Question:
The function \(f(x)=2x^3-9ax^2+12a^2x+1\), where \(a\gt 0\), attains its maximum and minimum at \(p\) and \(q\) respectively and \(p^2=q\). Then \(a=\)
The function \(f(x)=2x^3-9ax^2+12a^2x+1\), where \(a\gt 0\), attains its maximum and minimum at \(p\) and \(q\) respectively and \(p^2=q\). Then \(a=\)
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For finding local maximum and minimum, first solve \(f'(x)=0\), then use \(f''(x)\). If \(f''(x)\lt 0\), the point is of maximum, and if \(f''(x)\gt 0\), the point is of minimum.
Updated On: Jun 26, 2026
- \(1\)
- \(2\)
- \(\dfrac{1}{2}\)
- \(3\)
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The Correct Option is B
Solution and Explanation
Step 1: Differentiate the given function.
Given, \[ f(x)=2x^3-9ax^2+12a^2x+1 \] Differentiate with respect to \(x\): \[ f'(x)=6x^2-18ax+12a^2 \] Taking \(6\) common, \[ f'(x)=6(x^2-3ax+2a^2) \] Factorizing, \[ f'(x)=6(x-a)(x-2a) \]
Step 2: Find critical points.
For maximum or minimum, \[ f'(x)=0 \] So, \[ 6(x-a)(x-2a)=0 \] Hence, \[ x=a \quad \text{or} \quad x=2a \]
Step 3: Identify maximum and minimum points.
Now, \[ f''(x)=12x-18a \] At \(x=a\), \[ f''(a)=12a-18a=-6a \] Since \(a\gt 0\), \[ f''(a)\lt 0 \] Therefore, \(x=a\) is the point of maximum. Hence, \[ p=a \] At \(x=2a\), \[ f''(2a)=24a-18a=6a \] Since \(a\gt 0\), \[ f''(2a)\gt 0 \] Therefore, \(x=2a\) is the point of minimum. Hence, \[ q=2a \]
Step 4: Use the given condition.
It is given that \[ p^2=q \] Substitute \[ p=a \] and \[ q=2a \] So, \[ a^2=2a \] Since \(a\gt 0\), divide by \(a\): \[ a=2 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{2} \]
Given, \[ f(x)=2x^3-9ax^2+12a^2x+1 \] Differentiate with respect to \(x\): \[ f'(x)=6x^2-18ax+12a^2 \] Taking \(6\) common, \[ f'(x)=6(x^2-3ax+2a^2) \] Factorizing, \[ f'(x)=6(x-a)(x-2a) \]
Step 2: Find critical points.
For maximum or minimum, \[ f'(x)=0 \] So, \[ 6(x-a)(x-2a)=0 \] Hence, \[ x=a \quad \text{or} \quad x=2a \]
Step 3: Identify maximum and minimum points.
Now, \[ f''(x)=12x-18a \] At \(x=a\), \[ f''(a)=12a-18a=-6a \] Since \(a\gt 0\), \[ f''(a)\lt 0 \] Therefore, \(x=a\) is the point of maximum. Hence, \[ p=a \] At \(x=2a\), \[ f''(2a)=24a-18a=6a \] Since \(a\gt 0\), \[ f''(2a)\gt 0 \] Therefore, \(x=2a\) is the point of minimum. Hence, \[ q=2a \]
Step 4: Use the given condition.
It is given that \[ p^2=q \] Substitute \[ p=a \] and \[ q=2a \] So, \[ a^2=2a \] Since \(a\gt 0\), divide by \(a\): \[ a=2 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{2} \]
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