Question

If ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}$ and $f\left(1\right)=2,f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,y\in R$. Then $x^{f\left(1\right)}+y^{f\left(2\right)}+z^{f\left(3\right)}-\frac{x+y+z}{x^{f\left(1\right)}+y^{f\left(2\right)}+z^{f\left(3\right)}}$ is equal to

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

We know that $-\frac{\pi }{2}\le {\sin }^{-1}x,{\sin }^{-1}y,{\sin }^{-1}z\le \frac{\pi }{2}$
$\therefore {\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}$
$\Rightarrow {\sin }^{-1}x={\sin }^{-1}y={\sin }^{-1}z=\frac{\pi }{2}$
$\Rightarrow x+y+z=1$
It is given that $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,y\in R$
$\therefore f\left(x\right)={\left[f\right(1\left)\right]}^x$ for all $x\in R$
$\Rightarrow f\left(x\right)=2^x$ for all $x\in R$
$\Rightarrow f\left(1\right)=2,f\left(2\right)=4,f\left(3\right)=8$
$\therefore x^{f\left(1\right)}+y^{f\left(2\right)}+z^{f\left(3\right)}=\frac{x+y+z}{x^{f\left(1\right)}+y^{f\left(2\right)}+z^{f\left(3\right)}}$
$1+1+1-\frac{1+1+1}{1+1+1}=3-1=2$