Let `f(x) =` Maximum `{sin x, cos x} AA x in R`. minimum value of `f (x)` is

Options

A. `(1-sqrt2)`
B. `-1`
C. `(1-1/sqrt2)`
D. `(-1)/(sqrt2)`

Answer

To find the minimum value of the function \( f(x) = \max(\sin x, \cos x) \) for \( x \in \mathbb{R} \), we will analyze the behavior of the sine and cosine functions.

### Step-by-Step Solution:

1. **Understand the Functions**:
The functions \( \sin x \) and \( \cos x \) oscillate between -1 and 1. The maximum function will take the greater value between \( \sin x \) and \( \cos x \) at any point \( x \).

2. **Find Points of Intersection**:
To find where \( \sin x = \cos x \), we set:
\[ \sin x = \cos x \]
This occurs when:
\[ \tan x = 1 \quad \Rightarrow \quad x = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z} \]

3. **Evaluate the Function in Intervals**:
- For \( x \) in the interval \( \left[0, \frac{\pi}{4}\right] \):
- Here, \( \sin x \leq \cos x \), so \( f(x) = \cos x \).
- For \( x \) in the interval \( \left[\frac{\pi}{4}, \frac{5\pi}{4}\right] \):
- Here, \( \sin x \geq \cos x \), so \( f(x) = \sin x \).
- For \( x \) in the interval \( \left[\frac{5\pi}{4}, 2\pi\right] \):
- Here, \( \sin x \leq \cos x \), so \( f(x) = \cos x \).

4. **Evaluate the Function at Critical Points**:
- At \( x = 0 \): \( f(0) = \max(\sin 0, \cos 0) = \max(0, 1) = 1 \)
- At \( x = \frac{\pi}{4} \): \( f\left(\frac{\pi}{4}\right) = \max\left(\sin\frac{\pi}{4}, \cos\frac{\pi}{4}\right) = \max\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}} \)
- At \( x = \frac{5\pi}{4} \): \( f\left(\frac{5\pi}{4}\right) = \max\left(\sin\frac{5\pi}{4}, \cos\frac{5\pi}{4}\right) = \max\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right) = -\frac{1}{\sqrt{2}} \)
- At \( x = 2\pi \): \( f(2\pi) = \max(\sin 2\pi, \cos 2\pi) = \max(0, 1) = 1 \)

5. **Determine the Minimum Value**:
From the evaluations, we see that the minimum value occurs at \( x = \frac{5\pi}{4} \):
\[ \text{Minimum value of } f(x) = -\frac{1}{\sqrt{2}} \]

### Conclusion:
The minimum value of the function \( f(x) = \max(\sin x, \cos x) \) is \( -\frac{1}{\sqrt{2}} \).