# Inter 2nd Year Maths 2A Partial Fractions Solutions Ex 7(c)

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)
x2 = (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
2x3 = (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),
x3 = (x – a)(x – b) (x – c) + A(x – b) (x – c) + B(x – a) (x – c) + C(x – a) (x – b)
Put x = a,
a3 = A(a – b) (a – c)
⇒ A = $$\frac{a^3}{(a-b)(a-c)}$$
Put x = b,
b3 = B(b – a) (b – c)
⇒ B = $$\frac{b^3}{(b-a)(b-c)}$$
Put x = c, c3 = C(c – a) (c – b)
⇒ C = $$\frac{c^3}{(c-a)(c-b)}$$
∴ $$\frac{x^3}{(x-a)(x-b)(x-c)}$$ = $$1+\frac{a^3}{(a-b)(a-c)(x-a)}+\frac{b^3}{(b-a)(b-c)(x-b)}$$ + $$\frac{c^3}{(c-a)(c-b)(x-c)}$$