The points on the curve `f(x)=(x)/(1-x^(2))`, where the tangent to it has slope equal to unity, are `(x_(1), y_(1)), (x_(2), y_(2)) and (x_(3), y_(3))`. Then, `x_(1)+x_(2)+x_(3)` is equal to

Options

A. 0
B. `(sqrt3)/(2)`
C. `(sqrt3)/(4)`
D. `sqrt3`

Answer

To solve the problem, we need to find the points on the curve \( f(x) = \frac{x}{1 - x^2} \) where the slope of the tangent is equal to unity (1). We will do this by following these steps:

### Step 1: Find the derivative of the function
The slope of the tangent line to the curve at any point is given by the derivative of the function. We will differentiate \( f(x) \).

\[ f(x) = \frac{x}{1 - x^2} \]

Using the quotient rule, the derivative \( f'(x) \) is given by:

\[ f'(x) = \frac{(1)(1 - x^2) - (x)(-2x)}{(1 - x^2)^2} \]

### Step 2: Simplify the derivative
Now we simplify the expression we obtained for \( f'(x) \):

\[ f'(x) = \frac{1 - x^2 + 2x^2}{(1 - x^2)^2} = \frac{1 + x^2}{(1 - x^2)^2} \]

### Step 3: Set the derivative equal to 1
Since we are looking for points where the slope is equal to 1, we set the derivative equal to 1:

\[ \frac{1 + x^2}{(1 - x^2)^2} = 1 \]

### Step 4: Cross-multiply and simplify
Cross-multiplying gives us:

\[ 1 + x^2 = (1 - x^2)^2 \]

Expanding the right-hand side:

\[ 1 + x^2 = 1 - 2x^2 + x^4 \]

### Step 5: Rearranging the equation
Rearranging the equation leads to:

\[ x^4 - 3x^2 = 0 \]

### Step 6: Factor the equation
Factoring out \( x^2 \):

\[ x^2(x^2 - 3) = 0 \]

### Step 7: Solve for \( x \)
Setting each factor equal to zero gives us:

1. \( x^2 = 0 \) → \( x = 0 \)
2. \( x^2 - 3 = 0 \) → \( x = \sqrt{3} \) and \( x = -\sqrt{3} \)

Thus, the values of \( x \) are \( x_1 = 0 \), \( x_2 = \sqrt{3} \), and \( x_3 = -\sqrt{3} \).

### Step 8: Calculate \( x_1 + x_2 + x_3 \)
Now we can find the sum:

\[ x_1 + x_2 + x_3 = 0 + \sqrt{3} - \sqrt{3} = 0 \]

### Final Answer
Thus, \( x_1 + x_2 + x_3 = 0 \).

---