If sin−1⁡x+sin−1⁡y+sin−1⁡z=3π2 and f(1)=2,f(p+q)=f(p)⋅f(q),∀p,q∈R then xf(1)+yf(2)+zf(3)−(x+y+z)xf(1)+yf(2)+zf(3) is equal to

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

1. A

0

2. B

1

3. C

2

4. D

3

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### Solution:

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

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

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

which is possible only when

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