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1.Relation and Function
hard
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
$3$
(JEE MAIN-2021)
Solution
$f(x)=y=\frac{x-2}{x-3}$
$\therefore x=\frac{3 y-2}{y-1}$
$\therefore f^{-1}(x)=\frac{3 x-2}{x-1}$
$ g(x)=y=2 x-3$
$\therefore x=\frac{y+3}{2}$
$\therefore g^{-1}(x)=\frac{x+3}{2}$
$\because f^{-1}(x)+g^{-1}(x)=\frac{13}{2}$
$\therefore x^{2}-5 x+6=0$
$\therefore$ sum of roots
$x_{1}+x_{2}=5$
Standard 12
Mathematics
