The value of p so that the function `f(x) = sin x - cos x - px + q` decreases for all real values of x is given by

Options

A. `p ge sqrt(2)`
B. `p lt sqrt(2)`
C. `p ge 1 `
D. `p lt 1 `

Answer

To find the value of \( p \) such that the function \( f(x) = \sin x - \cos x - px + q \) decreases for all real values of \( x \), we need to follow these steps:

### Step 1: Find the derivative of \( f(x) \)
The first step is to differentiate the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(\sin x - \cos x - px + q) \]

Using the derivatives of sine and cosine, we have:

\[ f'(x) = \cos x + \sin x - p \]

### Step 2: Set the derivative less than or equal to zero
For the function to be decreasing for all \( x \), we need:

\[ f'(x) \leq 0 \]

This implies:

\[ \cos x + \sin x - p \leq 0 \]

Rearranging gives:

\[ \cos x + \sin x \leq p \]

### Step 3: Determine the maximum value of \( \cos x + \sin x \)
Next, we need to find the maximum value of \( \cos x + \sin x \).

Using the identity:

\[ \cos x + \sin x = \sqrt{2} \left( \sin\left(x + \frac{\pi}{4}\right) \right) \]

The maximum value of \( \sin \) function is 1, so:

\[ \cos x + \sin x \leq \sqrt{2} \]

### Step 4: Set the inequality for \( p \)
To ensure that \( f'(x) \leq 0 \) for all \( x \), we must have:

\[ \sqrt{2} \leq p \]

This means:

\[ p \geq \sqrt{2} \]

### Conclusion
Thus, the value of \( p \) such that the function \( f(x) \) decreases for all real values of \( x \) is:

\[ p \geq \sqrt{2} \]