Number of ordered pair (x, y) satisfying `x^(2)+1=y` and `y^(2)+1=x` is

Options

A. 0
B. 1
C. 2
D. 4

Answer

To find the number of ordered pairs \((x, y)\) that satisfy the equations \(x^2 + 1 = y\) and \(y^2 + 1 = x\), we can follow these steps:

### Step 1: Rewrite the equations
We have two equations:
1. \(y = x^2 + 1\)
2. \(x = y^2 + 1\)

### Step 2: Substitute \(y\) from the first equation into the second equation
Substituting \(y\) from the first equation into the second:
\[ x = (x^2 + 1)^2 + 1 \]

### Step 3: Expand the equation
Now, we will expand the equation:
\[ x = (x^2 + 1)^2 + 1 \]
Expanding \((x^2 + 1)^2\):
\[ (x^2 + 1)^2 = x^4 + 2x^2 + 1 \]
So, we have:
\[ x = x^4 + 2x^2 + 1 + 1 \]
This simplifies to:
\[ x = x^4 + 2x^2 + 2 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ x^4 + 2x^2 - x + 2 = 0 \]

### Step 5: Analyze the polynomial
We need to find the roots of the polynomial \(x^4 + 2x^2 - x + 2 = 0\). We can analyze this polynomial to find out how many real roots it has.

### Step 6: Use Descartes' Rule of Signs
We can apply Descartes' Rule of Signs to determine the number of positive and negative roots.

1. For \(f(x) = x^4 + 2x^2 - x + 2\):
- The signs of the coefficients are: \(+, +, -, +\) (2 sign changes)
- This indicates there could be 2 or 0 positive roots.

2. For \(f(-x) = x^4 + 2x^2 + x + 2\):
- The signs of the coefficients are: \(+, +, +, +\) (0 sign changes)
- This indicates there are no negative roots.

### Step 7: Check for real roots
To find the exact number of real roots, we can check the function values at some points:
- \(f(0) = 2\) (positive)
- \(f(1) = 1 + 2 - 1 + 2 = 4\) (positive)
- \(f(2) = 16 + 8 - 2 + 2 = 24\) (positive)

Since the polynomial is positive at these points and does not change signs, we conclude that there are no real roots.

### Conclusion
Since there are no real roots for the polynomial \(x^4 + 2x^2 - x + 2 = 0\), there are no ordered pairs \((x, y)\) that satisfy both equations.

Thus, the number of ordered pairs \((x, y)\) satisfying the given equations is **0**.

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