Question

Let $f\left(x\right)={\int }_0^{{\sin }^{2}x}{\sin }^{-1}\left(\sqrt{t}\right)dt+{\int }_0^{{\cos }^{2}x}{\cos }^{-1}\left(\sqrt{t}\right)dt$, then

• A. $f\left(x\right)$ is a constant function
• B. $f\left(\frac{\pi }{4}\right)$ = 0
• C. $f\left(\frac{\pi }{3}\right)=\frac{\pi }{4}$
• D. $f\left(\frac{\pi }{4}\right)=\frac{\pi }{4}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $f\left(x\right)$ is a constant function
C $f\left(\frac{\pi }{3}\right)=\frac{\pi }{4}$
D $f\left(\frac{\pi }{4}\right)=\frac{\pi }{4}$

Given, $f\left(x\right)={\int }_0^{{\sin }^{2}x}{\sin }^{-1}\left(\sqrt{t}\right)dt+{\int }_0^{{\cos }^{2}x}{\cos }^{-1}\left(\sqrt{t}\right)dt$

$\Rightarrow f^{\prime }\left(x\right)=x\cdot 2\sin x\cos x+x\cdot 2\cos x(-\sin x)$

$\Rightarrow f^{\prime }\left(x\right)=0\Rightarrow f\left(x\right)=\text{Constant}$

But $f\left(\frac{\pi }{4}\right)={\int }_0^{\frac{1}{2}}({\sin }^{-1}\sqrt{t}+{\cos }^{-1}\sqrt{t})dt$

= ${\int }_0^{\frac{1}{2}}\left(\frac{\pi }{2}\right)dt=\frac{\pi }{4}$

$\Rightarrow f\left(\frac{\pi }{3}\right)=\frac{\pi }{4},f\left(\frac{\pi }{4}\right)=\frac{\pi }{4}$