Question

Show that a function \( f : \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = \frac{2x{1 + x^2} \) is neither one-one nor onto. Further, find set \( A \) so that the given function \( f : \mathbb{R} \to A \) becomes an onto function.}

Hint:

For rational functions, analyze injectivity and surjectivity using algebraic methods and domain-range analysis.

Answer 1

Step 1: Test if \( f(x) \) is one-one.
For a function to be one-one, \( f(x_1) = f(x_2) \) should imply \( x_1 = x_2 \). Consider: \[ f(x) = \frac{2x}{1 + x^2}. \] Let \( f(x_1) = f(x_2) \): \[ \frac{2x_1}{1 + x_1^2} = \frac{2x_2}{1 + x_2^2}. \] Cross-multiply: \[ 2x_1 (1 + x_2^2) = 2x_2 (1 + x_1^2). \] Simplify: \[ x_1 + x_1x_2^2 = x_2 + x_2x_1^2. \] Rearranging terms: \[ x_1 - x_2 = x_2x_1^2 - x_1x_2^2. \] Factorize: \[ (x_1 - x_2)(1 + x_1x_2) = 0. \] Thus, either \( x_1 = x_2 \) or \( 1 + x_1x_2 = 0 \). The second case implies \( x_1x_2 = -1 \), so \( f(x) \) is not one-one. Step 2: Test if \( f(x) \) is onto.
For \( f(x) \) to be onto, every real number \( y \) should have a corresponding \( x \) such that: \[ y = \frac{2x}{1 + x^2}. \] Rearrange for \( x \): \[ y (1 + x^2) = 2x \quad \Rightarrow \quad y + yx^2 = 2x. \] Rearrange into a quadratic equation: \[ yx^2 - 2x + y = 0. \] The discriminant of this quadratic is: \[ \Delta = (-2)^2 - 4(y)(y) = 4 - 4y^2 = 4(1 - y^2). \] For \( x \) to exist, \( \Delta \geq 0 \), which implies: \[ 1 - y^2 \geq 0 \quad \Rightarrow \quad -1 \leq y \leq 1. \] Thus, \( f(x) \) is not onto because its range is \( [-1, 1] \), not \( \mathbb{R} \). Step 3: Find set \( A \) to make \( f(x) \) onto.
To make \( f(x) \) onto, let \( A = [-1, 1] \). Then, for every \( y \in A \), there exists an \( x \in \mathbb{R} \) such that: \[ y = \frac{2x}{1 + x^2}. \] Conclusion:
The function \( f(x) = \frac{2x}{1 + x^2} \) is: \[ \boxed{\text{Neither one-one nor onto.}} \] To make \( f(x) \) onto, let \( A = [-1, 1] \).