If $$f(x) = \begin{cases} \int_0^x (5 + |1 - t|) \, dt, & x > 2 \\ 5x + 1, & x \leq 2 \end{cases}$$, then

We begin by writing the definition of the function clearly:

$$ f(x)= \begin{cases} \displaystyle\int_{0}^{x}\bigl(5+\lvert 1-t\rvert\bigr)\,dt, & x>2\\[6pt] 5x+1, & x\le 2 \end{cases} $$

The questions to check are continuity and differentiability at the points where the formula changes or where the integrand changes its algebraic form. The possible “trouble spots’’ are therefore $$x=2$$ (where the definition changes) and $$x=1$$ (where $$\lvert 1-t\rvert$$ changes its expression).

Step 1 : Continuity at $$x=2$$

We must compare the left-hand limit, the right-hand limit and the actual value.

Left value (or left limit) comes from the polynomial part because for $$x\le 2$$ we have $$f(x)=5x+1$$. Substituting $$x=2$$, we get

$$f(2^{-})=5\cdot 2+1=10+1=11.$$

Right value is obtained from the integral formula with the upper limit $$x=2$$:

$$ f(2^{+})=\int_{0}^{2}\bigl(5+\lvert 1-t\rvert\bigr)\,dt. $$

To compute the integral, we split the interval at the point where the absolute-value expression changes sign, namely $$t=1$$.

For $$0\le t\le 1$$, we have $$\lvert 1-t\rvert=1-t$$, so

$$5+\lvert 1-t\rvert = 5+(1-t)=6-t.$$

For $$1\le t\le 2$$, we have $$\lvert 1-t\rvert=t-1$$, so

$$5+\lvert 1-t\rvert = 5+(t-1)=4+t.$$

Hence,

$$ \begin{aligned} f(2^{+}) &= \int_{0}^{1}(6-t)\,dt \;+\;\int_{1}^{2}(4+t)\,dt\\[6pt] &=\Bigl[6t-\tfrac{t^{2}}{2}\Bigr]_{0}^{1}\;+\;\Bigl[4t+\tfrac{t^{2}}{2}\Bigr]_{1}^{2}. \end{aligned} $$

Evaluating each bracket,

$$ \begin{aligned} \Bigl[6t-\tfrac{t^{2}}{2}\Bigr]_{0}^{1}&=\left(6\cdot1-\tfrac{1^{2}}{2}\right)-0=6-\tfrac12=\tfrac{11}{2},\\[6pt] \Bigl[4t+\tfrac{t^{2}}{2}\Bigr]_{1}^{2}&=\left(4\cdot2+\tfrac{2^{2}}{2}\right)-\left(4\cdot1+\tfrac{1^{2}}{2}\right)\\ &=(8+2)-(4+\tfrac12)=10-4.5=\tfrac{11}{2}. \end{aligned} $$

Adding the two halves,

$$f(2^{+})=\tfrac{11}{2}+\tfrac{11}{2}=11.$$

Because $$f(2^{-})=11$$ and $$f(2^{+})=11$$, we have

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

Therefore $$f(x)$$ is continuous at $$x=2$$.

Step 2 : Differentiability at $$x=2$$

For $$x\lt 2$$ we use the simple polynomial part, so

$$f'(x)=\dfrac{d}{dx}(5x+1)=5.$$

Hence the left derivative at 2 is

$$f'(2^{-})=5.$$

For $$x\gt 2$$ we invoke the Fundamental Theorem of Calculus, which states that if

$$F(x)=\int_{a}^{x}g(t)\,dt,$$

then $$F'(x)=g(x)$$ provided $$g$$ is continuous at $$x$$. Here $$g(t)=5+\lvert 1-t\rvert$$ is continuous for all real $$t$$, so

$$f'(x)=5+\lvert 1-x\rvert,\qquad x\gt 2.$$

Substituting $$x=2$$ from the right,

$$f'(2^{+})=5+\lvert 1-2\rvert=5+1=6.$$

The left derivative equals 5 while the right derivative equals 6, i.e.

$$f'(2^{-})\ne f'(2^{+}).$$

Because the two one-sided derivatives are unequal, $$f(x)$$ is not differentiable at $$x=2$$.

Step 3 : Differentiability at $$x=1$$

The point $$x=1$$ lies in the region $$x\le 2$$, so in a neighbourhood of 1 the definition is simply $$f(x)=5x+1$$. Since a linear function has a constant derivative 5 everywhere, $$f(x)$$ is certainly differentiable at $$x=1$$.

Step 4 : Collecting the facts

We have shown that

• $$f(x)$$ is continuous at $$x=2$$,
• but it is not differentiable at $$x=2$$ (the derivatives from the two sides differ),
• and $$f(x)$$ is differentiable at $$x=1$$.

Thus the function is “continuous but not differentiable at $$x=2$$.’’ This matches Option C.

Hence, the correct answer is Option C.