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#### $\mathrm{\text { Let } f(x) \text { be a function defined as }}$$\mathrm{f(x)=\left\{\begin{array}{ll} \int_0^x(3+|t-2|) d t, \text { if } x>4 \\ 2 x+8 \quad, \text { if } x \leq 4 \end{array} .\right.}$$\mathrm{\text { Then, } f(x) \text { is }}$Option: 1 discontinuous at x=4Option: 2  neither continuous nor differentiable at x=4Option: 3 everywhere continuous but not differentiable at x=4Option: 4  everywhere continuous and differentiable

For x > 4, we have

\mathrm{\begin{aligned} & f(x)=\int_0^x(3+|t-2|) d t \\ & \Rightarrow f(x)=\int_0^2(3-(t-2)) d t+\int_2^x(3+(t-2)) d t \end{aligned}}

\mathrm{\begin{aligned} & \Rightarrow f(x)=\int_0^2(5-t) d t+\int_2^x(1+t) d t \\ & \Rightarrow f(x)=\left[5 t-\frac{t^2}{2}\right]_0^2+\left[t+\frac{t^2}{2}\right]_2^x \Rightarrow f(x)=\frac{x^2}{2}+x+4 \end{aligned}}

Thus, we have

$\mathrm{f(x)= \begin{cases}\frac{x^2}{2}+x+4, & \text { if } x>4 \\ 2 x+8 & \text {,if } x \leq 4\end{cases}}$

Clearly, f(x) is continuous at x = 4.We have,

$\mathrm{\text { LHD of } f(x)(\text { at } x=4)=\left\{\frac{d}{d x}(2 x+8)\right\}_{x=4}=2}$

$\mathrm{\text { RHD of } f(x)(\text { at } x=4)=\left\{\frac{d}{d x}\left(\frac{x^2}{2}+x+4\right)\right\}_{x=4}=5}$

$\mathrm{\text { Clearly, LHD of } f(x) \text { (at } x=4) \neq \text { RHD of } f(x) \text { (at } x=4 \text { ) }}$

So,f(x) is not differentiable atx=4.

Thus,f(x) is everywhere continuous but not differentiable atx=4.