Question

If $f\left(x\right)=\frac{2x-1}{x+5}$, $x\ne -5$, then $f^{-1}\left(x\right)=$

• A. $\frac{x+5}{2x-1},x\ne \frac{1}{2}$
• B. $\frac{5x+1}{2-x},x\ne 2$
• C. $\frac{x-5}{2x-1},x\ne \frac{1}{2}$
• D. $\frac{5x-1}{2-x},x\ne 2$

Answer 1

The correct option is B

$\frac{5x+1}{2-x},x\ne 2$

Explanation for the correct option:

Find the value of $f^{-1}\left(x\right)$:

Let, $f\left(x\right)=y$, so $x=f^{-1}\left(y\right)$. Now,

$\begin{array}{rcl}y& =& \frac{2x-1}{x+5}\\ \Rightarrow xy+5y& =& 2x-1\\ \Rightarrow 2x-xy& =& 5y+1\\ & \Rightarrow & x\left(2-y\right)=5y+1\\ \Rightarrow x& =& \frac{5y+1}{2-y}\\ \Rightarrow f^{-1}\left(y\right)& =& \frac{5y+1}{2-y}\end{array}$

Therefore, $f^{-1}\left(x\right)=\frac{5x+1}{2-x},x\ne 2$

Hence, option B is correct.