Problem 146 - A Complete Resource Book in Mathematics for JEE Main

Question

Assertion: If \(f(x)=\frac{a^{x}}{a^{x}+\sqrt{a}}(a>0)\), then \(\sum_{r=1}^{2 n-1} 2 f\left(\frac{r}{2 n}\right)=2 n-1\) Reason: \(f(x)+f(1-x)=1 \forall x\)

Step-by-Step Solution

Answer

The sum equals \( 2n - 1 \), confirming that the assertion is correct.

Step 1: Understanding the assertion

The problem provides a function \( f(x) = \frac{a^x}{a^x + \sqrt{a}} \) where \( a > 0 \). The task is to evaluate the sum \( \sum_{r=1}^{2n-1} 2 f\left(\frac{r}{2n}\right) \) and show that it equals \( 2n-1 \).

Step 2: Understanding the reason

The reason given is \( f(x) + f(1-x) = 1 \) for all \( x \). This property of the function can be useful in simplifying the sum for our assertion.

Step 3: Simplifying the sum

Consider the transformation using the property \( f\left(\frac{r}{2n}\right) + f\left(\frac{2n-r}{2n}\right) = 1 \). Tranform the summation into pairs: \( r + (2n-r) = 2n \), which leads to this identity \( f\left(\frac{r}{2n}\right) + f\left(1-\frac{r}{2n}\right) = 1 \) for \( r = 1, 2, \ldots, n-1 \).

Step 4: Evaluating the sum using pair property

The pairs for \( r \) and \( 2n-r \) exist for each \( r \) from 1 to \( n-1 \). For these \( n-1 \) distinct pairs, the contributions sum up to \( n-1 \) because each pair sums to 2 due to the doubling factor in the sum: \( 2(f\left(\frac{r}{2n}\right) + f\left(1-\frac{r}{2n}\right)) = 2\).

Step 5: Consider the midpoint and edge case

When \( r = n \), \( f\left(\frac{n}{2n}\right) = f\left(\frac{1}{2}\right) \). For the middle term, the contribution is naturally 1 due to symmetry, as \( f\left(\frac{r}{2n}\right) + f\left(1-\frac{r}{2n}\right) = 1 \), implying \( 2f\left(\frac{1}{2}\right) = 1 \), giving an extra contribution of 1.

Step 6: Final result

Thus, summing the pairs contributes \( n-1 \times 2 = 2(n-1) \) and adding the middle term contributes additional 1, yielding \( 2(n-1) + 1 = 2n-1 \).

Step 7: Final Calculation

Calculate: \(2(n-1) + 1 = 2n - 1\) confirming that the assertion is correct.

Key Concepts

Real Functions

In mathematics, a **real function** is a function that associates a real number to each element in its domain, which is also a set of real numbers. One way to represent such a function is through its formula, as we see in the problem where \( f(x) = \frac{a^x}{a^x + \sqrt{a}} \). Here, \( a \) is a positive real number, and \( x \) can take any real value.
Real functions are fundamental because they help us model and solve real-world problems where inputs and outputs are both real numbers.
**Key Properties of Real Functions:**

• **Continuity:** A real function can be continuous, meaning that small changes in the input \( x \) result in small changes in the output \( f(x) \).
• **Domain and Range:** For the given function, the domain is \( \mathbb{R} \), the set of all real numbers, and the range, intuitively, is between 0 and 1 since both the numerator and denominator are positive, and the denominator is greater than the numerator.
• **Symmetric Properties:** For some functions, special properties such as \( f(x) + f(1-x) = 1 \) can simplify calculations.

These concepts are critical to understanding how to work with real functions and apply them, like finding sums or solving equations.

Summation

**Summation** refers to the addition of a sequence of numbers, and it is denoted by the sigma (\( \Sigma \)) symbol. In this exercise, we deal with a specific kind of summation over terms derived from a real function.
The summation given is \( \sum_{r=1}^{2n-1} 2 f\left(\frac{r}{2n}\right) \). This expression requires us to evaluate the function repeatedly at different points determined by \( \frac{r}{2n} \) for \( r = 1, 2, ..., 2n-1 \).
**Strategies for Simplifying Summations:**

• **Symmetrical Pairing:** The problem exploits a unique symmetry in the function values such that \( f\left(\frac{r}{2n}\right) + f\left(1-\frac{r}{2n}\right) = 1 \).
• **Pairwise Addition:** By grouping terms symmetrically, we get consistent results, simplifying the summation significantly. Each pair sums to 2, reducing the need to calculate each term individually.
• **Handling Edge Cases:** Special attention is given to the middle term \( r = n \). Evaluating the middle separately ensures we account for every term correctly.

By understanding and leveraging these strategies, complex summations become more manageable, and results can be confirmed confidently.

Symmetry in Mathematics

**Symmetry** in mathematics conveys a form of balance and predictability in structures. In this exercise, symmetry plays a crucial role in simplifying the summation.
The function \( f(x) = \frac{a^x}{a^x + \sqrt{a}} \) possesses a symmetric property: \( f(x) + f(1-x) = 1 \). This is known as **reflective symmetry** about the line \( x = \frac{1}{2} \).
**Benefits of Symmetry in Mathematical Problems:**

• **Reduction of Complexity:** Symmetry allows us to pair terms and simplify calculations, significantly reducing the work needed to compute large sums.
• **Predictable Outcomes:** With symmetry, we can predict that certain properties will hold throughout calculations, giving us confidence in the correctness of results.
• **Optimization:** Symmetry can lead to optimizations in computations, avoiding unnecessary work by grouping similar or equivalent terms.

Understanding symmetry not only aids in solving mathematical problems more efficiently but also provides deeper insight into how systems and functions behave.