If $f: R \rightarrow R$ is a continuous function satisfying $f(0)=1$ and $f(2 x)-f(x)=x$ for all $x \in R$ and $\lim\limits_{x \rightarrow \infty}\left\{f(x)-f\left(\frac{x}{2^n}\right)\right\}=P(x)$, then $P(x)$, is
Answer & explanation
Correct answer: option 2
We have,
$f(2 x)-f(x)=x$
$\Rightarrow f(x)-f\left(\frac{x}{2}\right)=\frac{x}{2}$ [Replacing x by $\frac{x}{2}$]
$\Rightarrow f\left(\frac{x}{2}\right)-f\left(\frac{x}{4}\right)=\frac{x}{4}$
$\Rightarrow f\left(\frac{x}{4}\right)-f\left(\frac{x}{8}\right)=\frac{x}{8}$
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$f\left(\frac{x}{2^{n-1}}\right)-f\left(\frac{x}{2^n}\right)=\frac{x}{2^n}$
Adding up all these equations, we obtain
$f(x)-f\left(\frac{x}{2^n}\right) =\frac{x}{2}+\frac{x}{4}+\frac{x}{8}+...+\frac{x}{2^n}$
$\Rightarrow f(x)-f\left(\frac{x}{2^n}\right) =\frac{x}{2} \frac{\left(1-\frac{1}{2^n}\right)}{\left(1-\frac{1}{2}\right)}$
$\Rightarrow f(x)-f\left(\frac{x}{2^n}\right)=x\left(1-\frac{1}{2^n}\right)$
$\Rightarrow \lim\limits_{n \rightarrow \infty}\left\{f(x)-f\left(\frac{x}{2^n}\right)\right\}=x$
$\Rightarrow P(x)=x$, which is a linear polynomial in x.