91Ó°ÊÓ

A function is considered one-one (or injective) if it maps every distinct input to a distinct output. This means there are no two different inputs in the domain that produce the same output in the range.
For our function \( f(x) = \tan^{-1} \left( \frac{2x}{1-x^2} \right) \), its injectivity implies that each unique input from \((-1, 1)\) should correspond to a unique output in our range.
This is crucial when proving a function is one-one:

• For a function with its domain restricted to \( (-1, 1) \), the term \( \frac{2x}{1-x^2} \) behaves uniquely for different values of \( x \) in the given interval.
• The essence of a one-one function for our given function means that as \( x \) increases from \(-1\) to \( 1\), \( f(x) \) properly increases without repeating any values, making it strictly increasing.

This specific behavior ensures that if \( x_1 eq x_2 \), then \( f(x_1) eq f(x_2) \).
In simple terms, for this tangent inverse function to be a one-one, it means every input leads to a new, unique point in the range of \((-\pi/2, \pi/2)\).

A function is considered onto (or surjective) if every element in the codomain (the potential output set) has at least one pre-image in the domain.
In the context of our function \( f(x) = \tan^{-1} \left( \frac{2x}{1-x^2} \right) \), confirming it is onto involves ensuring that for every possible value in the range \( (-\pi/2, \pi/2) \), there is a corresponding \( x \) value in the interval \((-1, 1)\) that maps to it.
Understanding onto functions involves ensuring:

• The entire range \((-\pi/2, \pi/2)\) is covered by working through values approached by the main functional form \( \frac{2x}{1-x^2} \).
• The inverse tangent expression covers values smoothly and completely within its conventional range without exceeding or missing any values.

This "covering" ensures every real number in the form of \((-\pi/2, \pi/2)\) as an output is possible, thereby showing the function \( f(x) \) is surjective onto this range. Essentially, no "target" value is left unmapped in the chosen set.

The range of a function is all the possible output values it can produce, given its domain. For \( f(x) = \tan^{-1} \left( \frac{2x}{1-x^2} \right) \), determining the range is pivotal when contemplating options like \((-\pi/2, \pi/2)\), \((0, \pi/2)\), and others provided in the exercise.
The core process for identifying the function range includes:

• Identifying the behavior of \( \frac{2x}{1-x^2} \) over \((-1, 1)\) supports realizing how it spans all real values negatively and positively.
• The \( \tan^{-1} \, \) or inverse tangent of this expression takes any real input and naturally limits itself to the interval \((-\pi/2, \pi/2)\).

This is rooted in the general property of the arctangent function—it inherently bounds outputs within \(-\pi/2\) and \(\pi/2\).
Thus, for our particular function \( f(x) \), the range assessment confirms it naturally and wholly covers \((-\pi/2, \pi/2)\), satisfying the condition for option (c) as the legitimate range occupancy.