The range of values of m for which the line y = mx and the curve `y=(x)/(x^(2)+1)` enclose a region, is

Options

A. `(-1, 1)`
B. `(0,1)`
C. `[0,1]`
D. `(1,oo)`

Answer

To find the range of values of \( m \) for which the line \( y = mx \) and the curve \( y = \frac{x}{x^2 + 1} \) enclose a region, we will follow these steps:

### Step 1: Find the derivative of the curve
We start with the curve given by the equation:
\[ y = \frac{x}{x^2 + 1} \]
To find the slope of the tangent to the curve, we need to compute the derivative \( \frac{dy}{dx} \).

Using the quotient rule:
\[ \frac{dy}{dx} = \frac{(x^2 + 1)(1) - x(2x)}{(x^2 + 1)^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2} \]

### Step 2: Evaluate the derivative at \( x = 0 \)
Next, we evaluate the derivative at \( x = 0 \):
\[ \frac{dy}{dx}\bigg|_{x=0} = \frac{1 - 0^2}{(0^2 + 1)^2} = \frac{1}{1} = 1 \]

### Step 3: Determine the slope conditions
The slope of the tangent line at \( x = 0 \) is 1. For the line \( y = mx \) to intersect the curve and enclose a region, the slope \( m \) must be less than the slope of the tangent at that point and greater than the slope of the curve at points where it intersects the x-axis.

Since the curve \( y = \frac{x}{x^2 + 1} \) is increasing in the interval \( (-\infty, 0) \) and decreasing in \( (0, \infty) \), we need to find the maximum value of the curve.

### Step 4: Find the maximum value of the curve
To find the maximum value of \( y \), we set the derivative equal to zero:
\[ 1 - x^2 = 0 \implies x^2 = 1 \implies x = \pm 1 \]
Evaluating the curve at these points:
\[ y(1) = \frac{1}{1^2 + 1} = \frac{1}{2}, \quad y(-1) = \frac{-1}{(-1)^2 + 1} = -\frac{1}{2} \]
The maximum value occurs at \( x = 1 \) where \( y = \frac{1}{2} \).

### Step 5: Determine the range of \( m \)
For the line \( y = mx \) to enclose a region with the curve, \( m \) must satisfy:
\[ 0 < m < \frac{1}{2} \]

### Conclusion
Thus, the range of values of \( m \) for which the line \( y = mx \) and the curve \( y = \frac{x}{x^2 + 1} \) enclose a region is:
\[ \boxed{(0, \frac{1}{2})} \]