1. Given function

$$f(x)= \begin{cases} \displaystyle \int_0^x \left(5+|1-t|\right)\,dt, & x>2 \\ 5x+1, & x\le 2 \end{cases}$$

We must check continuity and differentiability, especially at the junction point $x=2$ and also at $x=1$.

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2. Simplify the integrand

Since

$$|1-t|= \begin{cases} 1-t, & t\le 1 \\ t-1, & t\ge 1 \end{cases}$$

we get

$$5+|1-t|= \begin{cases} 6-t, & t\le 1 \\ 4+t, & t\ge 1 \end{cases}$$

For $x>2$, the interval $[0,x]$ crosses $t=1$, so

$$f(x)=\int_0^1 (6-t)\,dt+\int_1^x (4+t)\,dt.$$

Compute:

$$\int_0^1 (6-t)\,dt=\left[6t-\frac{t^2}{2}\right]_0^1=6-\frac12=\frac{11}{2}.$$

Also,

$$\int_1^x (4+t)\,dt=\left[4t+\frac{t^2}{2}\right]_1^x =4x+\frac{x^2}{2}-\left(4+\frac12\right) =4x+\frac{x^2}{2}-\frac92.$$

Thus for $x>2$,

$$f(x)=\frac{11}{2}+4x+\frac{x^2}{2}-\frac92 =1+4x+\frac{x^2}{2}.$$

So the function is

$$f(x)= \begin{cases} 1+4x+\dfrac{x^2}{2}, & x>2 \\ 5x+1, & x\le 2 \end{cases}$$

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3. Check continuity at $x=2$

For $x\le 2$,

$$f(2)=5(2)+1=11.$$

Right-hand limit:

$$\lim_{x\to 2^+} f(x)=1+4(2)+\frac{2^2}{2}=1+8+2=11.$$

Left-hand limit:

$$\lim_{x\to 2^-} f(x)=5(2)+1=11.$$

Hence,

$$\lim_{x\to 2}f(x)=f(2)=11,$$

so $f$ is continuous at $x=2$.

Therefore, Option A is false.

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4. Check differentiability at $x=2$

For $x<2$,

$$f(x)=5x+1 \implies f'(x)=5.$$

So the left derivative at $x=2$ is

$$f'_-(2)=5.$$

For $x>2$,

$$f(x)=1+4x+\frac{x^2}{2} \implies f'(x)=4+x.$$

So the right derivative at $x=2$ is

$$f'_+(2)=4+2=6.$$

Since

$$f'_-(2)\ne f'_+(2),$$

$f$ is not differentiable at $x=2$.

Thus, Option C is true.

And since it is not differentiable at $x=2$, Option B is false.

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5. Check differentiability at $x=1$

Note that for all $x\le 2$, the function is simply

$$f(x)=5x+1,$$

which is a linear polynomial.

Hence it is differentiable at every point $x\le 2$, including $x=1$.

So Option D is false.

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6. Conclusion

Only one statement is correct:

$$\boxed{\text{C: } f(x) \text{ is continuous but not differentiable at } x=2}$$

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7. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So the derived answer agrees with the stored answer.