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

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

Resolve the following into partial fractions.

Question 1.

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

Solution:

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

Multiplying with (x – 1) (x – 2)

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

Put x = 1, 1 = A(-1) ⇒ A = -1

Put x = 2, 4 = B(1) ⇒ B = 4

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

Question 2.

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

Solution:

Question 3.

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

Solution:

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

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

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

Put x = \(\frac{1}{2}\),

⇒ 2(\(\frac{1}{8}\)) = 2A(\(\frac{1}{4}\))

⇒ A = \(\frac{1}{2}\)

Put x = 1,

⇒ 2(1) = 2C(1)

⇒ C = 1

Put x = 0,

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

⇒ 2A + 2B – 2C = 1

⇒ 2B = 1 + 2C – 2A

⇒ 2B = 1 + 2 – 1 = 2

⇒ B = 1

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

Question 4.

\(\frac{x^3}{(x-a)(x-b)(x-c)}\)

Solution:

Let \(\frac{x^3}{(x-a)(x-b)(x-c)}\) = \(1+\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}\)

Multiplying with (x – a)(x – b) (x – c),

x^{3} = (x – a)(x – b) (x – c) + A(x – b) (x – c) + B(x – a) (x – c) + C(x – a) (x – b)

Put x = a,

a^{3} = A(a – b) (a – c)

⇒ A = \(\frac{a^3}{(a-b)(a-c)}\)

Put x = b,

b^{3} = B(b – a) (b – c)

⇒ B = \(\frac{b^3}{(b-a)(b-c)}\)

Put x = c, c^{3} = C(c – a) (c – b)

⇒ C = \(\frac{c^3}{(c-a)(c-b)}\)

∴ \(\frac{x^3}{(x-a)(x-b)(x-c)}\) = \(1+\frac{a^3}{(a-b)(a-c)(x-a)}+\frac{b^3}{(b-a)(b-c)(x-b)}\) + \(\frac{c^3}{(c-a)(c-b)(x-c)}\)