JEE Main 2025 — 4 April, Morning Shift — Question 39

$f(0)=1, f(2 x)-f(x)=x$

Replace $x \rightarrow \frac{x}{2}$

$$\begin{gathered} f(x)-f\left(\frac{x}{2}\right)=\frac{x}{2} \end{gathered}$$

Again, Replace $x \rightarrow \frac{x}{2}$

$$\begin{gathered} f\left(\frac{x}{2}\right)-f\left(\frac{x}{2^{2}}\right)=\frac{x}{2^{2}} \end{gathered}$$ $$\begin{gathered} f\left(\frac{x}{2^{n-1}}\right)-f\left(\frac{x}{2^{n}}\right)=\frac{x}{2^{n}} \end{gathered}$$

Adding $(1)+(2)+(3)+\ldots+(n)$

We get $f(x)-f\left(\frac{x}{2^{n}}\right)=\frac{x}{2}+\frac{x}{2^{2}}+\ldots .+\frac{x}{2^{n}}$

$\lim _{n \rightarrow \infty}\left(f(x)-f\left(\frac{x}{2^{n}}\right)\right)=\lim _{n \rightarrow \infty}\left(\frac{x}{2}+\frac{x}{2^{n}}+\ldots+\frac{x}{2^{n}}\right)$

$f(x)-f(0)=\frac{\frac{x}{2}}{\frac{1}{2}}$

$\Rightarrow \quad G(x)=x$

$\Rightarrow G\left(r^{2}\right)=r^{2}$

$\Rightarrow \quad \sum_{r=1}^{10} G\left(r^{2}\right)=\sum_{r=1}^{10} G\left(r^{2}\right)$

$=\frac{(10)(11)(21)}{6}=(55) 7$

$\Rightarrow 385$