Let the parabolas y= `x^(2)+ ax +b `and y =x(c-x) touch cach other at point (1,0). Then

Options

A. a=-3
B. b=2
C. c=2
D. b+c=3

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that the parabolas \( y = x^2 + ax + b \) and \( y = x(c - x) \) touch each other at the point \( (1, 0) \).

### Step 1: Substitute the point (1, 0) into both equations
Since the point \( (1, 0) \) lies on both parabolas, we can substitute \( x = 1 \) and \( y = 0 \) into both equations.

1. For the first parabola:
\[ 0 = 1^2 + a(1) + b \implies 0 = 1 + a + b \]
This gives us our first equation:
\[ a + b + 1 = 0 \quad \text{(Equation 1)} \]

2. For the second parabola:
\[ 0 = 1(c - 1) \implies 0 = c - 1 \implies c = 1 \]

### Step 2: Find the derivatives of both equations
Next, we need to find the derivatives of both equations to ensure that they have the same slope at the point of tangency.

1. For the first parabola:
\[ y = x^2 + ax + b \implies \frac{dy}{dx} = 2x + a \]
At \( x = 1 \):
\[ \frac{dy}{dx} = 2(1) + a = 2 + a \quad \text{(Slope 1)} \]

2. For the second parabola:
\[ y = x(c - x) = cx - x^2 \implies \frac{dy}{dx} = c - 2x \]
At \( x = 1 \):
\[ \frac{dy}{dx} = c - 2(1) = c - 2 \quad \text{(Slope 2)} \]

### Step 3: Set the slopes equal
Since the two parabolas touch each other, their slopes at the point of tangency must be equal:
\[ 2 + a = c - 2 \]
Substituting \( c = 1 \):
\[ 2 + a = 1 - 2 \implies 2 + a = -1 \]
Solving for \( a \):
\[ a = -1 - 2 = -3 \quad \text{(Equation 2)} \]

### Step 4: Substitute \( a \) back into Equation 1
Now, we substitute \( a = -3 \) back into Equation 1 to find \( b \):
\[ -3 + b + 1 = 0 \implies b - 2 = 0 \implies b = 2 \quad \text{(Equation 3)} \]

### Conclusion
The values of \( a \) and \( b \) are:
\[ a = -3, \quad b = 2 \]