The least value of `n (n in N)`, such that the function `f(n, x)=int n cos (nx) dx` satisfies `f(n, (pi)/(2))=-1`, is (given, `f(n, 0)=0`)

Options

A. 3
B. 4
C. 5
D. 6

Answer

To solve the problem, we need to find the least value of \( n \) (where \( n \in \mathbb{N} \)) such that the function \( f(n, x) = \int n \cos(nx) \, dx \) satisfies \( f(n, \frac{\pi}{2}) = -1 \) and \( f(n, 0) = 0 \).

### Step 1: Find the indefinite integral of \( f(n, x) \)

We start with the function:

\[ f(n, x) = \int n \cos(nx) \, dx \]

Since \( n \) is a constant with respect to \( x \), we can factor it out of the integral:

\[ f(n, x) = n \int \cos(nx) \, dx \]

The integral of \( \cos(nx) \) is:

\[ \int \cos(nx) \, dx = \frac{1}{n} \sin(nx) + C \]

Thus, we have:

\[ f(n, x) = n \left( \frac{1}{n} \sin(nx) + C \right) = \sin(nx) + C \]

### Step 2: Use the condition \( f(n, 0) = 0 \)

We know that:

\[ f(n, 0) = \sin(n \cdot 0) + C = \sin(0) + C = 0 + C = C \]

Since \( f(n, 0) = 0 \), we find that:

\[ C = 0 \]

Thus, the function simplifies to:

\[ f(n, x) = \sin(nx) \]

### Step 3: Set up the equation \( f(n, \frac{\pi}{2}) = -1 \)

Now we need to evaluate \( f(n, \frac{\pi}{2}) \):

\[ f(n, \frac{\pi}{2}) = \sin\left(n \cdot \frac{\pi}{2}\right) = \sin\left(\frac{n\pi}{2}\right) \]

We want this to equal -1:

\[ \sin\left(\frac{n\pi}{2}\right) = -1 \]

### Step 4: Determine the values of \( n \)

The sine function equals -1 at:

\[ \frac{n\pi}{2} = \frac{3\pi}{2} + 2k\pi \quad \text{for } k \in \mathbb{Z} \]

This simplifies to:

\[ n = 3 + 4k \]

Since \( n \) must be a natural number, we start with \( k = 0 \):

\[ n = 3 \]

For \( k = 1 \):

\[ n = 7 \]

And so on. The least value of \( n \) that satisfies the condition is:

\[ n = 3 \]

### Final Answer

Thus, the least value of \( n \) such that \( f(n, \frac{\pi}{2}) = -1 \) is:

\[ \boxed{3} \]