1. Understand the definition for $0\le x\le \pi$

We have $f(x)=\max\{\sin t:0\le t\le x\}, \qquad 0\le x\le \pi.$

So for each $x$, we need the maximum value attained by $\sin t$ on the interval $[0,x]$.

Recall:

• On $[0,\pi/2]$, $\sin t$ is increasing.
• On $[\pi/2,\pi]$, $\sin t$ is decreasing.
• The maximum of $\sin t$ on $[0,\pi]$ is $1$, attained at $t=\pi/2$.

Therefore:

• If $0\le x\le \pi/2$, then the maximum on $[0,x]$ occurs at $t=x$, so $f(x)=\sin x.$
• If $\pi/2\le x\le \pi$, then $\pi/2\in[0,x]$, so the maximum is already $1$, hence $f(x)=1.$

Thus the function can be rewritten as

2. Check continuity

We only need to inspect the junction points $x=\pi/2$ and $x=\pi$ since each piece is continuous on its own interval.

At $x=\pi/2$

Left value: $\lim_{x\to (\pi/2)^-} f(x)=\sin\frac{\pi}{2}=1.$ Right value: $\lim_{x\to (\pi/2)^+} f(x)=1.$ And $f\left(\frac{\pi}{2}\right)=1.$ So $f$ is continuous at $x=\pi/2$.

At $x=\pi$

From the left: $\lim_{x\to \pi^-} f(x)=1.$ From the right: $\lim_{x\to \pi^+} f(x)=2+\cos\pi=2-1=1.$ Also $f(\pi)=1.$ So $f$ is continuous at $x=\pi$.

Hence, $f$ is continuous everywhere on $[0,\infty)$.

3. Check differentiability in $(0,\infty)$

Possible trouble points are again $x=\pi/2$ and $x=\pi$.

For $0<x<\pi/2$

$f(x)=\sin x \implies f'(x)=\cos x.$ So differentiable here.

For $\pi/2<x<\pi$

$f(x)=1 \implies f'(x)=0.$ So differentiable here.

For $x>\pi$

$f(x)=2+\cos x \implies f'(x)=-\sin x.$ So differentiable here.

At $x=\pi/2$

Left derivative: $f'_-(\pi/2)=\cos\frac{\pi}{2}=0.$ Right derivative: $f'_+(\pi/2)=0.$ So $f$ is differentiable at $x=\pi/2$.

At $x=\pi$

Left derivative: Since $f(x)=1$ for $\pi/2<x\le \pi$, $f'_-(\pi)=0.$ Right derivative: Since $f(x)=2+\cos x$ for $x>\pi$, $f'_+(\pi)=-\sin\pi=0.$ So $f$ is differentiable at $x=\pi$.

Therefore, $f$ is differentiable everywhere in $(0,\infty)$.

4. Evaluate the options

• A: False, because $f$ is differentiable everywhere in $(0,\infty)$.
• B: True.
• C: False, because $f$ is continuous everywhere.
• D: False, because there are no points in $(0,\infty)$ where differentiability fails.

5. Final answer

The correct option is $\boxed{\text{B}}.$