# Let [x] denote the greatest integer less than or equal to x.

Question:

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) $h o f o g \neq h o g o f$

Solution:

(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 $=0$