Question

Let $f:R-\{3\}\to R-\{1\}$ be defined by $f(x)=\frac{x-2}{x-3}$. Let $g:R\to R$ be given as $g(x)=2x-3.$ 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)

$f(x)=y=\frac{x-2}{x-3}$
$\therefore x=\frac{3y-2}{y-1}$
$\therefore f^{-1}(x)=\frac{3x-2}{x-1}$
& $g(x)=y=2x-3$
$\therefore x=\frac{y+3}{2}$
$\therefore g^{-1}(x)=\frac{x+3}{2}$
$\because f^{-1}(x)+g^{-1}(x)=\frac{13}{2}$
$\therefore x^2-5x+6=0{<}_{x_2}^{x_1}$
$\therefore$ sum of roots
$x_1+x_2=5$