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If $f(x)=\left\{\begin{array}{l}\tan ^{-1} x \text {, when }|x| \leq 1 \\ \frac{1}{2}(|x|-1) \text { when }|x|>1\end{array}\right.$, then the domain of $\frac{d}{d x} f(x)$ is
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$\mathrm{R}-\{-1,1\}$
$f(x)= \begin{cases}\frac{1}{2}(-x-) & \text { if } x < -1 \\ \tan ^{-1} x & \text { if }-1 \leq x \leq 1 \\ \frac{1}{2}(x-1) & \text { if } x>1\end{cases}$
Since $\int(-1)=\frac{-\pi}{4} ; \int(1)=\frac{\pi}{4}$
$\lim _{x \rightarrow-1-} \int(x)=0$ and $\lim _{x \rightarrow 1+} \int(x)=0$
So, if is not continuous at $-1,1$, hence not differentiable at $-1,1$ Also
$f^{\prime}(x)= \begin{cases}\frac{-1}{2} & \text { if } x < -1 \\ \frac{1}{1+x^2} & \text { if }-1 < x < 1 \\ \frac{1}{2} & \text { if } x>1\end{cases}$
Thus the domain of $\int(x)$ is $R-\{-1,1\}$
Since $\int(-1)=\frac{-\pi}{4} ; \int(1)=\frac{\pi}{4}$
$\lim _{x \rightarrow-1-} \int(x)=0$ and $\lim _{x \rightarrow 1+} \int(x)=0$
So, if is not continuous at $-1,1$, hence not differentiable at $-1,1$ Also
$f^{\prime}(x)= \begin{cases}\frac{-1}{2} & \text { if } x < -1 \\ \frac{1}{1+x^2} & \text { if }-1 < x < 1 \\ \frac{1}{2} & \text { if } x>1\end{cases}$
Thus the domain of $\int(x)$ is $R-\{-1,1\}$
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