Let $[x]$ denote the greatest integer less than or equal to $x$. If $f(x)=\sin ^{-1} x, g(x)=\left[x^{2}\right]$ and $h(x)=2 x, \frac{1}{2} \leq x \leq \frac{1}{\sqrt{2}}$, then
(a) $f \circ g \circ h(x)=\frac{\pi}{2}$
(b) $\operatorname{fogoh}(x)=\pi$
(c) hofog $=$ hogof
(d) $h o f o g \neq h o g o f$
(c) hofog $=$ hogof
We have,
$g(x)=\left[x^{2}\right]$
$=0 \quad\left(A s \frac{1}{2} \leq x \leq \frac{1}{\sqrt{2}}\right.$
$\left.\therefore \frac{1}{4} \leq x^{2} \leq \frac{1}{2}\right)$
$f o g(x)=f(g(x))=\sin ^{-1}(0)$
$=0$
$h o f o g(x)=h(f(g(x)))=2 \times 0=0$
And
$f(x)=\sin ^{-1} x$
Now,
for, $x \in\left[\frac{1}{2}, \frac{1}{\sqrt{2}}\right]$
$f(x) \in\left[\frac{\pi}{6}, \frac{\pi}{4}\right]$
$f(x) \in[0.52,0.78]$
$g o f(x)=0 \quad(\mathrm{As}, f(x) \in[0.52,0.78])$
$=0$
$\operatorname{hogof}(x)=h(g(f(x)))=2 \times 0=0$
$\therefore$ hofog $=$ hogof $=0$