The sum of solutions of `sin pi x+cos pi x=0` in `[0, 100]` is (a) 4375 (b) 4975 (c) 5000 (d) 5025

Options

A. 4375
B. 4975
C. 5000
D. 5025

Explanation

To solve the equation \( \sin(\pi x) + \cos(\pi x) = 0 \) in the interval \([0, 100]\), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ \sin(\pi x) + \cos(\pi x) = 0 \]
This can be rearranged to:
\[ \sin(\pi x) = -\cos(\pi x) \]

### Step 2: Dividing by \(\cos(\pi x)\)
Since \(\cos(\pi x) \neq 0\) (as discussed in the video), we can divide both sides by \(\cos(\pi x)\):
\[ \tan(\pi x) = -1 \]

### Step 3: Finding General Solutions
The equation \(\tan(\pi x) = -1\) implies:
\[ \pi x = n\pi - \frac{\pi}{4} \quad \text{for } n \in \mathbb{Z} \]
This simplifies to:
\[ x = n - \frac{1}{4} \]

### Step 4: Determining the Range of \(n\)
We need to find the values of \(n\) such that \(x\) lies within the interval \([0, 100]\):
\[ 0 \leq n - \frac{1}{4} \leq 100 \]
This gives us two inequalities:
1. \(n - \frac{1}{4} \geq 0 \Rightarrow n \geq \frac{1}{4} \Rightarrow n \geq 1\) (since \(n\) is an integer)
2. \(n - \frac{1}{4} \leq 100 \Rightarrow n \leq 100 + \frac{1}{4} \Rightarrow n \leq 100\)

Thus, \(n\) can take integer values from \(1\) to \(100\).

### Step 5: Finding the Sum of Solutions
The solutions for \(x\) are:
\[ x = n - \frac{1}{4} \quad \text{for } n = 1, 2, \ldots, 100 \]
The sum of the solutions is:
\[ \sum_{n=1}^{100} \left(n - \frac{1}{4}\right) = \sum_{n=1}^{100} n - \sum_{n=1}^{100} \frac{1}{4} \]
Calculating each part:
1. The sum of the first \(100\) natural numbers:
\[ \sum_{n=1}^{100} n = \frac{100 \times 101}{2} = 5050 \]
2. The sum of \(\frac{1}{4}\) for \(100\) terms:
\[ \sum_{n=1}^{100} \frac{1}{4} = 100 \times \frac{1}{4} = 25 \]

Thus, the total sum becomes:
\[ 5050 - 25 = 5025 \]

### Final Answer
The sum of the solutions of \( \sin(\pi x) + \cos(\pi x) = 0 \) in the interval \([0, 100]\) is:
\[ \boxed{5025} \]