Question

If for non-zero x, $af\left(x\right)+bf\left(\frac{1}{x}\right)=\frac{1}{x},-5$, where $a\ne b$, then find $f\left(x\right)$.

Answer 1

We have,
$af\left(x\right)+bf\left(\frac{1}{x}\right)=\frac{1}{x}-5\dots \dots \left(i\right)$
$\Rightarrow af\left(\frac{1}{x}\right)+bf\left(x\right)=\frac{1}{\frac{1}{x}}-5$
$=x-5\dots \dots \left(ii\right)$
$\Rightarrow af\left(\frac{1}{x}\right)+bf\left(x\right)=x-5$
Adding equation (i) and (ii), we get
$af\left(x\right)+bf\left(x\right)+bf\left(\frac{1}{x}\right)+af\left(\frac{1}{x}\right)=\frac{1}{x}-5+x-5$
$\Rightarrow (a+b)f\left(x\right)+f\left(\frac{1}{x}\right)(a+b)=\frac{1}{x}+x-10$
$\Rightarrow f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{1}{a+b}[\frac{1}{x}+x-10]\dots \dots \left(iii\right)$
Subtracting equation (ii) from equation (i), we get
$af\left(x\right)-bf\left(x\right)+bf\left(\frac{1}{x}\right)-af\left(\frac{1}{x}\right)=\frac{1}{x}-5-x+5$
$\Rightarrow (a-b)f\left(x\right)-f\left(\frac{1}{x}\right)(a-b)=\frac{1}{x}-x$
$\Rightarrow f\left(x\right)-f\left(\frac{1}{x}\right)=\frac{1}{a-b}[\frac{1}{x}-x]$
Adding equation (iii) and (iv), we get
$2f\left(x\right)=\frac{1}{a+b}[\frac{1}{x}+x-10]+\frac{1}{a-b}[\frac{1}{x}-x]$
$\Rightarrow 2f\left(x\right)=\frac{(a-b)[\frac{1}{x}+x-10]+(a+b)[\frac{1}{x}-x]}{(a+b)(a-b)}$
$\Rightarrow 2f\left(x\right)$
$=\frac{\frac{a}{x}+ax-10a-\frac{b}{x}-bx+10b+\frac{a}{x}-ax+\frac{b}{x}-bx}{a^2-b^2}$
$\Rightarrow 2f\left(x\right)=\frac{\frac{2a}{x}-10a+10b-2bx}{a^2-b^2}$
$\Rightarrow 2f\left(x\right)$
$=\frac{1}{a^2-b^2}\times \frac{1}{2}[\frac{2a}{x}-10a+10b-2bx]$
$=\frac{1}{a^2-b^2}[\frac{a}{x}-5a+5b-bx]$
$=\frac{1}{a^2-b^2}[\frac{a}{x}-bx-5(a-b\left)\right]$