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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

1. $f$ is not continuous at $x=2$
2. $f$ is continuous and differentiable everywhere
3. $f$ is continuous everywhere but not differentiable at $x=1$
4. $f$ is continuous everywhere but not differentiable at $x=2$

$$f(x)=\begin{cases} \displaystyle{\int_{0}^{x}5+|1-y| \ \ \mathrm{d}y} & \text { if } x>2 \\[2ex] 5 x+{2} & \text { if } x \leq 2 \end{cases}$$

$$f'(x)=\begin{cases} 5+|1-x| & \text { if } x>2 \\[2ex] 5 & \text { if } x \leq 2 \end{cases}$$
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 ($\because f'(2^+)\neq f'(2^-)$) .
by

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

It is not continuous, after solving for x>2 condition we get  $((x^2)/2)+4x+1$  then f($2^+$) =11,

f($2^-$) =$5x+2$=12,

f($2^+$)$≠$f($2^-$)

So it is not continuous.

It would have been continuous if the f($2^-$)=$5x+1$, but it is not