Question

If a real value of function $f$of a real variable $x$is such that $\frac{1}{(1+x)(1+x^2)}=\frac{A}{(1+x)}+\frac{f\left(x\right)}{(1+x^2)}$then $f\left(x\right)=$

• A. $\frac{(1-x)}{2}$
• B. $\frac{(x^2+1)}{2}$
• C. $1-x$
• D. None of these

Answer 1

The correct option is C

$1-x$

Explanation for the correct options:

Find the function $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$:

Given that a real value of function $f$of a real variable $x$is such that $\frac{1}{(1+x)(1+x^2)}=\frac{A}{(1+x)}+\frac{f\left(x\right)}{(1+x^2)}$

$\Rightarrow 1=(1+x^2)A+f\left(x\right)(1+x)$

Put $x=-1$.we get

$$\begin{gathered} \Rightarrow 1=2A \\ \Rightarrow A=\frac{1}{2} \end{gathered}$$

Now, $\frac{1}{(1+x)(1+x^2)}-\frac{1}{2(1+x)}=\frac{f\left(x\right)}{(1+x^2)}$

$$\begin{gathered} \Rightarrow \frac{1}{(1+x)}\left[\frac{2-1-x^2}{(1+x^2)}\right]=\frac{f\left(x\right)}{(1+x^2)} \\ \Rightarrow \frac{1-x^2}{(1+x)}=f\left(x\right) \\ \Rightarrow f\left(x\right)=\frac{(1-x)(1+x)}{(1+x)} \\ \Rightarrow f\left(x\right)=(1-x) \end{gathered}$$

Hence, the correct option is (C).