Problem 18 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(f: R \rightarrow R\) 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 (a) \(\frac{7 n(n+1)}{2}\) (b) \(\frac{7 n}{2}\) (c) \(\frac{7(n+1)}{2}\) (d) \(7 n+(n+1)\)

Step-by-Step Solution

Answer

Option (a): \(\frac{7n(n+1)}{2}\).

Step 1: Understanding Function Property

We are given the functional equation \( f(x+y) = f(x) + f(y) \). This is known as Cauchy's functional equation. The given solution for \( f(1) = 7 \) can be used to find \( f(n) \) for any integer.

Step 2: Determine Function Form for Integers

Since \( f(1) = 7 \) and the equation holds additively, we can deduce that \( f(n) = n \cdot f(1) = 7n \) for all integers \( n \) by induction or by recognizing the homogeneity of linear functions.

Step 3: Setup the Summation

We want to find \( \sum_{r=1}^{n} f(r) \). Knowing that \( f(r) = 7r \), this becomes \( \sum_{r=1}^{n} 7r \).

Step 4: Simplify the Summation

Factor out the constant: \( 7\sum_{r=1}^{n} r \). Recognize that \( \sum_{r=1}^{n} r \) is the formula for the sum of the first \( n \) integers, which is \( \frac{n(n+1)}{2} \).

Step 5: Calculate Final Expression

Substitute back: \( 7 \cdot \frac{n(n+1)}{2} = \frac{7n(n+1)}{2} \). This represents the value of the summation.

Key Concepts

Additive Functions

Additive functions are a special class of functions characterized by the property that the function value of a sum is equal to the sum of the function values: \(f(x+y) = f(x) + f(y)\). This property is a central feature of Cauchy's functional equation. When a function satisfies this equation, it's said to be 'additive'.

For an additive function, it's important to realize that the property holds for all real numbers. This means that the output of the function conserves how inputs are added or subtracted. It's like the function respects the arithmetic operations we do between input numbers. In our exercise case, knowing that \(f(1)=7\) allows us to derive all values of \(f(n)\) using the additive property. Since \(f(n)\) is systematically built by adding \(f(1)\) repeatedly, \(f(n)\) simplifies to \(7n\). This realization is critical in fully utilizing the nature of additive functions.

Homogeneity of Linear Functions

The homogeneity of linear functions means that for any scalar multiple, the output of a function can simply be the scalar times the image of the function. In the context of our exercise, after recognizing that \(f(x)\) follows the additive rule, we can claim it adheres to homogeneity of a linear function.

This means if \( f(1) = 7 \), then \( f(n) \) for any integer \( n \) would be \( f(n) = n \cdot f(1) = 7n \). This is a key insight that stems from the homogeneity principle. Here, the constant \(7\) is a scalar multiplier that adjusts \( n \), maintaining the linearity—the straight proportionality of the function. This understanding simplifies our function evaluation processes across various inputs.

Summation of Integers

The summation of integers is a concept frequently encountered in mathematics, where we aim to find the total sum of a sequence of numbers. A common formula used for this is the sum of the first \( n \) natural numbers, given by \( \sum_{r=1}^{n}r = \frac{n(n+1)}{2} \).

In the original problem, we used this summation concept to find \( \sum_{r=1}^{n} f(r) \), knowing that \( f(r) = 7r \). We factored out the constant \(7\) and applied the formula for the sum of the first \( n \) integers, leading us to \( 7 \times \frac{n(n+1)}{2} \). This result, \( \frac{7n(n+1)}{2} \), elegantly utilizes both the linear nature of the function \(f\) and the classical summation of natural numbers. Recognizing this formula is crucial as it effectively summarizes the sum of outputs from an additive function across a specified range.