If the parabola `y=x^(2)+bx+c`, touches the straight line x=y at the point (1,1) then the value of b+c is

Options

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

Answer

To solve the problem, we need to find the values of \( b \) and \( c \) such that the parabola \( y = x^2 + bx + c \) touches the line \( x = y \) at the point \( (1, 1) \).

### Step-by-step Solution:

1. **Substituting the Point into the Parabola Equation:**
Since the parabola touches the line at the point \( (1, 1) \), this point must satisfy the equation of the parabola. Therefore, we substitute \( x = 1 \) into the parabola's equation:
\[ y = 1^2 + b(1) + c \]
Simplifying this gives:
\[ 1 = 1 + b + c \]
From this, we can rearrange to find:
\[ b + c = 0 \quad \text{(Equation 1)} \]

2. **Finding the Derivative of the Parabola:**
Next, we need to ensure that the parabola touches the line \( x = y \) at the point \( (1, 1) \). This means that the slopes of the parabola and the line must be equal at this point. The slope of the line \( x = y \) is \( 1 \).

To find the slope of the parabola, we differentiate \( y = x^2 + bx + c \):
\[ \frac{dy}{dx} = 2x + b \]
Now, we substitute \( x = 1 \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{x=1} = 2(1) + b = 2 + b \]

3. **Setting the Slopes Equal:**
Since the slope of the line is \( 1 \), we set the derivative equal to \( 1 \):
\[ 2 + b = 1 \]
Solving for \( b \):
\[ b = 1 - 2 = -1 \quad \text{(Equation 2)} \]

4. **Finding \( c \):**
Now that we have \( b = -1 \), we can substitute this value back into Equation 1 to find \( c \):
\[ -1 + c = 0 \]
Thus, we find:
\[ c = 1 \]

5. **Calculating \( b + c \):**
Finally, we calculate \( b + c \):
\[ b + c = -1 + 1 = 0 \]

### Conclusion:
The value of \( b + c \) is \( 0 \).