Given. $f(x)=\left\{\begin{aligned} \int_{0}^{x}|1-t| d t, & x>1 \\ x-\frac{1}{2}, & x \leq 1 \end{aligned}\right.$
Now, for $x>1$, $\int_{0}^{x}|1-t| d t$
$$
\begin{aligned}
&=\int_{0}^{1}(1-t) d t+\int_{1}^{x}(t-1) d t \\
&=\left[t-\frac{t^{2}}{2}\right]_{0}^{1}+\left[\frac{t^{2}}{2}-t\right]_{1}^{x} \\
&=\left[1-\frac{1}{2}-0\right]+\left[\frac{x^{2}}{2}-x-\frac{1}{2}+1\right] \\
&=\frac{1}{2}+\frac{x^{2}}{2}-x-\frac{1}{2}+1 \\
\Rightarrow \quad \frac{x^{2}}{2}-x+1
\end{aligned}
$$
$\therefore$ Eq. (i) becomes, $f(x)=\left\{\begin{array}{ll}\frac{x^{2}}{2}-x+1, & x>1 \\ x-\frac{1}{2}, & x \leq 1\end{array}\right.$
Continuity at $x=1$
and $\begin{aligned} \mathrm{LHL} &=\lim (f(x)\\ &=\lim _{x \rightarrow 1}\left(x-\frac{1}{2}\right)=1-\frac{1}{2} \\ \mathrm{RHL} &=\lim _{x \rightarrow 1^{+}} f(x) \\ &=\lim _{x \rightarrow 1}\left(\frac{x^{2}}{2}-x+1\right) \\ &=\frac{1}{2}-1+1=\frac{1}{2} \\ f(1) &=1-\frac{1}{2}=\frac{1}{2} \end{aligned}$
$\therefore \quad f(1)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)$
Hence, $f(x)$ is continuous at $x=1$
Differentiability at $x=1$
$\begin{aligned} L H D &=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \\ &=\lim _{h \rightarrow 0} \frac{(1-h)-\frac{1}{2}-\left(1-\frac{1}{2}\right)}{-h}=\lim _{h \rightarrow 0} \frac{-h}{-h}=1 \end{aligned}$
$\begin{aligned} \mathrm{RHD} &=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\frac{(1+h)^{2}}{2}-(1+h)+1-\left(1-\frac{1}{2}\right)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\frac{1+h^{2}+2 h}{2}-1-h+1-\frac{1}{2}}{h} \\ &=\lim _{h \rightarrow 0} \frac{\frac{1}{2}+\frac{h^{2}}{2}+h-h-\frac{1}{2}}{h} \\ &=\lim _{h \rightarrow 0} \frac{h^{2}}{2 h}=0 \end{aligned}$
$\mathrm{LHD} \neq \mathrm{RHD}$
Hence, $f(x)$ is not differentiable at $x=1$.