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

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

The derivative \(f'(x)\) is (B) \(\frac{1}{a^2-b^2}\left(\frac{a}{x^2}+b\right)\).

Step 1: Understand the given equation

The problem provides the equation: \[ a f(x) + b f\left(\frac{1}{x}\right) = \frac{1}{x} - 5 \]This is a function equation involving an expression in terms of \(f(x)\) and its reciprocal \(f\left(\frac{1}{x}\right)\). We need to find the derivative \(f'(x)\).

Step 2: Express \(f(x)\) in terms of \(f\left(\frac{1}{x}\right)\)

Rearrange the equation to isolate \(f\left(\frac{1}{x}\right)\) and express it in terms of \(f(x)\):\[ f\left(\frac{1}{x}\right) = \frac{1}{b}\left(\frac{1}{x} - 5 - a f(x)\right) \]

Step 3: Differentiate both sides of the original equation

Differentiate the entire expression with respect to \(x\). Use the chain rule for \(f\left(\frac{1}{x}\right)\): For differentiation of \(f\left(\frac{1}{x}\right)\), let \(u = \frac{1}{x}\), then \(\frac{d}{dx}f(u)\) = \(-\frac{1}{x^2} f'(u)\). Differentiation gives:\[ a f'(x) + b\left( -\frac{1}{x^2} f'\left(\frac{1}{x}\right) \right) = -\frac{1}{x^2} \]

Step 4: Solve for \(f'(x)\)

Substitute the expression derived in Step 2 into the differentiated equation from Step 3. Rearrange to solve for \(f'(x)\): After manipulation, you find:\[ f'(x) = \frac{1}{a^2 - b^2}\left(\frac{a}{x^2} + b\right) \]

Step 5: Verify your solution

Compare the answer you obtained with the provided options. The answer is in Option (B): \[ \frac{1}{a^2 - b^2}\left(\frac{a}{x^2} + b\right) \]

Key Concepts

Differential Calculus

Differential calculus is a branch of mathematics that focuses on how functions change. A primary tool in differential calculus is the concept of a derivative. The derivative of a function gives you the rate at which it changes for a small change in its input value. In other words, it tells us the slope or the steepness of the function's graph at any given point.

Differential calculus is widely used in various fields such as physics, engineering, and economics to optimize systems and predict behavior. It helps understand motion, growth trends, and how changing one variable affects another.

In this context, finding the derivative of a function is essential to solving many real-world problems. It involves utilizing rules like the power rule, product rule, quotient rule, and chain rule to handle different types of functions.

Function Derivative

A derivative represents the rate of change of a function concerning its input. If you have a function like \( f(x) \), its derivative is often denoted as \( f'(x) \) or \( \frac{df}{dx} \). This tells us how much \( f(x) \) changes for a small change in \( x \).

When you calculate the derivative, you're essentially looking for the function that describes the slope of the tangent line at each point on the graph of the original function. Knowing this helps in understanding the behavior of the function, such as increasing or decreasing trends, and identifying maximum and minimum values.

• If \( f(x) \) is a linear function, its derivative is constant.
• If \( f(x) \) is a polynomial, apply rules to give you a new polynomial.
• If \( f(x) \) involves more complex forms like trigonometric or exponential functions, specific differentiation rules apply.

For instance, the exercise requires finding the derivative of a composite function using advanced techniques such as the chain rule due to variable substitutions and transformations involved.

Chain Rule

The chain rule is a crucial tool in differential calculus used to differentiate composite functions. A composite function is a function formed when one function is applied to the result of another function; for instance, \( f(g(x)) \).

The principle behind the chain rule is that if you have two functions in the form \( h(x) = f(g(x)) \), then the derivative \( h'(x) \) is given by \( f'(g(x)) \cdot g'(x) \). Therefore, you first find the derivative of the outer function evaluated at the inner function, then multiply it by the derivative of the inner function.

• Helps in tackling problems where variables are nested.
• Efficient for functions expressed as combinations of different parts, each depending on the other.

In the case presented, the chain rule is applied when differentiating \( f\left(\frac{1}{x}\right) \) because \( u = \frac{1}{x} \) is a nested function inside \( f \). The key here is understanding how the change in \( x \) affects both the outer function and the transformation inside it, enabling the accurate calculation of the derivative.