Question

If ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}$ and $f\left(2\right)=2,f(a+b)=f\left(a\right)f\left(b\right),\forall a,bϵR$, then $x^{f\left(2\right)},y^{f\left(4\right)},z^{f\left(6\right)}$ are in

• A. A.P.
• B. G.P
• C. H.P
• D. None

Answer 1

Answer: (A), (B), (C)

The correct options are
A A.P.
B G.P
C H.P

we know if, $si{n}^{-1}x+si{n}^{-1}y+si{n}^{-1}z=\frac{3\pi }{2}$
$\Rightarrow x=y=z=1$, because maximum value of $si{n}^{-1}x$ is $\pi /2$ $\Rightarrow$ $x=1$
Now,
$f\left(4\right)=f(2+2)=f\left(2\right).f\left(2\right)=2.2=4$
and $f\left(6\right)=f(4+2)=f\left(4\right).f\left(2\right)=4.2=8$
Therefore, $x^{f\left(2\right)}=1,y^{f\left(4\right)}=1,z^{f\left(6\right)}=1$
Hence, numbers are both in A.P and in G.P and H.P.