Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function satisfying $f(0) = 1$ and $f(2x) - f(x) = x$ for all $x \in \mathbb{R}$. If $\lim_{n \to \infty} \left\{ f(x) - f\left( \frac{x}{2^n} \right) \right\} = G(x)$, then $\sum_{r=1}^{10} G(r^2)$ is equal to

• A. 540
• B. 385
• C. 420
• D. 215

Hint:

Use the sum of squares formula to calculate the sum.

Answer 1

The Correct Option is B

1. Use the given functional equation: \[ f(2x) - f(x) = x \]
2. Express $f(x)$ in terms of $f\left( \frac{x}{2^n} \right)$: \[ f(x) - f\left( \frac{x}{2} \right) = \frac{x}{2} \] \[ f\left( \frac{x}{2} \right) - f\left( \frac{x}{4} \right) = \frac{x}{4} \] \[ f\left( \frac{x}{4} \right) - f\left( \frac{x}{8} \right) = \frac{x}{8} \] \[ \vdots \] \[ f\left( \frac{x}{2^{n-1}} \right) - f\left( \frac{x}{2^n} \right) = \frac{x}{2^n} \]
3. Sum the series: \[ f(x) - f\left( \frac{x}{2^n} \right) = x \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{2^n} \right) \] \[ f(x) - f\left( \frac{x}{2^n} \right) = x \left( 1 - \frac{1}{2^n} \right) \]
4. Take the limit as $n \to \infty$: \[ G(x) = \lim_{n \to \infty} \left( f(x) - f\left( \frac{x}{2^n} \right) \right) = x \] 5. Calculate $\sum_{r=1}^{10} G(r^2)$: \[ \sum_{r=1}^{10} G(r^2) = \sum_{r=1}^{10} r^2 = 1^2 + 2^2 + 3^2 + \cdots + 10^2 \] \[ \sum_{r=1}^{10} r^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385 \] Therefore, the correct answer is (2) 385.

Answer 2

We are given a continuous function \( f:\mathbb{R}\to\mathbb{R} \) with \( f(0)=1 \) satisfying \( f(2x)-f(x)=x \) for all \( x \in \mathbb{R} \). We must find \( G(x)=\lim_{n\to\infty}\{f(x)-f(x/2^n)\} \) and then evaluate \( \sum_{r=1}^{10} G(r^2) \).

Concept Used:

Use the functional equation iteratively at the arguments \( x/2^{k+1} \) to obtain a telescoping sum for \( f(x)-f(x/2^n) \). Then pass to the limit as \( n\to\infty \). Finally, compute the required finite sum.

Step-by-Step Solution:

Step 1: Apply the given relation at \( x/2^{k+1} \):

\[ f\!\left(\frac{x}{2^{k}}\right)-f\!\left(\frac{x}{2^{k+1}}\right) = f\!\left(2\cdot\frac{x}{2^{k+1}}\right)-f\!\left(\frac{x}{2^{k+1}}\right) = \frac{x}{2^{k+1}},\quad k=0,1,\dots,n-1. \]

Step 2: Sum from \( k=0 \) to \( n-1 \) (telescoping):

\[ f(x)-f\!\left(\frac{x}{2^{n}}\right)=\sum_{k=0}^{n-1}\frac{x}{2^{k+1}} = x\!\sum_{k=0}^{n-1}\frac{1}{2^{k+1}} = x\!\left(1-\frac{1}{2^{n}}\right). \]

Step 3: Pass to the limit to get \( G(x) \):

\[ G(x)=\lim_{n\to\infty}\left\{f(x)-f\!\left(\frac{x}{2^{n}}\right)\right\} = \lim_{n\to\infty} x\!\left(1-\frac{1}{2^{n}}\right)=x. \]

(Equivalently, continuity gives \( f(x/2^n)\to f(0)=1 \), and the above identity also implies \( f(x)-1=x \Rightarrow f(x)=1+x \), consistent with the relation.)

Final Computation & Result

Thus \( G(x)=x \). Therefore,

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

Answer: \(385\).