Let `f(x)=min{1,cos x,1-sinx}, -pi le x le pi`, Then, f(x) is

Options

A. f(x) is differentiable at 0
B. f(x) is differentiable at `(pi)/(2)`
C. f(x) has local maxima at=0
D. none of the above

Answer

To solve the problem, we need to analyze the function \( f(x) = \min\{1, \cos x, 1 - \sin x\} \) for the interval \( -\pi \leq x \leq \pi \). We will determine the behavior of the function and check the given options.

### Step 1: Identify the functions involved
The function \( f(x) \) is defined as the minimum of three functions:
1. \( g_1(x) = 1 \)
2. \( g_2(x) = \cos x \)
3. \( g_3(x) = 1 - \sin x \)

### Step 2: Analyze the functions
- The function \( g_1(x) = 1 \) is a constant function.
- The function \( g_2(x) = \cos x \) oscillates between -1 and 1.
- The function \( g_3(x) = 1 - \sin x \) oscillates between 0 and 2.

### Step 3: Find the intersection points
To find where these functions intersect, we need to solve the following equations:
1. \( \cos x = 1 \)
2. \( \cos x = 1 - \sin x \)

#### For \( \cos x = 1 \):
This occurs at \( x = 0 \).

#### For \( \cos x = 1 - \sin x \):
Rearranging gives:
\[ \cos x + \sin x = 1 \]
Squaring both sides:
\[ \cos^2 x + 2\cos x \sin x + \sin^2 x = 1 \]
Using \( \cos^2 x + \sin^2 x = 1 \):
\[ 1 + 2\cos x \sin x = 1 \implies 2\cos x \sin x = 0 \]
This gives us \( \sin x = 0 \) or \( \cos x = 0 \).

- \( \sin x = 0 \) at \( x = 0, \pm \pi \)
- \( \cos x = 0 \) at \( x = \pm \frac{\pi}{2} \)

### Step 4: Evaluate \( f(x) \) at critical points
Now we evaluate \( f(x) \) at the critical points:
- At \( x = 0 \):
\[ f(0) = \min\{1, \cos(0), 1 - \sin(0)\} = \min\{1, 1, 1\} = 1 \]
- At \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = \min\{1, \cos\left(\frac{\pi}{2}\right), 1 - \sin\left(\frac{\pi}{2}\right)\} = \min\{1, 0, 0\} = 0 \]
- At \( x = -\frac{\pi}{2} \):
\[ f\left(-\frac{\pi}{2}\right) = \min\{1, \cos\left(-\frac{\pi}{2}\right), 1 - \sin\left(-\frac{\pi}{2}\right)\} = \min\{1, 0, 2\} = 0 \]
- At \( x = \pi \):
\[ f(\pi) = \min\{1, \cos(\pi), 1 - \sin(\pi)\} = \min\{1, -1, 1\} = -1 \]

### Step 5: Determine differentiability and local extrema
- **At \( x = 0 \)**: \( f(x) \) is continuous and differentiable since it is constant around this point.
- **At \( x = \frac{\pi}{2} \)**: \( f(x) \) is continuous but not differentiable because the minimum function switches from \( \cos x \) to \( 1 - \sin x \).
- **Local maxima/minima**: \( f(x) \) does not have a local maximum at \( x = 0 \) since \( f(x) \) decreases at \( x = \frac{\pi}{2} \).

### Conclusion
- \( f(x) \) is differentiable at \( x = 0 \) (Option 1 is correct).
- \( f(x) \) is not differentiable at \( x = \frac{\pi}{2} \) (Option 2 is incorrect).
- \( f(x) \) does not have a local maximum at \( x = 0 \) (Option 3 is incorrect).
- Therefore, the answer is **Option 1**.