Let $f: \mathbb{R} \to \mathbb{R}$ be defined as follows: $$f(x) = \left(\frac{|x|}{2} - x\right)\left(x - \frac{|x|}{2}\right)$$ Which of the following statements is/are true?

We simplify the function $f(x)$ by expanding:

Evaluating for the two cases of $x$:

• For $x \ge 0$, $|x| = x$:

  $$f(x) = -\left(x - \frac{x}{2}\right)^2 = -\left(\frac{x}{2}\right)^2 = -\frac{x^2}{4}$$

• For $x < 0$, $|x| = -x$:

  $$f(x) = -\left(x - \left(-\frac{x}{2}\right)\right)^2 = -\left(\frac{3x}{2}\right)^2 = -\frac{9x^2}{4}$$

1. Local Extrema (Options A and B):

   • For all $x

e 0$, $f(x) < 0$, and at $x = 0$, $f(0) = 0$.- Thus, $f(x) \le 0$ for all $x \in \mathbb{R}$, which means $f$ has a global (and local) maximum at $x = 0$. - $f(x) \to -\infty$ as $x \to \pm\infty$, so $f$ has no local minimum. - Therefore, (A) is true and (B) is false.

2. Continuity of $f^{\prime}$ (Option C):

   • For $x > 0$: $f^{\prime}(x) = -\frac{x}{2}$

   • For $x < 0$: $f^{\prime}(x) = -\frac{9x}{2}$

   • At $x = 0$:

     $$f^{\prime}(0) = \lim_{h \to 0^+} \frac{-h^2/4 - 0}{h} = 0 = \lim_{h \to 0^-} \frac{-9h^2/4 - 0}{h}$$

   • Since $\lim_{x \to 0^+} f^{\prime}(x) = 0$ and $\lim_{x \to 0^-} f^{\prime}(x) = 0$, $f^{\prime}(x)$ is continuous everywhere on $\mathbb{R}$.

   • Therefore, (C) is true.

3. Differentiability of $f^{\prime}$ (Option D):

   • The right-hand derivative of $f^{\prime}$ at $0$ is $\lim_{x \to 0^+} \frac{-x/2 - 0}{x} = -\frac{1}{2}$.

   • The left-hand derivative of $f^{\prime}$ at $0$ is $\lim_{x \to 0^-} \frac{-9x/2 - 0}{x} = -\frac{9}{2}$.

   • Since the left-hand and right-hand derivatives do not match at $x = 0$, $f^{\prime}$ is not differentiable at $x = 0$.

   • Therefore, $f^{\prime}$ is not differentiable over $\mathbb{R}$, making (D) true.

Thus, statements A, C, and D are correct.