If `(x+c)/(1+x^2)` where c is a constant , then when y is stationary , xy is equal to

Options

A. `1//2`
B. `3//4`
C. `5//8`
D. 1

Answer

To solve the problem, we need to find the value of \( xy \) when \( y \) is stationary. The function given is:

\[ y = \frac{x + c}{1 + x^2} \]

### Step 1: Understand the condition for stationary points
A function \( y \) is stationary when its derivative with respect to \( x \) is zero, i.e., \( \frac{dy}{dx} = 0 \).

### Step 2: Differentiate \( y \)
We will differentiate \( y \) using the quotient rule. The quotient rule states that if \( y = \frac{u}{v} \), then:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Here, \( u = x + c \) and \( v = 1 + x^2 \).

- \( \frac{du}{dx} = 1 \)
- \( \frac{dv}{dx} = 2x \)

Applying the quotient rule:

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

### Step 3: Simplify the derivative
Now we simplify the numerator:

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

### Step 4: Set the derivative to zero
To find the stationary points, we set the derivative equal to zero:

\[ 1 - x^2 - 2cx = 0 \]

Rearranging gives:

\[ x^2 + 2cx - 1 = 0 \]

### Step 5: Solve for \( x \)
This is a quadratic equation in \( x \). We can use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = 2c \), and \( c = -1 \):

\[ x = \frac{-2c \pm \sqrt{(2c)^2 - 4(1)(-1)}}{2(1)} = \frac{-2c \pm \sqrt{4c^2 + 4}}{2} \]

This simplifies to:

\[ x = -c \pm \sqrt{c^2 + 1} \]

### Step 6: Find \( y \) at stationary points
Now we substitute \( x \) back into the equation for \( y \):

\[ y = \frac{x + c}{1 + x^2} \]

Using \( x = -c + \sqrt{c^2 + 1} \):

\[ y = \frac{(-c + \sqrt{c^2 + 1}) + c}{1 + (-c + \sqrt{c^2 + 1})^2} \]

Calculating \( y \):

\[ y = \frac{\sqrt{c^2 + 1}}{1 + (-c + \sqrt{c^2 + 1})^2} \]

### Step 7: Calculate \( xy \)
Now we find \( xy \):

\[ xy = x \cdot y \]

Substituting the values of \( x \) and \( y \) into this expression gives:

\[ xy = \left(-c + \sqrt{c^2 + 1}\right) \cdot \frac{\sqrt{c^2 + 1}}{1 + (-c + \sqrt{c^2 + 1})^2} \]

After simplification, we find that:

\[ xy = \frac{1}{2} \]

### Final Answer
Thus, when \( y \) is stationary, \( xy \) is equal to:

\[ \boxed{\frac{1}{2}} \]