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

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Limits XII

Let

f : R \to R

be a continuous function satisfying

f (0) = 1

and

f (2 x) - f (x) = x

for all

x \in R

. If

lim_{n \to \infty} {f (x) - f (\frac{x}{2 ^{n}})} = G (x)

, then

\sum_{r = 1}^{10} G (r^{2})

is equal to

Answer: (B)

From

f (2 x) - f (x) = x

comes

f (x) - f (\frac{x}{2}) = \frac{x}{2}

, and telescoping down to

\frac{x}{2 ^{n}}

gives

f (x) - f (\frac{x}{2 ^{n}}) = x (1 - \frac{1}{2 ^{n}})

. Letting

n \to \infty

, with continuity ensuring

f (\frac{x}{2 ^{n}}) \to f (0)

, gives

G (x) = x

. Therefore

\sum_{r = 1}^{10} G (r^{2}) = \sum_{r = 1}^{10} r^{2} = \frac{10 \cdot 11 \cdot 21}{6} = 385

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option