Question

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

• A. $\frac{2a+b}{(a^2-b^2)}$
• B. $\frac{a}{a^2-b^2}$
• C. $\frac{a+2b}{a^2-b^2}$
• D. ab

Answer 1

The correct option is A $\frac{2a+b}{(a^2-b^2)}$

$af(x+1)+bf(\frac{1}{x+1})=x$

$putt=x+1⟹x=t-1$

$af\left(t\right)+bf(\frac{1}{t})=t-\mathrm{1....}\left(1\right)$

$af(\frac{1}{t})+bf\left(t\right)=\frac{1-t}{t}...\left(2\right)$

$\left(1\right)+\left(2\right)⟹(a+b)(f\left(t\right)+f(\frac{1}{t}))=\frac{(t-1{)}^2}{t}$

$\left(1\right)-\left(2\right)⟹(a-b)(f\left(t\right)+f(\frac{1}{t}))=\frac{t^2-1}{t}$

$f\left(t\right)=\frac{(t-1{)}^2}{t(a+b)}+\frac{t^2-1}{t(a-b)}$

$f\left(2\right)=\frac{1}{2}(\frac{1}{a+b}+\frac{3}{a-b})=\frac{1}{2}(\frac{a-b+3a+3b}{a^2-b^2}=\frac{2a+b}{a^2-b^2}$