Question

If $f\left(x\right)$ is a polynomial satisfying $f\left(x\right)f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$ and $f\left(2\right)>1,$ then $\underset{x\to 1}{\lim }f\left(x\right)$ is

Answer 1

Answer: 2

As given $f\left(x\right).f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right),f\left(2\right),$
$\Rightarrow f(x)f(\frac{1}{x})-f(x)-f(\frac{1}{x})=0$
$\Rightarrow (f(x)-1)(f(\frac{1}{x})-1)=1$
$\Rightarrow h(x)h(\frac{1}{x})=1$ as it $f(x)$ is a polynomial,hence $h(x)$ will also be a polynomial
$\Rightarrow h(x)=\pm x^n$
$\Rightarrow f(x)=\pm x^n+1$
Hence $f(x)=\pm x^n+1$
Now $f(2)>1$
$\Rightarrow f(x)=x^n+1$
hence $\underset{x\to 1}{\lim }f(x)=\underset{x\to 1}{\lim }(x^n+1)=2$