Show that the function `f:RrarrR` defined by `f(x)=(x)/(x^(2)+1),AAx inR` is neither one-one nor onto. Also, if `g:RrarrR` is defined as `g(x)=2x-1`, find fog(x).

Answer

To show that the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = \frac{x}{x^2 + 1} \) is neither one-one nor onto, we will follow these steps:

### Step 1: Check if the function is one-one

A function is one-one (injective) if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).

Assume \( f(x_1) = f(x_2) \):
\[ \frac{x_1}{x_1^2 + 1} = \frac{x_2}{x_2^2 + 1} \]

Cross-multiplying gives:
\[ x_1 (x_2^2 + 1) = x_2 (x_1^2 + 1) \]

Expanding both sides:
\[ x_1 x_2^2 + x_1 = x_2 x_1^2 + x_2 \]

Rearranging this gives:
\[ x_1 x_2^2 - x_2 x_1^2 + x_1 - x_2 = 0 \]

Factoring out \( (x_1 - x_2) \):
\[ (x_1 - x_2)(x_2 + x_1) = 0 \]

This implies either \( x_1 = x_2 \) or \( x_1 + x_2 = 0 \). The second condition shows that there are distinct values \( x_1 \) and \( x_2 \) such that \( f(x_1) = f(x_2) \) (for example, \( x_1 = 1 \) and \( x_2 = -1 \)). Therefore, \( f \) is not one-one.

### Step 2: Check if the function is onto

A function is onto (surjective) if for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} \) such that \( f(x) = y \).

Set \( y = f(x) \):
\[ y = \frac{x}{x^2 + 1} \]

Rearranging gives:
\[ y (x^2 + 1) = x \]
\[ yx^2 - x + y = 0 \]

This is a quadratic equation in \( x \). For \( x \) to have real solutions, the discriminant must be non-negative:
\[ D = (-1)^2 - 4y^2 = 1 - 4y^2 \]

Setting the discriminant \( D \geq 0 \):
\[ 1 - 4y^2 \geq 0 \]
\[ 4y^2 \leq 1 \]
\[ y^2 \leq \frac{1}{4} \]
\[ -\frac{1}{2} \leq y \leq \frac{1}{2} \]

This means the range of \( f(x) \) is \( \left[-\frac{1}{2}, \frac{1}{2}\right] \), which is not equal to the codomain \( \mathbb{R} \). Therefore, \( f \) is not onto.

### Conclusion

Since \( f \) is neither one-one nor onto, we conclude that:
\[ f: \mathbb{R} \to \mathbb{R} \text{ is neither one-one nor onto.} \]

### Step 3: Find \( f \circ g(x) \)

Given \( g(x) = 2x - 1 \), we need to find \( f(g(x)) \):
\[ f(g(x)) = f(2x - 1) = \frac{2x - 1}{(2x - 1)^2 + 1} \]

Calculating \( (2x - 1)^2 + 1 \):
\[ (2x - 1)^2 + 1 = 4x^2 - 4x + 1 + 1 = 4x^2 - 4x + 2 \]

Thus,
\[ f(g(x)) = \frac{2x - 1}{4x^2 - 4x + 2} \]

### Final Result
\[ f \circ g(x) = \frac{2x - 1}{4x^2 - 4x + 2} \]