Correct : a,c,d

First simplify f(x) by splitting into cases based on the sign of x.
For x ≥ 0: |x| = x, so f(x) = (x/2 − x)(x − x/2) = (−x/2)(x/2) = −x²/4
For x ≤ 0: |x| = −x, so f(x) = (−x/2 − x)(x + x/2) = (−3x/2)(3x/2) = −9x²/4
Both pieces give f(0) = 0, confirming continuity at the origin.
Option A — True: f has a local maximum
For all x ≠ 0, f(x) is negative (both −x²/4 and −9x²/4 are negative for x ≠ 0), while f(0) = 0. So f(0) = 0 is strictly greater than f(x) for all nearby x ≠ 0. Therefore x = 0 is a local maximum.
Option B — False: f has no local minimum
For x > 0, f(x) = −x²/4 is strictly decreasing toward −∞. For x < 0, f(x) = −9x²/4 also decreases toward −∞ as x moves away from 0. Neither branch has a turning point below zero. There is no local minimum anywhere.
Option C — True: f′ is continuous over ℝ
For x > 0: f′(x) = −x/2. For x < 0: f′(x) = −9x/2. At x = 0, the left-hand derivative = lim_x→0⁻(−9x/2) = 0 and the right-hand derivative = lim_x→0⁺(−x/2) = 0. Both equal zero, so f′(0) = 0 and f′ is continuous everywhere.
Option D — True: f′ is not differentiable over ℝ
f′(x) = −9x/2 for x < 0 and f′(x) = −x/2 for x > 0. The derivative of f′ from the left at 0 is −9/2 and from the right is −1/2. Since these differ, f′ is not differentiable at x = 0.
Correct answer: A, C, D ✓