Question

If $f(x)=2x^{3}-3x^{2}-\lambda x+1, x\in[0,3]$ attains a local minimum at $x=2$, then the value of $\lambda$ is equal to

• A. -12
• B. -36
• C. 12
• D. 36
• E. -6

Hint:

Logic Tip: Extrema always occur at critical points where $f'(x) = 0$. The moment you see "local minimum at $x=c$", immediately write down $f'(c) = 0$ to set up your primary equation.

Answer 1

The Correct Option is C

Concept:
Fermat's theorem states that if a function $f(x)$ has a local extremum (minimum or maximum) at a point $x=c$ and is differentiable there, then its first derivative must be zero at that point: $f'(c) = 0$.
Step 1: Find the first derivative of the function.
Given $f(x) = 2x^3 - 3x^2 - \lambda x + 1$. Differentiate with respect to $x$: $$f'(x) = \frac{d}{dx}(2x^3) - \frac{d}{dx}(3x^2) - \frac{d}{dx}(\lambda x) + \frac{d}{dx}(1)$$ $$f'(x) = 6x^2 - 6x - \lambda$$
Step 2: Apply the condition for a local minimum.
We are given that the local minimum occurs at $x = 2$. Therefore, substitute $x = 2$ into the derivative and set it equal to 0: $$f'(2) = 6(2)^2 - 6(2) - \lambda = 0$$
Step 3: Solve for $\lambda$.
$$6(4) - 12 - \lambda = 0$$ $$24 - 12 - \lambda = 0$$ $$12 - \lambda = 0$$ $$\lambda = 12$$
Step 4: Verify it's a minimum (Optional but good practice).
Find the second derivative $f''(x) = 12x - 6$. At $x = 2$, $f''(2) = 12(2) - 6 = 18$. Since $18>0$, the concavity is upwards, confirming it is indeed a local minimum.