If $af(x)+bf(\frac{1}{x})=x-1$, x ≠ 0 and a ≠ b, then f(2) =

$\frac{a}{a^2-b^2}$ $\frac{a+2b}{a^2-b^2}$ $\frac{a-2b}{a^2-b^2}$ none of these

none of these

Given, $af(x)+bf(\frac{1}{x})=x-1, x ≠ 0, a ≠ b$  …(i) $⇒af(\frac{1}{x})+bf(x)=\frac{1}{x}-1$   …(ii) a . (i) − b . (ii) $⇒ (a^2 − b^2) f(x) = a(x-1)-b(\frac{1}{x}-1)⇒(a^2 − b^2) f(2)=a+\frac{b}{2}=\frac{2a+b}{2}$ $⇒f(2)=\frac{2a+b}{2(a^2-b^2)}$