Question

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

• A. $\left[\frac{(2a+b)}{2(a^2+b^2)}\right]$
• B. $\left[\frac{(2a+b)}{2(a^2-b^2)}\right]$
• C. $\left[\frac{(a+2b)}{(a^2+b^2)}\right]$
• D. $\left[\frac{a}{(a^2-b^2)}\right]$

Answer 1

The correct option is B

$\left[\frac{(2a+b)}{2(a^2-b^2)}\right]$

Explanation for the correct option:

Step 1. Finding the value of $\mathit{f}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$:

Given, $af\left(x\right)+bf\left(\frac{1}{x}\right)\mathbf{=}x-1,x\ne 0$

Now, Put $x=2$in the given equation

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

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

Again put $x=\frac{1}{2}$in the given equation

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

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

Step 2. Multiply $\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{\times }\mathbf{\left(}\mathit{i}\mathbf{\right)}\mathbf{–}\mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{\times }\mathbf{\left(}\mathit{i}\mathit{i}\mathbf{\right)}$, we get

$$\begin{gathered} a^{2}f\left(2\right)+abf\left(\frac{1}{2}\right)=a \\ abf\left(\frac{1}{2}\right)+b^{2}f\left(2\right)=\frac{-b}{2} \end{gathered}$$

and

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

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

Hence, the correct option is $\mathbf{\left(}\mathit{B}\mathbf{\right)}$.