Problem 36 - A Complete Resource Book in Mathematics for JEE Main

Question

The minimum value of the function \(f(x)=\frac{x^{p}}{p}+\frac{x^{-q}}{q}\), where \(\frac{1}{p}+\frac{1}{q}=1, p>1\) is (A) 1 (B) 0 (C) 2 (D) None of these

Step-by-Step Solution

Answer

The minimum value of the function is 2, which corresponds to option (C).

Step 1: Understand the Problem

We are given a function of the form \(f(x)=\frac{x^{p}}{p}+\frac{x^{-q}}{q}\), where the exponents \(p\) and \(-q\) must satisfy the relationship \(\frac{1}{p} + \frac{1}{q} = 1\) with \(p>1\). We need to find the minimum value of this function.

Step 2: Use the Given Relationship

The condition \(\frac{1}{p} + \frac{1}{q} = 1\) can be rewritten as \(q = \frac{p}{p-1}\). This relationship will help us substitute \(q\) with a term involving \(p\) for our calculations.

Step 3: Express the Function

Substitute \(q = \frac{p}{p-1}\) into the function, resulting in \(f(x) = \frac{x^p}{p} + \frac{x^{-(\frac{p}{p-1})}}{\frac{p}{p-1}}\). Simplify the expression to \(f(x) = \frac{x^p}{p} + \frac{(p-1)x^{-rac{p}{p-1}}}{p}\).

Step 4: Differentiate the Function

To find the critical points, differentiate the function with respect to \(x\). Set \(f'(x) = 0\) to find points where the minimum could occur:\[f'(x) = x^{p-1} - (p-1)x^{-\frac{p}{p-1}-1} = x^{p-1} - (p-1)x^{-\frac{1}{p-1}}\].

Step 5: Solve for Critical Points

Set the derivative equal to zero: \[x^{p-1} - (p-1)x^{-\frac{1}{p-1}} = 0\].Simplify to find \(x\): \[x^{p-1} = (p-1)x^{-\frac{1}{p-1}}\].Multiply both sides by \(x^{\frac{1}{p-1}}\): \[x^{p-1 + \frac{1}{p-1}} = p-1\].Simplify the exponent to get \(x^p = p-1\). This gives \(x = (p-1)^{\frac{1}{p}}\).

Step 6: Evaluate the Function at Critical Point

Substitute \(x = (p-1)^{\frac{1}{p}}\) back into \(f(x)\):\[f((p-1)^{\frac{1}{p}}) = \frac{(p-1)}{p} + \frac{(p-1)^{\frac{1}{p}(-\frac{p}{p-1})}}{\frac{p}{p-1}}\].The second term simplifies to \(\frac{p-1}{p} = 1\). So, \(f(x) = 1 + 1 = 2\).

Step 7: Conclusion

Since the function \(f(x)\) reaches its minimum value of 2 at \(x = (p-1)^{\frac{1}{p}}\), the minimum value of the function \(f(x)\) is confirmed to be 2.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. The derivative represents the rate at which a function is changing at any given point. In our exercise, we have the function \[f(x)=\frac{x^{p}}{p}+\frac{x^{-q}}{q}\]To determine where the function attains its minimum value, we start by differentiating it with respect to the variable \(x\). This involves applying the power rule for differentiation, which states that the derivative of \(x^n\) is \(nx^{n-1}\).

• For the term \(\frac{x^p}{p}\), the derivative is \(x^{p-1}\), as the constant \(\frac{1}{p}\) remains unaffected by differentiation.
• For the term \(\frac{x^{-q}}{q}\), the derivative is \(-qx^{-q-1}\), but when rewritten with \(q = \frac{p}{p-1}\), it becomes \(-(p-1)x^{-{\frac{p}{p-1}}-1}\).

By setting the derived equation to zero, we find the critical points, which indicate where the minimum or maximum points may occur.

Critical Points

Critical points in a function occur where the derivative equals zero or is undefined. These points are crucial in calculus as they shed light on potential extrema (minimum or maximum points) of the function. In our function differentiation process, we reach the equation:\[x^{p-1} - (p-1)x^{-\frac{1}{p-1}} = 0\]Solving this equation helps us identify critical points. To simplify, we can isolate terms and adjust exponents:

1. First, bring terms to one side: \(x^{p-1} = (p-1)x^{-\frac{1}{p-1}}\).
2. Multiply both sides by \(x^{\frac{1}{p-1}}\) to simplify exponents: \(x^{p-1 + \frac{1}{p-1}} = p-1\).
3. This reduces the equation to \(x^p = p-1\), resulting in \(x = (p-1)^{\frac{1}{p}}\).

This calculated \(x\) is our critical point, where the function must be evaluated to confirm if it is indeed a minimum point.

Exponent Rules

Exponent rules are crucial when working with expressions like the function in our exercise. Understanding these rules allows us to manipulate and simplify expressions, leading to effective differentiation and critical point determination.Key exponent rules include:

• Product Rule: \(x^a \cdot x^b = x^{a+b}\) - When multiplying like bases, add the exponents.
• Power Rule: \((x^a)^b = x^{ab}\) - When raising a power to another power, multiply the exponents.
• Negative Exponent Rule: \(x^{-a} = \frac{1}{x^a}\) - A negative exponent indicates a reciprocal.

In the exercise, these rules assist in simplifying the function and its derivatives. For instance, converting \(x^{-\frac{p}{p-1}}\) into a manageable form for differentiation involved recognizing the negative exponent as a reciprocal. Understanding how these rules fit together facilitates solving calculus optimization problems efficiently.