If $f(x) = \begin{cases} -x - \frac{\pi}{2}, & x \leq -\frac{\pi}{2} \\ -\cos x, & -\frac{\pi}{2} < x \leq 0 \\ x - 1, & 0 < x \leq 1 \\ \ln x, & x > 1 \end{cases}$, then

If f(x) = -x - π/2, x ≤ -π/2 -cos x, -π/2 < x ≤ 0 x - 1, 0 < x ≤ 1 ln x, x > 1 , then $f(x) = \begin{cases} -x - \frac{\pi}{2} & x \leq -\frac{\pi}{2} \\[2pt] -\cos x & -\frac{\pi}{2} < x \leq 0 \\[2pt] x - 1 & 0 < x \leq 1 \\[2pt] \ln x & x > 1 \end{cases}$ $\text{Test continuity / differentiability at each break-point}$ $f(-\tfrac{\pi}{2}) = -(-\tfrac{\pi}{2}) - \tfrac{\pi}{2} = 0$ $\lim_{x \to -\pi/2^-} f = 0, \quad \lim_{x \to -\pi/2^+} f = -\cos\tfrac{\pi}{2} = 0$ $\Rightarrow \text{continuous at } -\tfrac{\pi}{2} \Rightarrow (1) \text{ TRUE}$ $f'(0^-) = \sin 0 = 0, \qquad f'(0^+) = \dfrac{d}{dx}(x-1) = 1$ $f'(0^-) \neq f'(0^+) \Rightarrow \text{not differentiable at } 0 \Rightarrow (2) \text{ TRUE}$ $f'(1^-) = 1, \qquad f'(1^+) = \left.\dfrac{1}{x}\right|_{1} = 1$ $\Rightarrow \text{differentiable at } 1 \Rightarrow (3) \text{ TRUE}$ $-\tfrac{3}{2} = -1.5 > -\tfrac{\pi}{2} \approx -1.571 \Rightarrow f = -\cos x$ $f(x) = -\cos x \text{ is smooth there} \Rightarrow \text{differentiable at } -\tfrac{3}{2}$ $\Rightarrow (4) \text{ TRUE}$ $\boxed{\text{All four are TRUE: } (1), (2), (3), (4)}$ $-\tfrac{\pi}{2}: \text{cont.};\quad 0: \text{not diff.};\quad 1: \text{diff.};\quad -\tfrac{3}{2}: \text{diff.}$