Question

Let $f\left(x\right)={\sin }^{-1}x+2{\tan }^{-1}x+x^5-5x^4+5x^3+1$. Then

• A. maximum value of $f\left(x\right)$ is $\pi +2$
• B. minimum value of $f\left(x\right)$ is $-\frac{3\pi }{2}-6$
• C. maximum value of $f\left(x\right)$ is $-\frac{3\pi }{2}+6$
• D. minimum value of $f\left(x\right)$ is $-\pi -10$

Answer 1

Answer: (A), (D)

The correct options are
A maximum value of $f\left(x\right)$ is $\pi +2$
D minimum value of $f\left(x\right)$ is $-\pi -10$

Domain of $f$ is $[-1,1]$
$f\left(x\right)={\sin }^{-1}x+2{\tan }^{-1}x+x^5-5x^4+5x^3+1$
$f^{\prime }\left(x\right)=\frac{1}{\sqrt{1-x^2}}+\frac{2}{1+x^2}+5x^4-20x^3+15x^2$
$=\frac{1}{\sqrt{1-x^2}}+\frac{2}{1+x^2}+5x^2(x-1)(x-3)>0,\forall x\in (-1,1)$

$f\left(1\right)=\pi +2$
and $f(-1)=-\pi -10$