Question

If $|{\sin }^{-1}x|+|{\cos }^{-1}x|=\frac{\pi }{2}$, then $x\in$

• A. $R$
• B. $[-1,1]$
• C. $[0,1]$
• D. $\varphi$

Answer 1

Answer: (C)

$[0,1]$

We have,

$\left|{\sin }^{-1}x\right|+\left|{\cos }^{-1}x\right|=\frac{\pi }{2}$

Let $x=0$

So,

$\left|{\sin }^{-1}0\right|+\left|{\cos }^{-1}0\right|=\frac{\pi }{2}$

$\Rightarrow 0+\frac{\pi }{2}=\frac{\pi }{2}$

$\Rightarrow \frac{\pi }{2}=\frac{\pi }{2}$

If put $x=1$

So,

$\left|{\sin }^{-1}1\right|+\left|{\cos }^{-1}1\right|=\frac{\pi }{2}$

$\Rightarrow \frac{\pi }{2}+0=\frac{\pi }{2}$

$\Rightarrow \frac{\pi }{2}=\frac{\pi }{2}$

$x\in [0,1]$

Hence, this is the answer.