Question

If \( f(x) = kx^3 - 3x^2 - 12x + 8 \) is strictly decreasing for all \( x \in \mathbb{R} \), then:

• A. \( k<-\frac{1}{4} \)
• B. \( k>-\frac{1}{4} \)
• C. \( k>\frac{1}{4} \)
• D. \( k<\frac{1}{4} \)

Hint:

A function is strictly decreasing if its first derivative is always negative. - Ensure the discriminant condition holds when working with quadratic inequalities.

Answer 1

The Correct Option is A

Step 1: Compute the first derivative
\[ f'(x) = 3kx^2 - 6x - 12. \] For \( f(x) \) to be strictly decreasing, we require: \[ f'(x)<0 \quad \forall x \in \mathbb{R}. \] Step 2: Find the condition for negativity
For a quadratic function to always be negative, its discriminant must be non-positive: \[ \Delta = (-6)^2 - 4(3k)(-12) = 36 + 144k \leq 0. \] Solving, \[ 144k \leq -36. \] \[ k \leq -\frac{1}{4}. \] Thus, the correct answer is \( \boxed{k<-\frac{1}{4}} \).