$|1-y| =\left\{\begin{matrix} y-1, & y>1\\ 1-y,&y \leq1 \end{matrix}\right.$
We have to evaluate $\int_{0}^x(5 + |1-y|)dy$ $=\int_{0}^1(5 + (1-y))dy + \int_{1}^x(5 + (y-1))dy$
$=\int_{0}^1(6-y)dy + \int_{1}^x(4+y)dy$
$=6\int_{0}^1 dy - \int_{0}^1y$ $dy + 4\int_{1}^xdy + \int_{1}^xy$ $dy$
$=6[$ $y$ $]_0^1 - \left[\dfrac{y^2}{2}\right]_0^1 + 4[$ $x$ $]_1^x + \left[\dfrac{y^2}{2}\right]_1^x$
$=(6-0) - \left[\dfrac{1}{2} - 0\right] + [4x -4] + \left[\dfrac{x^2}{2} - \dfrac{1}{2}\right]$
$= 6 - 4 -1 +4x +\dfrac{x^2}{2}$
$=\dfrac{x^2}{2} + 4x + 1$
Now $f(x) =\left\{\begin{matrix} \dfrac{x^2}{2} + 4x + 1, & x>2\\ 5x+2,&x \leq2 \end{matrix}\right.$
$ $
Continuity at $x =2$ :-
$f(2) = 5(2) + 2 = 12$
Left hand limit:- $\lim_{x \to 2^-} f(x) =\lim_{x \to 2^-} 5x + 2 = 5(2) + 2 = 12 = f(2)$
Right hand limt: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \dfrac{x^2}{2} + 4x + 1 =\dfrac{2^2}{2} + 4(2) + 1 = 2 + 8 + 1 = 11 \neq f(2)$
So $f(x)$ is not continuous at $x=0$, option A is correct one.