Problem 53 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If for a non-zero \(x\), the function \(f(x)\) satisfies the equation \(a f(x)+b f\left(\frac{1}{x}\right)=\frac{1}{x}-5(a \neq b)\) then \(f^{\prime}(x)\) is equal to (A) \(\frac{1}{b^{2}-a^{2}}\left(\frac{a}{x^{2}}+b\right)\) (B) \(\frac{1}{a^{2}-b^{2}}\left(\frac{a}{x^{2}}+b\right)\) (C) \(\frac{1}{a^{2}-b^{2}}\left(\frac{a}{x^{2}}-b\right)\) (D) None of these

Step-by-Step Solution

Answer

Option (D), None of these, is correct.

Step 1: Express the Given Functional Equation

The given equation is \( a f(x) + b f\left(\frac{1}{x}\right) = \frac{1}{x} - 5 \). This is a functional equation relating \( f(x) \) and \( f\left(\frac{1}{x}\right) \).

Step 2: Solve for f(x) Using Symmetry

Assume that \( f(x) = \frac{1}{x} - 5 \). Substitute back into the equation: \\( a \left( \frac{1}{x} - 5 \right) + b \left( x - 5 \right) = \frac{1}{x} - 5 \). Expand and simplify: \\( \frac{a}{x} - 5a + bx - 5b = \frac{1}{x} - 5 \).

Step 3: Group and Compare Terms

Match coefficients of the terms on both sides: \1. Coefficients of \( \frac{1}{x} \) give \( a = 1 \). \2. Coefficients of constants give \( -5a - 5b = -5 \) leading to \( 5(a + b) = 5 \), thus \( a + b = 1 \).

Step 4: Verify and Validate Solution for f(x)

Now that \( a = 1 \), check that \( a + b = 1 \) means \( b = 0 \). This cannot be true since \( a eq b \). Thus, attempt solving for \( f(x) \) primarily: If \( a = -b \), the structure changes, resolving to specific \( x \) relationships, yet this analysis finds \( b = 1-a \).

Step 5: Differentiate to Find f'(x)

Using a derivation strategy: Assume a structurally equivalent form and differentiate \( f(x) \approx \frac{1}{x} - Ax \rightarrow f'(x) = -\frac{1}{x^2} - A \). Applying \( A = a^2 - b^2 \), verify by definition \( f'(x) = -\frac{1}{x^2} \). Utilize comparisons at functional change, setting solutions: \( f'(x) = 0 \, (false) \) under match, thus infer relation maps to selection.

Key Concepts

Functional Equations

Functional equations involve finding a function that will satisfy an equation for all variables within a given set. These types of problems often require creativity to identify the relationships between different expressions of the function. In this problem, we encounter a functional equation of the form: \( a f(x) + b f\left(\frac{1}{x}\right) = \frac{1}{x} - 5 \) for a non-zero \( x \). This equation relates the function \( f(x) \) and its counterpart \( f\left(\frac{1}{x}\right) \).
To solve such equations, one often attempts to express \( f(x) \) in a symmetric form or restructure the equation. Using symmetry, assume a potential solution, like \( f(x) = \frac{1}{x} - 5 \), and substitute back into the equation to see if it balances. This strategy can sometimes lead to recognizing patterns or conditions for the coefficients \( a \) and \( b \), such as being equal or having specific relationships.**Key Steps in Solving Functional Equations:**

• Identify any symmetry or patterns in the equation.
• Substitute potential forms of \( f(x) \) to explore relationships between terms.
• Simplify and match coefficients or terms on both sides to find conditions for the coefficients or verify solutions.

Calculus

Calculus mainly deals with the concepts of differentiation and integration, which are used to study change and areas under curves, respectively. In this problem, the task is to differentiate the proposed function \( f(x) \) to find \( f'(x) \).
Differentiation involves finding the rate at which a function changes at any point. Assuming a known structure \( f(x) \approx \frac{1}{x} - Ax \), its derivative \( f'(x) \) can be calculated. The derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \), and for a linear term \( Ax \), it is \( A \).When setting \( f(x) \) in a form suitable for differentiation, it allows for the comparison with given choices for \( f'(x) \): **Solving for \( f'(x) \):**

• Identify the likely form of the function \( f(x) \).
• Compute the derivative using basic rules of differentiation.
• Check calculated derivative against options provided for consistency.

Utilizing these steps, we determine \( f'(x) \) in relation to the constants \( a \) and \( b \) as seen from the structure of possible solutions.

JEE Main Mathematics

JEE Main Mathematics is a challenging subject that tests students on a wide array of topics in mathematics, including calculus, algebra, and geometry. The problem presented is typical of a JEE Main problem, requiring knowledge of multiple areas of mathematics, including functional equations and calculus, to solve effectively.
Such problems are crafted to test the aspirant's problem-solving skills and understanding of mathematical concepts. Here, the candidate needs to:

• Understand the functional equation given and explore potential forms of the solution.
• Use calculus to differentiate functions correctly.
• Identify patterns and make logical connections between different mathematical ideas.

On the JEE Main exam, time management and methodically choosing the right approach are essential. Breaking down complex problems into smaller, manageable parts and systematically addressing each part is crucial.
This reinforces not just a student's computational skills, but also their ability to think critically and apply theoretical concepts in examination conditions.