It can be observed that area OABCO is symmetrical about x-axis.

∴ Area OABCO = 2 × Area OBC

Area OBCO = Area OMC + Area MBC

Question 16:
Area bounded by the curve y = x3, the x-axis and the ordinates x = –2 and x = 1 is
A. – 9
B.

Page 46 of 53

Class XII Chapter 8 – Application of Integrals Maths
C.
D.
Answer

Solve it yourself.
The correct option is D.

Question 17:

The area bounded by the curve , x-axis and the ordinates x = –1 and x = 1 is

given by

[Hint: y = x2 if x > 0 and y = –x2 if x < 0]

Page 47 of 53

Class XII Chapter 8 – Application of Integrals Maths
A. 0
B.
C.
D.
Answer

Thus, the correct answer is C.
Page 48 of 53

Class XII Chapter 8 – Application of Integrals Maths

Question 18:
The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is

A.

B.

C.

D.

Answer
The given equations are
x2 + y2 = 16 … (1)
y2 = 6x … (2)

Area bounded by the circle and parabola
Page 49 of 53

Class XII Chapter 8 – Application of Integrals Maths

Area of circle = π (r)2
= π (4)2
= 16π units

Thus, the correct answer is C.

Page 50 of 53

Class XII Chapter 8 – Application of Integrals Maths

Question 19:

The area bounded by the y-axis, y = cos x and y = sin x when
A.

B.

C.
D.
Answer
The given equations are
y = cos x … (1)
And, y = sin x … (2)

Required area = Area (ABLA) + area (OBLO)
Integrating by parts, we obtain

Page 51 of 53

Class XII Chapter 8 – Application of Integrals Maths

Thus, the correct answer is B.

Therefore, the required area is units
Page 52 of 53

Class XII Chapter 9 – Differential Equations Maths
Question 1: www.ncrtsolutions.blogspot.com
Exercise 9.1

Determine order and degree(if defined) of differential equation
Answer

The highest order derivative present in the differential equation is . Therefore, its
order is four.
The given differential equation is not a polynomial equation in its derivatives. Hence, its
degree is not defined.

Question 2:
Determine order and degree(if defined) of differential equation
Answer
The given differential equation is:

The highest order derivative present in the differential equation is . Therefore, its order
is one.
It is a polynomial equation in . The highest power raised to is 1. Hence, its degree is
one.

Question 3:

Determine order and degree(if defined) of differential equation
Answer

Page 1 of 120

Class XII Chapter 9 – Differential Equations Maths

The highest order derivative present in the given differential equation is . Therefore,
its order is two.

It is a polynomial equation in and . The power raised to is 1.
Hence, its degree is one.

Question 4:

Determine order and degree(if defined) of differential equation
Answer

The highest order derivative present in the given differential equation is . Therefore,
its order is 2.
The given differential equation is not a polynomial equation in its derivatives. Hence, its
degree is not defined.

Question 5:
Determine order and degree(if defined) of differential equation

Answer

The highest order derivative present in the differential equation is . Therefore, its
order is two.

Page 2 of 120

Class XII Chapter 9 – Differential Equations Maths

It is a polynomial equation in and the power raised to is 1.
Hence, its degree is one.

Question 6:
Determine order and degree(if defined) of differential equation

Answer

The highest order derivative present in the differential equation is . Therefore, its
order is three.

The given differential equation is a polynomial equation in .

The highest power raised to is 2. Hence, its degree is 2.

Question 7:
Determine order and degree(if defined) of differential equation
Answer

The highest order derivative present in the differential equation is . Therefore, its
order is three.

It is a polynomial equation in . The highest power raised to is 1. Hence, its
degree is 1.

Question 8:

Determine order and degree(if defined) of differential equation
Answer

Page 3 of 120

Class XII Chapter 9 – Differential Equations Maths

The highest order derivative present in the differential equation is . Therefore, its order
is one.
The given differential equation is a polynomial equation in and the highest power
raised to is one. Hence, its degree is one.

Question 9:
Determine order and degree(if defined) of differential equation
Answer

The highest order derivative present in the differential equation is . Therefore, its
order is two.
The given differential equation is a polynomial equation in and and the highest
power raised to is one.
Hence, its degree is one.

Question 10:
Determine order and degree(if defined) of differential equation
Answer

The highest order derivative present in the differential equation is . Therefore, its
order is two.
This is a polynomial equation in and and the highest power raised to is one.
Hence, its degree is one.

Question 11:
The degree of the differential equation

is

Page 4 of 120

Class XII Chapter 9 – Differential Equations Maths

(A) 3 (B) 2 (C) 1 (D) not defined
Answer

The given differential equation is not a polynomial equation in its derivatives. Therefore,
its degree is not defined.
Hence, the correct answer is D.

Question 12:
The order of the differential equation

is
(A) 2 (B) 1 (C) 0 (D) not defined
Answer

The highest order derivative present in the given differential equation is . Therefore,
its order is two.
Hence, the correct answer is A.

Page 5 of 120

Class XII Chapter 9 – Differential Equations Maths
Question 1: Exercise 9.2

Answer

Differentiating both sides of this equation with respect to x, we get:

Now, differentiating equation (1) with respect to x, we get:

Substituting the values of in the given differential equation, we get the L.H.S.
as:

Thus, the given function is the solution of the corresponding differential equation.
Question 2:

Answer

Differentiating both sides of this equation with respect to x, we get:

Substituting the value of in the given differential equation, we get:

L.H.S. = = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Page 6 of 120

Class XII Chapter 9 – Differential Equations Maths

Question 3:
Answer
Differentiating both sides of this equation with respect to x, we get:

Substituting the value of in the given differential equation, we get:

L.H.S. = = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Question 4:

Answer
Differentiating both sides of the equation with respect to x, we get:

Page 7 of 120

Class XII Chapter 9 – Differential Equations Maths

L.H.S. = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
Question 5:
Answer
Differentiating both sides with respect to x, we get:

Substituting the value of in the given differential equation, we get:
Hence, the given function is the solution of the corresponding differential equation.
Question 6:

Page 8 of 120

Class XII Chapter 9 – Differential Equations Maths
Answer

Differentiating both sides of this equation with respect to x, we get:

Substituting the value of in the given differential equation, we get:

Hence, the given function is the solution of the corresponding differential equation.
Question 7:

Answer
Differentiating both sides of this equation with respect to x, we get:

Page 9 of 120

Class XII Chapter 9 – Differential Equations Maths

L.H.S. = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
Question 8:
Answer
Differentiating both sides of the equation with respect to x, we get:

Substituting the value of in equation (1), we get:

Page 10 of 120

Class XII Chapter 9 – Differential Equations Maths

Hence, the given function is the solution of the corresponding differential equation.
Question 9:
Answer
Differentiating both sides of this equation with respect to x, we get:

Substituting the value of in the given differential equation, we get:
Page 11 of 120

Class XII Chapter 9 – Differential Equations Maths

Hence, the given function is the solution of the corresponding differential equation.
Question 10:

Answer
Differentiating both sides of this equation with respect to x, we get:

Substituting the value of in the given differential equation, we get:

Hence, the given function is the solution of the corresponding differential equation.
Page 12 of 120

Class XII Chapter 9 – Differential Equations Maths

Question 11:
The numbers of arbitrary constants in the general solution of a differential equation of
fourth order are:
(A) 0 (B) 2 (C) 3 (D) 4
Answer
We know that the number of constants in the general solution of a differential equation
of order n is equal to its order.
Therefore, the number of constants in the general equation of fourth order differential
equation is four.
Hence, the correct answer is D.

Question 12:
The numbers of arbitrary constants in the particular solution of a differential equation of
third order are:
(A) 3 (B) 2 (C) 1 (D) 0
Answer
In a particular solution of a differential equation, there are no arbitrary constants.
Hence, the correct answer is D.

Page 13 of 120

Class XII Chapter 9 – Differential Equations Maths
Question 1: Exercise 9.3

Answer

Differentiating both sides of the given equation with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Hence, the required differential equation of the given curve is
Question 2:
Answer
Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:
Page 14 of 120

Class XII Chapter 9 – Differential Equations Maths

Dividing equation (2) by equation (1), we get:

This is the required differential equation of the given curve.
Question 3:
Answer
Differentiating both sides with respect to x, we get:
Again, differentiating both sides with respect to x, we get:
Multiplying equation (1) with (2) and then adding it to equation (2), we get:

Now, multiplying equation (1) with equation (3) and subtracting equation (2) from it, we
get:

Substituting the values of in equation (3), we get:
Page 15 of 120

Class XII Chapter 9 – Differential Equations Maths

This is the required differential equation of the given curve.
Question 4:
Answer
Differentiating both sides with respect to x, we get:

Multiplying equation (1) with equation (2) and then subtracting it from equation (2), we
get:

Differentiating both sides with respect to x, we get:
Dividing equation (4) by equation (3), we get:

This is the required differential equation of the given curve.

Page 16 of 120

Class XII Chapter 9 – Differential Equations Maths
Question 5:

Answer

Differentiating both sides with respect to x, we get:

Again, differentiating with respect to x, we get:

Adding equations (1) and (3), we get:

This is the required differential equation of the given curve.

Question 6:
Form the differential equation of the family of circles touching the y-axis at the origin.
Answer
The centre of the circle touching the y-axis at origin lies on the x-axis.
Let (a, 0) be the centre of the circle.
Since it touches the y-axis at origin, its radius is a.
Now, the equation of the circle with centre (a, 0) and radius (a) is

Page 17 of 120

Class XII Chapter 9 – Differential Equations Maths

Differentiating equation (1) with respect to x, we get:
Now, on substituting the value of a in equation (1), we get:

This is the required differential equation.
Question 7:
Form the differential equation of the family of parabolas having vertex at origin and axis
along positive y-axis.
Answer
The equation of the parabola having the vertex at origin and the axis along the positive
y-axis is:

Page 18 of 120

Class XII Chapter 9 – Differential Equations Maths

Differentiating equation (1) with respect to x, we get:
Dividing equation (2) by equation (1), we get:

This is the required differential equation.
Question 8:
Form the differential equation of the family of ellipses having foci on y-axis and centre at
origin.
Answer
The equation of the family of ellipses having foci on the y-axis and the centre at origin is
as follows:

Page 19 of 120

Class XII Chapter 9 – Differential Equations Maths

Differentiating equation (1) with respect to x, we get:
Again, differentiating with respect to x, we get:
Substituting this value in equation (2), we get:
This is the required differential equation.

Page 20 of 120

Class XII Chapter 9 – Differential Equations Maths

Question 9:
Form the differential equation of the family of hyperbolas having foci on x-axis and
centre at origin.
Answer
The equation of the family of hyperbolas with the centre at origin and foci along the x-
axis is:

Differentiating both sides of equation (1) with respect to x, we get:
Again, differentiating both sides with respect to x, we get:
Substituting the value of in equation (2), we get:

Page 21 of 120

Class XII Chapter 9 – Differential Equations Maths

This is the required differential equation.

Question 10:
Form the differential equation of the family of circles having centre on y-axis and radius
3 units.
Answer
Let the centre of the circle on y-axis be (0, b).
The differential equation of the family of circles with centre at (0, b) and radius 3 is as
follows:

Differentiating equation (1) with respect to x, we get:
Substituting the value of (y – b) in equation (1), we get:

Page 22 of 120

Class XII Chapter 9 – Differential Equations Maths

This is the required differential equation. as the general solution?
Question 11:
Which of the following differential equations has
A.

B.
C.
D.
Answer
The given equation is:

Differentiating with respect to x, we get:

Again, differentiating with respect to x, we get:

Page 23 of 120

Class XII Chapter 9 – Differential Equations Maths

This is the required differential equation of the given equation of curve.
Hence, the correct answer is B.

Question 12: as one of its particular solution?
Which of the following differential equation has

A.

B.

C.

D.

Answer
The given equation of curve is y = x.
Differentiating with respect to x, we get:

Again, differentiating with respect to x, we get:

Now, on substituting the values of y, from equation (1) and (2) in each of

the given alternatives, we find that only the differential equation given in alternative C is
correct.

Page 24 of 120

Class XII Chapter 9 – Differential Equations Maths

Hence, the correct answer is C.

Page 25 of 120

Class XII Chapter 9 – Differential Equations Maths
Question 1: Exercise 9.4

Answer
The given differential equation is:

Now, integrating both sides of this equation, we get:

This is the required general solution of the given differential equation.
Question 2:
Answer
The given differential equation is:

Page 26 of 120

Class XII Chapter 9 – Differential Equations Maths

Now, integrating both sides of this equation, we get:

This is the required general solution of the given differential equation.
Question 3:

Answer
The given differential equation is:

Now, integrating both sides, we get:
Page 27 of 120

Class XII Chapter 9 – Differential Equations Maths

This is the required general solution of the given differential equation.
Question 4:
Answer
The given differential equation is:

Integrating both sides of this equation, we get:

Page 28 of 120

Class XII Chapter 9 – Differential Equations Maths

Substituting these values in equation (1), we get:

This is the required general solution of the given differential equation.
Question 5:
Answer
The given differential equation is:

Integrating both sides of this equation, we get:
Page 29 of 120

Class XII Chapter 9 – Differential Equations Maths

Let (ex + e–x) = t.
Differentiating both sides with respect to x, we get:

Substituting this value in equation (1), we get:

This is the required general solution of the given differential equation.
Question 6:
Answer
The given differential equation is:

Integrating both sides of this equation, we get:

Page 30 of 120

Class XII Chapter 9 – Differential Equations Maths

This is the required general solution of the given differential equation.
Question 7:
Answer
The given differential equation is:

Integrating both sides, we get:

Substituting this value in equation (1), we get:
Page 31 of 120

Class XII Chapter 9 – Differential Equations Maths

This is the required general solution of the given differential equation.
Question 8:
Answer
The given differential equation is:

Integrating both sides, we get:

This is the required general solution of the given differential equation.
Page 32 of 120