The number of values of the parameter `alpha in [0, 2pi]` for which the quadratic function `(sin alpha)x^(2)+(2cosalpha)x+(1)/(2)(cos alpha+sinalpha)` is the square of a linear funcion is

Options

A. 2
B. 4
C. 6
D. 8

Answer

To determine the number of values of the parameter \( \alpha \) in the interval \( [0, 2\pi] \) for which the quadratic function

\[ (\sin \alpha)x^2 + (2\cos \alpha)x + \frac{1}{2}(\cos \alpha + \sin \alpha) \]

is the square of a linear function, we need to analyze the conditions under which a quadratic can be expressed as a perfect square.

### Step 1: Identify the general form of a perfect square
A quadratic function \( ax^2 + bx + c \) can be expressed as the square of a linear function if it can be written in the form \( (px + q)^2 \). Expanding this gives:

\[ p^2x^2 + 2pqx + q^2 \]

From this, we can equate coefficients:

- \( a = p^2 \)
- \( b = 2pq \)
- \( c = q^2 \)

### Step 2: Set up the equations based on coefficients
For our quadratic function, we have:

- \( a = \sin \alpha \)
- \( b = 2 \cos \alpha \)
- \( c = \frac{1}{2}(\cos \alpha + \sin \alpha) \)

From the conditions for a perfect square, we can derive:

1. \( \sin \alpha = p^2 \)
2. \( 2\cos \alpha = 2pq \)
3. \( \frac{1}{2}(\cos \alpha + \sin \alpha) = q^2 \)

### Step 3: Solve for \( p \) and \( q \)
From the first equation, we have \( p = \sqrt{\sin \alpha} \) (valid if \( \sin \alpha \geq 0 \)).

From the second equation, substituting \( p \):

\[ 2\cos \alpha = 2\sqrt{\sin \alpha}q \implies \cos \alpha = \sqrt{\sin \alpha}q \]

From the third equation, substituting \( q \):

\[ \frac{1}{2}(\cos \alpha + \sin \alpha) = q^2 \]

### Step 4: Substitute and simplify
Substituting \( \cos \alpha = \sqrt{\sin \alpha}q \) into the third equation gives:

\[ \frac{1}{2}(\sqrt{\sin \alpha}q + \sin \alpha) = q^2 \]

Multiplying through by 2:

\[ \sqrt{\sin \alpha}q + \sin \alpha = 2q^2 \]

Rearranging gives:

\[ 2q^2 - \sqrt{\sin \alpha}q - \sin \alpha = 0 \]

### Step 5: Determine the discriminant
For \( q \) to have real solutions, the discriminant of this quadratic must be non-negative:

\[ D = (\sqrt{\sin \alpha})^2 - 4 \cdot 2 \cdot (-\sin \alpha) = \sin \alpha + 8\sin \alpha = 9\sin \alpha \]

Thus, we require:

\[ 9\sin \alpha \geq 0 \implies \sin \alpha \geq 0 \]

### Step 6: Find the range of \( \alpha \)
The condition \( \sin \alpha \geq 0 \) holds for:

\[ \alpha \in [0, \pi] \]

### Step 7: Count the values of \( \alpha \)
Within the interval \( [0, \pi] \), we need to check when \( \sin \alpha \) is non-negative. The values of \( \alpha \) that satisfy this condition are:

- \( \alpha = 0 \)
- \( \alpha = \frac{\pi}{2} \)
- \( \alpha = \pi \)

Thus, there are **3 values** of \( \alpha \) in the interval \( [0, 2\pi] \) where the quadratic is a perfect square.

### Final Answer
The number of values of the parameter \( \alpha \) in \( [0, 2\pi] \) for which the quadratic function is the square of a linear function is **3**.

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