If $f(x)=\left\{\begin{array}{ll}\int_{0}^{x}(5+|1-t|) d t, & x>2 \\ 5 x+1, & x \leq 2\end{array}\right.$, then
Correct Option: , 3
$f(\mathrm{x})=\int_{0}^{1}(5+(1-\mathrm{t})) \mathrm{dt}+\int_{1}^{\mathrm{x}}(5+(\mathrm{t}-1)) \mathrm{dt}$
$=6-\frac{1}{2}+\left.\left(4 \mathrm{t}+\frac{\mathrm{t}^{2}}{2}\right)\right|_{1} ^{\mathrm{x}}$
$=\frac{11}{2}+4 x+\frac{x^{2}}{2}-4-\frac{1}{2}$
$=\frac{x^{2}}{2}+4 x+1$
$f\left(2^{+}\right)=2+8+1=11$
$f(2)=f\left(2^{-}\right)=5 \times 2+1=11$
$\Rightarrow$ continuous at $x=2$
Clearly differentiable at $x=1$
$\operatorname{Lf}^{\prime}(2)=5$
$\operatorname{Rf}^{\prime}(2)=6$
$\Rightarrow$ not differentiable at $x=2$
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