Question

If $f\left(x\right)=\frac{x}{x-1}$, then $\frac{f\left(a\right)}{f(a+1)}$ is equal to

• A. $f(-a)$
• B. $f\left(1/a\right)$
• C. $f\left(a^2\right)$
• D. $f\left(\frac{-a}{a-1}\right)$

Answer 1

The correct option is C $f\left(a^2\right)$

Given,

$f\left(x\right)=\frac{x}{x-1}$

$f\left(a\right)=\frac{a}{a-1}$

$f(a+1)=\frac{a+1}{a+1-1}$

$=\frac{a+1}{a}$

Then,

$\frac{f\left(a\right)}{f(a+1)}=\frac{\frac{a}{a-1}}{\frac{a+1}{a}}$

$=\frac{a}{a-1}\times \frac{a}{a+1}=\frac{a^2}{a^2}$

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

$\text{Therefore option}C=f\left(a^2\right)$