Question

If $af(x+1)+bf\left(\frac{1}{x+1}\right)=x$ and $x\ne -1,a\ne b$ then $f\left(2\right)$ is equal to

• A. $\frac{2a+b}{2(a^2-b^2)}$
• B. $\frac{a}{a^2-b^2}$
• C. $\frac{a+2b}{a^2-b^2}$
• D. none of these

Answer 1

Answer: (A)

$\frac{2a+b}{2(a^2-b^2)}$

Substituting $x=1$ we get
$af\left(2\right)+bf\left(\frac{1}{2}\right)=1$ ...(i)
Substituting $x=-\frac{1}{2}$ we get
$af\left(\frac{1}{2}\right)+bf\left(2\right)=\frac{-1}{2}$ ...(ii)
$a\left(i\right)-b\left(ii\right)$ implies
$f\left(2\right)(a^2-b^2)=a-(-\frac{b}{2})$
$f\left(2\right)(a^2-b^2)=\frac{2a+b}{2}$
$f\left(2\right)=\frac{2a+b}{2(a^2-b^2)}$