Question

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

• A. $x=-1$
• B. $x=1$
• C. $x=\frac{1}{2}$
• D. $x=\frac{-1}{2}$

Answer 1

Answer: (A), (B)

The correct options are
A $x=-1$
B $x=1$

The following inequalities must be satisfied
Simultaneously,
$\cos \left(\sin x\right)\ge 0,\left|(1+x^2)/2x\right|\le 1.$For all real
$x,-1\le \sin x\le 1$
$\therefore \cos \left(\sin x\right)=\cos \left(y\right)$ where
$-1\le y\le 1$Since $\cos (-\theta )=\cos \theta$
$\therefore \cos \left(\sin x\right)=\cos \left(y\right)$ where $0\le y\le 1$ Ist quad.$\therefore \cos \left(\sin x\right)\ge 0\forall xϵR$
$\therefore D_1=R$
Let $D_2$ be the domain of ${\sin }^{-1}\left(\frac{1+x^2}{2x}\right)$
$\therefore \left|\frac{1+x^2}{2x}\right|\le 1$ or
$\frac{|1+x^2|}{2\left|x\right|}\le 1$But $1+x^2$ is always +ive$\therefore |1+x^2|=1+x^2$
$\therefore 1+x^2\le 2\left|x\right|or1+{\left|x\right|}^2\le 2\left|x\right|$or ${\left|x\right|}^2-2\left|x\right|+1\le 0$ or ${(\left|x\right|-1)}^2\le 0$
$\therefore \left|x\right|=1$ or $x=\pm 1$
$\therefore D_2=-1,1$ only$\therefore D=D_1\cap D_2=-1,1$ i.e. only two points$x=-1$ and $x=1$