Question

The set of real values of x for which $C = 2$is.

• A. $x^{2} + 3x - 7$
• B. $\frac{px + q}{(x - a)(x - b)} = \frac{pa + q}{(x - a)(a - b)} + \frac{pb + q}{(b - a)(x - b)}$
• C. $\frac{x}{(x - 1)(x^{2} + 1)^{2}} = \frac{1}{4}\left\lbrack \frac{1}{(x - 1)} - \frac{x + 1}{x^{2} + 1} \right\rbrack + y$
• D. None of these

Answer 1

Answer: (B)

$\frac{px + q}{(x - a)(x - b)} = \frac{pa + q}{(x - a)(a - b)} + \frac{pb + q}{(b - a)(x - b)}$

Answer 2

$\frac{x}{(x - 1)(x^{2} + 1)^{2}} = \frac{1}{4}\left\lbrack \frac{1}{(x - 1)} - \frac{x + 1}{x^{2} + 1} \right\rbrack + \frac{Ax + B}{(x^{2} + 1)^{2}}4x = (x^{2} + 1)^{2} - (x + 1)(x - 1)(x^{2} + 1) + 4(Ax + B)(x - 1)4A + 2 = 0$

$4B - 4A = 4$ $A = \frac{- 1}{2}B = \frac{1}{2}\therefore$

$y = \frac{Ax + B}{(x^{2} + 1)^{2}} = \frac{1}{2}\frac{(1 - x)}{(x^{2} + 1)^{2}}$ $\frac{5x + 6}{(2 + x)(1 - x)} = \frac{\frac{- 4}{3}}{2 + x} + \frac{\frac{11}{3}}{1 - x}$ $\Rightarrow$ $\frac{- 2}{3}\left\lbrack 1 - \frac{x}{2} + \frac{x^{2}}{4} - \frac{x^{3}}{8} + ...... + ( - 1)^{n}\frac{x^{n}}{2^{n}} + ...... \right\rbrack$ or $+ \frac{11}{3}\lbrack 1 + x + x^{2} + ....... + x^{n} + .....\rbrack$

But for $x^{n}$ to be defined, $\frac{- 2}{3}( - 1)^{n}\frac{1}{2^{n}} + \frac{11}{3}$ i.e., $x = 1$

$x = - r1 = A_{r}( - r)( - r + 1)( - r + 2),.....( - 1).1.2....( - r + n)$.