The range [ परिसर ] of the function `f(x)=(x)/(1+x^(2))` is

Options

A. `[0,1//2]`
B. `[-1//2,1//2]`
C. `[-1//2,0]`
D. `[-1//2,0) cup (0,1//2]`

Explanation

To find the range of the function \( f(x) = \frac{x}{1 + x^2} \), we will follow these steps:

### Step 1: Set up the equation
Let \( y = f(x) = \frac{x}{1 + x^2} \). Rearranging gives us:
\[ y(1 + x^2) = x \]
This can be rewritten as:
\[ yx^2 - x + y = 0 \]

### Step 2: Identify the quadratic equation
The equation \( yx^2 - x + y = 0 \) is a quadratic equation in \( x \). For \( x \) to have real solutions, the discriminant must be non-negative.

### Step 3: Calculate the discriminant
The discriminant \( D \) of the quadratic equation \( ax^2 + bx + c = 0 \) is given by \( D = b^2 - 4ac \). Here, \( a = y \), \( b = -1 \), and \( c = y \).
Thus, the discriminant is:
\[ D = (-1)^2 - 4(y)(y) = 1 - 4y^2 \]

### Step 4: Set the discriminant to be non-negative
For the quadratic to have real solutions, we need:
\[ 1 - 4y^2 \geq 0 \]
This simplifies to:
\[ 4y^2 \leq 1 \]
or
\[ y^2 \leq \frac{1}{4} \]

### Step 5: Solve for \( y \)
Taking the square root of both sides, we find:
\[ -\frac{1}{2} \leq y \leq \frac{1}{2} \]
Thus, the range of \( f(x) \) is:
\[ \text{Range of } f(x) = \left[-\frac{1}{2}, \frac{1}{2}\right] \]

### Final Answer
The range of the function \( f(x) = \frac{x}{1 + x^2} \) is \( \left[-\frac{1}{2}, \frac{1}{2}\right] \).

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