Question

The domain set of definition of the function $f(x)=\sqrt{\cos (\sin x)}+{\sin }^{-1}\left(\frac{1+x^2}{2x}\right)$ is

• A. $-1\le x\le 1$
• B. $x\ge 1$
• C. $x\le 1$
• D. $x=\pm 1$

Answer 1

Answer: (D)

$\cos (\sin x)\ge 0\forall x,x\ne 0$
Hence, $\cos (\sin x)>0\forall x$
$\frac{-\pi }{2}\le {\sin }^{-1}\left(\frac{1+x^2}{2x}\right)\le \frac{\pi }{2}$
i.e. $-1\le \frac{1+x^2}{2x}\le 1$. If $x<0$, then $\frac{1+x^2}{2x}<0$
i.e. $-1\le \frac{1+x^2}{2x}\Rightarrow -2x\ge \left(1+x^2\right)\Rightarrow \left(1+x^2\right)\le 0\Rightarrow x=-1$
The other side $(x>0),1+x^2\le 2x\Rightarrow (x-1{)}^2\le 0\Rightarrow x=1$.
The function is defined only for the two values $x=1$ and $x=-1$.