Problem 15 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(2 f(x)-3 f\left(\frac{1}{x}\right)=x^{2}, x\) is not equal to zero, then \(f(2)\) is equal to (A) \(-\frac{7}{4}\) (B) \(\frac{5}{2}\) (C) \(-1\) (D) None of these

Step-by-Step Solution

Answer

\(f(2) = -\frac{7}{4}\).

Step 1: Substitute and Rearrange

Given equation is: \(2f(x) - 3f\left(\frac{1}{x}\right) = x^2\). Let \(x = a\), so our equation is \(2f(a) - 3f\left(\frac{1}{a}\right) = a^2\). Let this be equation (1).

Step 2: Substitution with Reciprocal

Now, substitute \(x = \frac{1}{a}\) in the given equation, so we get: \[2f\left(\frac{1}{a}\right) - 3f(a) = \left(\frac{1}{a}\right)^2\] which simplifies to \[2f\left(\frac{1}{a}\right) - 3f(a) = \frac{1}{a^2}\]. Let this be equation (2).

Step 3: Solve Simultaneous Equations

We have two equations: \ (1) \: 2f(a) - 3f\left(\frac{1}{a}\right) = a^2 \ (2) \: 2f\left(\frac{1}{a}\right) - 3f(a) = \frac{1}{a^2} \ Multiply equation (1) by 2 and equation (2) by 3:\ \(4f(a) - 6f\left(\frac{1}{a}\right) = 2a^2\) \ \(6f\left(\frac{1}{a}\right) - 9f(a) = \frac{3}{a^2}\). \ Add these equations to eliminate \(f\left(\frac{1}{a}\right)\):\ \(-5f(a) = 2a^2 + \frac{3}{a^2}\) which simplifies to \(f(a) = -\frac{2a^4 + 3}{5a^2}\).

Step 4: Evaluate \(f(2)\)

Substitute \(a = 2\) into the expression for \(f(a)\): \[f(2) = -\frac{2(2)^4 + 3}{5(2)^2}\] \= -\frac{2(16) + 3}{5 * 4}\ \= -\frac{32 + 3}{20}\ \= -\frac{35}{20}\ \= -\frac{7}{4}\.

Key Concepts

Simultaneous Equations

In mathematics, simultaneous equations are a set of equations involving multiple variables, which are simultaneously satisfied. For instance, if you have two or more equations that both involve the same variables, solving them means finding the values for each variable that make all the equations true. This is a foundational concept used in algebra to solve complex problems.

In the provided example, we deal with two equations:

• Equation 1: \(2f(a) - 3f\left(\frac{1}{a}\right) = a^2\)
• Equation 2: \(2f\left(\frac{1}{a}\right) - 3f(a) = \frac{1}{a^2}\)

The goal is to solve for the function \(f(x)\) by simultaneously considering both equations. This means finding a value of \(f(a)\) and \(f\left(\frac{1}{a}\right)\) that satisfies both equations at the same time. This kind of problem-solving allows you to use relationships defined by multiple equations to determine unknowns.

Substitution Method

The substitution method is a technique used to solve simultaneous equations by expressing one variable in terms of another. This is done to simplify the equations so that we can solve them more easily. In the context of functional equations, substitution helps relate different forms of the function, such as \(f(x)\) and \(f\left(\frac{1}{x}\right)\), to each other.

In our exercise, the substitution method begins by substituting the reciprocal value in the original equation. By letting \(x = a\) and then \(x = \frac{1}{a}\), we end up with:

• For \(x = a\), you derive: \(2f(a) - 3f\left(\frac{1}{a}\right) = a^2\)
• For \(x = \frac{1}{a}\), you get: \(2f\left(\frac{1}{a}\right) - 3f(a) = \frac{1}{a^2}\)

This method is especially useful in functional equations where the goal is to find relationships between different instances of the function. In this case, substitution has enabled us to relate \(f(a)\) and \(f\left(\frac{1}{a}\right)\), making it possible to solve \(f(a)\).

Mathematical Problem Solving

Mathematical problem solving involves employing various strategies and methods to find solutions to problems. It requires a thorough understanding of mathematical concepts and the ability to apply them logically and creatively.

In this exercise on functional equations, problem solving involved multiple steps:

• Identifying the correct approach to handle the functional relation by setting up equations that relate different instances of \(f(x)\).
• Using substitution to manipulate and simplify equations by strategically choosing specific values, like \(x = a\) and \(x = \frac{1}{a}\), which expose underlying relationships.
• Applying algebraic techniques, such as multiplying equations and adding or subtracting them, to eliminate terms and isolate \(f(a)\).

Ultimately, we solved for \(f(2)\) by substituting back into our derived function \(f(a) = -\frac{2a^4 + 3}{5a^2}\). Mathematical problem solving is iterative and often requires trying different approaches until the desired result is obtained.