Question

If function $f\left(x\right)={\sin }^{-1}x+{\cos }^{-1}x$ & $g\left(x\right)$ are identical, then $g\left(x\right)$ can be equal to

• A. ${\sin }^{-1}\left|x\right|+\left|{\cos }^{-1}x\right|$
• B. ${\tan }^{-1}x+{\cot }^{-1}x$
• C. $\left|{\sin }^{-1}x\right|+{\cos }^{-1}\left|x\right|$
• D. ${\left(\sqrt{{\sin }^{-1}x}\right)}^2+{\left(\sqrt{{\cos }^{-1}x}\right)}^2$

Answer 1

The correct option is C $\left|{\sin }^{-1}x\right|+{\cos }^{-1}\left|x\right|$

$f\left(x\right)={\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$ $\forall x\in [-1,1]$

Clearly option B and D are not identical to $f$, since their domain are not same as $f$.

And option A is not correct since at $x=-1$. Its range is $\frac{3\pi }{2}\ne$ range of $f$.

Now check option C.

for $0\le x<1$

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

and for $-1\le x<0$

$g\left(x\right)=-{\sin }^{-1}x+{\cos }^{-1}(-x)=\frac{\pi }{2}$.