JEE Main 2003

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Animated Solution for Mathematics - Functions: If $f : R \to R$ satisfies $f (x + y) = f (x) + f (y)$ , for all $x, y \in R$ and $f (1) = 7$ , then $\sum_{r = 1}^{n} f (r)$ is

Q: Can I solve this by induction instead of using the general form?

Yes, you absolutely can! You can prove $f(n) = nf(1)$ for all natural numbers $n$ using induction. It's a fantastic way to verify the result. The general form $f(x) = cx$ is just a faster, more elegant way to get there.

Select Answer:

Progressive Hints

Hint 1

Look closely at the structure

f (x + y) = f (x) + f (y)

. This is not merely an algebraic relation; it is the hallmark of an additive function. Ask yourself: what kind of mathematical object preserves the operation of addition? This is the gateway to understanding the linearity of the function.

Hint 2

You are dealing with Cauchy's Functional Equation. For continuous functions defined on the real numbers, this equation forces the function to take the form

f (x) = c x

. Your task is to determine the constant

c

using the provided boundary condition

f (1) = 7

Hint 3

With the function identified as

f (x) = 7 x

, the summation

\sum_{r = 1}^{n} f (r)

transforms into

\sum_{r = 1}^{n} 7 r

. Use the linearity property of the summation operator to extract the constant

7

and apply the standard formula

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

to reach the solution.

Visualized Solution

Identify the Functional Equation

Given functional equation: $f (x + y) = f (x) + f (y)$
Domain and Range: $f : R \to R$
Given value: $f (1) = 7$
Objective: Find $\sum_{r = 1}^{n} f (r)$

Cauchy's Functional Equation

The relation $f (x + y) = f (x) + f (y)$ is known as Cauchy's Functional Equation.
For continuous functions on $R$ , the general solution is $f (x) = c x$ , where $c$ is a constant.
This represents a straight line passing through the origin $(0, 0)$ .

Visualizing the Function $f (x) = c x$

Since $f (x) = c x$ , the graph of this function is a straight line passing through the origin $(0, 0)$ with slope $c$ .
The constant $c$ determines the steepness of the line.

Finding the Constant $c$

We are given $f (1) = 7$ .
Substitute $x = 1$ into $f (x) = c x$ :
$f (1) = c \cdot 1 ⟹ 7 = c$ .

The Function $f (x) = 7 x$

Since $c = 7$ , our function is uniquely determined as:
$f (x) = 7 x$

Evaluating $f (r)$ for $r = 1, 2, 3, \dots$

Let's find the values of $f (r)$ for positive integers:
$f (1) = 7$
$f (2) = 14$
$f (3) = 21$
In general, $f (r) = 7 r$ .

Setting up the Summation $\sum_{r = 1}^{n} f (r)$

We need to find the sum: $S = \sum_{r = 1}^{n} f (r)$
Substitute $f (r) = 7 r$ into the summation:
$S = \sum_{r = 1}^{n} 7 r$

Factoring out the Constant

Using the linearity property of summation:
$\sum_{r = 1}^{n} 7 r = 7 \sum_{r = 1}^{n} r$

Sum of First $n$ Natural Numbers

Recall the standard formula for the sum of the first $n$ natural numbers:
$\sum_{r = 1}^{n} r = \frac{n ( n + 1 )}{2}$

Final Summation Formula

Substitute the sum formula back into our expression:
$S = 7 \cdot \frac{n ( n + 1 )}{2} = \frac{7 n ( n + 1 )}{2}$
This matches Option A.

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The Sigma Insight: Classification of Functions

Solution Diagram

Analyzing the Setup

The equation

f (x + y) = f (x) + f (y)

is known as Cauchy's Functional Equation. It defines an additive function, which, under the assumption of continuity or other mild conditions in the realm of real numbers, forces the function to take the linear form:

f (x) = c x

This equation represents a straight line passing through the origin

(0, 0)

, where

c

acts as a constant scaling factor or the slope of the line.

The Detective Work

Finding the Constant

We are provided with the specific condition

f (1) = 7

. This serves as our anchor point to determine the value of the constant

c

By substituting

x = 1

into our general form

f (x) = c x

, we obtain:

f (1) = c \cdot 1

Since we know

f (1) = 7

, it follows immediately that

c = 7

. Thus, the function is uniquely identified as:

f (x) = 7 x

The Grand Summation

Our objective is to calculate the sum

\sum_{r = 1}^{n} f (r)

. Substituting our derived function

f (r) = 7 r

into the summation, we get:

r = 1 \sum n 7 r

Using the linearity property of the summation operator, we can factor out the constant

7

7 \cdot r = 1 \sum n r

Recalling the standard formula for the sum of the first

n

natural numbers,

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

, we substitute this into our expression:

7 \cdot \frac{n ( n + 1 )}{2}

The final result of the summation is:

\frac{7n ( n + 1 )}{2}

Common Traps & FAQs

Wait, why can we just assume

f (x) = c x

? Is it possible there are other weird functions?

That is a brilliant question! For the general case of Cauchy's equation, if we do not assume continuity, there are indeed 'pathological' solutions that are not linear. However, in the context of JEE Advanced, we almost always work with continuous functions. For continuous functions,

f (x) = c x

is the unique solution. You are thinking like a mathematician!

Why can't we just use energy conservation or some other physics principle?

I love the enthusiasm! But remember, this is pure algebra. We aren't dealing with physical systems here, but with the 'DNA' of functions. The 'conservation' here is the conservation of the additive property, which leads us directly to the linear structure.

Can I solve this by induction instead of using the general form?

Yes, you absolutely can! You can prove

f (n) = n f (1)

for all natural numbers

n

using induction. It's a fantastic way to verify the result. The general form

f (x) = c x

is just a faster, more elegant way to get there.

Animated Solution for Mathematics - Functions: If $f : R \to R$ satisfies $f (x + y) = f (x) + f (y)$ , for all $x, y \in R$ and $f (1) = 7$ , then $\sum_{r = 1}^{n} f (r)$ is

Select Answer:

Visualized Solution

Identify the Functional Equation

Cauchy's Functional Equation

Visualizing the Function $f (x) = c x$

Finding the Constant $c$

The Function $f (x) = 7 x$

Evaluating $f (r)$ for $r = 1, 2, 3, \dots$

Setting up the Summation $\sum_{r = 1}^{n} f (r)$

Factoring out the Constant

Sum of First $n$ Natural Numbers

Final Summation Formula

The Sigma Insight: Classification of Functions

Analyzing the Setup

The Detective Work

The Grand Summation

Similar Questions

Let $f : R \to R$ be a function defined by $f (x) = (2 + 3 a) x^{2} + (\frac{a + 2}{a - 1}) x + b, a \neq = 1$ If $f (x + y) = f (x) + f (y) + 1 - \frac{2}{7} x y$ , then the value of $28 \sum_{i = 1}^{5} ∣ f (i) ∣$ is

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If a function $f$ satisfies $f (m + n) = f (m) + f (n)$ for all $m, n \in N$ and $f (1) = 1$ , then the largest natural number $λ$ such that $\sum_{k = 1}^{2022} f (λ + k) \leq (2022)^{2}$ is equal to _________

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A function $f (x)$ is given by $f (x) = \frac{5 ^{x}}{5 ^{x} + 5}$ , then the sum of the series $f (\frac{1}{20}) + f (\frac{2}{20}) + f (\frac{3}{20}) + \dots + f (\frac{39}{20})$ is equal to:

The number of functions $f : {1, 2, 3, 4} \to {a \in Z : ∣ a ∣ \leq 8}$ satisfying $f (n) + \frac{1}{n} f (n + 1) = 1, \forall n \in {1, 2, 3}$ is

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Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f:R→R satisfies f(x+y)=f(x)+f(y), for all x,y∈R and f(1)=7, then ∑r=1n​f(r) is

Select Answer:

Visualized Solution

Identify the Functional Equation

Cauchy's Functional Equation

Visualizing the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)=cx

Finding the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

The Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)=7x

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(r) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=1,2,3,…

Setting up the Summation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑r=1n​f(r)

Factoring out the Constant

Sum of First .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n Natural Numbers

Final Summation Formula

The Sigma Insight: Classification of Functions

Analyzing the Setup

The Detective Work

The Grand Summation

Similar Questions

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Animated Solution for Mathematics - Functions: If $f : R \to R$ satisfies $f (x + y) = f (x) + f (y)$ , for all $x, y \in R$ and $f (1) = 7$ , then $\sum_{r = 1}^{n} f (r)$ is

Visualizing the Function $f (x) = c x$

Finding the Constant $c$

The Function $f (x) = 7 x$

Evaluating $f (r)$ for $r = 1, 2, 3, \dots$

Setting up the Summation $\sum_{r = 1}^{n} f (r)$

Sum of First $n$ Natural Numbers

Let $f : R \to R$ be a function defined by $f (x) = (2 + 3 a) x^{2} + (\frac{a + 2}{a - 1}) x + b, a \neq = 1$ If $f (x + y) = f (x) + f (y) + 1 - \frac{2}{7} x y$ , then the value of $28 \sum_{i = 1}^{5} ∣ f (i) ∣$ is

Let $a, b, c \in R$ . If $f (x) = a x^{2} + b x + c$ is such that $a + b + c = 3$ and $f (x + y) = f (x) + f (y) + x y, \forall x, y \in R$ , then $\sum_{n = 1}^{10} f (n)$ is equal to:

Let $f : R - {0, 1} \to R$ be a function such that $f (x) + f (\frac{1}{1 - x}) = 1 + x$ . Then $f (2)$ is equal to :

Let $f : N \to R$ be a function such that $f (x + y) = 2 f (x) f (y)$ for natural numbers $x$ and $y$ . If $f (1) = 2$ , then the value of $α$ for which $\sum_{k = 1}^{10} f (α + k) = \frac{512}{3} (2^{10} - 1)$ holds, is

If a function $f$ satisfies $f (m + n) = f (m) + f (n)$ for all $m, n \in N$ and $f (1) = 1$ , then the largest natural number $λ$ such that $\sum_{k = 1}^{2022} f (λ + k) \leq (2022)^{2}$ is equal to _________

If $f (x) = \frac{2 ^{x}}{2 ^{x} + 2}, x \in R,$ then $\sum_{k = 1}^{81} f (\frac{k}{82})$ is equal to

Let $S = {1, 2, 3, 4, 5, 6, 7}$ . Then the number of possible functions $f : S \to S$ such that $f (m \cdot n) = f (m) \cdot f (n)$ for every $m, n \in S$ and $m \cdot n \in S$ is equal to

A function $f (x)$ is given by $f (x) = \frac{5 ^{x}}{5 ^{x} + 5}$ , then the sum of the series $f (\frac{1}{20}) + f (\frac{2}{20}) + f (\frac{3}{20}) + \dots + f (\frac{39}{20})$ is equal to:

The number of functions $f : {1, 2, 3, 4} \to {a \in Z : ∣ a ∣ \leq 8}$ satisfying $f (n) + \frac{1}{n} f (n + 1) = 1, \forall n \in {1, 2, 3}$ is

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