Suppose $f(x) = -x^2 + bx + 1$ and $g(x) = x^2 + 2x + c$, $b, c \in \mathbb{R}$, are such that maximum $f(x) \leq$ minimum $g(x)$ as $x$ varies over $\mathbb{R}$. Then possible values that $c$ can take is(are):
Step-by-Step Solution
Key Concept: The constraint requires finding the minimum value of $c$ such that there exists some real $b$ satisfying $c \geq \frac{b^2}{4} + 2$, which means $c \geq 2$ is necessary but any $c \geq 2$ is possible.
For $f(x) = -x^2 + bx + 1$, the maximum occurs at $x = \frac{b}{2}$ giving $f_{max} = \frac{b^2}{4} + 1$. For $g(x) = x^2 + 2x + c$, the minimum occurs at $x = -1$ giving $g_{min} = c - 1$. The constraint $f_{max} \leq g_{min}$ requires $\frac{b^2}{4} + 1 \leq c - 1$, which means $c \geq \frac{b^2}{4} + 2$. Since this must hold for all $b \in \mathbb{R}$, we need $c \geq 2 + \max(\frac{b^2}{4}) = \infty$ for all $b$. However, the condition must be satisfiable for some $b$, so $c$ must be at least $2$ (when $b = 0$). Testing the options: $c = 2$ works (with $b = 0$), and $c = \sqrt{5} \approx 2.236$ works (allowing $b^2 \leq 4(\sqrt{5}-2) \approx 1.71$).
Correct Answer: 3,4