Question

Let $f\left(x\right)=\sqrt{2-x-x^2}$ and $g\left(x\right)={\cos }^{-1}x$. The number of integral values of $x$ in the domain of $g\left[f^3\right(x\left)\right]$ is

Answer 1

$g\left[f\right(x\left)\right]=g\left(\sqrt{2-x-x^2}\right)$
$={\cos }^{-1}\sqrt{2-x-x^2}$

$f\left(x\right)$ is defined when $2-x-x^2\ge 0$
$\Rightarrow x^2+x-2\le 0$
$\Rightarrow (x+2)(x-1)\le 0$
$\Rightarrow x\in [-2,1]$

$g\left[f\right(x\left)\right]$ is defined when $-1\le \sqrt{2-x-x^2}\le 1$
$\Rightarrow 0\le 2-x-x^2\le 1$
$\Rightarrow 2-x-x^2\le 1$
$x\in (-\infty ,\frac{-1-\sqrt{5}}{2}]\cup [\frac{-1+\sqrt{5}}{2},\infty )$

Domain $g\left[f\right(x\left)\right]$ is $[-2,\frac{-1-\sqrt{5}}{2}]\cup [\frac{-1+\sqrt{5}}{2},1]$

Since, domain $g\left[f^3\right(x\left)\right]$ is same as domain of $g\left[f\right(x\left)\right]$.
So, the integers in the interval are $-2,1$.

Hence, number of integral value of $x$ is $2$.

Alternate Solution:
$g\left[f\right(x\left)\right]=g\left(\sqrt{2-x-x^2}\right)$
$={\cos }^{-1}\sqrt{2-x-x^2}$

$f\left(x\right)$ is defined when $2-x-x^2\ge 0$
$\Rightarrow x^2+x-2\le 0$
$\Rightarrow (x+2)(x-1)\le 0$
$\Rightarrow x\in [-2,1]$
The integrals values of $x$ in $[-2,1]$ are $-2,-1,0,1$.
Only for $x=-2,1;-1\le \sqrt{2-x-x^2}\le 1$
Hence, number of integral value of $x$ is $2$.