Let `f(x)= x sin pi x, x gt 0` then for all natural number `n, f'(x)` vanishes at

Options

A. a unique ponit in the interval `(n, n+(1)/(2))`
B. a unique point in the interval (`n+(1)/(2), n+1)`
C. a unique point in the interval `(n, n+1)`
D. two point in the interval `(n, n+1)`

Answer

To solve the problem, we need to find the points where the derivative \( f'(x) \) of the function \( f(x) = x \sin(\pi x) \) vanishes for all natural numbers \( n \).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = x \sin(\pi x) \]

2. **Differentiate the function**:
We will use the product rule for differentiation. The product rule states that if you have two functions \( u(x) \) and \( v(x) \), then the derivative \( (uv)' = u'v + uv' \).

Let:
- \( u = x \) and \( v = \sin(\pi x) \)

Then:
- \( u' = 1 \)
- \( v' = \pi \cos(\pi x) \)

Now applying the product rule:
\[ f'(x) = u'v + uv' = 1 \cdot \sin(\pi x) + x \cdot \pi \cos(\pi x) \]
Thus,
\[ f'(x) = \sin(\pi x) + \pi x \cos(\pi x) \]

3. **Set the derivative to zero**:
We want to find where \( f'(x) = 0 \):
\[ \sin(\pi x) + \pi x \cos(\pi x) = 0 \]

4. **Rearranging the equation**:
Rearranging gives:
\[ \sin(\pi x) = -\pi x \cos(\pi x) \]

5. **Divide both sides by \( \cos(\pi x) \)** (where \( \cos(\pi x) \neq 0 \)):
This gives:
\[ \tan(\pi x) = -\pi x \]

6. **Analyze the equation**:
The equation \( \tan(\pi x) = -\pi x \) suggests that we need to find the intersections of the functions \( y = \tan(\pi x) \) and \( y = -\pi x \).

7. **Finding the points of intersection**:
- The function \( \tan(\pi x) \) has vertical asymptotes at \( x = \frac{1}{2} + n \) for \( n \in \mathbb{Z} \).
- The line \( y = -\pi x \) is a straight line with a negative slope.

We can observe that:
- For \( n = 0 \), the first intersection occurs at \( x = 0 \).
- For \( n = 1 \), the next intersections occur between \( n \) and \( n + 1 \).

8. **Conclusion**:
Therefore, the derivative \( f'(x) \) vanishes at points \( x = n + \frac{1}{2} \) for all natural numbers \( n \).

### Final Answer:
The points where \( f'(x) \) vanishes for all natural numbers \( n \) are:
\[ x = n + \frac{1}{2}, \quad n \in \mathbb{N} \]