Given `A (0, 0)` and `B (x,y)`with `x epsilon (0,1)` and `y>0`. Let the slope of the line AB equals `m_1` Point C lies on the line `x= 1` such that the slope of BC equals `m_2` where `0< m_2< m_1` If the area of the triangle ABC can expressed as `(m_1- m_2)f(x)`, then largest possible value of `f(x)` is:

Options

A. 1
B. `1//2`
C. `1//4`
D. `1//8`

Explanation

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

### Step 1: Identify Points and Slopes
Given points:
- A = (0, 0)
- B = (x, y) where x ∈ (0, 1) and y > 0

The slope of line AB is given as \( m_1 \):
\[ m_1 = \frac{y - 0}{x - 0} = \frac{y}{x} \]
Thus, we can express \( y \) in terms of \( m_1 \) and \( x \):
\[ y = m_1 x \]

### Step 2: Determine Point C
Point C lies on the line \( x = 1 \), so we can denote it as \( C = (1, c) \). The slope of line BC is given as \( m_2 \):
\[ m_2 = \frac{c - y}{1 - x} \]
Substituting \( y = m_1 x \):
\[ m_2 = \frac{c - m_1 x}{1 - x} \]

### Step 3: Rearranging for c
Rearranging the equation for \( c \):
\[ c - m_1 x = m_2 (1 - x) \]
\[ c = m_2 (1 - x) + m_1 x \]

### Step 4: Calculate Area of Triangle ABC
The area \( A \) of triangle ABC can be calculated using the determinant formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
\[ A = \frac{1}{2} \left| 0(m_1 x - c) + x(c - 0) + 1(0 - m_1 x) \right| \]
This simplifies to:
\[ A = \frac{1}{2} \left| xc - m_1 x \right| = \frac{1}{2} x(c - m_1) \]

### Step 5: Substitute for c
Substituting \( c \) from step 3 into the area formula:
\[ A = \frac{1}{2} x \left( m_2(1 - x) + m_1 x - m_1 \right) \]
\[ = \frac{1}{2} x \left( m_2(1 - x) + (m_1 - m_1)x \right) \]
\[ = \frac{1}{2} x \left( m_2(1 - x) \right) \]

### Step 6: Express Area in Terms of \( m_1 \) and \( m_2 \)
We know from the problem statement that:
\[ A = (m_1 - m_2) f(x) \]
Equating the two expressions for area:
\[ \frac{1}{2} x m_2(1 - x) = (m_1 - m_2) f(x) \]

### Step 7: Solve for \( f(x) \)
Rearranging gives:
\[ f(x) = \frac{\frac{1}{2} x m_2(1 - x)}{m_1 - m_2} \]

### Step 8: Maximize \( f(x) \)
To find the maximum value of \( f(x) \), we can differentiate \( f(x) \) with respect to \( x \) and set the derivative to zero. This will give us the critical points.

### Step 9: Differentiate and Find Critical Points
Differentiating:
\[ f'(x) = \frac{1}{2} m_2 \left( (1 - 2x)(m_1 - m_2) \right) \]
Setting \( f'(x) = 0 \):
\[ 1 - 2x = 0 \implies x = \frac{1}{2} \]

### Step 10: Evaluate \( f(x) \) at Critical Point
Substituting \( x = \frac{1}{2} \) into \( f(x) \):
\[ f\left(\frac{1}{2}\right) = \frac{\frac{1}{2} \cdot \frac{1}{2} m_2(1 - \frac{1}{2})}{m_1 - m_2} \]
\[ = \frac{\frac{1}{2} \cdot \frac{1}{2} m_2 \cdot \frac{1}{2}}{m_1 - m_2} = \frac{m_2}{8(m_1 - m_2)} \]

### Conclusion
The largest possible value of \( f(x) \) is:
\[ \frac{1}{8} \]