Problem 109 - A Complete Resource Book in Mathematics for JEE Main

Question

Which of the following is not true? The function \(f(x)=x^{2}+\frac{\lambda}{x}\) has a (A) minimum at \(x=2\) if \(\lambda=16\) (B) maximum at \(x=2\) if \(\lambda=16\) (C) maximum for no real value of \(\lambda\) (D) point of inflexion at \(x=1\) if \(\lambda=-1\)

Step-by-Step Solution

Answer

The function does not have a maximum at \( x=2 \) if \( \lambda=16 \) (Option B is not true).

Step 1: Find the First Derivative

The given function is \( f(x) = x^2 + \frac{\lambda}{x} \). To find the critical points, we need its first derivative. Use the differentiation rules: the derivative of \( x^n \) is \( nx^{n-1} \), and the derivative of \( \frac{a}{x} \) is \( -\frac{a}{x^2} \). The first derivative, \( f'(x) \), is:\[ f'(x) = 2x - \frac{\lambda}{x^2} \]

Step 2: Evaluate Critical Points

Set the first derivative to zero to find critical points:\[ 2x - \frac{\lambda}{x^2} = 0 \]Rearrange to find \( \lambda \) in terms of \( x \):\[ 2x^3 = \lambda \]Substitute \( x = 2 \) into the equation:\[ \lambda = 2(2)^3 = 16 \]So, the function has a critical point at \( x=2 \) when \( \lambda = 16 \).

Step 3: Determine Nature of Critical Points

To determine if the function has a minimum or maximum at the critical point, we need the second derivative. Differentiate \( f'(x) \):\[ f''(x) = \frac{d}{dx}\left(2x - \frac{\lambda}{x^2}\right) = 2 + \frac{2\lambda}{x^3} \]Substitute \( x = 2 \) and \( \lambda = 16 \):\[ f''(2) = 2 + \frac{2(16)}{8} = 6 > 0 \]Since \( f''(2) > 0 \), the function has a minimum at \( x=2 \) when \( \lambda = 16 \), not a maximum.

Step 4: Check for Maximum for Any \( \lambda \)

The presence of a maximum at a critical point requires that the second derivative \( f''(x) \) is negative. The expression \( f''(x) = 2 + \frac{2\lambda}{x^3} \) can be zero or positive but cannot be negative for any real \( \lambda \) at any \( x \), hence never a maximum.

Step 5: Check for Point of Inflexion

For a point of inflexion, the second derivative should change sign at a point, which means you need to evaluate the third derivative:\[ f'''(x) = \frac{d}{dx} \left(2 + \frac{2\lambda}{x^3}\right) = -\frac{6\lambda}{x^4} \]If \( \lambda = -1 \) and \( x=1 \):\[ f'''(1) = -\frac{6(-1)}{1^4} = 6 eq 0 \]Thus, no inflection point at \( x=1 \) for \( \lambda = -1 \).

Step 6: Conclusion

Based on the calculations:- The function has a minimum, not a maximum, at \( x=2 \) if \( \lambda=16 \).- There is no maximum for any real \( \lambda \).- No point of inflexion for \( \lambda=-1 \) at \( x=1 \).Therefore, the statement that the function has a maximum at \( x=2 \) if \( \lambda=16 \) is not true.

Key Concepts

Critical Points

In calculus, finding the critical points of a function is a key step in understanding its behavior. A critical point occurs where the first derivative of a function is equal to zero or is undefined. In the problem's context, we are given the function:

• \( f(x) = x^2 + \frac{\lambda}{x} \)

To find critical points, we first calculate the first derivative:

• \( f'(x) = 2x - \frac{\lambda}{x^2} \)

Next, set this derivative to zero to find values of \( x \) that make it true:

• \( 2x - \frac{\lambda}{x^2} = 0 \)
• \( 2x^3 = \lambda \)

For \( x = 2 \) and \( \lambda = 16 \), a critical point is located. Critical points are vital as they can indicate potential maximum, minimum, or saddle points. Remember, further analysis is needed to determine the nature of these points.

Second Derivative Test

Once critical points are located, it's crucial to determine their nature—whether they represent a maximum, minimum, or something else. This is where the Second Derivative Test comes into play.The second derivative of a function helps reveal the concavity of the graph:

• \( f''(x) = \frac{d}{dx} \left(2x - \frac{\lambda}{x^2} \right) = 2 + \frac{2\lambda}{x^3} \)

For \( x = 2 \) and \( \lambda = 16 \), substituting into the second derivative gives:

• \( f''(2) = 2 + \frac{2(16)}{8} = 6 \)

Here, \( f''(2) > 0 \) indicating that the function is concave up and has a local minimum at \( x=2 \) when \( \lambda=16 \). If the second derivative had been less than zero, it would imply a local maximum, but since this is not the case, there is no maximum for these values.

Inflection Point

An inflection point on a graph is where the concavity changes direction. To find potential inflection points, we need to examine the function's second and third derivatives.In this particular problem, for \( \lambda = -1 \) and \( x=1 \), first ensure the second derivative is zero, which may suggest an inflection point. However, we also need the third derivative to confirm a change in concavity:

• \( f'''(x) = \frac{d}{dx} \left(2 + \frac{2\lambda}{x^3} \right) = -\frac{6\lambda}{x^4} \)

When \( \lambda = -1 \) and \( x=1 \):

• \( f'''(1) = -\frac{6(-1)}{1^4} = 6 \)

Since \( f'''(1) eq 0 \), it confirms there is no inflection point because the concavity does not change. Inflection points are significant as they mark where the graph switches from being concave up to concave down, or vice versa.