The curve `y=(lambda+1)x^2+2` intersects the curve `y=lambdax+3` in exactly one point, if `lambda` equals `{-2,2}` b. `{1}` c. `{-2}` d. `{2}`

Options

A. `{-2, 2}`
B. `{1}`
C. `{-2}`
D. `{2}`

Explanation

To solve the problem, we need to find the value of \( \lambda \) such that the curves \( y = (\lambda + 1)x^2 + 2 \) and \( y = \lambda x + 3 \) intersect at exactly one point. This means that the equations must be equal at one point, which leads us to set them equal to each other and analyze the resulting quadratic equation.

### Step-by-Step Solution:

1. **Set the equations equal to each other**:
\[ (\lambda + 1)x^2 + 2 = \lambda x + 3 \]

2. **Rearrange the equation**:
Move all terms to one side to form a standard quadratic equation:
\[ (\lambda + 1)x^2 - \lambda x + 2 - 3 = 0 \]
Simplifying this gives:
\[ (\lambda + 1)x^2 - \lambda x - 1 = 0 \]

3. **Identify coefficients**:
The quadratic equation is of the form \( ax^2 + bx + c = 0 \), where:
- \( a = \lambda + 1 \)
- \( b = -\lambda \)
- \( c = -1 \)

4. **Condition for exactly one intersection point**:
For the quadratic equation to have exactly one solution, the discriminant must be zero:
\[ D = b^2 - 4ac = 0 \]
Substituting the coefficients:
\[ (-\lambda)^2 - 4(\lambda + 1)(-1) = 0 \]

5. **Simplify the discriminant**:
\[ \lambda^2 + 4(\lambda + 1) = 0 \]
Expanding this gives:
\[ \lambda^2 + 4\lambda + 4 = 0 \]

6. **Factor the quadratic**:
The equation can be factored as:
\[ (\lambda + 2)^2 = 0 \]

7. **Solve for \( \lambda \)**:
Setting the factor equal to zero gives:
\[ \lambda + 2 = 0 \implies \lambda = -2 \]

### Conclusion:
The value of \( \lambda \) for which the curves intersect at exactly one point is \( \lambda = -2 \).

### Answer:
The correct option is **C. \{-2\}**.