For the function `f(x)=sinx+2cosx,AA"x"in[0,2pi]` we obtain

Options

A. a local point of maxima at `x = alpha`, where `alpha` is in `1^(st)` quadrant
B. a local point of maxima at `x = alpha`, where `alpha` is in `3^(rd)` quadrant
C. a local point of minima at `x = alpha`, where `alpha` is in `1^(st)` quadrant
D. a local point of minima at `x = alpha`, where `alpha` is in `2^(nd)` quadrant

Answer

To solve the problem, we need to analyze the function \( f(x) = \sin x + 2 \cos x \) over the interval \( [0, 2\pi] \) and determine the nature of its critical points.

### Step 1: Differentiate the function
First, we find the first derivative of the function \( f(x) \):

\[ f'(x) = \frac{d}{dx}(\sin x + 2 \cos x) = \cos x - 2 \sin x \]

**Hint:** Use the basic differentiation rules for sine and cosine functions.

### Step 2: Set the first derivative to zero
Next, we find the critical points by setting the first derivative equal to zero:

\[ \cos x - 2 \sin x = 0 \]

Rearranging gives:

\[ \cos x = 2 \sin x \]

**Hint:** This equation can be solved using trigonometric identities or by expressing one function in terms of the other.

### Step 3: Solve for \( x \)
Dividing both sides by \( \cos x \) (where \( \cos x \neq 0 \)), we get:

\[ 1 = 2 \tan x \implies \tan x = \frac{1}{2} \]

Now, we find the values of \( x \) in the interval \( [0, 2\pi] \):

\[ x = \tan^{-1}\left(\frac{1}{2}\right) \quad \text{and} \quad x = \tan^{-1}\left(\frac{1}{2}\right) + \pi \]

**Hint:** Remember that the tangent function is periodic with a period of \( \pi \).

### Step 4: Find the second derivative
To determine the nature of the critical points, we compute the second derivative:

\[ f''(x) = \frac{d}{dx}(\cos x - 2 \sin x) = -\sin x - 2 \cos x \]

**Hint:** Differentiate the first derivative again to find the second derivative.

### Step 5: Evaluate the second derivative at critical points
Now we evaluate \( f''(x) \) at the critical points found earlier.

1. For \( x = \tan^{-1}\left(\frac{1}{2}\right) \):
- Determine the signs of \( \sin x \) and \( \cos x \) in the first quadrant.
- Since both \( \sin x \) and \( \cos x \) are positive, \( f''(x) = -\sin x - 2\cos x < 0 \). This indicates a local maximum.

2. For \( x = \tan^{-1}\left(\frac{1}{2}\right) + \pi \):
- Determine the signs of \( \sin x \) and \( \cos x \) in the third quadrant.
- Here, both \( \sin x \) and \( \cos x \) are negative, so \( f''(x) = -\sin x - 2\cos x > 0 \). This indicates a local minimum.

**Hint:** Use the signs of the sine and cosine functions in different quadrants to evaluate the second derivative.

### Step 6: Conclusion
From our analysis:
- There is a local maximum at \( x = \tan^{-1}\left(\frac{1}{2}\right) \) (first quadrant).
- There is a local minimum at \( x = \tan^{-1}\left(\frac{1}{2}\right) + \pi \) (third quadrant).

Thus, the correct statement is that there is a local point of maxima at \( x = \alpha \) where \( \alpha \) belongs to the first quadrant.

**Final Answer:** The only correct option is option A: "A local point of maxima at \( x = \alpha \) where \( \alpha \) belongs to the first quadrant."