Question

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

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

Answer: (D)

$6$

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