Question

If $si{n}^{-1}x+si{n}^{-1}y+si{n}^{-1}z=\frac{3\pi }{2}$ and $f\left(1\right)=2,f(a+b)=f\left(a\right)f\left(b\right)$ for all $a,b,ϵR$, 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. 10
• B. 11
• C. 2
• D. 3

Answer 1

The correct option is C 2

Since $\frac{-\pi }{2}\le si{n}^{-1}x,si{n}^{-1}y,si{n}^{-1}z\le \frac{\pi }{2}$
$si{n}^{-1}x+si{n}^{-1}y+si{n}^{-1}z=\frac{3\pi }{2}$
$\Rightarrow$ $si{n}^{-1}x=si{n}^{-1}y=si{n}^{-1}z=\frac{\pi }{2}$
$\Rightarrow x=y=z=1$
$\therefore$ $x^{f\left(1\right)}+y^{f\left(2\right)}+z^{f\left(3\right)}-\frac{x+y+z}{x^{f\left(1\right)}+y^{f\left(2\right)}+z^{f\left(3\right)}}$
$=1+1+1-\left(\frac{1+1+1}{1+1+1}\right)=3-1=2$