`f(x)=1+[cosx]x,"in "0ltxle(pi)/(2)`

Options

A. has a minimum value 0
B. has a maximum value 2
C. is continuos in `[0,(pi)/(2)]`
D. is not differentiable at `x=(pi)/(2)`

Answer

To solve the problem step by step, we will analyze the function \( f(x) = 1 + \cos(x) \cdot x \) for \( 0 < x \leq \frac{\pi}{2} \).

### Step 1: Evaluate \( f(0) \)
Since the function is defined for \( 0 < x \), we will evaluate the limit as \( x \) approaches 0.

\[ f(0) = 1 + \cos(0) \cdot 0 = 1 + 1 \cdot 0 = 1 \]

**Hint:** Remember that \(\cos(0) = 1\).

### Step 2: Evaluate \( f\left(\frac{\pi}{2}\right) \)
Now, we will evaluate the function at the upper limit of the interval.

\[ f\left(\frac{\pi}{2}\right) = 1 + \cos\left(\frac{\pi}{2}\right) \cdot \frac{\pi}{2} = 1 + 0 \cdot \frac{\pi}{2} = 1 \]

**Hint:** Recall that \(\cos\left(\frac{\pi}{2}\right) = 0\).

### Step 3: Analyze the function for monotonicity
To find the maximum and minimum values, we need to check the behavior of \( f(x) \) within the interval \( (0, \frac{\pi}{2}) \).

1. **Find the derivative \( f'(x) \)**:
\[ f'(x) = -\sin(x) \cdot x + \cos(x) \]
This can be simplified as:
\[ f'(x) = \cos(x) - x \sin(x) \]

2. **Set the derivative to zero to find critical points**:
\[ \cos(x) - x \sin(x) = 0 \]
Rearranging gives:
\[ \cos(x) = x \sin(x) \]

**Hint:** Use trigonometric identities to analyze the behavior of the function.

### Step 4: Check the endpoints and critical points
From our evaluations:
- \( f(0) = 1 \)
- \( f\left(\frac{\pi}{2}\right) = 1 \)

Since \( f(x) \) is continuous and differentiable in the interval \( (0, \frac{\pi}{2}) \), we can conclude that the function does not exceed 1 in this interval.

### Step 5: Conclusion about the maximum and minimum
- The minimum value of \( f(x) \) is \( 1 \) at both endpoints.
- The maximum value of \( f(x) \) is also \( 1 \).

Thus, the function is constant and equal to 1 throughout the interval \( (0, \frac{\pi}{2}) \).

**Final Answer:** The function \( f(x) \) is continuous in the interval \( [0, \frac{\pi}{2}] \) and has a constant value of 1.