If the curve `y=x^(2)+bx +c ` touches the line y = x at the point (1,1), then the set of values of x for which the curve has a negative gradient is

Options

A. `(-oo,1//2)`
B. `(1//2,oo)`
C. `(-oo,-1//2)`
D. `(-1//2,oo)`

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Understand the curves
We have the curve given by the equation:
\[ y = x^2 + bx + c \]
and the line:
\[ y = x \]

### Step 2: Determine the point of tangency
The curve touches the line at the point (1, 1). This means that at \( x = 1 \), both the curve and the line must have the same value of \( y \) and the same slope.

### Step 3: Set up the equations
1. Substitute \( x = 1 \) into the curve equation:
\[ y = 1^2 + b(1) + c = 1 + b + c \]
Since this must equal 1 (the y-coordinate of the point of tangency), we have:
\[ 1 + b + c = 1 \]
Simplifying gives:
\[ b + c = 0 \quad \text{(Equation 1)} \]

2. Differentiate the curve to find the slope:
\[ \frac{dy}{dx} = 2x + b \]
For the line \( y = x \), the slope is:
\[ \frac{dy}{dx} = 1 \]

### Step 4: Set the slopes equal at the point of tangency
At \( x = 1 \):
\[ 2(1) + b = 1 \]
This simplifies to:
\[ 2 + b = 1 \]
Thus:
\[ b = 1 - 2 = -1 \quad \text{(Equation 2)} \]

### Step 5: Substitute \( b \) back into Equation 1
From Equation 1:
\[ -1 + c = 0 \]
Thus:
\[ c = 1 \quad \text{(Equation 3)} \]

### Step 6: Write the final equation of the curve
Now substituting \( b \) and \( c \) back into the curve equation:
\[ y = x^2 - x + 1 \]

### Step 7: Find the gradient of the curve
Differentiate the curve again:
\[ \frac{dy}{dx} = 2x - 1 \]

### Step 8: Determine where the gradient is negative
We want to find where the gradient is less than zero:
\[ 2x - 1 < 0 \]
Solving this inequality:
\[ 2x < 1 \]
\[ x < \frac{1}{2} \]

### Step 9: Write the solution in interval notation
The set of values of \( x \) for which the curve has a negative gradient is:
\[ (-\infty, \frac{1}{2}) \]

### Final Answer
The set of values of \( x \) for which the curve has a negative gradient is:
\[ (-\infty, \frac{1}{2}) \]

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