Question

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

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

Answer 1

The correct option is A

$f\left(a^2\right)$

Explanation for the correct option.

Step 1. Find the value of $f\left(a\right)$ and $f\left(a+1\right)$.

In the function $f\left(x\right)=\frac{x}{x-1}$, substitute $a$ for $x$ .

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

In the function $f\left(x\right)=\frac{x}{x-1}$, substitute $a+1$ for $x$ .

$\begin{array}{rcl}f\left(a+1\right)& =& \frac{a+1}{a+1-1}\\ & =& \frac{a+1}{a}\end{array}$

Step 2. Find the value of $\frac{f\left(a\right)}{f\left(a+1\right)}$.

The value of $\frac{f\left(a\right)}{f\left(a+1\right)}$ is given as:

$\begin{array}{rcl}\frac{f\left(a\right)}{f\left(a+1\right)}& =& \frac{{\displaystyle \frac{a}{a-1}}}{{\displaystyle \frac{a+1}{a}}}\\ & =& \frac{a\times a}{\left(a+1\right)\left(a-1\right)}\\ & =& \frac{a^2}{a^2-1}\end{array}$

The value $\frac{a^2}{a^2-1}$ is the function value of $f\left(x\right)=\frac{x}{x-1}$ when $x=a^2$.

So, $\frac{f\left(a\right)}{f\left(a+1\right)}=f\left(a^2\right)$.

Hence, the correct option is A.