Relations and Functions — Class 12 Maths Solution

$f:[ - 1,\;1] \to R$ is given by $f(x) = \cfrac{x}{{x + 2}},\;x \ne 2$
Let ${x_1},\;{x_2} \in [ - 1,\;1] \Rightarrow f({x_1}) = \cfrac{{{x_1}}}{{{x_1} + 2}}\;\;and\;\;f({x_2}) = \cfrac{{{x_2}}}{{{x_2} + 2}}$

$f({x_1}) = f({x_2})$

$\Rightarrow$ $\cfrac{{{x_1}}}{{{x_1} + 2}} = \cfrac{{{x_2}}}{{{x_2} + 2}} \Rightarrow {x_1}{x_2} + 2{x_1} = {x_1}{x_2} + 2{x_2}$

$\Rightarrow$ $2{x_1} = 2{x_2} \Rightarrow {x_1} = {x_2}.$ Thus, f is one-one.

Now, $f:[ - 1,\;1] \to R,$ be given for every $y \in R$ (co-domam of f), there exist $x \in [ - 1,\;1]$ (domain of f) such that f(x)$=$ y

$\Rightarrow$ $y = f(x) = \cfrac{x}{{x + 2}}\;\;for\;\;some\;\;x\;\;in\;\;[ - 1,\;1]$

As, $y = \cfrac{x}{{x + 2}} \Rightarrow yx + 2y = x \Rightarrow 2y = x(1 - y) \Rightarrow x = \cfrac{{2y}}{{1 - y}}$
$\Rightarrow$ $f\left[ {\cfrac{{2y}}{{1 - y}}} \right] = \cfrac{{\cfrac{{2y}}{{1 - y}}}}{{\cfrac{{2y}}{{1 - y}} + 2}} = \cfrac{{2y}}{{2y + 2 - 2y}} = y$

$\Rightarrow$ $f(x)$ is onto $\Rightarrow$ f is bijective and hence invertible

Now, $x = \cfrac{{2y}}{{1 - y}} \Rightarrow {f^{ - 1}}(y) = \cfrac{{2y}}{{1 - y}} \Rightarrow {f^{ - 1}}(x) = \cfrac{{2x}}{{1 - x}}$