If `f(x)=2sinx-x^(2)`, then in `x in [0, pi]`

Options

A. `f(x)` has no local maximum
B. `f(x)` has one local minimum
C. `f(x)` has 2 local maxima
D. `f(x)` has one local maximum

Answer

To solve the problem, we need to analyze the function \( f(x) = 2\sin x - x^2 \) over the interval \( [0, \pi] \) to determine the number of local maxima.

### Step-by-Step Solution:

1. **Find the First Derivative:**
We start by differentiating the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}(2\sin x - x^2) = 2\cos x - 2x \]
This simplifies to:
\[ f'(x) = 2(\cos x - x) \]

**Hint:** The first derivative helps us find the critical points where the function may have local maxima or minima.

2. **Set the First Derivative to Zero:**
To find the critical points, we set the first derivative equal to zero:
\[ 2(\cos x - x) = 0 \implies \cos x = x \]

**Hint:** The equation \( \cos x = x \) will help us find the points where the slope of the function is zero.

3. **Graphical Interpretation:**
We can analyze the functions \( y = \cos x \) and \( y = x \) on the interval \( [0, \pi] \):
- The graph of \( y = \cos x \) starts at \( (0, 1) \) and decreases to \( (pi, -1) \).
- The graph of \( y = x \) is a straight line starting from \( (0, 0) \) to \( (pi, pi) \).

**Hint:** The intersection points of these two graphs will give us the solutions to \( \cos x = x \).

4. **Determine the Number of Solutions:**
By plotting the graphs, we observe that they intersect at exactly one point in the interval \( [0, \pi] \). Therefore, there is one critical point.

**Hint:** Use the Intermediate Value Theorem to confirm that there is exactly one intersection point.

5. **Find the Second Derivative:**
Next, we find the second derivative to determine the nature of the critical point:
\[ f''(x) = \frac{d}{dx}(2\cos x - 2x) = -2\sin x - 2 \]

**Hint:** The second derivative test will help us determine if the critical point is a maximum or minimum.

6. **Evaluate the Second Derivative:**
Since \( -2\sin x - 2 \) is always negative (as \( \sin x \) ranges from 0 to 1 in the interval), we have:
\[ f''(x) < 0 \quad \text{for all } x \in [0, \pi] \]

**Hint:** A negative second derivative indicates that the function is concave down, confirming a local maximum.

7. **Conclusion:**
Since there is one critical point and the second derivative is negative at that point, we conclude that the function \( f(x) \) has exactly one local maximum in the interval \( [0, \pi] \).

**Final Answer:** The function \( f(x) = 2\sin x - x^2 \) has **one local maximum** in the interval \( [0, \pi] \).