Question

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

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is C 2

Since $-\frac{\pi }{2}\le {\sin }^{-1}x\le \frac{\pi }{2}$

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

${\sin }^{-1}x={\sin }^{-1}y={\sin }^{-1}z=\frac{\pi }{2}$

$\Rightarrow x=y=z=1$

Also $f(p+q)=f\left(p\right).f\left(q\right)\forall p,q\in R$ ....(1)

Given $f\left(1\right)=1$

From (1)

$f(1+1)=f\left(1\right).f\left(1\right)\Rightarrow f\left(2\right)=1^2=1$ ....(2)

From (2)

$f(2+1)=f\left(2\right)f\left(1\right)\Rightarrow f\left(2\right)=1^2.1=1$

Now $x^{f\left(1\right)}+y^{f\left(2\right)}+z^{f\left(3\right)}-\frac{x+y+z}{x^{f\left(1\right)}+y^{f\left(2\right)}+z^{f\left(3\right)}}=3-\frac{3}{3}=2$