Question

If $f\left(x\right)$ is a polynomial in $x(>0)$ satisfying the equation $f\left(x\right)+f\left(\frac{1}{x}\right)=f\left(x\right).f\left(\frac{1}{x}\right)$ and $f\left(2\right)=-7$, then $f\left(3\right)$ is equal to

• A. $-26$
• B. $-27$
• C. $-28$
• D. $-29$

Answer 1

The correct option is A $-26$

Whenever, $f\left(x\right)+f\left(\frac{1}{x}\right)=f\left(x\right).f\left(\frac{1}{x}\right)$, then $f\left(x\right)=1+x^n$ or $f\left(x\right)=1-x^n$ keeping in mind that $n$ must be greater than $0$.

So here, $f\left(2\right)=-7$.

So we choose $f\left(x\right)=1-x^n$.

So, $f\left(2\right)=1-2^n$

$\Rightarrow 2^n=8$

$\Rightarrow n=3$

So, the function is $f\left(x\right)=1-x^3$.

Now,$f\left(3\right)=1-3^3$

$\Rightarrow f\left(3\right)=-26$