Problem 23 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f(x)=x^{n}, n\) being a non-negative integer. The value of \(n\) for which equality \(f^{\prime}(a+b)=f^{\prime}(a)+f^{\prime}(b)\) is valid for all \(a, b>0\) is (A) 5 (B) 1 (C) 2 (D) 4

Step-by-Step Solution

Answer

The correct value of \(n\) is 2, so the answer is (C).

Step 1: Understand the Problem

We need to find the value of \(n\) such that the derivative of \(f(x) = x^n\) evaluated at \(a+b\) is equal to the sum of the derivatives evaluated separately at \(a\) and \(b\). In equation form: \(f'(a+b) = f'(a) + f'(b)\).

Step 2: Find the Derivative

To solve this, we first find the derivative of the function \(f(x) = x^n\). Using the power rule, the derivative is \(f'(x) = nx^{n-1}\).

Step 3: Substitute into the Equality

Using the derivative from Step 2, substitute into the equation \(f'(a+b) = f'(a) + f'(b)\). This becomes:\[n(a+b)^{n-1} = na^{n-1} + nb^{n-1}\]

Step 4: Simplify the Equation

Divide both sides of the equation by \(n\):\[(a+b)^{n-1} = a^{n-1} + b^{n-1}\]

Step 5: Test Values for n

To find the correct value of \(n\), we test the values. - For \(n=1\): \((a+b)^0 = a^0 + b^0\) which simplifies to \(1 = 1 + 1\), a contradiction.- For \(n=2\): \((a+b)^1 = a^1 + b^1\) simplifies to \(a+b = a + b\), a true statement.- For \(n=3, 4, 5\): Following similar logic, none satisfy the equation for all \(a, b > 0\).Therefore, \(n = 2\) satisfies the equation.

Key Concepts

Power Rule

The Power Rule is a fundamental tool in calculus that simplifies the process of finding derivatives for polynomial functions. When you have a function of the form \( f(x) = x^n \), where \( n \) is a real number, the Power Rule provides a straightforward way to determine the derivative. This rule states that the derivative of \( x^n \) is \( nx^{n-1} \).
To apply the Power Rule:

• Identify the exponent \( n \) in the function \( x^n \).
• Multiply the whole term by \( n \). This becomes your initial coefficient.
• Decrease the original exponent by 1. The new exponent after differentiation becomes \( n-1 \).

This simple rule greatly reduces the complexity when working with polynomial expressions, making it easier for students and professionals alike to handle a wide range of functions with varying degrees. Technique mastery in the Power Rule allows further exploration and understanding of calculus concepts, as seen in differential equations and more advanced topics.

Polynomial Function

Polynomial functions are mathematical expressions involving a sum of powers of one or more variables multiplied by coefficients. For example, a typical polynomial in a single variable \( x \) might look like: \( f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \).
Let's break it down:

• Terms: Each component, such as \( a_nx^n \), is called a term.
• Coefficients: The constants \( a_n, a_{n-1}, \ldots \) are coefficients, which are multiplied by each power of \( x \).
• Degree: The highest power of \( x \) (in this case \( n \)) is known as the degree of the polynomial.

Polynomials are simple yet powerful tools in mathematics, commonly used in algebra, calculus, numerical analysis, and more.
Understanding polynomials is crucial as they form the foundation for more complex mathematical models and calculations. Whether solving equations or analyzing functions, knowing how to manipulate polynomials enables discoveries in both theoretical and applied mathematics.

Equation Simplification

Simplifying equations is a central skill in mathematics, allowing us to solve problems more efficiently by reducing complex problems into a more manageable form. In the context of the exercise, equation simplification was essential after substituting derivatives into the equation from the original problem.
To simplify the expression \( n(a+b)^{n-1} = na^{n-1} + nb^{n-1} \):

• First, divide the entire equation by \( n \) to eliminate the \( n \) from both sides, making the equation more straightforward: \( (a+b)^{n-1} = a^{n-1} + b^{n-1} \).
• This step makes it easier to check various values for \( n \), as it helps in visualizing if and when both sides of the equation are equal, simplifying the testing process greatly.
• Finally, compare both sides to look for consistency or contradictions, which enables finding the right value of \( n \).

Simplifying expressions and equations is an invaluable mathematical technique. It provides a clearer view of relationships and behaviors of functions, leading ultimately to more effective problem-solving in algebra, calculus, and beyond.