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:

When analyzing rational functions, verify injectivity (one-one) by testing if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), and surjectivity (onto) by checking the range of the function.

Answer 1

Step 1: Check if \( f(x) \) is one-one. 
A function is one-one if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). Consider: \[ f(x) = \frac{2x}{1 + x^2}. \] Assume \( f(x_1) = f(x_2) \): \[ \frac{2x_1}{1 + x_1^2} = \frac{2x_2}{1 + x_2^2}. \] Cross-multiplying gives: \[ 2x_1 (1 + x_2^2) = 2x_2 (1 + x_1^2). \] Simplify: \[ x_1 + x_1x_2^2 = x_2 + x_2x_1^2. \] Rearranging terms: \[ x_1 - x_2 = x_2x_1^2 - x_1x_2^2. \] Factorizing: \[ (x_1 - x_2)(1 + x_1x_2) = 0. \] This implies either \( x_1 = x_2 \) or \( 1 + x_1x_2 = 0 \). The second case \( 1 + x_1x_2 = 0 \) implies \( x_1x_2 = -1 \). Therefore, \( f(x) \) is not one-one. 
Step 2: Check if \( f(x) \) is onto. 
A function is onto if every real number \( y \) has a corresponding \( x \) such that: \[ y = \frac{2x}{1 + x^2}. \] Rearranging for \( x \), we get: \[ y (1 + x^2) = 2x \quad \Rightarrow \quad y + yx^2 = 2x. \] This simplifies to 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 limited to \( [-1, 1] \), not all real numbers \( \mathbb{R} \). 
Step 3: Modify 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, restrict the codomain to \( A = [-1, 1] \).