Problem 165 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(x)+2 f\left(\frac{1}{x}\right)=3 x, x \neq 0\), and \(S=\\{x \in R: f(x)=f(-x)\\} ;\) then \(\mathrm{S}:\) (A) contains more than two elements. (B) is an empty set. (C) contains exactly one element (D) contains exactly two elements

Step-by-Step Solution

Answer

(C) contains exactly one element.

Step 1: Understanding the problem

We need to find the symmetric set \( S \) of real numbers such that \( f(x) = f(-x) \) based on the function equation \( f(x) + 2f\left(\frac{1}{x}\right) = 3x \). Determine the elements that satisfy this condition.

Step 2: Substitute and reorganize the function

Given the equation \( f(x) + 2f\left(\frac{1}{x}\right) = 3x \), let’s reverse the substitution: consider replacing \( x \) with \( \frac{1}{x} \) to find another equation.\[ f\left(\frac{1}{x}\right) + 2f(x) = \frac{3}{x} \]

Step 3: Solve simultaneous equations

You now have two equations:1. \( f(x) + 2f\left(\frac{1}{x}\right) = 3x \)2. \( f\left(\frac{1}{x}\right) + 2f(x) = \frac{3}{x} \).Subtract the second equation from the first equation:\[ (f(x) + 2f\left(\frac{1}{x}\right)) - (f\left(\frac{1}{x}\right) + 2f(x)) = 3x - \frac{3}{x} \]This simplifies to:\[ -3f(x) + f\left( \frac{1}{x} \right) = 3x - \frac{3}{x} \]

Step 4: Substitute and solve for \( f(x) \)

Rearrange and solve for \( f(x) \):\[ f(x) = \frac{x(x+1)}{x^2 + 1} \]Note that this function formula helps provide solutions if \( f(x) = f(-x) \).

Step 5: Check symmetry condition

For \( f(x) = f(-x) \), we want:\[ \frac{x(x+1)}{x^2 + 1} = \frac{(-x)(-x+1)}{(-x)^2 + 1} \]This can be simplified as \( x = 0 \), meaning \( f(x) = f(-x) \) only holds when \( x = 0 \).

Step 6: Conclusion on symmetric set \( S \)

The calculation shows that \( f(x) = f(-x) \) holds only at \( x = 0 \). Thus, set \( S \) contains exactly one element.

Key Concepts

Function Equation

Grasping the idea of a function equation is crucial to understand the scenario in the problem. A function equation involves an equation which places a functional relationship between variables like in our exercise. Here, we are working with the function equation:

• \( f(x) + 2f\left(\frac{1}{x}\right) = 3x \)

This equation is telling us that if you input any rational number (except zero), the relationship between the function values and this input should satisfy the equation.

To break it down further, we see that the equation includes two expressions: \( f(x) \) and another function evaluated at its reciprocal, \( f\left(\frac{1}{x}\right) \). This complexity requires considering how each part responds to various inputs and hints at the complexity of finding a solution that satisfies all inputs.

Regardless, once comprehended properly, solving a function equation often involves substitution, algebraic manipulation, or even setting up new equations as illustrated in our solution.

Simultaneous Equations

Simultaneous equations come into play when two or more equations have to be solved together. This scenario usually happens when two variables appear in more than one equation. It is a key part in solving this problem because both equations share similar unknowns. The function equation provided:

• \( f(x) + 2f\left(\frac{1}{x}\right) = 3x \)
• \( f\left(\frac{1}{x}\right) + 2f(x) = \frac{3}{x} \)

These create a system of simultaneous equations that need solutions for the same function, \( f \). By solving such systems, you find common solutions that apply to both, often through techniques like substitution or elimination.

In our exercise, subtraction of one equation from another led to a simpler form, allowing us to solve for \( f(x) \). This process tends to help in determining exact values or expressions for unknowns.

Symmetry Condition

The symmetry condition in math generally involves conditions where a function behaves identically when certain transformations are applied. In this exercise, the symmetry condition is defined through:

• \( f(x) = f(-x) \)

This means that the function should produce identical outputs whether you input \( x \) or \( -x \), showcasing a reflective symmetry along the y-axis in a graph.

For our problem, this involves substituting \( x \) with \( -x \) in the expression for \( f(x) \) obtained from previous steps, confirming

• \( \frac{x(x+1)}{x^2 + 1} = \frac{(-x)(-x+1)}{(-x)^2 + 1} \)

This equation simplifies under certain conditions, pointing out where true symmetry exists. In this instance, symmetry occurs only when \( x = 0 \). Recognizing and testing these conditions is a critical part of solving function-related problems where symmetry is explored.