JEE Main 2025 Apr 4 Shift 1, Mathematics Q10: Methods to Evaluate Limits

JEE Main2025Apr 4, Shift 1Mathematics

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 $n \to \infty lim {f (x) - f (\frac{x}{2 ^{n}})} = G (x)$ , then $r = 1 \sum 10 G (r^{2})$ is equal to

A
540
B
385
C
420
D
215

Solution

Replace $x$ by $x /2$ in the functional equation: $f (x) - f (x /2) = x /2$ . Iterating, $f (x / 2^{k - 1}) - f (x / 2^{k}) = x / 2^{k}$ .

Telescoping from $k = 1$ to $n$ : $f (x) - f (x / 2^{n}) = k = 1 \sum n \frac{x}{2 ^{k}} = x (1 - \frac{1}{2 ^{n}})$ .

Letting $n \to \infty$ : $G (x) = x$ .

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

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JEE Main 2025 Apr 4 Shift 1, Mathematics Q10: Methods to Evaluate Limits

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