Points on the curve `f(x)=x/(1-x^2)` where the tangent is inclined at an angle of `pi/4` to the x-axis are (a) (0,0) (b) `(sqrt(3),-(sqrt(3))/2)` (c) `(-2,2/3)` (d) `(-sqrt(3),(sqrt(3))/2)`

Options

A. `(0,0)`
B. `(sqrt(3),-(sqrt3)/(2))`
C. `(-2,(2)/(3))`
D. `(sqrt3,-(sqrt3)/(2))`

Answer

To find the points on the curve \( f(x) = \frac{x}{1 - x^2} \) where the tangent is inclined at an angle of \( \frac{\pi}{4} \) (or 45 degrees) to the x-axis, we can follow these steps:

### Step 1: Determine the slope of the tangent
The slope of the tangent line at any point on the curve is given by the derivative \( \frac{dy}{dx} \). Since the tangent is inclined at an angle of \( \frac{\pi}{4} \), the slope is:
\[ m = \tan\left(\frac{\pi}{4}\right) = 1 \]

### Step 2: Find the derivative of the function
We need to find the derivative of the function \( f(x) \):
\[ f(x) = \frac{x}{1 - x^2} \]
Using the quotient rule, where if \( u = x \) and \( v = 1 - x^2 \), then:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Calculating \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = -2x \):
\[ \frac{dy}{dx} = \frac{(1 - x^2)(1) - (x)(-2x)}{(1 - x^2)^2} = \frac{1 - x^2 + 2x^2}{(1 - x^2)^2} = \frac{1 + x^2}{(1 - x^2)^2} \]

### Step 3: Set the derivative equal to the slope
We set the derivative equal to the slope:
\[ \frac{1 + x^2}{(1 - x^2)^2} = 1 \]

### Step 4: Solve for \( x \)
Cross-multiplying gives:
\[ 1 + x^2 = (1 - x^2)^2 \]
Expanding the right side:
\[ 1 + x^2 = 1 - 2x^2 + x^4 \]
Rearranging gives:
\[ x^4 - 3x^2 = 0 \]
Factoring out \( x^2 \):
\[ x^2(x^2 - 3) = 0 \]
This gives us:
\[ x^2 = 0 \quad \text{or} \quad x^2 = 3 \]
Thus, \( x = 0, \sqrt{3}, -\sqrt{3} \).

### Step 5: Find corresponding \( y \) values
Now we substitute these \( x \) values back into the original function to find \( y \):

1. For \( x = 0 \):
\[ y = f(0) = \frac{0}{1 - 0^2} = 0 \quad \Rightarrow \quad (0, 0) \]

2. For \( x = \sqrt{3} \):
\[ y = f(\sqrt{3}) = \frac{\sqrt{3}}{1 - 3} = \frac{\sqrt{3}}{-2} = -\frac{\sqrt{3}}{2} \quad \Rightarrow \quad \left(\sqrt{3}, -\frac{\sqrt{3}}{2}\right) \]

3. For \( x = -\sqrt{3} \):
\[ y = f(-\sqrt{3}) = \frac{-\sqrt{3}}{1 - 3} = \frac{-\sqrt{3}}{-2} = \frac{\sqrt{3}}{2} \quad \Rightarrow \quad \left(-\sqrt{3}, \frac{\sqrt{3}}{2}\right) \]

### Final Points
The points on the curve where the tangent is inclined at an angle of \( \frac{\pi}{4} \) to the x-axis are:
- \( (0, 0) \)
- \( \left(\sqrt{3}, -\frac{\sqrt{3}}{2}\right) \)
- \( \left(-\sqrt{3}, \frac{\sqrt{3}}{2}\right) \)

### Conclusion
Thus, the correct options are:
- (a) \( (0, 0) \)
- (b) \( \left(\sqrt{3}, -\frac{\sqrt{3}}{2}\right) \)
- (d) \( \left(-\sqrt{3}, \frac{\sqrt{3}}{2}\right) \)