Problem 11 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

Let \(f:(-1,1) \rightarrow B\), be a function defined by \(f(x)=\tan ^{-1} \frac{2 x}{1-x^{2}}\) Then \(f\) is both one-one and onto when \(B\) is the interval (a) \([0, \pi / 2)\) (b) \((0, \pi / 2)\) (c) \((-\pi / 2, \pi / 2)\) (d) \([-\pi / 2, \pi / 2]\)

Step-by-Step Solution

Answer

Option (c) \((-\pi/2, \pi/2)\) is correct.

Step 1: Understand the function

The function given is \( f(x) = \tan^{-1} \left( \frac{2x}{1-x^2} \right) \). This is an inverse tangent function, which typically has a range of \((-\frac{\pi}{2}, \frac{\pi}{2})\). Thus, we initially suspect option (c) or (d) might be correct as these include intervals touching near these end points.

Step 2: Determine the identity

Recognize the input form \( \frac{2x}{1-x^2} \) as related to the trigonometric identity for tangent, specifically related to \( \tan(2\theta) = \frac{2 \tan(\theta)}{1 - \tan^2(\theta)} \). Thus, we might expect \( \frac{2x}{1-x^2} \) to simulate the tangent function doubling angle scenarios.

Step 3: Analyze domain and range

The function \( f(x) \) is defined for \( x \in (-1, 1) \). Within this interval, \( \frac{2x}{1-x^2} \) can achieve any real number, leading \( \tan^{-1} \left( \frac{2x}{1-x^2} \right) \) to span \( (-\pi/2, \pi/2) \).

Step 4: Verify one-one and onto condition

For the function to be one-one, different inputs \( x_1 \) and \( x_2 \) within \((-1, 1)\) must lead to different outputs, which holds as the function \( \tan^{-1} \left( \frac{2x}{1-x^2} \right) \) is strictly increasing in this interval. The function being onto means every value in the output interval \((-\pi/2, \pi/2)\) is achievable, which is met here.

Step 5: Conclude the correct option

Given the analysis, the function is one-one and onto if the range is \((-\pi/2, \pi/2)\), matching option (c).

Key Concepts

One-One Function

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 \((-\pi/2, \pi/2)\).

Onto Function

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.

Function Range

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.