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: Use the condition for maximum and minimum.
If a function attains maximum or minimum at a point, then its first derivative at that point must be zero. Since the function has maximum at \(x=1\) and minimum at \(x=2\), we must have:
\[ f'(1)=0 \quad \text{and} \quad f'(2)=0 \]

Step 2: Differentiate the given function.
The given function is:
\[ f(x)=ax^3-9x^2+6ax+6 \] Differentiating with respect to \(x\), we get:
\[ f'(x)=3ax^2-18x+6a \]

Step 3: Apply the condition \(f'(1)=0\).
Substitute \(x=1\) into the derivative:
\[ f'(1)=3a(1)^2-18(1)+6a=0 \] \[ 3a-18+6a=0 \] \[ 9a-18=0 \] \[ 9a=18 \] \[ a=2 \]

Step 4: Apply the condition \(f'(2)=0\).
Now substitute \(x=2\):
\[ f'(2)=3a(2)^2-18(2)+6a=0 \] \[ 12a-36+6a=0 \] \[ 18a-36=0 \] \[ 18a=36 \] \[ a=2 \] So both conditions consistently give:
\[ a=2 \]

Step 5: Verify the nature of extrema using second derivative.
Differentiate again:
\[ f''(x)=6ax-18 \] Now put \(a=2\):
\[ f''(x)=12x-18 \]

Step 6: Check at \(x=1\) and \(x=2\).
At \(x=1\):
\[ f''(1)=12-18=-6<0 \] So \(x=1\) is a point of maximum.
At \(x=2\):
\[ f''(2)=24-18=6>0 \] So \(x=2\) is a point of minimum. This confirms the result.

Step 7: State the final answer.
Hence, the required value of \(a\) is:
\[ \boxed{2} \] which matches option \((5)\).