Let's go through each part of the question:

(i) Formula used for $f(x)$

The problem asks for the formula of the half-range sine series for the given function $f(x) = \pi x - x^2$ over $(0, \pi)$.

The half-range sine series only includes sine terms, which means that $f(x)$ is represented as:

$f(x) = \sum_{n=1}^{\infty} b_n \sin(nx)$

So, the correct option is (d).

(ii) Formula used for $b_n$

For the sine series, the coefficients $b_n$ are given by:

$b_n = \frac{2}{\pi} \int_0^{\pi} f(x) \sin(nx) \, dx$

Therefore, the correct option is (b).

(iii) Value of $b_n$

To determine the value of $b_n$, we use the formula:

$b_n = \frac{2}{\pi} \int_0^{\pi} (\pi x - x^2) \sin(nx) \, dx$

This integral evaluates to $\frac{4}{n^3} [1 + (-1)^n]$. Therefore, the correct option is (b).

Final Answers:

• (i) Formula used for $f(x)$: (d)
• (ii) Formula used for $b_n$: (b)
• (iii) Value of $b_n$: (b)

Would you like a more detailed explanation for any of these parts?

Related Questions:

1. What is the full Fourier series representation of $f(x)$?
2. How is the half-range cosine series derived?
3. What is the general form for the coefficients $a_n$ in a cosine series?
4. Can we compute $b_n$ for a more complex function over $(0, \pi)$?
5. How do you determine if a function can be represented by a half-range sine series?

Tip: Always pay attention to the range of the function when selecting a half-range sine or cosine series!