JEE Main 2025 — Functions Question with Solution
JEE Main 2025 (4 Apr Shift 1)
Question
Let be a continuous function satisfying and for all . If , then is equal to
Choose an option
Show full solutionCorrect option: B
Correct answer
B385
Step-by-step explanation
$\begin{aligned}
& f(2 x)-f(x)=x \\ & f(x)-f\left(\frac{x}{2}\right)=\frac{x}{2} \\ & \mathrm{f}\left(\frac{\mathrm{x}}{2}\right)-\mathrm{f}\left(\frac{\mathrm{x}}{4}\right)=\frac{\mathrm{x}}{4} \\ & \mathrm{f}\left(\frac{\mathrm{x}}{4}\right)-\mathrm{f}\left(\frac{\mathrm{x}}{8}\right)=\frac{\mathrm{x}}{8} \\ & f\left(\frac{x}{2^{n-1}}\right)-f\left(\frac{x}{2^n}\right)=\frac{x}{2^n} \\ & f(2 x)-f\left(\frac{x}{2^n}\right)=x\left\{\frac{1-\left(\frac{1}{2}\right)^{n-1}}{1-\frac{1}{2}}\right\} \\ & f(x)-f\left(\frac{x}{2^n}\right)=2 x\left(1-\left(\frac{1}{2}\right)^{n+1}\right) \\ & \mathrm{f}(\mathrm{x})+\mathrm{x}-\mathrm{f}\left(\frac{\mathrm{x}}{2^{\mathrm{n}}}\right)=2 \mathrm{x}\left(1-\left(\frac{1}{2}\right)^{\mathrm{n}+1}\right) \\ & \lim _{n \rightarrow \infty}\left(f(x)-f\left(\frac{x}{2^n}\right)\right)=\lim _{n \rightarrow \infty}\left(2 x\left(1-\left(\frac{1}{2}\right)^{n+1}\right)-x\right) \\ & G(x)=x \\ & \sum_{r=1}^{10} G\left(r^2\right)=\sum_{r=1}^{10} r^2=385
\end{aligned}$
& 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
\end{aligned}$
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