The differentiable function `y= f(x)` has a property that the chord joining any two points `A (x _(1), f (x_(1)) and B (x_(2), f (x _(2)))` always intersects y-axis at `(0,2 x _(1) x _(2)).` Given that `f (1) =-1.` then:
`int _(0)^(1//2) f (x) dx` is equal to : (a)`1/6` (b) `1/8` (c) `1/12` (d) `1/24`

Options

A. `1/6`
B. `1/8`
C. `1/12`
D. `1/24`

Answer

To solve the problem, we need to find the integral \( \int_{0}^{\frac{1}{2}} f(x) \, dx \) given that the function \( f(x) \) has a specific property related to the chords joining two points on its graph.

### Step 1: Understand the property of the function
The function \( f(x) \) has the property that the chord joining any two points \( A(x_1, f(x_1)) \) and \( B(x_2, f(x_2)) \) intersects the y-axis at the point \( (0, 2x_1x_2) \). This means that when we find the equation of the line joining points \( A \) and \( B \), it should pass through the point \( (0, 2x_1x_2) \).

### Step 2: Find the equation of the chord
The slope \( m \) of the line joining points \( A \) and \( B \) is given by:
\[ m = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \]
Using the point-slope form of the line, we can write the equation of the line as:
\[ y - f(x_1) = m(x - x_1) \]
Substituting \( m \) into the equation, we have:
\[ y - f(x_1) = \frac{f(x_2) - f(x_1)}{x_2 - x_1}(x - x_1) \]

### Step 3: Substitute \( x = 0 \) to find the y-intercept
Setting \( x = 0 \) in the equation gives:
\[ y - f(x_1) = \frac{f(x_2) - f(x_1)}{x_2 - x_1}(0 - x_1) \]
This simplifies to:
\[ y = f(x_1) - \frac{f(x_2) - f(x_1)}{x_2 - x_1} x_1 \]
According to the problem, this y-intercept should equal \( 2x_1x_2 \):
\[ f(x_1) - \frac{f(x_2) - f(x_1)}{x_2 - x_1} x_1 = 2x_1x_2 \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 2x_1x_2 = f(x_1) - \frac{f(x_2) - f(x_1)}{x_2 - x_1} x_1 \]
Multiplying through by \( (x_2 - x_1) \) and simplifying leads us to a functional equation.

### Step 5: Substitute \( x_2 = 1 \)
Let’s set \( x_2 = 1 \) and substitute \( f(1) = -1 \):
\[ 2x_1 \cdot 1 = f(x_1) - \frac{-1 - f(x_1)}{1 - x_1} x_1 \]
This simplifies to:
\[ 2x_1 = f(x_1) + \frac{1 + f(x_1)}{1 - x_1} x_1 \]

### Step 6: Solve for \( f(x) \)
After rearranging and simplifying, we find:
\[ f(x) = x - 2x^2 \]

### Step 7: Evaluate the integral
Now we can evaluate the integral:
\[ \int_{0}^{\frac{1}{2}} f(x) \, dx = \int_{0}^{\frac{1}{2}} (x - 2x^2) \, dx \]
Calculating this integral:
\[ = \left[ \frac{x^2}{2} - \frac{2x^3}{3} \right]_{0}^{\frac{1}{2}} \]
Substituting the limits:
\[ = \left( \frac{(\frac{1}{2})^2}{2} - \frac{2(\frac{1}{2})^3}{3} \right) - \left( 0 - 0 \right) \]
\[ = \left( \frac{1/4}{2} - \frac{2/8}{3} \right) = \left( \frac{1}{8} - \frac{1}{12} \right) \]

### Step 8: Find a common denominator
Finding a common denominator (24):
\[ = \frac{3}{24} - \frac{2}{24} = \frac{1}{24} \]

### Conclusion
Thus, the value of the integral \( \int_{0}^{\frac{1}{2}} f(x) \, dx \) is \( \frac{1}{24} \).

### Final Answer
The answer is \( \frac{1}{24} \) (option d).