Consider the cubic $f(x) = 8x^3 + 4ax^2 + 2bx + a$ where $a, b \in \mathbb{R}$. For $a = 1$ if $y = f(x)$ is strictly increasing $\forall x \in \mathbb{R}$ then maximum range of values of $b$ is:
Step-by-Step Solution
Key Concept: A cubic function is strictly increasing when its derivative (a quadratic) is non-negative everywhere, requiring non-positive discriminant.
For $f(x) = 8x^3 + 4ax^2 + 2bx + a$ with $a = 1$, we have $f(x) = 8x^3 + 4x^2 + 2bx + 1$. For strict monotonicity, we need $f'(x) \geq 0$ for all $x \in \mathbb{R}$. Computing the derivative: $f'(x) = 24x^2 + 8x + 2b$. For this quadratic to be non-negative for all $x$, its discriminant must be non-positive: $\Delta = 64 - 4(24)(2b) = 64 - 192b \leq 0$. This gives $192b \geq 64$, so $b \geq \frac{1}{3}$. Therefore, the maximum range of values of $b$ is $\left[\frac{1}{3}, \infty\right)$.
Correct Answer: 3