Which of the following equations have no real solutions ?

Options

A. `x^(2) - 2x+ 5 + pi^(2) = 0`
B. `
C. `x^(4) - 2x^(2)sin^(2)'(pix)/(2)+1 = 0`
D. all of these

Answer

To determine which of the given equations have no real solutions, we will analyze each equation using the discriminant method for quadratic equations and other relevant properties for higher degree polynomials.

### Step-by-Step Solution:

1. **Identify the First Equation**:
The first equation given is:
\[ x^2 - 2x + 5 + \pi^2 = 0 \]
We can rewrite this as:
\[ x^2 - 2x + (5 + \pi^2) = 0 \]

2. **Identify Coefficients**:
Here, we have:
- \( a = 1 \)
- \( b = -2 \)
- \( c = 5 + \pi^2 \)

3. **Calculate the Discriminant**:
The discriminant \( D \) is given by the formula:
\[ D = b^2 - 4ac \]
Plugging in the values:
\[ D = (-2)^2 - 4 \cdot 1 \cdot (5 + \pi^2) \]
\[ D = 4 - 4(5 + \pi^2) \]
\[ D = 4 - 20 - 4\pi^2 \]
\[ D = -16 - 4\pi^2 \]

4. **Analyze the Discriminant**:
Since \( \pi^2 \) is a positive number, \( -16 - 4\pi^2 \) is definitely less than zero. Thus, the first equation has no real solutions.

5. **Identify the Second Equation**:
The second equation is:
\[ x^4 - 2x^4 \sin^2\left(\frac{\pi x}{2}\right) + 1 = 0 \]
Rearranging gives:
\[ x^4 + 1 = 2x^4 \sin^2\left(\frac{\pi x}{2}\right) \]

6. **Analyze the Second Equation**:
We can express it as:
\[ \frac{x^4 + 1}{x^4} = 2 \sin^2\left(\frac{\pi x}{2}\right) \]
The left side simplifies to:
\[ 1 + \frac{1}{x^4} \]
The expression \( 1 + \frac{1}{x^4} \) is always greater than or equal to 2 for all \( x \neq 0 \).

7. **Conclusion for the Second Equation**:
Since \( 2 \sin^2\left(\frac{\pi x}{2}\right) \) can take values from 0 to 2, the equation can have real solutions. Therefore, this equation does have real solutions.

### Final Answer:
The first equation \( x^2 - 2x + (5 + \pi^2) = 0 \) has no real solutions.