Problem 129 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f(x)=\frac{9^{x}}{9^{x}+3}\). Then, (A) \(f(x)+f(1-x)=1\) (B) \(f(x)+f(1-x)=-1\) (C) \(f\left(\frac{1}{1996}\right)+f\left(\frac{2}{1996}\right)\) \(+f\left(\frac{3}{1996}\right)+\ldots+f\left(\frac{1995}{1996}\right)=998\) (D) \(f\left(\frac{1}{1996}\right)+f\left(\frac{2}{1996}\right)\) \(+f\left(\frac{3}{1996}\right)+\ldots+f\left(\frac{1995}{1996}\right)=997 \frac{1}{2}\)

Step-by-Step Solution

Answer

Answer: (A) is correct; (C) the sum = 998.

Step 1: Substituting the Equation

Let's consider the function \(f(x) = \frac{9^x}{9^x + 3}\). To solve part (A), we first express \(f(1-x)\) as \(\frac{9^{1-x}}{9^{1-x} + 3}\).

Step 2: Simplifying f(1-x)

The expression for \(9^{1-x}\) can be rewritten as \(\frac{9}{9^x}\). Substituting this in, \(f(1-x) = \frac{9/9^x}{9/9^x + 3}\). This simplifies to \(\frac{9}{9 + 3 \times 9^x}\).

Step 3: Showing f(x) + f(1-x)

Now, compute \(f(x) + f(1-x)\): \[f(x) + f(1-x) = \frac{9^x}{9^x + 3} + \frac{9}{9^x + 27}\]. The common denominator is \((9^x + 3)(9^x + 27)\).

Step 4: Finding the Common Denominator

Calculate the sum: \[\frac{9^x (9^x + 27) + 9 (9^x + 3)}{(9^x + 3)(9^x + 27)} = \frac{9^{2x} + 27 \times 9^x + 9^{x+1} + 27}{(9^x + 3)(9^x + 27)}\]. Simplify this expression.

Step 5: Simplifying Numerator Expression

Combine like terms: the numerator becomes \(9^{2x} + 9^{x+1} + 27 \times 9^x + 27 = (9^x + 3)(9^x + 27)\). As they are equal, the expression simplifies to 1.

Step 6: Conclusion on Part A and Validity of Other Parts

Since \(f(x) + f(1-x) = 1\), part (A) is correct. Other parts (B and D) cannot be correct, as the sum property is well-defined and is based on pairing terms as shown.

Key Concepts

Function Properties

A function is a special type of relation that assigns each input exactly one output. Understanding the properties of functions helps us solve complex equations with ease. Here are some important properties of the function involved in our exercise:

• **Domain and Range**: The domain of a function is the set of all possible input values, while the range is the set of possible outputs. For the function \( f(x) = \frac{9^x}{9^x+3} \), the domain is all real numbers because exponentiation and addition with positive constants are defined for all real inputs. The range is limited to \( 0 < f(x) < 1 \), since the function cannot equal zero or one within the real number system.
• **Symmetry**: Some functions exhibit symmetry properties, making them easier to solve. Our function shows symmetry with the transformation \( x \to 1-x \), as seen in our exercise.
• **Behavior**: Investigating how functions behave as \( x \to \infty \) or \( x \to -\infty \) can inform solutions. The function \( f(x) \) approaches 1 as \( x \to \infty \) and approaches 0 as \( x \to -\infty \).

By understanding these properties, solving problems with this function becomes simpler and intuitive.

Functional Equations

Functional equations require us to find the function or its properties given certain conditions. In this exercise, we are dealing with:

• **Transformation Properties**: We transformed \( f(1-x) \) by substituting and simplifying expressions to uncover symmetries. Starting with \( \frac{9^{1-x}}{9^{1-x}+3} \), we find that the expression simplifies and reveals a pattern.
• **Equation Setup**: When we added \( f(x) \) and \( f(1-x) \), we noticed a repeated structure, leading us to the correct simplification. This type of functional equation often exploits symmetrical properties.
• **Verification**: It's important to verify solutions by plugging them back into the initial conditions. This validates the transformation and ensures the result satisfies the equation entirely.

Mastering functional equations involves recognizing useful transformations and appropriately simplifying them, as demonstrated in the exercise.

Simplifying Expressions

Simplifying expressions is important for making complex calculations more manageable. In the exercise's solution, we simplify to identify patterns quickly.

• **Fraction Simplification**: Given \( f(x) \) and \( f(1-x) \), we simplified expressions by finding a common denominator. This technique makes them easier to add or compare, as seen during the solution of \( f(x) + f(1-x) \).
• **Combining Like Terms**: After distributing and expanding product terms, we learn to combine like terms. The formula we used consolidated terms neatly, leading to a cancellation that showed equality of denominator and numerator.
• **Elimination Techniques**: Tidying up expressions can be done by eliminating common denominators or numerators where possible. This results in a cleaner expression, facilitating the discovery of key insights, such as symmetry or consistent results.

Simplifying expressions not only aids in solving specific problems—like the exercise on \( f(x) \) here—but also forms a base skill for tackling more advanced algebraic topics.