Problem 34 - A Complete Resource Book in Mathematics for JEE Main

Question

A function \(f\) is such that \(f^{\prime}(4)=f^{\prime \prime}(4)=0\) and \(f\) has minimum value 10 at \(x=4\). Then \(f(x)=\) (A) \(4+(x-4)^{4}\) (B) \(10+(x-4)^{4}\) (C) \((x-4)^{4}\) (D) None of these

Step-by-Step Solution

Answer

Option (B) \(10 + (x-4)^4\) is the correct function.

Step 1: Analyze Given Information

The function \( f \) has its first derivative \( f'(x) \) equal to zero at \( x = 4 \), indicating that there is a critical point at \( x = 4 \). Additionally, the second derivative \( f''(x) \) is also zero at this point, suggesting the point could be a minimum, maximum, or point of inflection. The problem states that the function has a minimum value of 10 at \( x = 4 \).

Step 2: Determine the General Form of Function

A function with a flat (zero first and second derivative) minimum value typically includes terms starting from the third power or higher. Since the function has its minimum at \( x = 4 \) and exhibits zero slopes at that point for both the first and second derivatives, a likely form is \( f(x) = c + (x-4)^k \) where \( k\geq 2 \). For the simplest case, consider \( k=4 \), given the options presented.

Step 3: Evaluate given Options

Option (A) is \( f(x) = 4 + (x-4)^4 \), which does not match the minimum value of 10 because at \( x=4 \), \( f(x) = 4 + 0 = 4 \).Option (B) is \( f(x) = 10 + (x-4)^4 \). At \( x=4 \), this gives \( f(x) = 10 + 0 = 10 \), which matches the given minimum value.Option (C) is \( f(x) = (x-4)^4 \), which results in \( f(4) = 0 \), not matching the given minimum value of 10.

Step 4: Confirm the Correct Option

From Step 3, option (B) fits both conditions: the value of 10 at \( x = 4 \) and the nature of its derivative properties, without proposing unnecessary complexity, making it the correct choice.

Key Concepts

Critical Points

In calculus, critical points of a function are where its derivative is zero or undefined. At these points, the function might have a local maximum, a local minimum, or a saddle point. Finding these points is crucial for analyzing the function's behavior.
To determine a critical point, we calculate the first derivative of the function, denoted as \( f'(x) \). If \( f'(4) = 0 \), as in our exercise, it indicates a critical point at \( x = 4 \).

• If \( f'(x) \) changes sign around a critical point, it shows a potential extremum (maximum or minimum).
• If there is no sign change, the point might be a plateau or a point of inflection.

Identifying critical points is the first step in understanding how a function behaves at or around these values.

First Derivative Test

The First Derivative Test helps determine whether a critical point is a local minimum or maximum. This is done by observing the changes in sign of the first derivative, \( f'(x) \), around the critical points.
Here is how you use the test:

• Calculate \( f'(x) \) and identify critical points.
• Analyze the sign of \( f'(x) \) before and after the critical point.

If \( f'(x) \) changes from positive to negative, then the point is a local maximum.
If \( f'(x) \) shifts from negative to positive, there's a local minimum.
No change in sign results in neither a local max nor min.
In the given exercise, since \( f'(4) = 0 \), we know \( x = 4 \) is a critical point but the function's context defined it as a minimum value. This points towards further testing with the Second Derivative Test.

Second Derivative Test

The Second Derivative Test is a useful tool for determining the nature of critical points when they aren't clearly defined by the first derivative test.
This test involves the second derivative, \( f''(x) \):

• If \( f''(x) > 0 \) at a critical point, the point is a local minimum because the function is concave up.
• If \( f''(x) < 0 \), it's a local maximum since the function is concave down.
• With \( f''(x) = 0 \), the test isn't conclusive, indicating either a possible point of inflection or higher-order testing is needed.

In our problem, \( f''(4) = 0 \), hence the second derivative test alone could not determine the nature of \( x = 4 \). However, since the problem specifies \( x = 4 \) as a minimum, the function is likely aligned as in option B, adhering to higher powers \((x-4)^4\) to satisfy both conditions given initially.