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$\operatorname{Let} f(x)=\left\{\begin{array}{l}0, \text { if }-1 \leq x < 0 \\ 1, \text { if } x=0 \\ 2, \text { if } 0 < x \leq 1\end{array}\right.$ and let $F(x)=\int_{-1}^{x} f(t) d t,-1 \leq x \leq 1$, then
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The correct answers are:
$\mathrm{F}$ is continuous function in $[-1,1]$, $\mathrm{F}^{\prime}(\mathrm{x})$ does not exists at $\mathrm{x}=0$
$f(x)=\left\{\begin{array}{ll}0 & -1 \leq x < 0 \\ 1 & x=0 \\ 2 & 0 < x \leq 1\end{array}\right.$
$F(x)=\left\{\begin{array}{ll}0 & -1 \leq x \leq 0 \\ 2 x & 0 < x \leq 1\end{array}\right.$
$F(x)=\left\{\begin{array}{ll}0 & -1 \leq x \leq 0 \\ 2 x & 0 < x \leq 1\end{array}\right.$
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