Let be a continuous function satisfying f(0) = 1 and f(2x) – f(x) = x for all x, y R. If , then is equal to [2025]

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Solution

Q.
Let $f : R \to R$ be a continuous function satisfying f(0) = 1 and f(2x) – 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 [2025]

1 540

2 420

3 385

4 215

Ans.
(3)

We have, f(2x) – f(x) = x

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

$\Rightarrow f (\frac{x}{2}) - f (\frac{x}{4}) = \frac{x}{4}$

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

On adding all the above statements, we get

$f (2 x) - f (\frac{x}{2^{n}}) = x + \frac{x}{2} + \frac{x}{4} + . . . + \frac{x}{2^{n}}$

$= x {\frac{1 - {(\frac{1}{2})}^{n + 1}}{1 - \frac{1}{2}}} = 2 x [1 - {(\frac{1}{2})}^{n + 1}]$

$\Rightarrow f (x) + x - f (\frac{x}{2^{n}})$

$= 2 x [1 - {(\frac{1}{2})}^{n + 1}]$

$\Rightarrow \lim_{n \to \infty} [f (x) - f (\frac{x}{2^{n}})]$

$= \lim_{n \to \infty} [2 x (1 - {(\frac{1}{2})}^{n + 1}) - x]$

$\Rightarrow G (x) = x \Rightarrow \sum_{r = 1}^{10} r^{2} = 385$ .

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