If `sin""(pix)/(2)=x^(2)-2x+2`, then the value of x is

Options

A. 0
B. 1
C. `-1`
D. None of these

Answer

To solve the equation \( \sin(\pi x) = x^2 - 2x + 2 \), we will follow these steps:

### Step 1: Understand the equation
We have the equation:
\[ \sin(\pi x) = x^2 - 2x + 2 \]
We need to find the values of \( x \) that satisfy this equation.

### Step 2: Analyze the left-hand side (LHS)
The left-hand side, \( \sin(\pi x) \), can take values between -1 and 1 for all real \( x \). Therefore, we need to ensure that the right-hand side (RHS), \( x^2 - 2x + 2 \), also falls within this range.

### Step 3: Analyze the right-hand side (RHS)
The expression \( x^2 - 2x + 2 \) is a quadratic equation. We can rewrite it in vertex form:
\[ x^2 - 2x + 2 = (x-1)^2 + 1 \]
This shows that the minimum value of \( x^2 - 2x + 2 \) occurs at \( x = 1 \) and is equal to 1.

### Step 4: Set up the inequality
Since \( \sin(\pi x) \) can only take values between -1 and 1, we set up the inequality:
\[ 1 \leq x^2 - 2x + 2 \leq 1 \]
This implies that:
\[ x^2 - 2x + 2 = 1 \]

### Step 5: Solve the equation
Now we solve the equation:
\[ x^2 - 2x + 2 - 1 = 0 \]
which simplifies to:
\[ x^2 - 2x + 1 = 0 \]
Factoring gives:
\[ (x - 1)^2 = 0 \]
Thus, we find:
\[ x - 1 = 0 \implies x = 1 \]

### Step 6: Verify the solution
Now we need to check if \( x = 1 \) satisfies the original equation:
- LHS: \( \sin(\pi \cdot 1) = \sin(\pi) = 0 \)
- RHS: \( 1^2 - 2 \cdot 1 + 2 = 1 - 2 + 2 = 1 \)

Since LHS (0) does not equal RHS (1), we need to check other possible values.

### Step 7: Check other integer values
Let's check \( x = 0 \):
- LHS: \( \sin(\pi \cdot 0) = \sin(0) = 0 \)
- RHS: \( 0^2 - 2 \cdot 0 + 2 = 2 \)

Now check \( x = 2 \):
- LHS: \( \sin(\pi \cdot 2) = \sin(2\pi) = 0 \)
- RHS: \( 2^2 - 2 \cdot 2 + 2 = 2 \)

### Conclusion
After checking possible values, we find that the only solution that satisfies the equation is \( x = 1 \).

### Final Answer
The value of \( x \) is:
\[ \boxed{1} \]