JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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

Q: Can we just assume $f(x) = x + c$ and solve it?

While guessing $f(x) = x + c$ is a great heuristic to check your answer, it is not a rigorous proof. In JEE Advanced, you must demonstrate the derivation. The telescoping sum method is the formal way to show that $f(x) = x + 1$ is the only continuous solution satisfying the given conditions.

Select Answer:

Visualized Solution

Understanding the Functional Equation

Given: $f (2 x) - f (x) = x$
Target: $G (x) = lim_{n \to \infty} {f (x) - f (\frac{x}{2 ^{n}})}$
Constraint: $f (x)$ is continuous and $f (0) = 1$

The Telescoping Strategy

Strategy: Use the telescoping sum method.
We need terms that cancel each other out when added.
Replace $x$ with $\frac{x}{2}$ in the original equation.

Generating the First Link

Substitute $x \to \frac{x}{2}$ :
$f (2 \cdot \frac{x}{2}) - f (\frac{x}{2}) = \frac{x}{2}$
Simplified: $f (x) - f (\frac{x}{2}) = \frac{x}{2}$

Continuing the Chain

Substitute $x \to \frac{x}{4}$ :
$f (\frac{x}{2}) - f (\frac{x}{4}) = \frac{x}{4}$
The $n$ -th substitution ( $x \to \frac{x}{2 ^{n}}$ ):
$f (\frac{x}{2 ^{n - 1}}) - f (\frac{x}{2 ^{n}}) = \frac{x}{2 ^{n}}$

Summing the Telescoping Series

Add all $n$ equations together.
Observe the diagonal cancellation of terms.
Only the first and last terms on the LHS survive.

The Resulting Equation

LHS: $f (x) - f (\frac{x}{2 ^{n}})$
RHS: $\frac{x}{2} + \frac{x}{4} + \dots + \frac{x}{2 ^{n}}$
$f (x) - f (\frac{x}{2 ^{n}}) = x (\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2 ^{n}})$

Applying the Limit for $G (x)$

$G (x) = lim_{n \to \infty} {f (x) - f (\frac{x}{2 ^{n}})}$
Substitute the RHS into the limit.
$G (x) = lim_{n \to \infty} x (\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2 ^{n}})$

Evaluating the Infinite GP

Infinite GP sum formula: $S_{\infty} = \frac{a}{1 - r}$
Here, $a = \frac{1}{2}$ and $r = \frac{1}{2}$ .
$S_{\infty} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1$
Therefore, $G (x) = x \cdot 1 = x$ .

Setting up the Final Summation

We need to find: $\sum_{r = 1}^{10} G (r^{2})$
Since $G (x) = x$ , we substitute $x = r^{2}$ .
The expression becomes: $\sum_{r = 1}^{10} r^{2}$

Applying the Sum of Squares Formula

Formula: $\sum_{r = 1}^{n} r^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$
Substitute $n = 10$ :
Sum $= \frac{10 ( 10 + 1 ) ( 2 \cdot 10 + 1 )}{6}$

Final Calculation

Sum $= \frac{10 \times 11 \times 21}{6}$
Simplify: $\frac{2310}{6} = 385$
The final answer is 385.

00:00 / 00:00

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Solution Diagram

Analyzing the Setup

We are given a continuous function

f : R \to R

satisfying

f (0) = 1

and the functional equation:

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

Our objective is to determine the value of

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

, where

G (x)

is defined by the limit:

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

The Telescoping Strategy

To solve this, we observe the behavior of the function by substituting

x

with smaller values. Replacing

x

with

\frac{x}{2}

in the original equation yields:

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

Continuing this substitution process for

n

steps, we generate a sequence of equations:

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

Building the Chain

When we sum these equations from

k = 1

n

, the intermediate terms cancel out in a telescoping fashion. This leaves us with the following expression:

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

The Limit and the Identity

We now evaluate the limit

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

. Substituting our summation result, we obtain:

G (x) = x \cdot n \to \infty lim k = 1 \sum n (\frac{1}{2})^{k}

The summation is an infinite geometric series with first term

a = \frac{1}{2}

and common ratio

r = \frac{1}{2}

. Using the formula

S_{\infty} = \frac{a}{1 - r}

, we find:

S_{\infty} = \frac{1/2}{1 - 1/2} = 1

Thus, the function simplifies to the identity function:

G (x) = x

The Final Summation

With

G (x) = x

, the required sum becomes the sum of the squares of the first ten natural numbers:

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

Applying the standard formula

\sum_{r = 1}^{n} r^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

for

n = 10

r = 1 \sum 10 r^{2} = \frac{10 ( 11 ) ( 21 )}{6} = \frac{2310}{6}

The final result is:

385

Common Traps & FAQs

Wait, why does

f (0) = 1

not seem to matter in the final calculation?

That is a brilliant observation! In this specific problem, the continuity of

f

x = 0

ensures that

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

. However, because the telescoping sum

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

already isolates

f (x)

and

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

, the limit of the sum is determined by the geometric series. The value

f (0) = 1

is a constraint that guarantees the function is well-defined and continuous, but the functional equation itself dictates the behavior of

G (x)

Can we just assume

f (x) = x + c

and solve it?

While guessing

f (x) = x + c

is a great heuristic to check your answer, it is not a rigorous proof. In JEE Advanced, you must demonstrate the derivation. The telescoping sum method is the formal way to show that

f (x) = x + 1

is the only continuous solution satisfying the given conditions.

Why does the telescoping sum work here?

Telescoping sums work whenever you can express a term as a difference of two values,

a_{k} - a_{k + 1}

. When you sum these from

k = 1

n

, the intermediate terms like

a_{2}, a_{3}, \dots, a_{n}

all cancel out, leaving only the first and the last terms. Here, we transformed the functional equation into the form

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

, which is perfectly set up for this cancellation.

Select Answer:

Visualized Solution

Understanding the Functional Equation

The Telescoping Strategy

Generating the First Link

Continuing the Chain

Summing the Telescoping Series

The Resulting Equation

Applying the Limit for $G (x)$

Evaluating the Infinite GP

Setting up the Final Summation

Applying the Sum of Squares Formula

Final Calculation

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Telescoping Strategy

Building the Chain

The Limit and the Identity

The Final Summation

Similar Questions

Let $f (x) = lim_{n \to \infty} \sum_{r = 0}^{n} (\frac{t a n ( x / 2 ^{r + 1} ) + t a n ^{3} ( x / 2 ^{r + 1} )}{1 - t a n ^{2} ( x / 2 ^{r + 1} )}) .$ Then $lim_{x \to 0} \frac{e ^{x} - e ^{f (x)}}{x - f ( x )}$ is equal to

$lim_{n \to \infty} {(2^{\frac{1}{2}} - 2^{\frac{1}{3}}) (2^{\frac{1}{2}} - 2^{\frac{1}{5}}) \dots (2^{\frac{1}{2}} - 2^{\frac{1}{2 n + 1}})}$ is equal to

The value of $lim_{n \to \infty} \frac{[ r ] + [ 2 r ] + \dots + [ n r ]}{n ^{2}}$ , where $r$ is non-zero real number and $[r]$ denotes the greatest integer less than or equal to $r$ , is equal to :

$lim_{n \to \infty} (1 + \frac{1}{n ^{2}} + \frac{1}{n ^{2}} + ....... + \frac{1}{n})^{n}$ is equal to :

Let $f : R \to R$ be a differentiable function satisfying $f^{'} (3) + f^{'} (2) = 0$ . Then $lim_{x \to 0} (\frac{1 + f ( 3 + x ) - f ( 3 )}{1 + f ( 2 - x ) - f ( 2 )})^{1/ x}$ is equal to

$lim_{n \to \infty} \frac{( 1 ^{2} - 1 ) ( n - 1 ) + ( 2 ^{2} - 2 ) ( n - 2 ) + \dots + (( n - 1 ) ^{2} - ( n - 1 )) \cdot 1}{( 1 ^{3} + 2 ^{3} + \dots + n ^{3} ) - ( 1 ^{2} + 2 ^{2} + \dots + n ^{2} )}$ is equal to :

Let $f : (- \infty, \infty) - {0} \to R$ be a differentiable function such that $f^{'} (1) = lim_{a \to \infty} a^{2} f (\frac{1}{a})$ . Then $lim_{a \to \infty} \frac{a ( a + 1 )}{2} tan^{- 1} (\frac{1}{a}) + a^{2} - 2 lo g_{e} a$ is equal to

The value of $lim_{x \to 0} 2 (\frac{1 - c o s x c o s 2 x 3 c o s 3 x \dots 10 c o s 10 x}{x ^{2}})$ is

Let $f : [- \frac{π}{2}, \frac{π}{2}] \to R$ be a differentiable function such that $f (0) = \frac{1}{2}$ . If $lim_{x \to 0} \frac{x \int _{0}^{x} f ( t ) d t}{e ^{x^{2}} - 1} = α$ , then $8 α^{2}$ is equal to :

Let $f (x) = {x - 1, 2 x, x is even, x is odd, x \in N$ . If for some $a \in N, f (f (f (a))) = 21$ , then $lim_{x \to a^{-}} {\frac{∣ x ∣ ^{3}}{a} - [\frac{x}{a}]}$ , where $[t]$ denotes the greatest integer less than or equal to $t$ , is equal to :

Select Answer:

Visualized Solution

Understanding the Functional Equation

The Telescoping Strategy

Generating the First Link

Continuing the Chain

Summing the Telescoping Series

The Resulting Equation

Applying the Limit for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G(x)

Evaluating the Infinite GP

Setting up the Final Summation

Applying the Sum of Squares Formula

Final Calculation

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Telescoping Strategy

Building the Chain

The Limit and the Identity

The Final Summation

Similar Questions

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Applying the Limit for $G (x)$

Let $f (x) = lim_{n \to \infty} \sum_{r = 0}^{n} (\frac{t a n ( x / 2 ^{r + 1} ) + t a n ^{3} ( x / 2 ^{r + 1} )}{1 - t a n ^{2} ( x / 2 ^{r + 1} )}) .$ Then $lim_{x \to 0} \frac{e ^{x} - e ^{f (x)}}{x - f ( x )}$ is equal to

$lim_{n \to \infty} {(2^{\frac{1}{2}} - 2^{\frac{1}{3}}) (2^{\frac{1}{2}} - 2^{\frac{1}{5}}) \dots (2^{\frac{1}{2}} - 2^{\frac{1}{2 n + 1}})}$ is equal to

The value of $lim_{n \to \infty} \frac{[ r ] + [ 2 r ] + \dots + [ n r ]}{n ^{2}}$ , where $r$ is non-zero real number and $[r]$ denotes the greatest integer less than or equal to $r$ , is equal to :

$lim_{n \to \infty} (1 + \frac{1}{n ^{2}} + \frac{1}{n ^{2}} + ....... + \frac{1}{n})^{n}$ is equal to :

Let $f : R \to R$ be a differentiable function satisfying $f^{'} (3) + f^{'} (2) = 0$ . Then $lim_{x \to 0} (\frac{1 + f ( 3 + x ) - f ( 3 )}{1 + f ( 2 - x ) - f ( 2 )})^{1/ x}$ is equal to

$lim_{n \to \infty} \frac{( 1 ^{2} - 1 ) ( n - 1 ) + ( 2 ^{2} - 2 ) ( n - 2 ) + \dots + (( n - 1 ) ^{2} - ( n - 1 )) \cdot 1}{( 1 ^{3} + 2 ^{3} + \dots + n ^{3} ) - ( 1 ^{2} + 2 ^{2} + \dots + n ^{2} )}$ is equal to :

Let $f : (- \infty, \infty) - {0} \to R$ be a differentiable function such that $f^{'} (1) = lim_{a \to \infty} a^{2} f (\frac{1}{a})$ . Then $lim_{a \to \infty} \frac{a ( a + 1 )}{2} tan^{- 1} (\frac{1}{a}) + a^{2} - 2 lo g_{e} a$ is equal to

The value of $lim_{x \to 0} 2 (\frac{1 - c o s x c o s 2 x 3 c o s 3 x \dots 10 c o s 10 x}{x ^{2}})$ is

Let $f : [- \frac{π}{2}, \frac{π}{2}] \to R$ be a differentiable function such that $f (0) = \frac{1}{2}$ . If $lim_{x \to 0} \frac{x \int _{0}^{x} f ( t ) d t}{e ^{x^{2}} - 1} = α$ , then $8 α^{2}$ is equal to :