The straight line x+y= k touches the parabola `y=x-x^(2)` then k =

Options

A. a) 0
B. b)`-1`
C. c) 1
D. d) 2

Answer

To solve the problem, we need to find the value of \( k \) such that the line \( x + y = k \) touches the parabola \( y = x - x^2 \).

### Step-by-Step Solution:

1. **Rewrite the line equation**:
The equation of the line can be rewritten as:
\[ y = k - x \]
(This is our Equation 1).

2. **Set the equations equal**:
Since the line touches the parabola, we can substitute \( y \) from Equation 1 into the parabola's equation:
\[ k - x = x - x^2 \]

3. **Rearrange the equation**:
Rearranging gives us:
\[ x^2 - 2x + k = 0 \]
(This is our Equation 2).

4. **Identify coefficients**:
The quadratic equation is in the form \( ax^2 + bx + c = 0 \) where:
- \( a = 1 \)
- \( b = -2 \)
- \( c = k \)

5. **Use the condition for tangency**:
For the line to touch the parabola, the discriminant of the quadratic equation must be zero. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
Substituting the values:
\[ D = (-2)^2 - 4(1)(k) = 4 - 4k \]

6. **Set the discriminant to zero**:
For the line to touch the parabola, we set the discriminant equal to zero:
\[ 4 - 4k = 0 \]

7. **Solve for \( k \)**:
Rearranging gives:
\[ 4k = 4 \implies k = 1 \]

### Final Answer:
Thus, the value of \( k \) is:
\[ \boxed{1} \]