Question:

Step 1: Understand $f(x) = \pi\left\{\frac{x}{\pi}\right\}$. Since $\{t\} \in [0,1)$ for all real $t$, we have $f(x) = \pi \cdot \left\{\frac{x}{\pi}\right\} \in [0, \pi)$. So the range of $f(x)$ is $[0, \pi)$. Option (a) is correct. Step 2: Compute $\displaystyle\lim_{x \to \frac{\pi}{2}} f(x)$. As $x \to \frac{\pi}{2}$, $\frac{x}{\pi} \to \frac{1}{2}$, which is not an integer, so $\left\{\frac{x}{\pi}\right\} \to \frac{1}{2}$. Thus $f(x) \to \pi \cdot \frac{1}{2} = \frac{\pi}{2}$. Option (b) is correct. Step 3: Compute $\displaystyle\int_0^{2\pi} f(x)\, dx$. Substitute $t = \frac{x}{\pi}$, $dx = \pi\, dt$: $$\int_0^{2\pi} \pi\left\{\frac{x}{\pi}\right\} dx = \pi \int_0^{2} \pi\{t\} \cdot \pi\, dt = \pi^2 \int_0^2 \{t\}\, dt$$ Since $\int_0^2 \{t\}\, dt = 2 \int_0^1 t\, dt = 2 \cdot \frac{1}{2} = 1$: $$= \pi^2 \cdot 1 = \pi^2$$ Option (c) is correct. Step 4: Compute $f'\!\left(\frac{5\pi}{2}\right)$. For $x$ not at integer multiples of $\pi$, $f(x) = \pi\left\{\frac{x}{\pi}\right\} = \pi\left(\frac{x}{\pi} - \left\lfloor\frac{x}{\pi}\right\rfloor\right) = x - \pi\left\lfloor\frac{x}{\pi}\right\rfloor$. So $f'(x) = 1$ wherever $f$ is differentiable. At $x = \frac{5\pi}{2}$, $\frac{x}{\pi} = \frac{5}{2}$ which is not an integer, so $f'\!\left(\frac{5\pi}{2}\right) = 1$. Option (d) is correct.