Question

If the function $f(x)=ax^3-9x^2+6ax+6$ attains maximum at $x=1$ and minimum at $x=2$, then the value of $a$ is:

• A. \(6\)
• B. \(5\)
• C. \(4\)
• D. \(3\)
• E. \(2\)

Hint:

For maximum and minimum questions, first use \(f'(x)=0\) to find the unknown parameter, and then use \(f''(x)\) to verify whether the point is a maximum or minimum.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
For a differentiable function, local maxima and minima (extrema) occur at critical points, which are points where the first derivative is zero or undefined. Since the given function is a polynomial, its derivative is defined everywhere. Therefore, the locations of the maximum and minimum, \( x=1 \) and \( x=2 \), must be roots of the first derivative, \( f'(x) = 0 \).
Step 2: Key Formula or Approach:
1. Find the first derivative of the function, \( f'(x) \).
2. Since \( x=1 \) and \( x=2 \) are locations of extrema, set \( f'(1) = 0 \) and \( f'(2) = 0 \).
3. Solve the resulting equation(s) for the unknown parameter \( a \).
4. (Optional but recommended) Use the second derivative test to confirm that \( x=1 \) is a maximum and \( x=2 \) is a minimum.
Step 3: Detailed Explanation:
First, find the first derivative of \( f(x) \):
\[ f(x) = ax^3 - 9x^2 + 6ax + 6 \] \[ f'(x) = \frac{d}{dx}(ax^3 - 9x^2 + 6ax + 6) = 3ax^2 - 18x + 6a \] We are given that the function has extrema at \( x=1 \) and \( x=2 \). This means \( f'(1)=0 \) and \( f'(2)=0 \). We can use either condition to find \( a \). Let's use \( x=1 \):
\[ f'(1) = 3a(1)^2 - 18(1) + 6a = 0 \] \[ 3a - 18 + 6a = 0 \] \[ 9a - 18 = 0 \] \[ 9a = 18 \] \[ a = \frac{18}{9} = 2 \] Let's verify this using the second condition, \( f'(2)=0 \), with \( a=2 \):
\[ f'(x) = 3(2)x^2 - 18x + 6(2) = 6x^2 - 18x + 12 \] \[ f'(2) = 6(2)^2 - 18(2) + 12 = 6(4) - 36 + 12 = 24 - 36 + 12 = 0 \] The value \( a=2 \) satisfies both conditions.
To confirm the nature of the extrema, let's use the second derivative test.
\[ f''(x) = \frac{d}{dx}(6x^2 - 18x + 12) = 12x - 18 \] At \( x=1 \): \( f''(1) = 12(1) - 18 = -6 \). Since \( f''(1)<0 \), there is a local maximum at \( x=1 \).
At \( x=2 \): \( f''(2) = 12(2) - 18 = 24 - 18 = 6 \). Since \( f''(2)>0 \), there is a local minimum at \( x=2 \).
The conditions given in the problem are fully satisfied with \( a=2 \).
Step 4: Final Answer:
The value of \( a \) is 2. This corresponds to option (E).