Question

If the function \( f(x) = 2x^3 - 9ax^2 + 12a^2x + 1 \), where \( a>0 \), attains its local maximum and minimum at \( p \) and \( q \), respectively, such that \( p^2 = q \), then \( f(3) \) is equal to:

• A. 23
• B. 55
• C. 10
• D. 37

Hint:

Critical points from \( f'(x) = 0 \) help determine max/min values. Use given condition on them.

Answer 1

The Correct Option is D

The problem asks for the value of \( f(3) \) for the function \( f(x) = 2x^3 - 9ax^2 + 12a^2x + 1 \) (with \( a>0 \)), given that its local maximum and minimum occur at points \( p \) and \( q \) respectively, and that these points are related by the condition \( p^2 = q \).

Concept Used:

To find the local maximum and minimum of a differentiable function, we use the following steps:

1. First Derivative Test: The local extrema of a function occur at its critical points, which are the points where the first derivative is zero or undefined. For a polynomial function, we find the roots of \( f'(x) = 0 \).

2. Second Derivative Test: To classify the critical points, we use the second derivative. Let \( c \) be a critical point such that \( f'(c) = 0 \).

• If \( f''(c) < 0 \), then the function has a local maximum at \( x = c \).
• If \( f''(c) > 0 \), then the function has a local minimum at \( x = c \).

Step-by-Step Solution:

The given function is \( f(x) = 2x^3 - 9ax^2 + 12a^2x + 1 \).

First, we find the first derivative of \( f(x) \) to locate the critical points:

\[ f'(x) = \frac{d}{dx}(2x^3 - 9ax^2 + 12a^2x + 1) = 6x^2 - 18ax + 12a^2 \]

To find the critical points, we set \( f'(x) = 0 \):

\[ 6x^2 - 18ax + 12a^2 = 0 \]

We can simplify this quadratic equation by dividing the entire equation by 6:

\[ x^2 - 3ax + 2a^2 = 0 \]

This equation can be factored by finding two numbers that multiply to \( 2a^2 \) and add up to \( -3a \). These numbers are \( -a \) and \( -2a \).

\[ (x - a)(x - 2a) = 0 \]

The critical points are \( x = a \) and \( x = 2a \). These are the values of \( p \) and \( q \).

To determine which point corresponds to the maximum and which to the minimum, we use the second derivative test. We find the second derivative, \( f''(x) \):

\[ f''(x) = \frac{d}{dx}(6x^2 - 18ax + 12a^2) = 12x - 18a \]

Now, we evaluate \( f''(x) \) at each critical point:

At \( x = a \):

\[ f''(a) = 12(a) - 18a = -6a \]

Since it is given that \( a > 0 \), we have \( f''(a) = -6a < 0 \). Therefore, the function has a local maximum at \( x = a \). This means \( p = a \).

At \( x = 2a \):

\[ f''(2a) = 12(2a) - 18a = 24a - 18a = 6a \]

Since \( a > 0 \), we have \( f''(2a) = 6a > 0 \). Therefore, the function has a local minimum at \( x = 2a \). This means \( q = 2a \).

We are given the condition \( p^2 = q \). Substituting our values for \( p \) and \( q \):

\[ (a)^2 = 2a \implies a^2 - 2a = 0 \] \[ a(a - 2) = 0 \]

This gives two possible solutions: \( a = 0 \) or \( a = 2 \). Since the problem states that \( a > 0 \), we must have \( a = 2 \).

Now that we have the value of \( a \), we can write the specific function for \( f(x) \):

\[ f(x) = 2x^3 - 9(2)x^2 + 12(2)^2x + 1 = 2x^3 - 18x^2 + 48x + 1 \]

Final Computation & Result:

We are asked to find the value of \( f(3) \). We substitute \( x = 3 \) into the function:

\[ f(3) = 2(3)^3 - 18(3)^2 + 48(3) + 1 \] \[ f(3) = 2(27) - 18(9) + 144 + 1 \] \[ f(3) = 54 - 162 + 144 + 1 \] \[ f(3) = 199 - 162 = 37 \]

Hence, the value of \( f(3) \) is 37.

Answer 2

First derivative: \[ f'(x) = 6x^2 - 18ax + 12a^2 = 6(x^2 - 3ax + 2a^2) \] Roots of \( f'(x) = 0 \) are \( x = a, 2a \). Given: \( p^2 = q \Rightarrow a^2 = 2a \Rightarrow a = 2 \) Now, \[ f(3) = 2(3)^3 - 9(2)(3)^2 + 12(2)^2(3) + 1 = 54 - 162 + 144 + 1 = 37 \]