The most general values of 'x' for which ` sin x + cos x ="min"_(a in R)[1,a^(2)-4a+6]` are given by

Options

A. `2 n pi `
B. `2 n pi +(pi)/(2)`
C. `n pi +(-1)^(n)(pi)/(4)-(pi)/(4)`
D. None of these

Answer

To solve the problem, we need to find the most general values of \( x \) for which

\[ \sin x + \cos x = \min_{a \in \mathbb{R}} [1, a^2 - 4a + 6]. \]

### Step 1: Find the minimum value of \( a^2 - 4a + 6 \)

The expression \( a^2 - 4a + 6 \) is a quadratic function. We can complete the square to find its minimum value.

\[ a^2 - 4a + 6 = (a^2 - 4a + 4) + 2 = (a - 2)^2 + 2. \]

The minimum value of \( (a - 2)^2 \) occurs when \( a = 2 \), which gives \( (2 - 2)^2 = 0 \). Therefore, the minimum value of \( a^2 - 4a + 6 \) is:

\[ 0 + 2 = 2. \]

### Step 2: Compare the minimum values

Now we have:

\[ \min_{a \in \mathbb{R}} [1, a^2 - 4a + 6] = \min[1, 2] = 1. \]

### Step 3: Set up the equation

Now we set the equation:

\[ \sin x + \cos x = 1. \]

### Step 4: Rewrite the left-hand side

We can rewrite \( \sin x + \cos x \) using the identity:

\[ \sin x + \cos x = \sqrt{2} \left( \sin x \cdot \frac{1}{\sqrt{2}} + \cos x \cdot \frac{1}{\sqrt{2}} \right) = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right). \]

### Step 5: Solve the equation

Thus, we have:

\[ \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) = 1. \]

Dividing both sides by \( \sqrt{2} \):

\[ \sin \left( x + \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}. \]

### Step 6: Find the general solutions

The general solutions for \( \sin \theta = \frac{1}{\sqrt{2}} \) are:

\[ \theta = n\pi + (-1)^n \frac{\pi}{4}, \quad n \in \mathbb{Z}. \]

Substituting \( \theta = x + \frac{\pi}{4} \):

\[ x + \frac{\pi}{4} = n\pi + (-1)^n \frac{\pi}{4}. \]

### Step 7: Isolate \( x \)

Now, we can isolate \( x \):

1. For \( n \) even:

\[ x + \frac{\pi}{4} = n\pi + \frac{\pi}{4} \implies x = n\pi. \]

2. For \( n \) odd:

\[ x + \frac{\pi}{4} = n\pi - \frac{\pi}{4} \implies x = n\pi - \frac{\pi}{2}. \]

### Final Answer

Thus, the most general values of \( x \) are:

\[ x = n\pi \quad \text{and} \quad x = n\pi - \frac{\pi}{2}, \quad n \in \mathbb{Z}. \]