Question

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

• A. $\frac{1-x}{2}$
• B. $\frac{x+1}{2}$
• C. $1-x$
• D. none of these

Answer 1

Answer: (A)

$\frac{1-x}{2}$

Given, $\frac{1}{(1+x)(1+x^2)}=\frac{A}{1+x}+\frac{f\left(x\right)}{1+x^2}$
Consider,$\frac{1}{(1+x)(1+x^2)}=\frac{A}{1+x}+\frac{Bx+C}{1+x^2}$ ....(1)
$\Rightarrow 1=A(1+x^2)+(Bx+C)(1+x)$
$\Rightarrow 1=(A+B)x^2+(B+C)x+(A+C)$
$\Rightarrow A+B=0,B+C=0,A+C=1$
Solving these, we get
$A=C=\frac{1}{2},B=-\frac{1}{2}$
Put these values in (1), we get
$\frac{1}{(1+x)(1+x^2)}=\frac{1}{2(1+x)}+\frac{-x+1}{2(1+x^2)}$
Comparing this with (1), we get
$f\left(x\right)=\frac{1-x}{2}$