Question

If ${\cos }^{-1}\left(a\right)+{\cos }^{-1}\left(b\right)+{\cos }^{-1}\left(c\right)=3\pi$ and $f$ be a function such that $f\left(1\right)=2$ and $f(x+y)=f\left(x\right)\cdot f\left(y\right)$ for all $x,y\in R$, then the value of $a^{2f\left(1\right)}+b^{2f\left(2\right)}+c^{2f\left(3\right)}+\frac{a+b+c}{a^{2f\left(1\right)}+b^{2f\left(2\right)}+c^{2f\left(3\right)}}$ is

• A. $0$
• B. $2$
• C. $4$
• D. $3$

Answer 1

The correct option is B $2$

We know that ${\cos }^{-1}x\in [0,\pi ]$
$\therefore {\cos }^{-1}\left(a\right)+{\cos }^{-1}\left(b\right)+{\cos }^{-1}\left(c\right)=3\pi$ is possible iff $a=b=c=-1$

Now, $f\left(1\right)=2$ and $f(x+y)=f\left(x\right)\cdot f\left(y\right)$
Let, $f\left(x\right)=a^{kx}$
$\therefore a^k=2$
Hence, $f\left(x\right)=2^x$
$\Rightarrow f\left(2\right)=4,f\left(3\right)=8$

$\therefore a^{2f\left(1\right)}+b^{2f\left(2\right)}+c^{2f\left(3\right)}+\frac{a+b+c}{a^{2f\left(1\right)}+b^{2f\left(2\right)}+c^{2f\left(3\right)}}$
$=1+1+1-\frac{3}{1+1+1}=2$