The number of values of the pair (a, b) for which `a(x+1)^2 + b(-x^2 – 3x - 2) + x + 1 = 0` is an identity in x, is

Options

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

Explanation

To find the number of values of the pair (a, b) for which the expression

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

is an identity in \( x \), we need to ensure that this equation holds true for all values of \( x \). This means that the coefficients of \( x^2 \), \( x \), and the constant term must all equal zero.

### Step 1: Expand the expression

First, we expand the expression:

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

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

Now, substituting these into the original equation gives:

\[ ax^2 + 2ax + a - bx^2 - 3bx - 2b + x + 1 = 0 \]

### Step 2: Combine like terms

Combining the like terms, we have:

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

This simplifies to:

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

### Step 3: Set coefficients to zero

For the equation to be an identity, each coefficient must be zero:

1. Coefficient of \( x^2 \):
\[ a - b = 0 \quad \Rightarrow \quad a = b \]

2. Coefficient of \( x \):
\[ 2a - 3b + 1 = 0 \]

3. Constant term:
\[ a - 2b + 1 = 0 \]

### Step 4: Substitute \( a = b \) into the equations

Substituting \( a = b \) into the second equation:

\[ 2a - 3a + 1 = 0 \quad \Rightarrow \quad -a + 1 = 0 \quad \Rightarrow \quad a = 1 \]

Since \( a = b \), we also have:

\[ b = 1 \]

### Step 5: Check the constant term

Now substituting \( a = 1 \) into the constant term equation:

\[ 1 - 2(1) + 1 = 0 \quad \Rightarrow \quad 1 - 2 + 1 = 0 \quad \Rightarrow \quad 0 = 0 \]

This is satisfied.

### Conclusion

Thus, the only pair \( (a, b) \) that satisfies the identity is \( (1, 1) \). Therefore, there is only **one** value for the pair \( (a, b) \).

### Final Answer

The number of values of the pair \( (a, b) \) is **1**.

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