Question

If $f\left(x\right)=\{\begin{array}{c}-\sqrt{1-x^2}0\le x\le 1\\ -x,x>1\end{array}$ then

• A. Maximum of $f\left(x\right)$ exist at $x=1$
• B. Maximum of $f\left(x\right)$ doesn't exist
• C. Maximum of $f^{-1}\left(x\right)$ exist at $x=1$
• D. minimum of $f^{-1}\left(x\right)$ exist at $x=1$

Answer 1

Answer: (A)

Maximum of $f\left(x\right)$ exist at $x=1$

for $xϵ[0,1]:x^2ϵ[0,1]$

$\Rightarrow 1-x^2ϵ[0,1]$

$\Rightarrow -\sqrt{1-x^2}ϵ[-1,0]$

For $x>1:-xϵ(-\infty ,-1)$

Hence, maximum value of $f\left(x\right)$ is $0$, and it is attained at

$-\sqrt{1-x^2}=0\Rightarrow x=1$

This can also be seen from graph.