Question

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

Answer 1

Given:
$af\left(x\right)+bf\left(\frac{1}{x}\right)=\frac{1}{x}-5$ ...(i)
$\Rightarrow af\left({\displaystyle \frac{1}{x}}\right)+bf\left(x\right)=\frac{1}{{\displaystyle \frac{1}{x}}}-5$
$\Rightarrow af\left({\displaystyle \frac{1}{x}}\right)+bf\left(x\right)=x-5$ ...(ii)

On adding equations (i) and (ii), we get:

$af\left(x\right)+bf\left(x\right)+bf\left(\frac{1}{x}\right)+af\left(\frac{1}{x}\right)=\frac{1}{x}-5+x-5$
$\Rightarrow \left(a+b\right)f\left(x\right)+\left(a+b\right)f\left(\frac{1}{x}\right)=\frac{1}{x}+x-10$
$\Rightarrow f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{1}{\left(a+b\right)}\left[\frac{1}{x}+x-10\right]$ ...(iii)

On subtracting (ii) from (i), we get:

$af\left(x\right)-bf\left(x\right)+bf\left(\frac{1}{x}\right)-af\left(\frac{1}{x}\right)=\frac{1}{x}-5-x+5$
$\Rightarrow \left(a-b\right)f\left(x\right)-f\left(\frac{1}{x}\right)\left(a-b\right)=\frac{1}{x}-x$
$\Rightarrow f\left(x\right)-f\left(\frac{1}{x}\right)=\frac{1}{\left(a-b\right)}\left[\frac{1}{x}-x\right]$ ...(iv)

On adding equations (iii) and (iv), we get:

$2f\left(x\right)=\frac{1}{a+b}\left[\frac{1}{x}+x-10\right]+\frac{1}{a-b}\left[\frac{1}{x}-x\right]$
$\Rightarrow 2f\left(x\right)=\frac{\left(a-b\right)\left[\frac{1}{x}+x-10\right]+\left(a+b\right)\left[\frac{1}{x}-x\right]}{\left(a+b\right)\left(a-b\right)}$
$\Rightarrow 2f\left(x\right)=\frac{{\displaystyle \frac{a}{x}}+ax-10a-{\displaystyle \frac{b}{x}}-bx+10b+{\displaystyle \frac{a}{x}}-ax+{\displaystyle \frac{b}{x}}-bx}{a^2-b^2}$
$\Rightarrow 2f\left(x\right)=\frac{{\displaystyle \frac{2a}{x}}-10a+10b-2bx}{a^2-b^2}$
$\Rightarrow f\left(x\right)=\frac{1}{a^2-b^2}\times \frac{1}{2}\left[\frac{2a}{x}-10a+10b-2bx\right]$
$=\frac{1}{a^2-b^2}\left[\frac{a}{x}-5a+5b-bx\right]$

Therefore,
$f\left(x\right)=\frac{1}{a^2-b^2}\left[\frac{a}{x}-bx-5a+5b\right]$
$=\frac{1}{a^2-b^2}\left[\frac{a}{x}-bx\right]-\frac{5\left(a-b\right)}{a^2-b^2}$
$=\frac{1}{a^2-b^2}\left[\frac{a}{x}-bx\right]-\frac{5\left(a-b\right)}{\left(a-b\right)\left(a+b\right)}$
$=\frac{1}{a^2-b^2}\left[\frac{a}{x}-bx\right]-\frac{5}{\left(a+b\right)}$

Hence,
$f\left(x\right)=\frac{1}{a^2-b^2}\left[\frac{a}{x}-bx\right]-\frac{5}{\left(a+b\right)}$