Problem 36 - A Complete Resource Book in Mathematics for JEE Main 2017

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 1, option (A).

Step 1: Define Conditions

First, let's analyze the conditions given: \ \(\frac{1}{p} + \frac{1}{q} = 1\). This implies a relationship between \(p\) and \(q\) often associated with Holder's inequality where \(p > 1\).

Step 2: Express q in terms of p

From the condition \(\frac{1}{p} + \frac{1}{q} = 1\), we rearrange to express \(q\) in terms of \(p\). Thus, we get: \ \(\frac{1}{q} = 1 - \frac{1}{p} = \frac{p-1}{p}\) \ which implies \(q = \frac{p}{p-1}\).

Step 3: Compute Derivative

To find the minimum value, calculate the derivative of the function \(f(x)\) and set it to zero: \ \(f'(x) = \frac{d}{dx}\left(\frac{x^p}{p} + \frac{x^{-q}}{q}\right) = x^{p-1} - x^{-q-1}\).

Step 4: Solve for Critical Points

Set the derivative \(f'(x)\) to zero to find critical points: \ \(x^{p-1} = x^{-q-1}\) \ implies \(x^{p-1+q+1}=1\) \ Thus, \(x^{p+q-2} = 1\).

Step 5: Substitute q as a Function of p

We already have \(q = \frac{p}{p-1}\). Substitute this into the equation: \ \(x^{p+\frac{p}{p-1}-2} = 1\). \ Solving, notice this simplifies to \(x^0 = 1\), indicating \(x=1\) is a critical point.

Step 6: Calculate Function Value at Critical Point

Substitute \(x = 1\) into the original function: \ \(f(1) = \frac{1^p}{p} + \frac{1^{-q}}{q} = \frac{1}{p} + \frac{1}{q}\). \ From the condition \(\frac{1}{p} + \frac{1}{q} = 1\), so \(f(1)=1\).

Step 7: Verify the Minimum

To confirm \(f(x)\) has a minimum at \(x=1\), note that \(f(x) \to \infty\) as \(x \to 0^+\) and \(x \to \infty\), indicating \(x=1\), where \(f(1)=1\), is indeed the minimum point.

Key Concepts

Holder's Inequality

Holder's inequality is a fundamental concept in calculus and analysis. It describes a relationship between integrals and sums by generalizing the Cauchy-Schwarz inequality. The condition \[ \frac{1}{p} + \frac{1}{q} = 1 \] arises from this inequality and imposes a strict requirement on the values of \(p\) and \(q\). This relationship ensures the existence of conjugate exponents, which are often denoted as \(p\) and \(q\). This forms the basis for analyzing certain optimization problems.

• Understanding Conjugate Exponents: In Holder's inequality, if one exponent is known, the other can be determined using the relationship \(\frac{1}{p} + \frac{1}{q} = 1\).
• Utilizing Holder's in Optimization: It helps in characterizing functions and their behavior under given conditions.

Knowing how Holder’s inequality influences the structure of functions can enhance comprehension of problems like minimization or finding critical points.

Derivative Calculation

Derivative calculation is a crucial process in calculus used to determine the rate of change of a function with respect to a variable. In this problem, we want to find the derivative of \[ f(x) = \frac{x^p}{p} + \frac{x^{-q}}{q}. \]The derivative, represented as \(f'(x)\), captures how the function value changes with \(x\).

• Formula Application: For any term in the function, apply the rule for derivatives, essentially reducing the power by one.
• Example Application:
  • \(\frac{d}{dx}(x^p) = px^{p-1}\)
  • \(\frac{d}{dx}(x^{-q}) = -qx^{-q-1}\)

By setting \(f'(x) = 0\), we find points termed as 'critical points' where potential minima or maxima occur.

Critical Points Method

The critical points method is used to locate points on a function where the derivative equals zero, indicating potential minima, maxima, or points of inflection. Here's how we apply it:

• Find Derivative: Calculating the derivative as shown in the exercise, gives us \(f'(x) = x^{p-1} - x^{-q-1}\).
• Identify Critical Points: By solving \(f'(x) = 0\), we find that this simplifies to the criticality condition \(x^{p+q-2} = 1\).
• Solve for \(x\): Substituting \(q = \frac{p}{p-1}\), leads to \(x = 1\).

These points are significant as they simplify the process of determining function minima or maxima, guiding us in analyzing optimization problems effectively.

Function Minimization

Function minimization involves finding the smallest value of a function, often necessary in optimization problems. The steps typically involve:

• Identify Critical Points: Determine these from the derivative, where \(f'(x) = 0\).
• Substitute and Compare Values: Substituting critical points back into the original function, as with \(x = 1\), helps verify potential minima.
• Evaluate Function Behavior: Ensure the function's behavior at boundaries (\(x \to 0^+\) and \(x \to \infty\)) aligns with the expected extremes, confirming a minimum at chosen points.

Applying these steps ensures a systematic approach to solving for minimum values, guaranteeing accuracy and clarity in calculus problems.