Question

Let [x] denote the greatest integer less than or equal to x. If $f\left(x\right)={\sin }^{-1}x,g\left(x\right)=\left[x^2\right]\mathrm{and}h\left(x\right)=2x,\frac{1}{2}\le x\le \frac{1}{\sqrt{2}}$, then
(a) $fogoh\left(x\right)=\frac{\pi }{2}$
(b) $fogoh\left(x\right)=\pi$
(c) $hofog=hogof$
(d) $hofog\ne hogof$

Answer 1

(c) $hofog=hogof$

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ g\left(x\right)=\left[x^2\right] \\ =0\left(As\frac{1}{2}\le x\le \frac{1}{\sqrt{2}} \\ \therefore \frac{1}{4}\le x^2\le \frac{1}{2}\right) \\ fog\left(x\right)=f\left(g\left(x\right)\right)={\sin }^{-1}\left(0\right) \\ =0 \\ hofog\left(x\right)=h\left(f\left(g\left(x\right)\right)\right)=2\times 0=0 \end{gathered}$$

$$\begin{gathered} \mathrm{And} \\ f\left(x\right)={\sin }^{-1}x \\ \mathrm{Now}, \\ for,x\in \left[\frac{1}{2},\frac{1}{\sqrt{2}}\right] \\ f\left(x\right)\in \left[\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{4}\right] \\ f\left(x\right)\in \left[0.52,0.78\right] \\ gof\left(x\right)=0\left(\mathrm{As},f\left(x\right)\in \left[0.52,0.78\right]\right) \\ =0 \\ hogof\left(x\right)=h\left(g\left(f\left(x\right)\right)\right)=2\times 0=0 \end{gathered}$$

$\therefore hofog=hogof=0$