NCERT Maths solutions for 12th

Vector Algebra(Ex. 10.2)

2020-11-11T19:18:00.000+05:30

Exercise 10.2

1. Compute the magnitude of the following vectors:

Solun:- Given

We know that:-

The magnitude of a vector is

Given vector is:-

2. Write two different vectors having same magnitude.

Solun:- Let

and

We know that:-

The magnitude of a vector is

and

3. Write two different vectors having same direction.

Solun:- Let

and

We know that:-

The Direction cosine of vector

Then

and

4. Find the values of x and y so that the vectors and are equal.

Solun:- We know that for the equal vectors:-

=

Compare both sides:

x = 2 and y = 3.

5. Find the scalar and vector components of the vector with the initial point (2, 1) and terminal point (-5, 7).

Solun:- Let A(2, 1) and B(-5, 7) then

Scalar Components are:- -7 and 6

Vector Components are:-

6. Find the sum of vectors ,, and .

Solun:- Sum of Vectors = sum of components of vectors

7. Find the unit vector in the direction of the vector

Solun:- Given

We know that:-

8. Find the unit vector in the direction of vector , where P and Q are the points (1, 2, 3) and (4, 5, 6) respectively.

Solun:- Given P(1, 2, 3) and Q(4, 5, 6) then

We know that:-

9. For given vectors, and , find the unit vector in the direction of the vector .

Solun:- Given and

Let

We know that:-

The magnitude of a vector is

We know that:-

10. Find a vector in the direction of vector which has magnitude 8 units.

Solun:- Let given vector is

We have to find a vector in the direction of which has magnitude 8 units then

We know that:-

Let unknown vector is

11. Show that the vectors and are collinear.

Solun:- We know that:- if two vectors and are collinear if and only if there exists a non-zero scalar such that

Let = and =

Here = -2

Hence is a non-zero scalar then and are collinear vectors.

12. Find the direction cosine of the vector .

Solun:- Let =

Direction ratios in is:- a = 1, b = 2, and c = 3

We know that:- direction cosine in the form of direction ratio is:-

, , and

13. Find the direction cosine of the vector joining the points A(1, 2, -3) and

B(-1, -2, 1), directed from A to B.

Solun:- Given A(1, 2, -3) and B(-1, -2, 1)

Then

Direction ratios in is:- a = -2, b = -4, and c = 4

We know that:- direction cosine in the form of direction ratio is:-

, , and

Direction cosine:- -1/3, -2/3, and 2/3

14. Show that the vector is equally inclined to the axes OX, OY, and OZ.

Solun:- Let =

Direction ratios in is:- a = 1, b = 1, and c = 1

We know that:- direction cosine in the form of direction ratio is:-

, , and

We know that direction cosine is the cosine of direction angles and direction cosines of are equal then makes the same angle with all the axes.

Hence given vector is equally inclined to the axes OX, OY, and OZ.

15. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are and respectively, in the ratio 2:1.

(i) internally (ii) externally

Solun:- Let = and =

And point R() divides line joining by and

(i) internally in the ration of 2:1

m = 2 and n = 1

(ii) externally in the ration of 2:1

m = 2 and n = 1

16. Find the position vector of the mid point of the vector joining the points

P(2, 3, 4) and Q(4, 1, -2).

Solun:- Given points are P(2, 3, 4) and Q(4, 1, -2)

Then

We know that:-

For midpoint

17. Show that the points A, B, and C with position vectors, ,, and respectively form the vertices of a right angled triangle.

Solun:- Given Vectors:-

41 = 35 + 6

41 = 41

So,

Hence ABC is right angle triangle.

18. In triangle ABC, which of the following is not true:

Solun:- According to triangle law of addition:-

So option B is correct.

We know that:-

So option A is correct.

Also:-

So option D is also correct.

Hence option C is not true.

The correct answer is C.

19. If and are two collinear vectors, then which of the following are incorrect:

(D) both the vectors and have same direction, but different magnitude

Solun:- According to the definition of collinear vector:- Parallel vectors are said to be collinear vectors irrespective of their magnitude.

Such that , is non-zero scalar.

So, option A is true.

Here is -1 or 1 that is a non-zero scalar.

So, option B is correct.

And option C is also correct because and are proportional so components of and are also proportional.

Option D is incorrect because vectors having the opposite direction are also collinear.

The correct answer is D.

Download PDF of Exercise 10.2

See Also:-

Notes of Vector Algebra

Exercise 10.1

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Vector Algebra(Ex. 10.1)

2020-11-09T20:23:00.000+05:30

Exercise 10.1

1. Represent graphically a displacement of 40 km, 30o east of north.

Solun:-

Here has magnitude 40 km and direction is 30O east of north.

2. Classify the following measures as scalars and vectors.

(i) 10 kg ----> Scalar

(ii) 2 meters north-west ----> Vector

(iii) 40O ----> Scalar

(iv) 40 watt ----> Scalar

(v) 10-19 coulomb ----> Scalar

(vi) 20 m/s2 ----> Vector

3. Classify the following as scalars and vector quantities:

(i) time period ----> Scalar

(ii) distance ----> Scalar

(iii) force ----> Vector

(iv) velocity ----> Vector

(v) work done ----> Scalar

4. In Fig. 10.6 (a square), identify the following vectors.

(i) Coinitial ----> and

(ii) Equal ----> and

(iii) Collinear but not equal ----> and

5. Answer the following as true or false.

(i) and are collinear.

Solun:- Given statement is true.

(ii) Two collinear vector always equal in magnitude.

Solun:- Given statement is false because according to the definition of collinear vectors, vectors that are parallel to each other are said to be collinear, irrespective of their magnitude.

(iii) Two vectors having same magnitude are collinear.

Solun:- Given statement is false because according to the definition of collinear vectors, vectors that are parallel to each other are said to be collinear, irrespective of their magnitude.

(iv) Two collinear vectors having the same magnitude are equal.

Solun:- Given statement is false because according to the definition of equal vectors, vectors having the same magnitude as well as the same direction are called equal vectors.

Download PDF of Exercise 10.1

See Also:-

Notes of Vector Algebra

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Vector Algebra

2020-11-07T18:29:00.000+05:30

Definitions and Formulas

Scalars:- A quantity that involves only value (magnitude) or a real number.

Vectors:- A quantity that involves value (magnitude) as well as direction.

Representation:- or

Here the point A from where the vector starts is called its initial point, and the point B where it ends is called the terminal point. The distance between the initial and terminal points of a vector is called the magnitude (or length) of the vector, denoted as.

Position vector:- Position vector of a point P(x,y,z) is given as and its magnitude by .

Direction cosines:- Consider a position vector of a point P(x,y,z). Then the angles made by the vector with the positive directions of x, y, and z-axes respectively are called the direction angles. The cosine values of these angles cosα, cosβ, and cosγ are called direction cosine of the vector and usually denoted by l, m, and n.

cosα = x/r , cosβ = y/r, and cosγ = z/r

We know that:-

l = cosα, m= cosβ, and n= cosγ

Then lr = x, mr = y and nr = z

These numbers lr, mr, and nr are directly proportional to the direction cosines are called direction ratios of vector and denoted as a, b, and c respectively.

NOTE:- l2 + m2 + n2 = 1

Components of a vector:-

Let A(1,0,0), B(0,1,0), and C(0,0,1) respectively on the x-axis, y-axis, and z-axis

then

Each having magnitude 1 is called unit vectors along the axes OX, OY, and OZ respectively, and denoted by , and respectively.

NOTE:- (1) Two vectors and are called collinear if and only if there exists a non-zero scalar such that .

If the vectors and are given in component form then the two vectors are called collinear if and only if

Compare both sides:-

b1 = a1, b2 = a2, and b3 = a3

(2) If then a1, a2 and a3 are also called direction ratios of .

Types of Vectors

1. Zero Vector:- A vector whose initial and terminal points coincide or same, is called a zero vector (or null vector) and denoted as or and it may be having any direction because it has no definite direction.

2. Unit Vector:- A vector whose magnitude is unity is called a unit vector and denoted as .

For unit vector:-

3. Coinitial Vectors:- Two or more vectors having the same initial point are called coinitial vectors.

4. Collinear Vectors:- Two or more vectors are said to be collinear if they are parallel to the same line, irrespective of their magnitudes and directions.

5. Equal Vectors:- Two vectors are said to be equal if they have the same magnitude and direction regardless of the positions of their initial points.

If and

Then vector is called equal vector

6. Negative of a Vector:- A vector whose magnitude is the same as that of a given vector, but the direction is opposite to that of it, is called a negative of a given vector.

For example:- Vector is negative of the given vector , written as

Example:- Find the magnitude of .

Solun:-

Addition of Vectors:-

This is known as the triangle law of addition.

Properties of vector addition:-

1. Commutative Property:- For any two vectors and

2. Associative Property:- For any vectors , and

NOTE:- (1) For any vector

Here, the zero vector is called the additive identity for the vector addition.

(2) For any vector

Here, the vector is called the additive inverse or negative of vector .

Multiplication of a vector by a scalar:- Product of the vector by the scalar , is denoted as , is called the multiplication of a vector by the scalar .

Note that is also a vector, collinear to the vector so has the same direction as . But, the magnitude of is times the magnitude of vector .

Product of two vectors:- Multiplication of two vectors is also defined in two ways- scalar product and vector product. Scalar product where the result is scalar and vector product where the result is a vector.

1. Scalar (or dot) product of two vectors:-

The scalar product of two non-zero vectors and , denoted by . is defined as

where θ is the angle between and , .

Properties of scalar product:-

1. The scalar product is commutative i.e.

2. Distributivity of scalar product over addition:- Let ,andbe any three vectors, then

3. Let and be any two vectors, λ be any scalars. Then

4. Let and be two non-zero vectors, then

(i) if and only if both the vectors are perpendicular to each other or θ = 90o.

(ii) If θ = 0 or both the vectors are parallel to each other then

NOTE:-

(ii) If θ = π or both the vectors are parallel to each other and direction is opposite to each other then

5. The angle between two non-zero vectors and is

NOTE:- As conventionally , and are mutually perpendicular unit vectors then

Projection of a vector on a line:-

(1) Projection of vector on is

And denotes the projection of in the direction of .

(2) Projection of vector onis

And denotes the projection of in the direction of .

NOTE:- If is the unit vector along line l then the projection of on the line is given by:-

2. Vector (or cross) product of two vectors:-

The vector product of two non-zero vectors and , denoted by x is defined as

where θ is the angle between and , and is

a unit vector perpendicular to both and .

Properties of vector product:-

1. The scalar product is not commutative i.e.

Because in vector product direction of the resultant vector differs from the initial direction.

2. Distributivity of vector product over addition:- Let ,andbe any three vectors, then

3. Let and be any two vectors, λ be any scalars. Then

3. Let and be two non-zero vectors, then

(i) if both the vectors are parallel or collinear or θ = 0o.

(ii) If θ = 90o or both the vectors are perpendicular to each other then

4. The angle between two non-zero vectors and is

NOTE:- As conventionally , and are mutually perpendicular unit vectors then

5. Area of Triangle:-

Proof:-

5. Area of Parallelogram:-

Proof:-

6. If three points are collinear then

7. Find a unit vector

Vector joining two points:-

If P1(x1,y1,z1) and P2(x2,y2,z2) are two points, then the vector joining P1 and P2 is the vector .

According to the triangle law of addition:-

Section Formula:-

Case 1: When R divides PQ internally

Case 2: When R divides PQ externally

NOTE:- If R is the midpoint of PQ then m = n

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Differential Equations(Ex. 9.5)

2020-11-04T20:31:00.001+05:30

Exercise 9.5

In each of the Exercises 1 to 10, show that the given differential equation is

homogeneous and solve each of them.

1. (x2 + xy).dy = (x2 + y2).dx

Solun:- Given eq. is:- (x2 + xy).dy = (x2 + y2).dx

.....(1)

Let F(x,y)=

Replace F(x,y) with F(λx,λy):-

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

.......(2)

From eq. 1 and 2:-

Separating the variables:-

Integrating both sides:-

We know that:-

Put the value of v:-

Here e-C = c both are constants.

Solun:- Given eq. is:-

.....(1)

Let F(x,y)=

Replace F(x,y) with F(λx,λy):-

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

.......(2)

From eq. 1 and 2:-

Separating the variables:-

Integrating both sides:-

We know that:-

Put the value of v:-

y = xlog|x| + cx

3. (x - y).dy - (x + y).dx = 0

Solun:- Given eq. is:- (x - y).dy - (x + y).dx = 0

(x - y).dy = (x + y).dx

.....(1)

Let F(x,y)=

Replace F(x,y) with F(λx,λy):-

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

.......(2)

From eq. 1 and 2:-

Separating the variables:-

Integrating both sides:-

We know that:-

Put the value of v:-

4. (x2 - y2).dx + 2xy.dy = 0

Solun:- Given eq. is:- (x2 - y2).dx + 2xy.dy = 0

(x2 - y2).dx = -2xy.dy

.....(1)

Let F(x,y)=

Replace F(x,y) with F(λx,λy):-

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

.......(2)

From eq. 1 and 2:-

Separating the variables:-

Integrating both sides:-

We know that:-

Put the value of v:-

⇒ log |x2 + y2| = 2log |x| - log |x| + log |C|

⇒ log |x2 + y2| = log |x| + log |C|

⇒ log |x2 + y2| = log |C.x|

⇒ x2 + y2 = C.x

Solun:- Given eq. is:-

.....(1)

Let F(x,y)=

Replace F(x,y) with F(λx,λy):-

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

.......(2)

From eq. 1 and 2:-

Separating the variables:-

Integrating both sides:-

We know that:-

Put the value of v:-

Let -log |C| = c

Solun:- Given eq. is:-

.....(1)

Let F(x,y)=

Replace F(x,y) with F(λx,λy):-

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

.......(2)

From eq. 1 and 2:-

Separating the variables:-

Integrating both sides:-

We know that:-

Put the value of v:-

Solun:- Given eq. is:-

.....(1)

Let F(x,y)=

Replace F(x,y) with F(λx,λy):-

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

.......(2)

From eq. 1 and 2:-

Separating the variables:-

Integrating both sides:-

We know that:-

Put the value of v:-

Let 1/C = c

Solun:- Given eq. is:-

.....(1)

Let F(x,y)=

Replace F(x,y) with F(λx,λy):-

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

.......(2)

From eq. 1 and 2:-

Separating the variables:-

Integrating both sides:-

We know that:-

Put the value of v:-

Solun:- Given eq. is:-

.....(1)

Let F(x,y)=

Replace F(x,y) with F(λx,λy):-

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

.......(2)

From eq. 1 and 2:-

Separating the variables:-

Integrating both sides:-

We know that:-

Put the value of v:-

Let 1/C = c

Solun:- Given eq. is:-

.....(1)

Let F(y,x)=

Replace F(y,x) with F(λy,λx):-

Then F(y,x) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute x = vy

Differentiate with respect to y:-

.......(2)

From eq. 1 and 2:-

Separating the variables:-

Integrating both sides:-

We know that:-

Put the value of v:-

For each of the Differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition:

; y = 1 when x = 1

Solun:- Given eq. is:-

.....(1)

Let F(x,y)=

Replace F(x,y) with F(λx,λy):-

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

.......(2)

From eq. 1 and 2:-

Separating the variables:-

Integrating both sides:-

We know that:-

Put the value of v:-

......(3)

Given y = 1 and x = 1

⇒ tan-1(1) +1/2log |2| = log |C|

Put the value of log |C| in eq. 3:-

; y = 1 when x = 1

Solun:- Given eq. is:-

.....(1)

Let F(x,y)=

Replace F(x,y) with F(λx,λy):-

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

.......(2)

From eq. 1 and 2:-

Separating the variables:-

Integrating both sides:-

We know that:-

Put the value of v:-

......(3)

Given y = 1 and x = 1

Put the value of log |C| in eq. 3:-

⇒ 3x2y = y + 2x

; y = π/4 when x = 1

Solun:- Given eq. is:-

.....(1)

Let F(x,y)=

Replace F(x,y) with F(λx,λy):-

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

.......(2)

From eq. 1 and 2:-

Separating the variables:-

Integrating both sides:-

We know that:-

Put the value of v:-

......(3)

Given y = π/4 and x = 1

⇒ C = e

Put the value of C in eq. 3:-

; y = 0 when x = 1

Solun:- Given eq. is:-

.....(1)

Let F(x,y)=

Replace F(x,y) with F(λx,λy):-

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

.......(2)

From eq. 1 and 2:-

Separating the variables:-

Integrating both sides:-

We know that:-

⇒ cos v = log |x| + C

Put the value of v:-

......(3)

Given y = 0 and x = 1

⇒ cos (0) = log |1| + C

⇒ 1 = C

⇒ C = 1

Put the value of C in eq. 3:-

; y = 2 when x = 1

Solun:- Given eq. is:-

.....(1)

Let F(x,y)=

Replace F(x,y) with F(λx,λy):-

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

.......(2)

From eq. 1 and 2:-

Separating the variables:-

Integrating both sides:-

We know that:-

Put the value of v:-

......(3)

Given y = 2 and x = 1

⇒ C = -1

Put the value of log |C| in eq. 3:-

16. A homogeneous differential equation of the form can be solved by

making the substitution.

(A) y = vx (B) v = yx (C) x = vy (D) x = v

Solun:- Given eq. is:-

This type of homogeneous equation is solved by the substitution of x = vy.

The correct answer is C.

17. Which of the following is a homogeneous differential equation?

(A) (4x+6y+5).dy - (3y+2x+4).dx = 0

(B) (xy).dx - (x3+ y3).dy = 0

(C) (x3+2y2).dx + 2xy.dy = 0

(D) y2.dx + (x2-xy-y2).dy = 0

Solun:- The correct answer is D because power of all the terms all are same.

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See Also:-

Differential Equations(Ex. 9.4)

2020-11-02T22:47:00.000+05:30

Exercise 9.4

For each of the differential equations in Exercises 1 to 10 find the general equation:

Solun:- Given equation is:-

Separate variables of eq. 1:-

Integrating both sides:-

We know that:-

cos 2x = 2cos2x - 1

cos 2x = 1 - 2sin2x

We know that:-

, (-2 < y < 2)

Solun:- Given equation is:-

........(1)

Separate variables of eq. 1:-

Integrating both sides:-

We know that:-

⇒ y/2 = sin (x + C)

⇒ y = 2sin (x + C)

, (y ≠ 1)

Solun:- Given equation is:-

........(1)

Separate variables of eq. 1:-

Integrating both sides:-

We know that:-

Let -1/ec = A (A is constant)

⇒ y = 1 + Ae-x

4. sec2x.tany.dx + sec2y.tanx.dy = 0

Solun:- Given equation is:- sec2x.tany.dx + sec2y.tanx.dy = 0

sec2x.tany.dx = -sec2y.tanx.dy........(1)

Separate variables of eq. 1:-

Integrating both sides:-

We know that:-

⇒ log |tan x| = -log |tan y| + log C

⇒ log |tan x| + log |tan y| = log C

⇒ log |tan x.tan y| = log C

⇒ tan x.tan y = C

5. (ex + e-x).dy - (ex - e-x).dx = 0

Solun:- Given equation is:- (ex + e-x).dy - (ex - e-x).dx = 0

(ex + e-x).dy = (ex - e-x).dx........(1)

Separate variables of eq. 1:-

Integrating both sides:-

We know that:-

⇒ y = log |ex + e-x| + C

Solun:- Given equation is:-

........(1)

Separate variables of eq. 1:-

Integrating both sides:-

We know that:-

7. ylog y.dx - x.dy = 0

Solun:- Given equation is:- ylog y.dx - x.dy = 0

⇒ ylog y.dx = x.dy........(1)

Separate variables of eq. 1:-

Integrating both sides:-

We know that:-

⇒ log x = log |log y| + log C

⇒ log x - log C = log |log y|

⇒ log x + log c = log |log y|

⇒ log |c.x| = log |log y|

⇒ log y = c.x

⇒ y = ecx

Solun:- Given equation is:-

........(1)

Separate variables of eq. 1:-

⇒ y-5.dy = -x-5.dx

Integrating both sides:-

We know that:-

⇒ x-4 + y-4 = C

Solun:- Given equation is:-

........(1)

Separate variables of eq. 1:-

⇒ dy = sin-1x.dx

Integrating both sides:-

According to ILATE:- u = sin-1x and v = 1

We know that:-

Let 1 - x2 = t2

Differentitate w.r.t. to x:-

⇒ -2x.dx = 2t.dt

⇒ 2x.dx = -2t.dt

10. extan y.dx + (1 - ex).sec2y.dy = 0

Solun:- Given equation is:- extan y.dx + (1 - ex).sec2y.dy = 0

extan y.dx = -(1 - ex).sec2y.dy........(1)

Separate variables of eq. 1:-

Integrating both sides:-

We know that:-

⇒ log |tan y| = log |1 - ex| + log C

⇒ log |tan y| = log |C.(1 - ex)|

⇒ tan y = C.(1 - ex)

For each of the differential equations in Exercises 11 to 14, Find a particular solution satisfying the given condition:

; y = 1 when x = 0

Solun:- Given equation is:-

........(1)

Separate variables of eq. 1:-

Integrating both sides:-

By partial fractions:-

⇒ 1 = A(x2 + 1) + (Bx+C)(x + 1)......(1)

Put x = -1 in eq. 1:-

⇒ 1 = A(1 + 1)

⇒ A = 1/2

Put x = 0 in eq. 1:-

⇒ 1 = A(0 + 1) + (C)(1)

⇒ 1 = A + C

⇒ C = 1/2

Put x = 1 in eq. 1:-

⇒ 1 = A(1 + 1) + (B + C)(2)

⇒ 1 = 2(1/2) + (B + 1/2)(2)

⇒ 1 = 1 + 2B + 1

⇒ 1 = 2 + 2B

⇒ B = -1/2

We know that:-

Given y = 1 and x = 0

⇒ 1 = 1/2(log |1|) - 1/2tan-1(0) + 3/4(log |1|) + C

⇒ 1 = (1/2)x0 -(1/2)x0 + (3/4)x0 + C

⇒ C = 1

; y = 0 when x = 2

Solun:- Given equation is:-

........(1)

Separate variables of eq. 1:-

Integrating both sides:-

By partial fractions:-

⇒ 1 = A(x2 - 1) + B(x)(x + 1) + C(x)(x - 1)......(1)

Put x = 0 in eq. 1:-

⇒ 1 = A(-1)

⇒ A = -1

Put x = 1 in eq. 1:-

⇒ 1 = B(1)(2)

⇒ B = 1/2

Put x = -1 in eq. 1:-

⇒ 1 = C(-1)(-2)

⇒ C = 1/2

We know that:-

Given y = 0 and x = 2

; y = 1 when x = 0

Solun:- Given equation is:-

........(1)

Separate variables of eq. 1:-

Integrating both sides:-

Given y = 1 and x = 0

⇒ C = 1

; y = 1 when x = 0

Solun:- Given equation is:-

........(1)

Separate variables of eq. 1:-

Integrating both sides:-

We know that:-

Given y = 1 and x = 0

⇒ log |1| = log |sec 0| + C

⇒ C = 0

⇒ y = sec x

15. Find the equation of a curve passing through the point (0,0) and whose differential equation is y’ = exsin x.

Solun:- Given equation is:- y’ = exsin x

........(1)

Separate variables of eq. 1:-

Integrating both sides:-

......(2)

Let

According to ILATE:- u = sin x and v = ex

We know that:-

⇒ 2I = ex(sin x - cos x) + C

Put the value of I in eq. 2:-

Given y = 0 and x = 0 because curve passing through (0,0)

⇒ C = 1/2

⇒ 2y - 1 = ex(sin x - cos x)

16. For the differential equation , find the solution curve passing through the point (1,-1).

Solun:- Given equation is:-

........(1)

Separate variables of eq. 1:-

Integrating both sides:-

We know that:-

Given y = -1 and x = 1 because curve passing through (1,-1)

⇒ -1 - 2log |-1+2| = 1 + 2log |1| + C

⇒ -1 - 2log |1| = 1 + C

⇒ -1 = 1 + C

⇒ C = -2

⇒ y - 2log |y + 2| = x + 2log |x| - 2

⇒ y - log |(y + 2)2| = x + log |x2| - 2

⇒ y - x = log |(y + 2)2| + log |x2| - 2

⇒ y - x + 2 = log |x2(y + 2)2|

17. Find the equation of a curve passing through the point (0,-2) given that at any point (x,y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Solun:- Given:- y(slope of its tangent) = x

........(1)

Separate variables of eq. 1:-

Integrating both sides:-

We know that:-

Given x = 0 and y = -2 because curve passing through (0,-2)

⇒ 2 = 0 + C

⇒ C = 2

⇒ y2 - x2 = 4

18. At any point (x,y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4,-3). Find the equation of the curve given that it passes through (-2,1).

Solun:- Slope of the line segment joining the point of contact of the curve and the point (-4,-3) is:-

We know that:-

Slope =

Given:- slope of its tangent = 2(Slope of joining line)

........(1)

Separate variables of eq. 1:-

Integrating both sides:-

We know that:-

Given x = -2 and y = 1 because curve passing through (-2,1)

⇒ log |4| = 2log |2| + C

⇒ log |4| = log |22| + C

⇒ log |4| = log |4| + C

⇒ C = 0

⇒ log |y+3| = log |(x+4)2| + 0

⇒ log |y+3| - log |(x+4)2| = 0

⇒ y + 3 = (x + 4)2

⇒ y = (x + 4)2 - 3

19. The volume of spherical balloon being inflated changes at a constant rate. If initially, its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Solun:- Given that volume of the spherical balloon being inflated changes at a constant rate.

We know that volume of a sphere is (4/3)πr3.

........(1)

Separate variables of eq. 1:-

Integrating both sides:-

We know that:-

……(2)

Given t = 0 and r = 3 units

⇒ (4π)x(9) = k(0) + C

⇒ C = 36π……….(3)

Given t = 3 and r = 6 units

⇒ 288π = k(3) + C

⇒ 288π = 3k + C

From eq. 3:-

⇒ 288π = 3k + 36π

⇒ 252π = 3k

⇒ k = 84π

From eq. 2:-

Put the value of k and C:-

⇒ r3 = 3(21t + 9)

⇒ r3 = 9(7t + 3)

⇒ r = [9(7t + 3)]1/3

20. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 doubles itself in 10 years (loge2 = 0.6931).

Solun:- Let p,t, and r represent principal, time, and rate.

........(1)

Separate variables of eq. 1:-

Integrating both sides:-

We know that:-

⇒ log |p| = rt/100 + C……(2)

Given t = 0 and p = 100 Rs

⇒ log |100| = r(0)/100 + C

⇒ C = log |100|……….(3)

Given t = 10 and p = 200 Rs

⇒ log |200| = r(10)/100 + C

⇒ log |200| = r/10 + C

From eq. 3:-

⇒ log |200| = r/10 + log |100|

⇒ log |200| - log |100| = r/10

⇒ log |2| = r/10

⇒ r = 10log |2|

Given log |2| = 0.6931

⇒ r = 10x0.6931

⇒ r = 6.93%

21. In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648).

Solun:- Let p, and t represent principal and time.

........(1)

Separate variables of eq. 1:-

Integrating both sides:-

We know that:-

⇒ log |p| = 5t/100 + C……(2)

Given t = 0 and p = 1000 Rs

⇒ log |1000| = 5(0)/100 + C

⇒ C = log |1000|……….(3)

Given t = 10, after 10 years

⇒ log |p| = 5(10)/100 + C

⇒ log |p| = 1/2 + C

From eq. 3:-

⇒ log |p| = 1/2 + log |1000|

⇒ log |p| - log |1000| = 0.5

⇒ p/1000 = e0.5

⇒ p = 1000e0.5

Given e0.5 = 1.648

⇒ p = 1000x1.648

⇒ p = 1648.

22. In culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Solun:- Let n, and t represent the count of bacteria and time.

........(1)

Separate variables of eq. 1:-

Integrating both sides:-

We know that:-

⇒ log |n| = kt + C……(2)

Given t = 0 and let n = n0 bacteria

⇒ log |n0| =k(0) + C

⇒ C = log |n0|……….(3)

Given t = 2 hour, after 2 hours

From eq. 3:-

From eq. 2 and eq. 3:-

n0 = 1,00,000 at t = 0

At t = t and n = 2,00,000

23. The general solution of the differential equation

Solun:- Given equation is
........(1)
Separate variables of eq. 1:-
⇒ e-y.dy = ex.dx
Integrating both sides:-
We know that:-
The correct answer is A.

Download PDF of Exercise 9.4

See Also:-
Notes of Differential Equations
Exercise 9.1
Exercise 9.2

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Differential Equations(Ex. 9.2)

2020-10-31T18:52:00.000+05:30

Exercise 9.2

In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) are a solution of the corresponding differential equation:

1. y = ex + 1 : y’’ - y’ = 0

Solun:- Given eq. is:- y = ex + 1

Differentiate w.r.t. to x:-

⇒ y’ = ex……....(1)

Again differentiate w.r.t. to x:-

⇒ y’’ = ex……....(2)

Eq. 2 - Eq. 1:-

⇒ y’’ - y’ = 0 this is the given differential equation.

Hence Proved……

2. y = x2 + 2x + C : y’ - 2x - 2 = 0

Solun:- Given eq. is:- y = x2 + 2x + C

We know that:-

Order of differential equation = No. of arbitrary constants

Differentiate w.r.t. to x:-

⇒ y’ = 2x + 2

⇒ y’ - 2x - 2 = 0 this is the given differential equation.

Hence Proved……

3. y = cos x + C : y’ + sin x = 0

Solun:- Given eq. is:- y = cos x + C

We know that:-

Order of differential equation = No. of arbitrary constants

Differentiate w.r.t. to x:-

⇒ y’ = - sin x

⇒ y’ + sin x = 0 this is the given differential equation.

Hence Proved……

:

Solun:- Given eq. is:-

Differentiate w.r.t. to x:-

From given equation:-

this is the given differential equation.

Hence Proved……

5. y = Ax : xy’ = y (x ≠ 0)

Solun:- Given eq. is:- y = Ax

We know that:-

Order of differential equation = No. of arbitrary constants

Differentiate w.r.t. to x:-

⇒ y’ = A

From given eq.:-

⇒ A = y/x

⇒ y’ = y/x, x ≠ 0

⇒ xy’ = y, (x ≠ 0) this is the given differential equation.

Hence Proved……

6. y = x.sin x : (x ≠ 0 and x > y or x < -y)

Solun:- Given eq. is:- y = x.sinx

Differentiate w.r.t. to x:-

⇒ y’ = x.cos x + sin x

From given eq.:-

⇒ sin x = y/x and cos x = Here x ≠ 0 and x > y or x < -y

, (x ≠ 0 and x > y or x < -y)

This is the given differential equation.

Hence Proved……

7. xy = log y + C : (xy ≠ 1)

Solun:- Given eq. is:- xy = log y + C

Differentiate w.r.t. to x:-

⇒ xyy’ + y2 = y’

⇒ y2 = y’ - xyy’

⇒ y2 = y’(1 - xy)

, (xy ≠ 1)

This is the given differential equation.

Hence Proved……

8. y - cos y = x : (ysin y + cos y + x).y’ = y

Solun:- Given eq. is:- y - cos y = x

Differentiate w.r.t. to x:-

⇒ y’ + (sin y).y’ = 1

Given differential equation is: (ysin y + cos y + x).y’ = y

Taking L.H.S.

= (ysin y + cos y + x).y’

From given eq.:-

⇒ y = x + cos y

= (ysin y + y).y’

Put the value of y’:-

= R.H.S.

Hence Proved……

9. x + y = tan-1y : y2y’ + y2 + 1 = 0

Solun:- Given eq. is:- x + y = tan-1y

Differentiate w.r.t. to x:-

⇒ (1 + y’).(1 + y2) = y’

⇒ 1 + y2 + y’ + y’y2 = y’

⇒ y2y’ + y2 + 1 = 0 this is the given differential equation

Hence Proved……

10. , x ∈ (-a, a) : (y ≠ 0)

Solun:- Given eq. is:-

Differentiate w.r.t. to x:-

From given eq.:-

this is the given differential equation

Hence Proved……

11. The number of arbitrary constants in the general solution of a differential equation of fourth order are:

(A) 0 (B) 2 (C) 3 (D) 4

Solun:- Given the order of differential equation is 4

We know that:-

Order of differential equation = No. of arbitrary constants

Then no. of arbitrary constants are also 4.

The correct answer is D.

12. The number of arbitrary constants in the particular solution of a differential equation of third order are:

(A) 3 (B) 2 (C) 1 (D) 0

Solun:- Given the order of differential equation is 3

But according to the definition of a particular solution:- no. of arbitrary constants are zero because a particular solution is obtained by putting the values to the arbitrary constant in the general solution.

The correct answer is D.

Download PDF of Exercise 9.2

See Also:-

Notes of Differential Equations

Exercise 9.1

Differential Equations(Ex. 9.1)

2020-10-29T18:19:00.000+05:30

Exercise 9.1

Determine order and degree (if defined) of differential equations given in Exercises 1 to 10.

Solun:- Given eq. is:-

Order of the given differential equation is 4.

But the degree of the given differential equation is not defined because the given equation is not a polynomial equation of derivatives.

2. y’ + 5y = 0

Solun:- Given eq. is:- y’ + 5y = 0

Order of the given differential equation is 1.

Degree of the given differential equation is also 1 because the degree of the differential equation is the power of the highest order derivative.

Solun:- Given eq. is:-

Order of the given differential equation is 2.

Degree of the given differential equation is 1 because the degree of the differential equation is the power of the highest order derivative.

Solun:- Given eq. is:-

Order of the given differential equation is 2.

But the degree of the given differential equation is not defined because the given equation is not a polynomial equation of derivatives.

Solun:- Given eq. is:-

Order of the given differential equation is 2.

Degree of the given differential equation is 1 because the degree of the differential equation is the power of the highest order derivative.

Solun:- Given eq. is:-

Order of the given differential equation is 3.

Degree of the given differential equation is 2 because the degree of the differential equation is the power of the highest order derivative.

Solun:- Given eq. is:-

Order of the given differential equation is 3.

Degree of the given differential equation is 1 because the degree of the differential equation is the power of the highest order derivative.

Solun:- Given eq. is:-

Order of the given differential equation is 1.

Degree of the given differential equation is 1 because the degree of the differential equation is the power of the highest order derivative.

Solun:- Given eq. is:-

Order of the given differential equation is 2.

Degree of the given differential equation is 1 because the degree of the differential equation is the power of the highest order derivative.

Solun:- Given eq. is:-

Order of the given differential equation is 2.

Degree of the given differential equation is 1 because the degree of the differential equation is the power of the highest order derivative.

11. The degree of the differential equation

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

Solun:- Given eq. is:-

Degree of the given differential equation is not defined because the given equation is not a polynomial equation of derivatives.

The correct answer is D.

12. The order of the differential equation

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

Solun:- Given eq. is:-

Order of the given differential equation is 2.

The correct answer is A.

Download PDF of Exercise 9.1

See Also:-

Notes of Differential Equations

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Differential Equations

2020-10-27T18:42:00.000+05:30

Definitions and Formulas

Dependent variable:- Dependent variable is a variable whose value depends on another variable (Generally it is denoted by y).

Independent variable:- Independent variable is a variable whose value does not depend on another variable (Generally it is denoted by x).

For example:- x + y = 7

In this equation, y is dependent and x is the independent variable.

Differential Equation:- Differential equation is an equation that contains dependent, independent variables and derivative of the dependent variable with respect to the independent variable x. For example:-

In general, an equation involving derivatives of the dependent variable with respect to the independent variable x is called a differential equation.

NOTE:- Notations for derivatives:

Order of differential equation:- The order of the differential equation is the order of the highest derivative present in the given equation.

Order of differential equation is the order of the highest order derivative of the dependent variable with respect to the independent variable involved in the given differential equation. For example:-

Order of differential eq. 1,2 and 3 respectively are 1, 2, 3.

Degree of a differential equation:- For the study of the degree of a differential equation it is necessary that the equation must be polynomial of derivatives then the degree of that differential equation is the power of highest order derivative.

For example:- Degree of the eq. 1, 2 and 3 respectively are 1, 1, 1.

Order of differential eq. 4 is 1 but the degree of this differential eq. is not defined because eq. 4 is not a polynomial of derivatives.

General solutions of a differential equation:- The solution which contains arbitrary constants is called the general(primitive) solution of the differential equation.

Particular solutions of a differential equation:- The solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution of the differential equation.

Formation of differential equation whose general equation is given:-

The required differential equation is obtained by eliminating constants from equations.

NOTE:- The order of differential equation is the same as the number of arbitrary constants present in the given equation.

Methods of Solving First Order, First Degree Differential Equations

(1) Differential equations with variable separable:-

(i) Separate both independent and dependent variables.

(ii) Integrate both sides then calculating the general solution of the given differential equation.

Example:- Find the general solution of the differential equation , (y≠2).

Solun:- Given

Separating the variables:-

Integrating both sides:-

x2 + y2 + 2x - 4y + c = 0

(2) Homogeneous differential equations:-

If x and y are replaced by λx and λy respectively and then F(λx, λy) = λnF(x, y) then the given equation is called a homogeneous equation if it is a homogeneous function of degree zero where λ is a non-zero constant.

Solving Homogeneous Differential Equation:- If the given equation is a homogeneous equation then

Type-1: ......(1)

(i) Substitute y = vx and then differentiate with respect to x

..........(2)

(ii) Replace dy/dx in given equation 1 with eq. 2

(iii) Separate variables and then integrating both sides and put the value of v in integrating result

(iv) Calculate the general solution of the given equation

Type-2: ......(1)

(i) Substitute x = vy and then differentiate with respect to y

..........(2)

(ii) Replace dx/dx in given equation 1 with eq. 2

(iii) Separate variables and then integrating both sides and put the value of v in integrating result

(iv) Calculate the general solution of the given equation

Example:- Show that the differential equation is a homogeneous equation and solve it.

Solun:- Given

.....(1)

Let F(x,y)=

Replace F(x,y) with F(λx,λy):-

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

.......(2)

From eq. 1 and 2:-

Separating the variables:-

Integrating both sides:-

We know that:-

Numerator = A.d(Denominator)/dx+B

⇒ 1-v = A.d(1+v+v2)/dv + B

⇒ 1-v = A(2v+1) + B

⇒ 1-v = 2Av + A + B

Compare both sides:-

⇒ 2A = -1

⇒ A = -1/2

⇒ A + B = 1

⇒ -1/2 + B = 1

⇒ B = 3/2

Then 1-v = -1/2(2v+1)+3/2

We know that:-

Put the value of v:-

Here -C = c both are constants.

(3) Linear differential equations:-

A differential of the form

Type-1: ......(1)

(i) In eq. 1, P and Q are constants or functions of x only.

(ii) Find the Integrating factor(I.F.) =

(iii) Then the solution of the given differential equation is:

Type-2: ......(2)

(i) In eq. 2, P and Q are constants or functions of y only.

(ii) Find the Integrating factor(I.F.) =

(iii) Then the solution of the given differential equation is:

Example:- Find the general solution of the differential equation .

Solun:- Given

.....(1)

Given equation is a linear differential equation of type

......(2) then

Compare equation 1 and 2:-

P = -1 and Q = cos x

I.F. =

The solution of equation 1 is:-

......(3)

Let

According to ILATE:- u = cos x and v = e-x

We know that:-

From eq. A:-

Put the value of I in equation 3:-

Download PDF notes of Differential Equations

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Application of Integrals(Ex. 8.1)

2020-10-24T19:01:00.000+05:30

Exercise 8.1

1. Find the area of the region bounded by the curve y2 = x and the line x = 1, x = 4, and the x-axis in the first quadrant.

Solun:- Given eq. is:- y2 = x

⇒ y = √x

Area of given curve from x = 1 to x = 4 is:-

We know that:-

2. Find the area of the region bounded by y2 = 9x, x = 2, x = 4, and the x-axis in the first quadrant.

Solun:- Given eq. is:- y2 = 9x

⇒ y = √9x = 3√x

Area of given curve from x = 2 to x = 4 is:-

We know that:-

3. Find the area of the region bounded by x2 = 4y, y = 2, y = 4, and the y-axis in the first quadrant.

Solun:- Given eq. is:- x2 = 4y

⇒ x = √4y = 2√y

Area of the given curve from y = 2 to y = 4 is:-

We know that:-

4. Find the area of the region bounded by the ellipse .

Solun:- Given eq. is:-

Area of given curve from x = 0 to x = 4 is:-

Area of given ellipse = 4(Area of region AOBA)

Area of region AOBA:-

We know that:-

Then Area of given ellipse = 4(3π) = 12π.

5. Find the area of the region bounded by the ellipse .

Solun:- Given eq. is:-

Area of given curve from x = 0 to x = 2 is:-

Area of given ellipse = 4(Area of region AOBA)

Area of region AOBA:-

We know that:-

Then Area of given ellipse = 4(3π/2) = 6π.

6. Find the area of the region in the first quadrant enclosed by x-axis, line x = (√3)y, and the circle x2 + y2 = 4.

Solun:- Given eq. is:-

x2 + y2 = 4…....(1)

x = (√3)y…....(2)

Solving eq. 1 and 2:-

⇒ 3y2 + y2 = 4

⇒ 4y2 = 4

⇒ y2 = 1

⇒ y = 1,-1 and x = √3,-√3

Area of given curve from x = 0 to x = 2 is:-

A = (Area of region from x = 0 to x = √3) +

(Area of region from x = √3 to x = 2)

We know that:-

7. Find the area of smaller part of circle x2 + y2 = a2 cut off by line x = a/√2.

Solun:- Given eq. is:-

x2 + y2 = a2…....(1)

x = a/√2

Area of given curve from x = a/√2 to x = a is:-

A = 2(Area of region ADCA)

Area of region ADCA:-

We know that:-

8. The area between x = y2 and x = 4 is divided into two equal parts by the line x = a, find the value of a.

Solun:- Given eq. is:-

y = √x…....(1)

Given:-

Area of region AEOCA = Area of region ACDBFEA

2(Area of region ACOA) = 2(Area of region ABDCA)

Area of region ACOA = Area of region ABDCA…...(1)

Area of region ACOA:-

We know that:-

Area of region ABDCA:-

We know that:-

From eq. 1:-

9. Find the area of the region bounded by the parabola y = x2 and y = |x|.

Solun:- Given eq. is:-

y = x2…....(1)

y = |x|

If x > 0 then y = x…....(2)

If x < 0 then y = -x…....(3)

From eq. 1 and 2:-

⇒ x = x2

⇒ x2 - x = 0

⇒ x(x - 1) = 0

⇒ x = 0 and y = 0 and

⇒ x = 1 and y = 1

From eq. 1 and 3:-

⇒ x = 0 and y = 0 and

⇒ x = -1 and y = 1

Area of bounded region between (y = x2) and (y = |x|) = 2(Area of shaded region)....(1)

Area of region ADOA = (area of ΔACO) - area of region ADOCA…....(2)

Area of ΔACO = 1/2

Area of region ADOCA:-

We know that:-

From eq. 2:-

Area of region ADOA = 1/2 - 1/3 = 1/6

From eq. 1:-

Area of bounded region between (y = x2) and (y = |x|) = 2(1/6)

Area of bounded region between (y = x2) and (y = |x|) = 1/3

10. Find the area bounded by the curve x2 = 4y and the line x = 4y - 2.

Solun:- Given eq. is:-

x2 = 4y…....(1)

x = 4y - 2…....(2)

From eq. 1 and 2:-

⇒ x = -1 and y = 1/4 and

⇒ x = 2 and y = 1

Area of shaded region = area(AEOA)+area(EBOE)...(1)

Area of region AEOA = area(ACOEA)-area(ACOA)

Area of AEOA:-

We know that:-

Area of EBOE:-

We know that:-

From eq. 1:-

Area of shaded region = 5/6 + 7/24 = 9/8

11. Find the area of the region bounded by the curve y2 = 4x and the line x = 3.

Solun:- Given eq. is:-

y2 = 4x

Area of region ABOA = 2(Area of region ACOA)....(1)

Area of region ACOA:-

We know that:-

From eq. 1:-

Area of ABOA = 2(4√3) = 8√3.

Choose the correct answer in the following Exercises 12 and 13.

12. Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x = 2 is

(A) π (B) π/2 (C) π/3 (D) π/4

Solun:- Given eq. is:- x2 + y2 = 4

Area of the shaded region:-

We know that:-

The correct answer is A.

13. Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3 is

(A) 2 (B) 9/4 (C) 9/3 (D) 9/2

Solun:- Given eq. is:- y2 = 4x

Area of the shaded region:-

We know that:-

The correct answer is B.

Download PDF of Exercise 8.1

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Notes of Application of Integrals

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Application of Integrals

2020-10-22T18:16:00.000+05:30

Definitions and Formulas

Application of Integrals:- Application of Integrals is finding the area of curves.

Area Under Simple Curves:-

(1) The area A of the region bounded by the curve y = f(x), the x-axis and line x = a, x = b is given by:-

But Below x-axis:-

(2) The area A of the region bounded by the curve x = g(y), the y-axis and line y = c, y = d is given by:-

But in -y-axis:-

Area between two curves:-

If y = f(x) and y = g(x)

where f(x) > g(x) in [a,b]

then area between curves f(x) and g(x)

Or A = [area bounded by y = f(x), x-axis and line x = a, x = b]

- [area bounded by y = g(x), x-axis and line x = a, x = b]

where f(x) > g(x)

Example:- Find the area enclosed by the circle x2 + y2 = a2.

Solun:- Area of given circle is:- 4(area of the region AOBA)........(1)

Area of the region AOBA:-

We know that:-

Area of given circle is = 4(area of the region AOBA) = 4(πa2/4)

Area of the given circle is = πa2.

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Integrals(Miscellaneous Ex.)

2020-10-20T17:59:00.001+05:30

Miscellaneous Exercise

Integrate the functions in Exercises 1 to 24.

Solun:- Let f(x) =

Integrate f(x):-

By Partial Fraction:-

⇒ 1 = A(1 - x2) + B(x)(1 + x) + C(x)(1 - x)......(1)

Put x = 0 in eq. 1:-

⇒ 1 = A(1)

⇒ A = 1

Put x = 1 in eq. 1:-

⇒ 1 = B(1)(2)

⇒ B = 1/2

Put x = - 1 in eq. 1:-

⇒ 1 = C(-1)(2)

⇒ C = - 1/2

We know that:-

⇒ log m - log n = log (m/n)

⇒ log mn = n.log m

⇒ log m + log n = log (mn)

Solun:- Let f(x) =

Integrate f(x):-

We know that:-

Solun:- Let f(x) =

Integrate f(x):-

Let x = a/t

Differentiate w.r.t. to t:-

⇒ dx = - a/t2.dt

We know that:-

Put the value of t:-

Solun:- Let f(x) =

Integrate f(x):-

Let 1 + 1/x4 = t4

Differentiate w.r.t. to t:-

⇒ (0 + (-4/x5)).dx = 4t3.dt

⇒ 1/x5.dx = (-4t3/4).dt

⇒ 1/x5.dx = -t3.dt

Put the value of t:-

Solun:- Let f(x) =

Integrate f(x):-

Let x = t6

Differentiate w.r.t. to t:-

⇒ dx = 6t5.dt

On Dividing:-

Put the value of t:-

Solun:- Let f(x) =

Integrate f(x):-

By Partial Fraction:-

⇒ 5x = A(x2 + 9) + (Bx + C)(x + 1).....(1)

Put x = -1 in eq. 1:-

⇒ -5 = A(10)

⇒ A = -1/2

Put x = 0 in eq. 1:-

⇒ 0 = A(9) + C(1)

⇒ 0 =(-1/2)(9) + C

⇒ 0 =(-9/2) + C

⇒ C = 9/2

Put x = 1 in eq. 1:-

⇒ 5 = A(10) + (B + C)(2)

⇒ 5 = (-1/2)(10) + (B + 9/2)(2)

⇒ 5 = -5 + 2B + 9

⇒ 1 = 2B

⇒ B = 1/2

We know that:-

Solun:- Let f(x) =

Integrate f(x):-

Let x - a = t

Differentiate w.r.t. to t:-

⇒ dx = dt

We know that:-

Put the value of t:-

Solun:- Let f(x) =

Integrate f(x):-

We know that:-

Solun:- Let f(x) =

Integrate f(x):-

Let sin x = t

Differentiate w.r.t. to t:-

⇒ (cos x).dx = dt

We know that:-

Put the value of t:-

Solun:- Let f(x) =

Integrate f(x):-

We know that:-

⇒ cos2x - sin2x = cos 2x

We know that:-

Solun:- Let f(x) =

Integrate f(x):-

We know that:-

Solun:- Let f(x) =

Integrate f(x):-

Let x4 = t

Differentiate w.r.t. to t:-

⇒ 4x3.dx = dt

⇒ x3.dx = (1/4).dt

We know that:-

Put the value of t:-

Solun:- Let f(x) =

Integrate f(x):-

Let ex = t

Differentiate w.r.t. to t:-

⇒ ex.dx = dt

By Partial Fraction:-

⇒ 1 = A(2 + t) + B(1 + t)....(1)

Put t = - 1 in eq. 1:-

⇒ 1 = A(1)

⇒ A = 1

Put t = - 2 in eq. 1:-

⇒ 1 = B(-1)

⇒ B = - 1

We know that:-

Put the value of t:-

Solun:- Let f(x) =

Integrate f(x):-

Put x2 = t

By Partial Fraction:-

⇒ 1 = A(t + 4) + B(t + 1)....(1)

Put t = - 1 in eq. 1:-

⇒ 1 = A(3)

⇒ A = 1/3

Put t = - 4 in eq. 1:-

⇒ 1 = B(-3)

⇒ B = - 1/3

We know that:-

Solun:- Let f(x) = cos3x.elog sinx

Integrate f(x):-

Let cos x = t

Differentiate w.r.t. to t:-

⇒ - sin x.dx = dt

⇒ sin x.dx = - dt

We know that:-

Put the value of t:-

Solun:- Let f(x) =

Integrate f(x):-

Let x4 + 1 = t

Differentiate w.r.t. to t:-

⇒ 4x3.dx = dt

⇒ x3.dx = (1/4).dt

We know that:-

Put the value of t:-

Solun:- Let f(x) =

Integrate f(x):-

Let f(ax + b) = t

Differentiate w.r.t. to t:-

⇒ a.f’(ax + b).dx = dt

⇒ f’(ax + b).dx = (1/a).dt

We know that:-

Put the value of t:-

Solun:- Let f(x) =

Integrate f(x):-

Let cos α + cot x.sin α = t

Differentiate w.r.t. to t:-

⇒ - sin α.cosec2x.dx = dt

⇒ cosec2x.dx = (-1/sin α).dt

We know that:-

Put the value of t:-

Solun:- Let f(x) =

Integrate f(x):-

We know that:-

sin-1x + cos-1x = π/2

Let sin-1√x = t

⇒ √x = sin t

⇒ x = sin2 t

Differentiate w.r.t. to t:-

⇒ dx = 2sint.cost.dt

⇒ dx = sin 2t.dt

We know that:-

Put the value of t:-

⇒ cos t = √(1-x)

⇒ sin 2t = 2√(x-x2)

⇒ cos 2t = 1 - 2x

Solun:- Let f(x) =

Integrate f(x):-

Let x = cos2y

Differentiate w.r.t. to y:-

⇒ dx = 2cosy.(-siny).dy

⇒ dx = - sin 2y.dy

We know that:-

⇒ cos 2y = 2cos2y - 1

⇒ cos 2y = 1 - 2sin2y

We know that:-

Put the value of y:-

⇒ cos y = √x

⇒ sin y = √(1-x)

⇒ sin 2y = 2√(x - x2)

Solun:- Let f(x) =

Integrate f(x):-

We know that:-

Let f(x) = tan x

⇒ f’(x) = sec2x

Solun:- Let f(x) =

Integrate f(x):-

By Partial Fraction:-

⇒ x2+x+1 = A(x + 1)(x + 2) + B(x + 2) + C(x + 1)2......(1)

Put x = -1 in eq. 1:-

⇒ 1 = B(1)

⇒ B = 1

Put x = -2 in eq. 1:-

⇒ 3 = C(1)

⇒ C = 3

Put x = 0 in eq. 1:-

⇒ 1 = 2A + 2B + C

⇒ 1 = 2A + 2 + 3

⇒ 2A = - 4

⇒ A = -2

We know that:-

Solun:- Let f(x) =

Integrate f(x):-

Let x = cos y

Differentiate w.r.t. to y:-

⇒ dx = -siny.dy

We know that:-

According to ILATE:- u = y and v = sin y

We know that:-

Put the value of y:-

⇒ cos y = x

⇒ sin y = √(1-x2)

Solun:- Let f(x) =

Integrate f(x):-

Let 1+1/x2 = t

Differentiate w.r.t. to t:-

⇒ -2/x3.dx = dt

⇒ 1/x3.dx = (-1/2).dt

We know that:-

According to ILATE:- u = log t and v = √t

Put the value of t:-

Evaluate the definite integrals in Exercises 25 to 33.

Solun:- Let f(x) =

Integrate f(x):-

We know that:-

Let f(x) = - cot (x/2)

Then

Solun:- Let f(x) =

Integrate f(x):-

Let tan2x = t…....(1)

Differentiate w.r.t. to t:-

⇒ 2tan x.sec2x.dx = dt

⇒ tan x.sec2x.dx = (1/2).dt

Limits Change:-

Put x = 0 in eq. 1:-

⇒ t = 0

Put x = π/4 in eq. 1:-

⇒ t = 1

We know that:-

Solun:- Let f(x) =

Integrate f(x):-

Let tan x = t…....(1)

Differentiate w.r.t. to t:-

⇒ sec2x.dx = dt

Limits change:-

Put x = 0 in eq. 1:-

⇒ t = 0

Put x = π/2 in eq. 1:-

⇒ t = ∞

We know that:-

Solun:- Let f(x) =

Integrate f(x):-

Let sin x - cos x = t…....(1)

Differentiate w.r.t. to t:-

⇒ (cos x + sin x).dx = dt

Limits change:-

Put x = π/6 in eq. 1:-

⇒ t = (1 - √3)/2

Put x = π/3 in eq. 1:-

⇒ t = (√3 - 1)/2

We know that:-

sin-1(-x) = -sin-1(x)

Solun:- Let f(x) =

Integrate f(x):-

We know that:-

Solun:- Let f(x) =

Integrate f(x):-

Let sin x - cos x = t…....(1)

Differentiate w.r.t. to t:-

⇒ (cos x + sin x).dx = dt

Limits change:-

Put x = 0 in eq. 1:-

⇒ t = - 1

Put x = π/4 in eq. 1:-

⇒ t = 0

We know that:-

Solun:- Let f(x) = sin 2x.tan-1(sin x)

Integrate f(x):-

Let sin x = t…....(1)

Differentiate w.r.t. to t:-

⇒ cos x.dx = dt

Limits change:-

Put x = 0 in eq. 1:-

⇒ t = 0

Put x = π/2 in eq. 1:-

⇒ t = 1

We know that:-

According to ILATE:- u = tan-1t and v = t

We know that:-

tan-1(-x) = tan-1x

Solun:- Let f(x) =

Integrate f(x):-

By Property:-

Add eq. 1 and 2:-

We know that:-

Solun:- Let f(x) = [|x - 1| + |x - 2| + |x - 3|]

Integrate f(x):-

Calculating I1:-

Calculating I2:-

Calculating I3:-

Put the value of I1, I2, and I3 in eq. 1:-

Prove the following (Exercises 34 to 39)

Solun:- Let f(x) =

Integrate f(x):-

By Partial Fraction:-

⇒ 1 = A(x)(x + 1) + B(x + 1) + C(x2)......(1)

Put x = 0 in eq. 1:-

⇒ 1 = B(1)

⇒ B = 1

Put x = -1 in eq. 1:-

⇒ 1 = C(1)

⇒ C = 1

Put x = 1 in eq. 1:-

⇒ 1 = A(1)(2) + B(2) + C(1)

⇒ 1 = A(1)(2) + 1(2) + 1(1)

⇒ 1 - 3 = A(1)(2)

⇒ A = - 1

We know that:-

Hence Proved...

Solun:- Let f(x) = x.ex

Integrate f(x):-

We know that:-

According to ILATE:- u = x and v = ex

Hence Proved…....

Solun:- Let f(x) = x17.cos4x

Integrate f(x):-

By Property:-

f(-x) = (-x)17.[cos (-x)]4 = - x17.cos4x

f(-x) = - f(x)

Hence Proved….

Solun:- Let f(x) = sin3x

Integrate f(x):-

We know that:-

sin 3x = 3sinx - 4sin3x

We know that:-

Hence Proved…

Solun:- Let f(x) = 2tan3x

Integrate f(x):-

Let tan x = t…....(1)

Differentiate w.r.t. to t:-

⇒ sec2x.dx = dt

Limits Change:-

Put x = 0 in eq. 1:-

⇒ t = 0

Put x = π/4 in eq. 1:-

⇒ t = 1

We know that:-

Hence Proved…

Solun:- Let f(x) = sin-1x

Integrate f(x):-

Let sin-1x = t and x = sin t

Differentiate w.r.t. to t:-

⇒ dx = cos t.dt

Limits Change:-

Put x = 0 in eq. 1:-

⇒ t = 0

Put x = 1 in eq. 1:-

⇒ t = π/2

We know that:-

According to ILATE:- u = t and v = cos t

We know that:-

Hence Proved…

Choose the correct answer in Exercises 41 to 44.

is equal to

Solun:- Let f(x) =

Integrate f(x):-

Let ex = t

Differentiate w.r.t. to t:-

⇒ ex.dx = dt

We know that:-

Put the value of t:-

The correct answer is A.

is equal to

Solun:- Let f(x) =

Integrate f(x):-

We know that:-

cos 2x = cos2x - sin2x

Let sin x + cos x = t

Differentiate w.r.t. to t:-

⇒ (cos x - sin x).dx = dt

We know that:-

Put the value of t:-

The correct answer is B.

43. If f(a+b-x) = f(x), then is equal to

Solun:- Given

By Property:-

Given f(a+b-x) = f(x)

Add eq. 1 and 2:-

The correct answer is D.

44. The value of is

Solun:- Given

We know that:-

By Property:-

Add eq. 1 and 2:-

The correct answer is B.

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