Let A = {x &epsis; R | −1 ≤ x ≤ 1} and let f : A → A, g : A → A be two functions defined by


Let $A=\{x$ \&epsis; $R \mid-1 \leq x \leq 1\}$ and let $f: A \rightarrow A, g: A \rightarrow A$ be two functions defined by $f(x)=x^{2}$ and $g(x)=\sin (\pi x / 2)$. Show that $g^{-1}$ exists but $f^{-1}$ does not exist. Also, find $g-1$.


 f is not one-one because


and $f(1)=1^{2}=1$

$\Rightarrow-1$ and 1 have the same image under $f$.

$\Rightarrow f$ is not a bijection.

So, $f^{-1}$ does not exist.

Injectivity of g:

Let $x$ and $y$ be any two elements in the domain $(A)$, such that


$\Rightarrow \sin \left(\frac{\pi x}{2}\right)=\sin \left(\frac{\pi y}{2}\right)$

$\Rightarrow\left(\frac{\pi x}{2}\right)=\left(\frac{\pi y}{2}\right)$

$\Rightarrow x=y$

So, is one-one.

Surjectivity of g:

Range of $g=\left[\sin \left(\frac{\pi(-1)}{2}\right), \sin \left(\frac{\pi(1)}{2}\right)\right]$

$=\left[\sin \left(\frac{-\pi}{2}\right), \sin \left(\frac{\pi}{2}\right)\right]=[-1,1]=A($ co-domain of $g)$

$\Rightarrow g$ is onto.

$\Rightarrow g$ is a bijection.

So, $g^{-1}$ exists.


let $g^{-1}(x)=y$                                  $\ldots(1)$

$\Rightarrow g(y)=x$

$\Rightarrow \sin \left(\frac{\pi y}{2}\right)=x$

$\Rightarrow\left(\frac{\pi y}{2}\right)=\sin ^{-1} x$

$\Rightarrow y=\frac{2}{\pi} \sin ^{-1} x$

$\Rightarrow g^{-1}(x)=\frac{2}{\pi} \sin ^{-1} x$            $[$ from (1) $]$



Leave a comment