Q. 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 $\underset{x \rightarrow 1}{lim} f \left(x\right)$ is
NTA AbhyasNTA Abhyas 2020Limits and Derivatives
Solution:
As given $f\left(x\right) \, .f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right), \, f\left(2\right),$
$\Rightarrow f \left(\right. x \left.\right) f \left(\right. \frac{1}{x} \left.\right) - f \left(\right. x \left.\right) - f \left(\right. \frac{1}{x} \left.\right) = 0$
$\Rightarrow \left(\right. f \left(\right. x \left.\right) - 1 \left.\right)\left(\right. f \left(\right. \frac{1}{x} \left.\right) - 1 \left.\right)=1$
$\Rightarrow h\left(\right.x\left.\right)h\left(\right. \frac{1}{x} \left.\right)=1$ as it $f \left(\right. x \left.\right)$ is a polynomial,hence $h \left(\right. x \left.\right)$ will also be a polynomial
$\Rightarrow h \left(\right. x \left.\right) = \pm x^{n}$
$\Rightarrow f \left(\right. x \left.\right) = \pm x^{n} + 1$
Hence $f \left(\right. x \left.\right) = \pm x^{n} + 1$
Now $f \left(\right. 2 \left.\right) > 1$
$\Rightarrow f \left(\right. x \left.\right) = x^{n} + 1$
hence $\underset{x \rightarrow 1}{lim} f \left(\right. x \left.\right) = \underset{x \rightarrow 1}{lim} \left(\right. x^{n} + 1 \left.\right) = 2$
$\Rightarrow f \left(\right. x \left.\right) = \pm x^{n} + 1$
Hence $f \left(\right. x \left.\right) = \pm x^{n} + 1$
Now $f \left(\right. 2 \left.\right) > 1$
$\Rightarrow f \left(\right. x \left.\right) = x^{n} + 1$
hence $\underset{x \rightarrow 1}{lim} f \left(\right. x \left.\right) = \underset{x \rightarrow 1}{lim} \left(\right. x^{n} + 1 \left.\right) = 2$