The function `f(x)=(x)/(x^(2)+1)` from R to R is

Options

A. one-one as well as onto
B. onto but not one-one
C. neither one-one nor onto
D. one-one but not onto

Explanation

To determine the properties of the function \( f(x) = \frac{x}{x^2 + 1} \), we will analyze whether the function is one-to-one (1-1) and onto (onto).

### Step 1: Check if the function is one-to-one (1-1)

A function is one-to-one if it is either monotonically increasing or monotonically decreasing throughout its domain. To check this, we will differentiate the function.

**Differentiation:**
\[ f'(x) = \frac{d}{dx} \left( \frac{x}{x^2 + 1} \right) \]
Using the quotient rule, where \( u = x \) and \( v = x^2 + 1 \):
\[ f'(x) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Calculating \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = 2x \):
\[ f'(x) = \frac{(x^2 + 1)(1) - (x)(2x)}{(x^2 + 1)^2} \]
\[ = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2} \]

### Step 2: Analyze the sign of \( f'(x) \)

The sign of \( f'(x) \) depends on the numerator \( 1 - x^2 \):
- \( 1 - x^2 > 0 \) when \( |x| < 1 \) (i.e., \( -1 < x < 1 \)) → \( f'(x) > 0 \) (increasing)
- \( 1 - x^2 < 0 \) when \( |x| > 1 \) (i.e., \( x < -1 \) or \( x > 1 \)) → \( f'(x) < 0 \) (decreasing)

Thus, the function is increasing on the interval \( (-1, 1) \) and decreasing on the intervals \( (-\infty, -1) \) and \( (1, \infty) \). Since it is not consistently increasing or decreasing over all real numbers, the function is not one-to-one.

### Step 3: Check if the function is onto (onto)

A function is onto if every element in the codomain (here, \( \mathbb{R} \)) has a pre-image in the domain. To check this, we need to find the range of \( f(x) \).

**Setting up the equation:**
Let \( y = f(x) = \frac{x}{x^2 + 1} \). Rearranging gives:
\[ y(x^2 + 1) = x \implies yx^2 - x + y = 0 \]
This is a quadratic equation in \( x \). For \( x \) to have real solutions, the discriminant must be non-negative:
\[ D = (-1)^2 - 4(y)(y) = 1 - 4y^2 \geq 0 \]
This simplifies to:
\[ 1 \geq 4y^2 \implies \frac{1}{4} \geq y^2 \implies -\frac{1}{2} \leq y \leq \frac{1}{2} \]

### Conclusion

The range of \( f(x) \) is \( \left[-\frac{1}{2}, \frac{1}{2}\right] \), which does not cover all real numbers. Therefore, the function is not onto.

### Final Answer
The function \( f(x) = \frac{x}{x^2 + 1} \) is neither one-to-one nor onto.

---