Question

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

• A. 7
• B. 2
• C. 5
• D. 3

Answer 1

Answer: (C)

$f(x)=y=\frac{x-2}{x-3}$
$\therefore x=\frac{3 y-2}{y-1}$
$\therefore f^{-1}(x)=\frac{3 x-2}{x-1}$
& $ g(x)=y=2 x-3$
$\therefore x=\frac{y+3}{2}$
$\therefore g ^{-1}( x )=\frac{ x +3}{2}$
$\because f^{-1}(x)+g^{-1}(x)=\frac{13}{2}$
$\therefore x^{2}-5 x+6=0 < ^{x_{1}} _{x_{2}}$
$\therefore $ sum of roots
$x _{1}+ x _{2}=5$