Let f: \mathbbR arrow \mathbbR be a continuous function satisfying f(0)=1 and f(2 x)-f(x)=x for all x \in \mathbbR . If \lim_n arrow \infty\f(x)-f(\fracx2^n)\=G(x) , then \sum_r=1^10 G(r^2) is equal to

Limits Continuity and Differentiability PYQ — JEE Main 2025 · 4 Apr · Shift 1 · Q26 with Solution

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Correct answer: C

We have $f(2x)-f(x)=x \qquad \forall x\in\mathbb R,$ with $f$ continuous and $f(0)=1$ .

We need $G(x)=\lim_{n\to\infty}\left[f(x)-f\left(\frac{x}{2^n}\right)\right].$

Apply the given relation with $x\mapsto \dfrac{x}{2}$ : $f(x)-f\left(\frac{x}{2}\right)=\frac{x}{2}.$

Similarly, with $x\mapsto \dfrac{x}{2^2}$ : $f\left(\frac{x}{2}\right)-f\left(\frac{x}{2^2}\right)=\frac{x}{2^2}.$

Continuing, in general, $f\left(\frac{x}{2^{k-1}}\right)-f\left(\frac{x}{2^k}\right)=\frac{x}{2^k} \qquad (k\ge 1).$

Now sum from $k=1$ to $n$ :

=\sum_{k=1}^n \frac{x}{2^k}.$$ The left side telescopes to $$f(x)-f\left(\frac{x}{2^n}\right).$$ Hence, $$f(x)-f\left(\frac{x}{2^n}\right)=x\sum_{k=1}^n \frac{1}{2^k}.$$ Since $$\sum_{k=1}^n \frac{1}{2^k}=1-\frac{1}{2^n},$$ we get $$f(x)-f\left(\frac{x}{2^n}\right)=x\left(1-\frac{1}{2^n}\right).$$ --- 3. **Take the limit** Therefore, $$G(x)=\lim_{n\to\infty}\left[f(x)-f\left(\frac{x}{2^n}\right)\right] =\lim_{n\to\infty} x\left(1-\frac{1}{2^n}\right)=x.$$ So, $$G(x)=x.$$ --- 4. **Compute the required sum** We need $$\sum_{r=1}^{10} G(r^2)=\sum_{r=1}^{10} r^2.$$ Using $$\sum_{r=1}^{n} r^2=\frac{n(n+1)(2n+1)}{6},$$ for $n=10$, $$\sum_{r=1}^{10} r^2=\frac{10\cdot 11\cdot 21}{6}=385.$$ --- 5. **Option check** - A: $215$ ❌ - B: $420$ ❌ - C: $385$ ✅ - D: $540$ ❌ Thus the correct answer is **Option C**.