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If the local maximum value of the function f(x)=x3−2λx2+λ2x,λ>0f(x)=x^3-2\lambda x^2+\lambda^2 x, \lambda>0 is 108, Then the local minimum value of the function occurs at

Solution

✅ Correct Option: 2

f′(x)=3x2−4λx+λ2=(3x−λ)(x−λ)f'(x)=3x^2-4\lambda x+\lambda^2=(3x-\lambda)(x-\lambda), so critical points are x=λ3x=\frac{\lambda}{3} and x=λx=\lambda.

Since f′′=6x−4λf''=6x-4\lambda, the maximum is at x=λ3x=\frac{\lambda}{3} and the minimum at x=λx=\lambda.

f(λ3)=4λ327=108f\left(\frac{\lambda}{3}\right)=\frac{4\lambda^3}{27}=108 gives λ3=729\lambda^3=729, so λ=9\lambda=9 and the minimum occurs at x=9x=9.

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