Question

If ${\sin }^{-1}a+{\sin }^{-1}b+{\sin }^{-1}c=\frac{3\pi }{2}$ and $f\left(2\right)=2,f(x+y)=f\left(x\right)f\left(y\right)\forall x,yϵR$ then $a^{f\left(2\right)}+b^{f\left(4\right)}+c^{f\left(6\right)}-\frac{3(a^{f\left(2\right)}.b^{f\left(4\right)}.c^{f\left(6\right)})}{a^{f\left(2\right)}+b^{f\left(4\right)}+c^{f\left(6\right)}}$ equals

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

The correct option is A 2

Since
$si{n}^{-1}\left(a\right)+si{n}^{-1}\left(b\right)+si{n}^{-1}\left(c\right)=\frac{3\pi }{2}$
Hence
$a=b=c=1$.
Now
$f(x+y)=f\left(x\right).f\left(y\right)$ implies
$f\left(x\right)=a^x$
Now
$f\left(2\right)=2$
Or
$a^2=2$
Or
$a=\sqrt{2}$.
Hence
$f\left(x\right)=2^{\frac{x}{2}}$.
Substituting the values of f(x) and a,b,c in the above expression, we get
$1^2+1^{2^2}+1^{2^3}-\frac{3\left(1^2{.1}^{2^2}{.1}^{2^3}\right)}{1^2+1^{2^2}+1^{2^3}}$
$=3-\frac{3.\left(1\right)}{3}$
$=2$.