If `0 le x le 2pi`, then `2^(cosec^(2) x) sqrt(1/2 y^(2) -y+1) le sqrt(2)`

Options

A. is satisfied by exactly one value of y
B. is satisfied by exactly two value of x
C. is satisfied by x for which `cos x=0`
D. is satisfied by x for which `sin x=0`

Explanation

To solve the inequality \( 2^{\csc^2 x} \sqrt{\frac{1}{2} y^2 - y + 1} \leq \sqrt{2} \) for \( 0 \leq x \leq 2\pi \), we will break it down step by step.

### Step 1: Analyze the inequality
We start with the inequality:
\[ 2^{\csc^2 x} \sqrt{\frac{1}{2} y^2 - y + 1} \leq \sqrt{2} \]

### Step 2: Simplify the expression
We can rewrite \( \sqrt{2} \) as \( 2^{1/2} \). Thus, we can express the inequality as:
\[ 2^{\csc^2 x} \sqrt{\frac{1}{2} y^2 - y + 1} \leq 2^{1/2} \]

### Step 3: Divide both sides by \( \sqrt{2} \)
To simplify, we divide both sides by \( \sqrt{2} \):
\[ 2^{\csc^2 x - \frac{1}{2}} \sqrt{\frac{1}{2} y^2 - y + 1} \leq 1 \]

### Step 4: Analyze \( 2^{\csc^2 x - \frac{1}{2}} \)
Since \( 2^{\csc^2 x - \frac{1}{2}} \) is always positive, we can isolate \( \sqrt{\frac{1}{2} y^2 - y + 1} \):
\[ \sqrt{\frac{1}{2} y^2 - y + 1} \leq 2^{\frac{1}{2} - \csc^2 x} \]

### Step 5: Square both sides
Squaring both sides gives:
\[ \frac{1}{2} y^2 - y + 1 \leq 2^{1 - 2\csc^2 x} \]

### Step 6: Analyze the right side
The term \( 2^{1 - 2\csc^2 x} \) can vary depending on \( \csc^2 x \). Since \( \csc^2 x \geq 1 \) for \( x \in [0, 2\pi] \), we know:
\[ 2^{1 - 2\csc^2 x} \leq 2^{1 - 2} = 2^{-1} = \frac{1}{2} \]

### Step 7: Set up the quadratic inequality
Now we can set up the quadratic inequality:
\[ \frac{1}{2} y^2 - y + 1 \leq \frac{1}{2} \]
This simplifies to:
\[ \frac{1}{2} y^2 - y + \frac{1}{2} \leq 0 \]
Multiplying through by 2 gives:
\[ y^2 - 2y + 1 \leq 0 \]
This factors to:
\[ (y - 1)^2 \leq 0 \]

### Step 8: Solve the quadratic inequality
The only solution to \( (y - 1)^2 = 0 \) is:
\[ y = 1 \]

### Step 9: Determine values of \( x \)
Now, we need to find \( x \) such that \( \csc^2 x \) is defined. Since \( \csc^2 x = \frac{1}{\sin^2 x} \), we have:
\[ \sin^2 x = 1 \implies \sin x = \pm 1 \]
This occurs at:
\[ x = \frac{\pi}{2}, \frac{3\pi}{2} \]

### Conclusion
Thus, the solution to the inequality \( 2^{\csc^2 x} \sqrt{\frac{1}{2} y^2 - y + 1} \leq \sqrt{2} \) under the given conditions is:
\[ y = 1 \quad \text{and} \quad x = \frac{\pi}{2}, \frac{3\pi}{2} \]