Q37 / 75 · JEE Main 2025 (4 Apr Shift 1) · Mathematics

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function satisfying $f(0)=1$ and $f(2 \mathrm{x})-f(\mathrm{x})=\mathrm{x}$ for all $\mathrm{x} \in \mathbb{R}$. If $\lim _{n \rightarrow \infty}\left\{f(x)-f\left(\frac{x}{2^n}\right)\right\}=G(x)$, then $\sum_{r=1}^{10} G\left(r^2\right)$ is equal to

  • A
    540
  • Correct answer
    B
    385
  • C
    420
  • D
    215

Solution

$\begin{aligned}<br/>& f(2 x)-f(x)=x \\ & f(x)-f\left(\frac{x}{2}\right)=\frac{x}{2} \\ & \mathrm{f}\left(\frac{\mathrm{x}}{2}\right)-\mathrm{f}\left(\frac{\mathrm{x}}{4}\right)=\frac{\mathrm{x}}{4} \\ & \mathrm{f}\left(\frac{\mathrm{x}}{4}\right)-\mathrm{f}\left(\frac{\mathrm{x}}{8}\right)=\frac{\mathrm{x}}{8} \\ & f\left(\frac{x}{2^{n-1}}\right)-f\left(\frac{x}{2^n}\right)=\frac{x}{2^n} \\ & f(2 x)-f\left(\frac{x}{2^n}\right)=x\left\{\frac{1-\left(\frac{1}{2}\right)^{n-1}}{1-\frac{1}{2}}\right\} \\ & f(x)-f\left(\frac{x}{2^n}\right)=2 x\left(1-\left(\frac{1}{2}\right)^{n+1}\right) \\ & \mathrm{f}(\mathrm{x})+\mathrm{x}-\mathrm{f}\left(\frac{\mathrm{x}}{2^{\mathrm{n}}}\right)=2 \mathrm{x}\left(1-\left(\frac{1}{2}\right)^{\mathrm{n}+1}\right) \\ & \lim _{n \rightarrow \infty}\left(f(x)-f\left(\frac{x}{2^n}\right)\right)=\lim _{n \rightarrow \infty}\left(2 x\left(1-\left(\frac{1}{2}\right)^{n+1}\right)-x\right) \\ & G(x)=x \\ & \sum_{r=1}^{10} G\left(r^2\right)=\sum_{r=1}^{10} r^2=385<br/>\end{aligned}$

Hence, the correct answer is option: 'B'.

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