The number of real solutions of `|2x-x^2-3|=1` is

Options

A. `0`
B. `2`
C. `3`
D. `4`

Explanation

To solve the equation \( |2x - x^2 - 3| = 1 \), we will break it down into two separate cases based on the definition of absolute value.

### Step 1: Set up the two cases
The equation \( |A| = B \) implies two cases:
1. \( A = B \)
2. \( A = -B \)

In our case, we have:
1. \( 2x - x^2 - 3 = 1 \)
2. \( 2x - x^2 - 3 = -1 \)

### Step 2: Solve the first case
**Case 1:** \( 2x - x^2 - 3 = 1 \)

Rearranging gives:
\[ -x^2 + 2x - 3 - 1 = 0 \implies -x^2 + 2x - 4 = 0 \]
Multiplying through by -1:
\[ x^2 - 2x + 4 = 0 \]

Now, we calculate the discriminant \( D \):
\[ D = b^2 - 4ac = (-2)^2 - 4(1)(4) = 4 - 16 = -12 \]

Since \( D < 0 \), there are no real solutions for this case.

### Step 3: Solve the second case
**Case 2:** \( 2x - x^2 - 3 = -1 \)

Rearranging gives:
\[ -x^2 + 2x - 3 + 1 = 0 \implies -x^2 + 2x - 2 = 0 \]
Multiplying through by -1:
\[ x^2 - 2x + 2 = 0 \]

Now, we calculate the discriminant \( D \):
\[ D = b^2 - 4ac = (-2)^2 - 4(1)(2) = 4 - 8 = -4 \]

Since \( D < 0 \), there are no real solutions for this case as well.

### Conclusion
Since both cases yield no real solutions, the total number of real solutions to the equation \( |2x - x^2 - 3| = 1 \) is **0**.

### Final Answer
The number of real solutions is **0**.

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