Question

Let $f:R-\left\{3\right\}\to R-\left\{1\right\}$ be defined by $f\left(x\right)=\frac{\left(x-2\right)}{\left(x-3\right)}$. Let $g:R$ be given as $g\left(x\right)=2x-3$. Then, the sum of all the values of $x$ for which $f^{-1}\left(x\right)+g^{-1}\left(x\right)=\frac{13}{2}$ is equal to?

• A. $7$
• B. $5$
• C. $2$
• D. $3$

Answer 1

The correct option is B

$5$

Explanation for the correct option:

Step 1:Finding $f^{-1}$and $g^{-1}$:

$f\left(x\right)=\frac{\left(x-2\right)}{\left(x-3\right)}$

Let $y=\frac{x-2}{x-3}$

$$\begin{gathered} \Rightarrow yx-3y=x-2 \\ \Rightarrow x(y-1)=3y-2 \\ \Rightarrow x=\frac{3y-2}{y-1} \end{gathered}$$

Therefore $f^{-1}\left(x\right)=\frac{3x-2}{x-1}$

$g\left(x\right)=2x-3$

Let $y=2x-3$

$\Rightarrow x=\frac{y+3}{2}$

Therefore $g^{-1}\left(x\right)=\frac{x+3}{2}$

Step 2: finding value of $x$:

We have,

$$\begin{gathered} \Rightarrow f^{-1}\left(x\right)+g^{-1}\left(x\right)=\frac{13}{2}\left[\mathrm{Given}\right] \\ \Rightarrow \frac{3x-2}{x-1}+\frac{x+3}{2}=\frac{13}{2} \\ \Rightarrow 2(3x-2)+(x-1)(x+3)=\frac{13}{2} \\ \Rightarrow x^2-5x+6=0 \\ \Rightarrow x=2or3 \end{gathered}$$

Sum: $2+3=5$

Therefore, the correct answer is option (B).