Question

If the range of the function $f(x)=\sqrt{{\tan }^{-1}x+1}+\sqrt{1-{\tan }^{-1}x}$ is $[a,b]$ then find the value of $\left(a^2+b^2\right)$.

Answer 1

Answer: 0006

$\text{ For }y\text{ to the real }-1\le {\tan }^{-1}x\le 1$
$\Rightarrow -\tan 1\le x\le \tan 1$
$y=\sqrt{t+1}+\sqrt{1-t}\Rightarrow y^2=2+2\sqrt{1-t^2}$
$\Rightarrow y^2\in [2,4]\Rightarrow y\in [\sqrt{2},2]$
$\Rightarrow \left(a^2+b^2\right)=6$