Question

The set of real values of x for which $B = \frac{1}{2}$is.

• A. $y = \frac{Ax + B}{(x^{2} + 1)^{2}} = \frac{1}{2}\frac{(1 - x)}{(x^{2} + 1)^{2}}$
• B. $\frac{5x + 6}{(2 + x)(1 - x)} = \frac{\frac{- 4}{3}}{2 + x} + \frac{\frac{11}{3}}{1 - x}$
• C. $\frac{\frac{- 4}{3}}{2\left( 1 + \frac{x}{2} \right)} + \frac{\frac{11}{3}}{1 - x} = \frac{- 2}{3}\left( 1 + \frac{x}{2} \right)^{- 1} + \frac{11}{3}(1 - x)^{- 1}$
• D. None of these

Answer 1

Answer: (A)

$y = \frac{Ax + B}{(x^{2} + 1)^{2}} = \frac{1}{2}\frac{(1 - x)}{(x^{2} + 1)^{2}}$

Answer 2

$2 = 9 - B$ …..(i)

For log to be defined,

$\Rightarrow$ $B = 73x + 4 = A(x - 1) - B(x - 2)$ or $\Rightarrow$

Now from (i), $\frac{1}{2} \leq \log_{0.1},x \leq 2$

$4 = - A + 2B \Rightarrow$ …..(ii)

Case (i) $A = 10,B = 7$

From (ii), $\therefore$

$\Rightarrow$ $\frac{f(x)}{x + 1} = \varphi_{1}(x) + \frac{6}{x + 1},\frac{f(x)}{x - 2} = \varphi_{2}(x) + \frac{3}{x - 2}$

$\Rightarrow$ $\frac{f(x)}{(x + 1)(x + 2)(x - 2)} = \varphi(x) + \frac{Q(x)}{(x + 1)(x + 2)(x - 2)}$.

Case (ii) $Q(x)$

From (ii), $Q(x)(x + 1)f(x)(x + 1)Q(x)$

$(x - 2)f(x)$; $x + 2,\therefore$.