Problem 72 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(x)=x^{m}, m\) being a non-negative integer, then the value of \(m\) for which \(f^{\prime}(\alpha+\beta)=f^{\prime}(\alpha)+f^{\prime}(\beta)\), for all \(\alpha, \beta>0\), is (A) 1 (B) 2 (C) 0 (D) None of these

Step-by-Step Solution

Answer

The value of \(m\) for which the condition holds is "None of these."

Step 1: Differentiate the function

The given function is \( f(x) = x^m \). To find the derivative, apply the power rule for differentiation: \( f'(x) = m \cdot x^{m-1} \). This is the derivative of the function with respect to \( x \).

Step 2: Apply the derivative to the condition

According to the problem, \( f'(\alpha+\beta) = f'(\alpha) + f'(\beta) \) for all \( \alpha, \beta > 0 \). Substitute the derivative expression into this equation: \( m (\alpha + \beta)^{m-1} = m \alpha^{m-1} + m \beta^{m-1} \).

Step 3: Simplify the equation

Factor out \( m \) from both sides, simplifying the equation to: \( (\alpha + \beta)^{m-1} = \alpha^{m-1} + \beta^{m-1} \).

Step 4: Analyze the simplified equation

The equality \( (\alpha + \beta)^{m-1} = \alpha^{m-1} + \beta^{m-1} \) must hold for all \( \alpha, \beta > 0 \). Consider specific cases to deduce \( m\).

Step 5: Test for specific value of m

Test \( m = 1 \): \((\alpha + \beta)^{1-1} = \alpha^{1-1} + \beta^{1-1}\) becomes \((\alpha + \beta)^0 = \alpha^0 + \beta^0\), which simplifies to \( 1 = 1 + 1 \). This is not true, so \( m eq 1\).

Step 6: Test for other possible values of m

Test \( m = 0 \): \((\alpha + \beta)^{0-1} = \alpha^{0-1} + \beta^{0-1}\) becomes \(1/(\alpha + \beta) = 1/\alpha + 1/\beta\), which is also not valid for all values. Thus, \( m eq 0 \). Similarly, test \( m = 2 \): \((\alpha + \beta)^{2-1} = \alpha^{2-1} + \beta^{2-1}\), does not hold either. Explore further conditions or conclude with last option.

Step 7: Solution Verification

Investigate further by considering unique behavior, restrictions, computational checks, or higher values. Since none satisfy all conditions without contradiction, determine \( m = \text{"None of these"} \) fits best given options.

Key Concepts

Power Rule for Differentiation

Differentiation is a key concept in calculus, which deals with determining the rate at which a function is changing. One common method used is the power rule for differentiation. This rule is particularly useful when differentiating expressions of the form \( x^m \), where \( m \) is any real number. The power rule states that if you have a function \( f(x) = x^m \), the derivative \( f'(x) \) is equal to \( m \cdot x^{m-1} \). This means you multiply the power by the variable itself reduced by one degree.

• The power rule is straightforward and quick to apply.
• It simplifies the process of finding the derivative of polynomial functions.

Becoming familiar with the power rule is essential for solving many differentiation problems. Remember, the power rule is only applicable when dealing with polynomial expressions. Each application of the rule provides the function's slope, or rate of change at any given point \( x \). This computational ease makes the power rule a cornerstone of calculus.

Function Derivative

A function derivative represents the rate at which a function's value changes as the input changes. It is essentially a function that gives the slope of the tangent line at any given point on the original function. The concept of a derivative is fundamental, as it provides insight into how a function behaves.When determined using the power rule, the derivative of a function like \( f(x) = x^m \) is achieved efficiently. The derivative is expressed as \( f'(x) = m \cdot x^{m-1} \). By differentiating, you understand:

• Where a function reaches its maximum or minimum values.
• The rate of increase or decrease of the function's value.
• The function's behavior over different intervals.

Derivatives allow you to find critical points and analyze function trends, such as increases or decreases, and can be used to solve complex mathematical problems. Understanding derivatives goes beyond calculation; it's about understanding the nature and behavior of functions over their domain.

Equation Simplification

Equation simplification is an important process that involves reducing an equation to its simplest form. In complex calculus problems, this often occurs after differentiation, such as when using the power rule. Simplification helps make equations easier to handle and solve, especially when dealing with multiple steps or components.For instance, in our problem, after finding the derivative using the power rule, we applied it to the condition given: \( f'(\alpha + \beta) = f'(\alpha) + f'(\beta) \). Simplifying the derived expressions leads to \( m(\alpha + \beta)^{m-1} = m \alpha^{m-1} + m \beta^{m-1} \), which further reduces by factoring out \( m \). This process reveals the core relationship that must hold among \( \alpha \), \( \beta \), and \( m \).

• Simplification helps in identifying errors early by ensuring expressions are in their correct form.
• It reduces the possibility of computational errors in later steps.

In simplification, you're breaking down complex expressions into manageable pieces. This often involves combining like terms, factoring, or distributing terms properly. The goal is to achieve clarity and accuracy, making complex equations much easier to interpret and solve.