Students must practice this TS Intermediate Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(d) to find a better approach to solving the problems.

## TS Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(d)

Question 1.

Find the coefficietil of x^{3} in the power series expansion of \(\frac{5 x+6}{(x+2)(1-x)}\) specifying the region in which the expansion is valid.

Solution:

Given rational fraction \(\frac{5 x+6}{(x+2)(1-x)}\)

Let \(\frac{5 x+6}{(x+2)(1-x)}\) = \(\frac{A}{x+2}+\frac{B}{1-x}\)

⇒ A (1 – x) + B (x + 2) = 5x + 6 …………..(1)

Substituting x = 1 in (1), we get

3B = 11

⇒ B = \(\frac{11}{3}\)

Substituting x = – 2 in (1), we get

3A = – 4

⇒ A = \(\frac{-4}{3}\)

Question 2.

Find the coefficient of x^{4} in the power series expansion of \(\frac{3 x^2+2 x}{\left(x^2+2\right)(x-3)}\) specifying the Interval in which the expansion is valid.

Solution:

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

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

Substituting x = 3 ¡n (1), we get

11A = 33

⇒ A = 3

Equating coefficient of x^{2} on both sides in (1), we get

A + B = 3

⇒ 3 + B = 3

⇒ B = 0

Substituting x = 0 in (1), we get

2A – 3C = 0

⇒ 3C = 2A

⇒ C = 2

The above expansions are valid for |x| < √2.

Question 3.

Find the coefficient of x^{n} in the power series expansion of \(\frac{x-4}{x^2-5 x+6}\) specifying the region in which the expansion is valid.

Solution:

We know that

\(\frac{x-4}{\left(x^2-5 x+6\right)}=\frac{x-4}{(x-2)(x-3)}\)

Let \(\frac{x-4}{x^2-5 x+6}=\frac{A}{(x-2)}+\frac{B}{(x-3)}\)

⇒ A(x – 3) + B(x – 2) = x – 4 …………(1)

Substituting x = 3 in (1), we get B = – 1

substituting x = 2 in (1), we get A = 2

Question 4.

Find the coefficient of x^{n} in the power series expansion of \(\frac{3 x}{(x-1)(x-2)^2}\)

Solution:

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

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

Substituting x = 1 in (1), we get A = 3

Substituting x = 2 in (2), we get C = 6

Equating coefficient of x^{2} in (1) we get

A + B = 0

⇒ B = – A

⇒ B = – 3

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

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

Now

(1 – x)^{-1} = 1 + x + x^{2} + …………. + x^{n} + ……….., if |x| < 1