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(A, D) Given $f(x) = \begin{cases} 0, & -1 \leq x < 0 \ 1, & x = 0 \ 2, & 0 < x \leq 1 \end{cases}$.For $-1 \leq x \leq 0$,$F(x) = \int_{-1}^{x} 0 \

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$F'(x)$ does not exist at $x = 0$

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(A, D) Given $f(x) = \begin{cases} 0, & -1 \leq x For $-1 \leq x \leq 0$,$F(x) = \int_{-1}^{x} 0 \, dt = 0$.For $0 So,$F(x) = \begin{cases} 0, & -1 \leq x \leq 0 \ 2x, & 0 Checking continuity at $x = 0$:$\lim_{x 	o 0^-} F(x) = 0$ and $\lim_{x 	o 0^+} F(x) = 2(0) = 0$.Since $\lim_{x 	o 0^-} F(x) = \lim_{x 	o 0^+} F(x) = F(0) = 0$,$F(x)$ is continuous in $[-1, 1]$.Checking differentiability at $x = 0$:Left-hand derivative $LHD = \lim_{h 	o 0^-} \frac{F(0+h) - F(0)}{h} = \lim_{h 	o 0^-} \frac{0 - 0}{h} = 0$.Right-hand derivative $RHD = \lim_{h 	o 0^+} \frac{F(0+h) - F(0)}{h} = \lim_{h 	o 0^+} \frac{2(0+h) - 0}{h} = \lim_{h 	o 0^+} \frac{2h}{h} = 2$.Since $LHD 
eq RHD$,$F'(x)$ does not exist at $x = 0$. Thus,options $A$ and $D$ are correct.

Let $f(x) = \begin{cases} 0, & \text{if } -1 \leq x < 0 \\ 1, & \text{if } x = 0 \\ 2, & \text{if } 0 < x \leq 1 \end{cases}$ and let $F(x) = \int_{-1}^{x} f(t) \, dt, -1 \leq x \leq 1$. Then:

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