The first task should have been to check whether the function is continuous. Indeed, it is continuous. Hence, **D **is the correct answer.

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Consider the function $$f(x) = \begin{cases} \int_0^x \{5+ \mid 1-y \mid \} dy & \text{ if } x>2 \\ 5x+2 & \text{ if } x \leq 2 \end{cases}$$ Then

- $f$ is not continuous at $x=2$
- $f$ is continuous and differentiable everywhere
- $f$ is continuous everywhere but not differentiable at $x=1$
- $f$ is continuous everywhere but not differentiable at $x=2$

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$$f(x)=\begin{cases}

\displaystyle{\int_{0}^{x}5+|1-y| \ \ \mathrm{d}y} & \text { if } x>2 \\[2ex]

5 x+{2} & \text { if } x \leq 2

\end{cases}$$

$$f'(x)=\begin{cases}

5+|1-x| & \text { if } x>2 \\[2ex]

5 & \text { if } x \leq 2

\end{cases}$$

Using Newton-Leibnitz rule for differentiation

for checking the differentiability: $$f'(2^+)=\lim_\limits{h \to 0} 5+|1-(2+h)|=6 $$

And

$$f'(2^-)=5$$

Hence,

the function is not differentiable ($\because f'(2^+)\neq f'(2^-) $) .

\displaystyle{\int_{0}^{x}5+|1-y| \ \ \mathrm{d}y} & \text { if } x>2 \\[2ex]

5 x+{2} & \text { if } x \leq 2

\end{cases}$$

$$f'(x)=\begin{cases}

5+|1-x| & \text { if } x>2 \\[2ex]

5 & \text { if } x \leq 2

\end{cases}$$

Using Newton-Leibnitz rule for differentiation

for checking the differentiability: $$f'(2^+)=\lim_\limits{h \to 0} 5+|1-(2+h)|=6 $$

And

$$f'(2^-)=5$$

Hence,

the function is not differentiable ($\because f'(2^+)\neq f'(2^-) $) .

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