Problem 109 - A Complete Resource Book in Mathematics for JEE Main 2017

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

Option (B) is not true.

Step 1: Find the First Derivative

The first step to analyzing the function is to find its first derivative, since it helps in determining the critical points where the function might have a maximum, minimum, or point of inflexion. The given function is:\[ f(x) = x^2 + \frac{\lambda}{x} \]The first derivative \( f'(x) \) is found using the rules of differentiation:\[ f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}\left(\frac{\lambda}{x}\right) = 2x - \frac{\lambda}{x^2} \]

Step 2: Find Critical Points

We find critical points by setting \( f'(x) = 0 \):\[ 2x - \frac{\lambda}{x^2} = 0 \]Solving for \( x \), we get:\[ 2x^3 = \lambda \quad \Rightarrow \quad \lambda = 2x^3 \]

Step 3: Use Second Derivative for Concavity

To establish whether the critical points are maxima, minima, or points of inflection, we take the second derivative of \( f(x) \):\[ f''(x) = \frac{d}{dx}(2x - \frac{\lambda}{x^2}) = 2 + \frac{2\lambda}{x^3} \]

Step 4: Check Options with \\(\lambda=16\\) and x=2

For option (A) and (B), use \( \lambda = 16 \) and \( x = 2 \):- Substitute in \( \, \lambda = 16 \, \) and set the equation \( 2x^3 = \lambda \, \rightarrow \, 2\times 2^3 = 16 \). So, \( x = 2 \, \) is a critical point.- Computing \( f''(x) \) at \( x = 2 \): \[ f''(2) = 2 + \frac{2\times 16}{2^3} = 2 + \frac{32}{8} = 2 + 4 = 6 \]The positive value at \( f''(2) = 6 \) indicates a minimum at \( x = 2 \), not a maximum.

Step 5: Verify Option C

The condition for a maximum requires \( f''(x) < 0 \). Since for any real \( \lambda \), the term \( 2 + \frac{2\lambda}{x^3} \) can never be less than zero for positive \( x \), there can be no maximum for any real \( \lambda \). Hence, option (C) is true.

Step 6: Check Point of Inflexion for Option D

A point of inflexion occurs where \( f''(x) = 0 \). Let's check for \( x = 1 \) with \( \lambda = -1 \):- Substitute \( x = 1 \), \( \lambda = -1 \) into \( f''(x) \):\[ f''(1) = 2 + \frac{2(-1)}{1^3} = 2 - 2 = 0 \]Hence, there is indeed a point of inflexion at \( x = 1 \), making option (D) true.

Key Concepts

Differentiation

Differentiation is a key technique in calculus used to find the rate at which a function is changing at any point. In simple terms, it helps you understand how steep the function is at various points. For the function given in the exercise, \( f(x) = x^2 + \frac{\lambda}{x} \), the process of differentiation involves applying the power rule to \( x^2 \) and the quotient rule to \( \frac{\lambda}{x} \).

• The derivative of \( x^2 \) is \( 2x \), highlighting the change in the slope as \( x \) moves.
• The derivative of \( \frac{\lambda}{x} \) is \( -\frac{\lambda}{x^2} \), which provides how \( \lambda \) affects the slope as \( x \) changes.

The resulting first derivative, \( f'(x) = 2x - \frac{\lambda}{x^2} \), gives us a tool to find critical points by setting it to zero, which tells us where the function changes its slope. Discovering these points is the first step in exploring if they correspond to a maximum, minimum, or other features like points of inflexion.

Critical Points

Critical points occur where the first derivative of a function equals zero, or where the derivative does not exist. These points are important because they provide potential locations of maximum, minimum, or inflection points on the function. For the given function, the critical points are obtained by solving:

• Setting the derivative to zero: \( 2x - \frac{\lambda}{x^2} = 0 \).
• From this equation, solve for \( x \) to find \( x^3 = \frac{\lambda}{2} \).
• Thus, \( x = \sqrt[3]{\frac{\lambda}{2}} \).

These critical points are where the change in the function's behavior could occur. If these points fall within the domain of the function, they help in determining the shape and turning points of the function, which is critical in determining the nature of the graph.

Second Derivative Test

The Second Derivative Test utilizes the second derivative of a function to determine the concavity of graphs at critical points. This test helps classify whether these points are maxima, minima, or points of inflection. For the function \( f(x) = x^2 + \frac{\lambda}{x} \), the second derivative is determined as:

• \( f''(x) = 2 + \frac{2\lambda}{x^3} \)
• This expression informs us about the concavity of the curve.

We evaluate this second derivative at the critical points. For instance, with \( \lambda = 16 \) and \( x = 2 \), substituting gives \( f''(2) = 6 \), indicating a positive value.

• If \( f''(x) > 0 \) at a critical point, the point is a local minimum because the curve is concave upwards.
• If \( f''(x) < 0 \), it indicates a local maximum as the curve is concave downwards.
• If \( f''(x) = 0 \), the test is inconclusive, possibly indicating an inflection point.

This test is a quick method to determine the nature of critical points without the need for further computation.

Maximum and Minimum

The concepts of maximum and minimum point to the high and low turning points on the function's graph. In analyzing the exercise function \( f(x) = x^2 + \frac{\lambda}{x} \):

• A local maximum is where the function reaches its highest value in some neighborhood, determined by \( f'(x) = 0 \) and \( f''(x) < 0 \).
• A local minimum occurs where the function takes its lowest value, identified by \( f'(x) = 0 \) and \( f''(x) > 0 \).

In this exercise:

• For \( \lambda = 16 \) and \( x = 2 \), \( f''(x) = 6 > 0 \), indicating a minimum.
• There is no \( \lambda \) ensuring \( f''(x) < 0 \) for \( x > 0 \), showing no real maximum exists for any real \( \lambda \).

This illustrates how differentiation can be utilized to explore the range of values a function can achieve, giving us insight into its behavior and critical points.