If ${\sin }^{-1}\mathrm{x}+{\sin }^{-1}\mathrm{y}+{\sin }^{-1}\mathrm{z}=\frac{3\mathrm{\pi }}{2}$ and $\mathrm{f}\left(1\right)=2,\mathrm{f}(\mathrm{p}+\mathrm{q})=\mathrm{f}\left(\mathrm{p}\right)\cdot \mathrm{f}\left(\mathrm{q}\right),\mathrm{\forall }\mathrm{p},\mathrm{q}\in \mathrm{R}$ then ${\mathrm{x}}^{\mathrm{f}\left(1\right)}+{\mathrm{y}}^{\mathrm{f}\left(2\right)}+{\mathrm{z}}^{\mathrm{f}\left(3\right)}-\frac{(\mathrm{x}+\mathrm{y}+\mathrm{z})}{{\mathrm{x}}^{\mathrm{f}\left(1\right)}+{\mathrm{y}}^{\mathrm{f}\left(2\right)}+{\mathrm{z}}^{\mathrm{f}\left(3\right)}}$ is equal to

Solution:

$\because -\frac{\mathrm{\pi }}{2}\le {\sin }^{-1}\mathrm{x}\le \frac{\mathrm{\pi }}{2},-\frac{\mathrm{\pi }}{2}\le {\sin }^{-1}\mathrm{y}\le \frac{\mathrm{\pi }}{2}$

and  $-\frac{\mathrm{\pi }}{2}\le {\sin }^{-1}\mathrm{z}\le \frac{\mathrm{\pi }}{2}$

Given that ${\sin }^{-1}\mathrm{x}+{\sin }^{-1}\mathrm{y}+{\sin }^{-1}\mathrm{z}=\frac{3\mathrm{\pi }}{2}$

which is possible only when

$\begin{array}{l}{\sin }^{-1}\mathrm{x}={\sin }^{-1}\mathrm{y}={\sin }^{-1}\mathrm{z}=\frac{\mathrm{\pi }}{2}\\ \Rightarrow \mathrm{x}=\mathrm{y}=\mathrm{z}=1\\ \mathrm{put} \mathrm{p}=\mathrm{q}=1\\ \mathrm{then}, \mathrm{f}\left(2\right)=\mathrm{f}\left(1\right)\mathrm{f}\left(1\right)=2\cdot 2=4\\ \mathrm{and} \mathrm{put} \mathrm{p}=1,\mathrm{q}=2\\ \mathrm{then} \mathrm{f}\left(3\right)=\mathrm{f}\left(1\right)\mathrm{f}\left(2\right)=2\cdot 2^2=8\end{array}$

$\begin{array}{r}\therefore {\mathrm{x}}^{\mathrm{f}\left(1\right)}+{\mathrm{y}}^{\mathrm{f}\left(2\right)}+{\mathrm{z}}^{\mathrm{f}\left(3\right)}-\frac{\mathrm{x}+\mathrm{y}+\mathrm{z}}{{\mathrm{x}}^{\mathrm{f}\left(1\right)}+{\mathrm{y}}^{\mathrm{f}\left(2\right)}+{\mathrm{z}}^{\mathrm{f}\left(3\right)}}\\ =1+1+1-\frac{3}{1+1+1}=3-1=2\end{array}$