Question:
If the local maximum value of the real valued function
\[
f(x)=2(x^3-1)-3ax(x+4a),\qquad a>0
\]
is \(54\), then \(a=\)
If the local maximum value of the real valued function
\[
f(x)=2(x^3-1)-3ax(x+4a),\qquad a>0
\]
is \(54\), then \(a=\)
Show Hint
To find the local maximum of a polynomial:
\[
\boxed{\text{Find }f'(x)=0,\text{ then use }f''(x)\text{ to classify the critical points.}}
\]
Updated On: Jul 18, 2026
- \(4\)
- \(2\)
- \(9\)
- \(27\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Find the critical points.
Given,
\[
f(x)=2x^3-2-3ax^2-12a^2x.
\]
Differentiating,
\[
f'(x)=6x^2-6ax-12a^2
=6(x-2a)(x+a).
\]
Hence, the critical points are
\[
x=2a,\qquad x=-a.
\]
Step 2: Identify the local maximum point. Again, \[ f''(x)=12x-6a. \] At \[ x=-a, \] \[ f''(-a)=-18a<0 \] since \[ a>0. \] Hence, the local maximum occurs at \[ x=-a. \]
Step 3: Use the given maximum value. Now, \[ f(-a) = 2(-a^3-1)-3a(-a)(3a) = -2a^3-2+9a^3 = 7a^3-2. \] Given, \[ 7a^3-2=54. \] Therefore, \[ 7a^3=56, \] \[ a^3=8, \] \[ a=2. \] Hence, \[ \boxed{2}. \] Thus, \[ \boxed{(B)} \] is the correct answer.
Step 2: Identify the local maximum point. Again, \[ f''(x)=12x-6a. \] At \[ x=-a, \] \[ f''(-a)=-18a<0 \] since \[ a>0. \] Hence, the local maximum occurs at \[ x=-a. \]
Step 3: Use the given maximum value. Now, \[ f(-a) = 2(-a^3-1)-3a(-a)(3a) = -2a^3-2+9a^3 = 7a^3-2. \] Given, \[ 7a^3-2=54. \] Therefore, \[ 7a^3=56, \] \[ a^3=8, \] \[ a=2. \] Hence, \[ \boxed{2}. \] Thus, \[ \boxed{(B)} \] is the correct answer.
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