Question

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

Answer 1

We have, $af\left(x\right)+bf\left(\frac{1}{x}\right)=\frac{1}{x}-5$ . . . (i)

x is replace by $\frac{1}{x}$, we get

$af\left(\frac{1}{x}\right)+bf\left(x\right)=x-5$ . . . . (ii)

Multiply by a in Eq. (i), we get

$a^{2}f\left(x\right)+abf\left(\frac{1}{x}\right)=a(\frac{1}{x}-5)$ . . . (iii)

Multiply by b in Eq. (ii), we get

$abf\left(\frac{1}{x}\right)+b^{2}f\left(x\right)=b(x-5)$ . . . (iv)

On subtracting Eq. (iv) from Eq. (iii), we get

$(a^2-b^2)f\left(x\right)=a(\frac{1}{x}-5)-b(x-5)$

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