If the local maximum value of the function \( f(x)=x^3-2\lambda x^2+\lambda^2x,\lambda\gt 0 \) is 108, then the local minimum value of the function occurs at
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( x=9 \)
The critical points come from \( f'(x)=0 \):
\( \displaystyle f'(x)=3x^2-4\lambda x+\lambda^2=(3x-\lambda)(x-\lambda) \)
\( \displaystyle f'(x)=0\ \Rightarrow\ x=\frac{\lambda}{3}\ \text{or}\ x=\lambda \)
\( \displaystyle f''(x)=6x-4\lambda \)
\( \displaystyle f''\left(\frac{\lambda}{3}\right)=-2\lambda\lt 0\ \text{(local maximum)} \)
\( \displaystyle f''(\lambda)=2\lambda\gt 0\ \text{(local minimum)} \)
The local maximum value is at \( x=\frac{\lambda}{3} \):
\( \displaystyle f\left(\frac{\lambda}{3}\right)=\frac{\lambda^3}{27}-\frac{2\lambda^3}{9}+\frac{\lambda^3}{3} \)
\( \displaystyle =\frac{\lambda^3-6\lambda^3+9\lambda^3}{27} \)
\( \displaystyle =\frac{4\lambda^3}{27} \)
\( \displaystyle \frac{4\lambda^3}{27}=108 \)
\( \displaystyle \lambda^3=729\ \Rightarrow\ \lambda=9 \)
So the local minimum occurs at \( x=\lambda=9 \).
Why the other options are incorrect
Option (1) \( x=3 \) — Incorrect. x = 3 = λ/3 is where the local maximum occurs, not the minimum.
Option (3) \( x=9\sqrt{3} \) — Incorrect. The minimum is at x = λ = 9, not \( 9\sqrt{3} \).
Option (4) \( x=3\sqrt{3} \) — Incorrect. The minimum is at x = 9, not \( 3\sqrt{3} \).