The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.

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

Discover the best professional documents and content resources in AnyFlip Document Base.
Search
Published by rajusingh79, 2019-07-31 10:20:56

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

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

Keywords: IIT JEE study material, NEET Study Material, JEE mains Study Material, JEE advanced study material, AIIMS Study Material, IIT JEE Foundation study material, NEET Foundation study material, CBSE Study Material, Test Series, Question Bank, Editable Study Material, School Exams study material, board exams study material, XII board exams Study Material, X board exams Study Material, Study Material, JEE mains, JEE advanced, Video Lectures, Study Innovations, online tuition, home tuition, online tutors, coaching & tutorials for English, Mathematics, Science, Physics, Chemistry, Biology, Soft copy study material, customized study material provider for coaching institutes, how to make study material for coaching institute, study material for coaching classes

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


Click to View FlipBook Version