Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(b) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(b)

Resolve the following into partial fractions.

Question 1.

\(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}\)

Solution:

Let \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{A}{x-1}+\frac{B x+C}{x^2+2}\)

Multiplying with (x – 1) (x^{2} + 2)

2x^{2} + 3x + 4 = A(x^{2} + 2) + (Bx + C) (x – 1)

x = 1

⇒ 2 + 3 + 4 = A(1 + 2)

⇒ 9 = 3A

⇒ A = 3

Equating the coefficients of x^{2}

2 = A + B

⇒ B = 2 – A = 2 – 3 = -1

Equating constants

4 = 2A – C

⇒ C = 2A – 4 = 6 – 4 = 2

∴ \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{3}{x-1}+\frac{-x+2}{x^2+2}\)

Question 2.

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

Solution:

Let \(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}=\frac{A}{2+x}+\frac{B x+C}{1-x+x^2}\)

Multiplying with (2 + x) (1 – x + x^{2})

3x – 1 = A(1 – x + x^{2}) (Bx + C) (2 + x)

x = -2

⇒ -7 = A(1 + 2 + 4) = 7A

⇒ A = -1

Equating the coefficients of x^{2}

0 = A + B ⇒ B = -A = 1

Equating the constants

-1 = A + 2C

⇒ 2C = -1 – A = -1 + 1 = 0

⇒ C = 0

∴ \(\frac{3 x-1}{\left(1-x+x^2\right)(2+x)}=-\frac{1}{2+x}+\frac{x}{1-x+x^2}\)

Question 3.

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

Solution:

Let \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{A}{x+2}+\frac{B x+C}{x^2+1}\)

Multiplying with (x + 2) (x^{2} + 1)

x^{2} – 3 = A(x^{2} + 1) + (Bx + C) (x + 2)

x = -2

⇒ 4 – 3 = A(4 + 1)

⇒ 1 = 5A

⇒ A = \(\frac{1}{5}\)

Equating the coefficients of x^{2}

1 = A + B

⇒ B = 1 – A = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)

Equating the constants

-3 = A + 2C

⇒ 2C = -3 – A

⇒ 2C = -3 – \(\frac{1}{5}\)

⇒ 2C = \(-\frac{16}{5}\)

⇒ C = \(-\frac{8}{5}\)

∴ \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{1}{5(x+2)}+\frac{4 x-8}{5\left(x^2+1\right)}\)

Question 4.

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

Solution:

Let \(\frac{x^2+1}{\left(x^2+x+1\right)^2}=\frac{A x+B}{x^2+x+1}+\frac{C x+D}{\left(x^2+x+1\right)^2}\)

Multiplying with (x^{2} + x + 1)^{2}

x^{2} + 1 = (Ax + B) (x^{2} + x + 1) + (Cx + D)

Equating the coefficients of x^{3},

A = 0

Equating the coefficients of x^{2},

A + B = 1 ⇒ B = 1

Equating the coefficients of x,

A + B + C = 0

⇒ 1 + C = 0

⇒ C = -1

Equating the constants,

B + D = 1

⇒ D = 1 – B = 1 – 1 = 0

∴ Ax + B = 1, Cx + D = -x

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

Question 5.

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

Solution:

∴ x^{3} + x^{2} + 1 = A(x – 1) (x^{2} + x + 1) + B(x^{2} + x + 1) + (Cx + D) (x – 1)^{2} …….(2)

Put x = 1 in (2)

1 + 1 + 1 = A(0) + B(1 + 1 + 1) + (C(1) + D) (0)

⇒ 3B = 3

⇒ B = 1

Equating the coefficients of x^{3} in (2)

1 = A + C ………(3)

Equating the coefficients of x^{2} in (2)

1 = A(1 – 1) + B(1) + C(-2) + D(1)

⇒ 1 = B – 2C + D

∵ B = 1,

⇒ 1 = 1 – 2C + D

⇒ 2C = D ………(4)

Put x = 0 in (2)

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

⇒ -A + B + D = 1

⇒ -A + 1 + D = 1

⇒ A = D ………(5)

From (3), (4) and (5)