The number of pairs (a,b) for which `a(x+1)^2+b(x^(2)-3x-2)+x+1=0 AA x in R ` is :

Options

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

Answer

To solve the problem of finding the number of pairs \((a, b)\) for which the equation

\[ a(x+1)^2 + b(x^2 - 3x - 2) + x + 1 = 0 \]

is satisfied for all \(x \in \mathbb{R}\), we will follow these steps:

### Step 1: Expand the equation

First, we expand the terms in the equation:

1. Expand \(a(x+1)^2\):
\[ a(x+1)^2 = a(x^2 + 2x + 1) = ax^2 + 2ax + a \]

2. Expand \(b(x^2 - 3x - 2)\):
\[ b(x^2 - 3x - 2) = bx^2 - 3bx - 2b \]

3. Combine all terms:
\[ ax^2 + 2ax + a + bx^2 - 3bx - 2b + x + 1 = 0 \]

### Step 2: Combine like terms

Now, we combine the coefficients of \(x^2\), \(x\), and the constant term:

1. Coefficient of \(x^2\):
\[ (a + b)x^2 \]

2. Coefficient of \(x\):
\[ (2a - 3b + 1)x \]

3. Constant term:
\[ (a - 2b + 1) \]

Thus, we rewrite the equation as:
\[ (a + b)x^2 + (2a - 3b + 1)x + (a - 2b + 1) = 0 \]

### Step 3: Set coefficients to zero

For the equation to hold for all \(x \in \mathbb{R}\), each coefficient must be equal to zero:

1. \(a + b = 0\) (Equation 1)
2. \(2a - 3b + 1 = 0\) (Equation 2)
3. \(a - 2b + 1 = 0\) (Equation 3)

### Step 4: Solve the system of equations

Now we will solve these equations:

1. From Equation 1, we can express \(b\) in terms of \(a\):
\[ b = -a \]

2. Substitute \(b = -a\) into Equation 2:
\[ 2a - 3(-a) + 1 = 0 \implies 2a + 3a + 1 = 0 \implies 5a + 1 = 0 \implies a = -\frac{1}{5} \]

3. Substitute \(a = -\frac{1}{5}\) back into Equation 1 to find \(b\):
\[ b = -(-\frac{1}{5}) = \frac{1}{5} \]

### Step 5: Verify the solution

Now we check if these values satisfy Equation 3:
\[ -\frac{1}{5} - 2\left(\frac{1}{5}\right) + 1 = -\frac{1}{5} - \frac{2}{5} + 1 = -\frac{3}{5} + 1 = \frac{2}{5} \neq 0 \]

Since Equation 3 does not hold, there are no values of \(a\) and \(b\) that satisfy all three equations simultaneously.

### Conclusion

Thus, the number of pairs \((a, b)\) for which the original equation holds for all \(x \in \mathbb{R}\) is:

\[ \text{Number of pairs } (a, b) = 0 \]