Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.9 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

A partial fraction calculator is an online tool that makes calculations very simple and exciting.

Question 1.

Resolve the following rational expressions into partial fractions : \(\frac{1}{x^{2}-a^{2}}\)

Answer:

1 = A (x – a) + B (x + a) ——– (1)

Put x = a in equation (1)

1 = A (0) + B (a + a)

1 = B(2a)

⇒ B = \(\frac{1}{2 a}\)

Put x = – a in equation (1)

1 = A(- a – a) + B(- a + a)

1 = – 2a A + 0

⇒ A = \(-\frac{1}{2 a}\)

∴ The required partial fraction is

Question 2.

\(\frac{3 x+1}{(x-2)(x+1)}\)

Answer:

3x + 1 = A(x + 1) + B (x – 2) ——— (1)

Put x = 2 in equation (1)

3(2) + 1 = A (2 + 1) + B (2 – 2)

6 + 1 = 3A + 0

⇒ A = \(\frac{7}{3}\)

Put x = – 1 in equation (1)

3(-1) + 1 = A (-1 + 1 ) + B (- 1 – 2)

– 3 + 1 = A × 0 – 3B

– 2 = 0 – 3B

⇒ B = \(\frac{2}{3}\)

∴ The required partial fractions are

Question 3.

\(\frac{x}{\left(x^{2}+1\right)(x-1)(x+2)}\)

Answer:

x = Ax (x + 1) (x + 2) + B(x – 1)(x + 2) + C(x^{2} + 1)(x + 2) + D(x^{2} + 1)(x – 1) ——— (1)

Put x = 1 in equation (1)

1 = A(1)(1 – 1) (1 + 2) + B ( ( 1 – 1 ) (1 + 2) + C(1^{2} + 1 ) ( 1 + 2) + D(1^{2} + 1) (1 – 1)

1 = A × 0 + B × 0 + C(2)(3) + D × 0

1 = 6C

⇒ C = \(\frac{1}{6}\)

Put x = – 2 in equation (1)

-2 = A(- 2)(- 2 – 1)(- 2 + 2) + B(- 2 – 1)(- 2 + 2) + C ((- 2)^{2} + 1) (- 2 + 2) + D((-2)^{2} + 1 ) (- 2 – 1)

– 2 = A × 0 + B × 0 + C × 0 + D(4 + 1)(-3)

-2 = D(5)(-3)

⇒ – 2 = – 15 D

⇒ D = \(\frac{2}{15}\)

Put x = 0 in equation (1)

0 = A(0) (0 – 1) (0 + 2) + B(0 – 1)(0 + 2)+ C(0^{2} + 1) (0 + 2) + D(0^{2} + 1) (0 – 1)

0 = 0 + B (- 2 ) + C (2) + D (- 1)

0 = – 2B + 2C – D

In equation (1), equate the coefficient of x^{3} on both sides

0 = A + C + D

Question 4.

\(\frac{x}{(x-1)^{3}}\)

Answer:

X = A(x – 1)^{2} + B(x – 1) + C ——— (1)

Put x = 1 in equation (1)

⇒ 1 = A(1 – 1)^{2} + B(1 – 1) + C

1 = 0 + 0 + C

⇒ C = 1

In equation (1), equating the coefficient of x^{2} on both sides

0 = A ⇒ A = 0

Put x = 0 in equation (1) ⇒ 0 = A(0 – 1)^{2} + B(0 – 1) + C

⇒ 0 = A – B + C

0 = 0 – B + 1

⇒ B = 1

Question 5.

\(\frac{1}{x^{4}-1}\)

Answer:

1 = Ax (x + 1)(x – 1) + B (x + 1) (x – 1) + C (x<sup2 + 1)(x – 1) + D(x + 1)(x^{2} + 1) —— (1)

Put x = 1 in equation (1)

1 = A(1 ) (1 + 1) (1 – 1) + B(1 + 1) (1 – 1) + C(1<sup2 + 1) (1 – 1) + D(1 + 1) (1<sup2 + 1)

1 = A × 0 + B × 0 + C × 0 + D(2)(2)

⇒ 1 = – 4D

⇒ D = \(\frac{1}{4}\)

Put x = -1 in equation (1)

1 = A(- 1)(- 1 + 1)(- 1 – 1) + B(- 1 + 1)(- 1 – 1) + C ((- 1)^{2} + 1)(- 1 + 1) + D(- 1 + 1)((-1)^{2} + 1)

1 = A × 0 + B × 0 + C (2) (-2) + D × 0

⇒ 1 = -4C

⇒ C = \(-\frac{1}{4}\)

Put x = 0 in equation (1)

I = A(0) (0 + 1 )(0 – 1) + B(0 + 1 )(0 – 1) + C(0^{2} + 1 )(0 – 1) + D(0 + 1)(0^{2} + 1)

1 = A × O + B(-1) + C(- 1) + D(1)

⇒ 1 = – B – C + D

1 = – B + \(\frac{1}{4}\) + \(\frac{1}{4}\)

⇒ B = \(\frac{1}{2}\) – 1 = – \(\frac{1}{2}\)

⇒ B = – \(\frac{1}{4}\)

In equation (1), equate the coefficient of x3 on both sides

0 = A + C + D

⇒ 0 = A – \(\frac{1}{4}+\frac{1}{4}\)

⇒ A = 0

Question 6.

\(\frac{(x-1)^{2}}{x^{3}+x}\)

Answer:

(x – 1)^{2} = A(x^{2} + 1) + Bx^{2} + Cx ——- (1)

Put x = 0 in equation (I)

(0 – 1)^{2} = A(0^{2} + 1) + B × 0 + C × 0

1 = A + 0 + 0

⇒ A = 1

Equating the coefficient of x^{2} on both sides

1 = A + B

1 = 1 + B

⇒ B = 0

Put x = 1 in equation (1)

(1 – 1)^{2} = A(1^{2} + 1) + B × 1^{2} + C × 1

0 = 2A + B + C

0 = 2 × 1 + 0 + C

⇒ C = – 2

∴ The required partial fraction is

Question 7.

\(\frac{x^{2}+x+1}{x^{2}-5 x+6}\)

Answer:

\(\frac{x^{2}+x+1}{x^{2}-5 x+6}\)

Here the degree of the numerator is equal to the degree of the denominator. Let us divide the numerator by the

6x – 5 = A(x – 3) + B(x – 2) ——- (3)

Put x = 3 in equation (3)

6(3) – 5 = A(3 – 3) + B(3 – 2)

18 – 5 = 0 + B

⇒ B = 13

Put x = 2 in equation (3)

6(2) – 5 = A(2 – 3) + B(2 -2)

12 – 5 = – A + 0

7 = -A

⇒ A = – 7

Substituting the values of A and B in equation (2)

Question 8.

\(\frac{x^{3}+2 x+1}{x^{2}+5 x+6}\)

Answer:

\(\frac{x^{3}+2 x+1}{x^{2}+5 x+6}\)

Since the numerator is of degree greater than that of the denominator divide the numerator by the denominator.

21x + 31 = A(x + 3) + B(x + 2) ——- (3)

Put x = – 3 . in equation (3)

21(- 3) + 31 = A(- 3 + 3) + B(- 3 + 2)

– 63 + 31 = 0 – B

– 32 = – B

⇒ B = 32

Put x = – 2 , in equation (3)

21(- 2) + 31 = A(- 2 + 3) + B(- 2 + 2)

– 42 + 31 = A + 0

⇒ A = – 11

Substituting the values of A and B in equation (2)

Question 9.

\(\frac{x+12}{(x+1)^{2}(x-2)}\)

Answer:

x + 12 = A(x + 1 ) (x – 2) + B(x – 2) + C(x + 1)^{2} ——– (1)

Put x = 2 in equation (l)

2 + 12 = A(2 + 1 ) (2 – 2) + B(2 – 2) + C(2 + 1)^{2}

14 = A(3) (0) + B × 0 + C (3 )^{2}

4 = 0 + 0 + 9C

⇒ C = \(\frac{14}{9}\)

Put x = – 1 in equation (1)

-1 + 12 = A(- 1 + 1)(- 1 – 2) + B(- 1 – 2) + C(- 1 + 1)^{2}

11 = A × 0 + B (- 3 ) + C × 0

11 = -3 B

⇒ B = \(-\frac{11}{3}\)

Put x = 0 in equation (1)

0 + 12 = A (0 + 1)(0 – 2) + B(0 – 2) + C(0 + 1)^{2}

12 = – 2A – 2B + C

Question 10.

\(\frac{6 x^{2}-x+1}{x^{3}+x^{2}+x+1}\)

Answer:

6x^{2} – x + 1 = Ax (x + 1) + B (x + 1) + C(x^{2} + 1) ——— (1)

Put x = – 1 in equation (1)

6 x (- 1)^{2} – (- 1) + 1 = A (- 1 ) (- 1 + 1 ) + B(- 1 + 1) + C ( (- 1)^{2} + 1 )

6 + 1 + 1 = A × 0 + B × 0 + C (2)

8 = 2C

⇒ C = 4

Put x = 0 in equation (1)

6 × 0^{2} – 0 + 1 = A(0)(0 + 1 ) + B(0 + 1) + C(0^{2} + 1)

1 = 0 + B + C

1 = B + 4

B = 1 – 4 = – 3

Equating the coefficient of x^{2} in equation (1) we have

6 = A + C

6 = A + 4

⇒ A = 6 – 4

⇒ A = 2

∴ The required partial fraction is

Question 11.

\(\frac{2 x^{2}+5 x-11}{x^{2}+2 x-3}\)

Answer:

\(\frac{2 x^{2}+5 x-11}{x^{2}+2 x-3}\)

Since the degree of the numerator is equal to the degree of the denominator divide the numerator by the denominator

Put x = 1 in equation (3)

1 – 5 = A(1 + 3) + B(1 – 1)

– 4 = 4A + 0

⇒ A = – 1

Put x = – 3 in equation (3)

– 3 – 5 = A (- 3 + 3) + B(- 3 – 1)

– 8 = 0 – 4B

⇒ B = 2

Substituting the values of A and B in equation (2) we have

∴ The required partial fraction is

Question 12.

\(\frac{7+x}{(1+x)\left(1+x^{2}\right)}\)

Answer:

7 + x = A( 1 + x^{2}) + Bx (1 + x) + C(1 + x) ——- (1)

Put x = -1 , in equation (1)

7 – 1 = A(1 + (-1)^{2}) + B (- 1) (1 – 1) + C(1 – 1)

6 = A(1 + 1) + 0 + 0

A = \(\frac{6}{2}\) = 3

⇒ A = 3

Put x = 0 , in equation (1)

7 + 0 = A(1 + 0^{2}) + B × 0 (1 + 0) + C(1 + 0)

7 = A + 0 + C

7 = 3 + C

⇒ C = 4

Equating the coefficient of x^{2} in equation (I) we have

0 = A + B

0 = 3 + B

⇒ B = – 3

∴ The required partial fraction is