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JEE Mathematics 2025 Question with Solution

Let f:R→Rf: \mathbb{R} \to \mathbb{R} be a continuous function satisfying f(0)=1f(0) = 1 and f(2x)−f(x)=xf(2x) - f(x) = x for all x∈Rx \in \mathbb{R}. If lim⁡n→∞{f(x)−f(x2n)}=G(x)\lim_{n \to \infty} \left\{ f(x) - f\left( \frac{x}{2^n} \right) \right\} = G(x), then ∑r=110G(r2)\sum_{r=1}^{10} G(r^2) is equal to

  • A

    540540

  • B

    385385

  • C

    420420

  • D

    215215

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: f:R→Rf: \mathbb{R} \to \mathbb{R} is continuous, f(0)=1f(0)=1, and f(2x)−f(x)=xf(2x)-f(x)=x for all x∈Rx \in \mathbb{R}.

Find: ∑r=110G(r2)\sum_{r=1}^{10} G(r^2) where

G(x)=lim⁡n→∞{f(x)−f(x2n)}.G(x)=\lim_{n\to\infty}\left\{f(x)-f\left(\frac{x}{2^n}\right)\right\}.

Use the functional equation at successive arguments:

f(x)−f(x2)=x2f(x)-f\left(\frac{x}{2}\right)=\frac{x}{2} f(x2)−f(x4)=x4f\left(\frac{x}{2}\right)-f\left(\frac{x}{4}\right)=\frac{x}{4} f(x4)−f(x8)=x8f\left(\frac{x}{4}\right)-f\left(\frac{x}{8}\right)=\frac{x}{8}

Continuing up to the nnth step,

f(x2n−1)−f(x2n)=x2n.f\left(\frac{x}{2^{n-1}}\right)-f\left(\frac{x}{2^n}\right)=\frac{x}{2^n}.

Adding these equations gives a telescoping sum:

f(x)−f(x2n)=x(12+14+18+⋯+12n).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).

Hence,

f(x)−f(x2n)=x(1−12n).f(x)-f\left(\frac{x}{2^n}\right)=x\left(1-\frac{1}{2^n}\right).

Now take the limit as n→∞n\to\infty:

G(x)=lim⁡n→∞(f(x)−f(x2n))=lim⁡n→∞x(1−12n)=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.

Therefore,

∑r=110G(r2)=∑r=110r2.\sum_{r=1}^{10} G(r^2)=\sum_{r=1}^{10} r^2.

Using the sum of squares formula,

∑r=110r2=10⋅11⋅216=385.\sum_{r=1}^{10} r^2=\frac{10\cdot 11\cdot 21}{6}=385.

Therefore, the correct option is B.

Telescoping Sum with Limit

Given: f(2x)−f(x)=xf(2x)-f(x)=x and f(0)=1f(0)=1.

Find: G(x)G(x) and then evaluate ∑r=110G(r2)\sum_{r=1}^{10} G(r^2).

Apply the relation at x2k+1\frac{x}{2^{k+1}}:

f(x2k)−f(x2k+1)=x2k+1,k=0,1,…,n−1.f\left(\frac{x}{2^k}\right)-f\left(\frac{x}{2^{k+1}}\right)=\frac{x}{2^{k+1}}, \quad k=0,1,\dots,n-1.

Summing from k=0k=0 to n−1n-1,

f(x)−f(x2n)=∑k=0n−1x2k+1=x∑k=0n−112k+1=x(1−12n).\begin{aligned} 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). \end{aligned}

Now let n→∞n\to\infty:

G(x)=lim⁡n→∞{f(x)−f(x2n)}=x.G(x)=\lim_{n\to\infty}\left\{f(x)-f\left(\frac{x}{2^n}\right)\right\}=x.

So,

∑r=110G(r2)=∑r=110r2=12+22+32+⋯+102.\sum_{r=1}^{10} G(r^2)=\sum_{r=1}^{10} r^2=1^2+2^2+3^2+\cdots+10^2.

Using the formula,

∑r=110r2=10⋅11⋅216=385.\sum_{r=1}^{10} r^2=\frac{10\cdot 11\cdot 21}{6}=385.

Thus, the required value is 385385, so the correct option is B.

Common mistakes

  • A common mistake is to write f(x)−f(x2n)=x(12+14+⋯+12n−1)f(x)-f\left(\frac{x}{2^n}\right)=x\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n-1}}\right) and miss the last term. This gives the wrong finite geometric sum. Include all terms up to x2n\frac{x}{2^n} before taking the limit.

  • Some students incorrectly use continuity to conclude immediately that f(x2n)→0f\left(\frac{x}{2^n}\right)\to 0. Continuity at 00 gives f(x2n)→f(0)=1f\left(\frac{x}{2^n}\right)\to f(0)=1, not 00. Use the telescoping identity carefully.

  • Another mistake is to compute ∑r=110r2\sum_{r=1}^{10} r^2 using the formula for ∑r\sum r instead of the sum of squares formula. Here the correct formula is ∑r=1nr2=n(n+1)(2n+1)6\sum_{r=1}^{n} r^2=\frac{n(n+1)(2n+1)}{6}.

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