Question:

Step 1: Determine the condition for f(x) to be increasing. For a function to be increasing for all $x \in \mathbb{R}$, its derivative must be non-negative for all $x \in \mathbb{R}$. Therefore, we need: $$f'(x) \geq 0 \text{ for all } x \in \mathbb{R}$$ Step 2: Find the derivative of f(x). Given $f(x) = p^2x - \int 2^{4-x^2}\,dx$, we differentiate with respect to $x$: $$f'(x) = p^2 - 2^{4-x^2}$$ Step 3: Establish the inequality that must hold. For $f(x)$ to be increasing for all $x \in \mathbb{R}$, we require: $$p^2 - 2^{4-x^2} \geq 0 \text{ for all } x \in \mathbb{R}$$ Rearranging: $$p^2 \geq 2^{4-x^2} \text{ for all } x \in \mathbb{R}$$ Step 4: Find the maximum value of $2^{4-x^2}$. The expression $2^{4-x^2}$ is maximized when the exponent $4-x^2$ is maximized. Since $x^2 \geq 0$ for all real $x$, the exponent $4-x^2$ is maximized when $x^2 = 0$, i.e., when $x = 0$. Therefore: $$\max(2^{4-x^2}) = 2^{4-0} = 2^4 = 16$$ Step 5: Determine the condition on p. For the inequality $p^2 \geq 2^{4-x^2}$ to hold for all $x \in \mathbb{R}$, we need: $$p^2 \geq \max(2^{4-x^2}) = 16$$ This gives us: $$p^2 \geq 16$$ Step 6: Solve for p. Taking square roots of both sides: $$|p| \geq 4$$ This means: $$p \leq -4 \text{ or } p \geq 4$$ In interval notation: $$p \in (-\infty, -4] \cup [4, \infty)$$ **Final Answer:** The set of values of $p$ for which $f(x)$ is increasing for all $x \in \mathbb{R}$ is $(-\infty, -4] \cup [4, \infty)$, which corresponds to **Option 3**.