Problem 50 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

If the range of the function \(f(x)=\cot ^{-1}\left(\frac{x^{2}}{x^{2}+1}\right)\) is \((a, b)\), find the value of \(\left(\frac{b}{a}+2\right)\).

Step-by-Step Solution

Answer

The value of \(\left(\frac{b}{a}+2\right)\) is infinity (∞).

Step 1: Identify the Inner Function

Before proceeding with the calculations, let's denote \(z = \frac{x^{2}}{x^{2}+1}\). So, the function \(f(x)\) now can be recast as \(f(x)=\cot ^{-1}(z)\).

Step 2: Express \(\cot^{-1}(z)\) in Terms of \(\tan^{-1}(z)\)

Since \(\cot ^{-1}(z)\) can also be written as \(\frac{\pi}{2}-\tan ^{-1}(z)\), \(f(x)\) becomes \(f(x)=\frac{\pi}{2}-\tan ^{-1}(z)\). The range of \(\tan ^{-1}(z)\) function is \((-π/2, π/2)\). Therefore, the range of \(f(x)\) considering \(a\) as the smallest value and \(b\) as the largest, is \((a, b) = (\frac{\pi}{2} - \frac{\pi}{2}, \frac{\pi}{2} - (-\frac{\pi}{2})) = (0, \pi)\).

Step 3: Find the Value of \((\frac{b}{a}+2)\)

Substituting the values of \(a\) and \(b\), we get \(\frac{b}{a}+2 = \frac{\pi}{0} + 2\). However, division by zero is not defined in mathematics, so we can't proceed with this. However, looking at the range, we see that \(a\) can be any very small positive number, as the function never attains the zero value. This means, we can assume \(a\) to be minutely greater than zero. In that case, \(\frac{\pi}{a}\) would be a very large number, practically serving as infinity (∞). Hence, \(\frac{b}{a}+2 = ∞ + 2 = ∞\).

Key Concepts

Cotangent Function

The cotangent function, often denoted as \( \cot(x) \), is one of the six fundamental trigonometric functions. It is defined as the reciprocal of the tangent function. In simpler terms, \( \cot(x) = \frac{1}{\tan(x)} \). The cotangent function is related to right triangles where it represents the ratio of the adjacent side to the opposite side of a given angle. In the context of inverse trigonometric functions, like in the equation \( \cot^{-1}(z) \), we are dealing with the principal value of angles whose cotangent is \( z \). This is very useful for finding angles in trigonometric contexts.

When working with the inverse cotangent function, it's important to understand that its range, by convention, is \((0, \pi)\). This means that the angles returned by \( \cot^{-1} \) will always be between these two values, providing a clear limit to the possible outputs.

Arctangent Function

The arctangent function, represented by \( \tan^{-1}(x) \) or \( \text{arctan}(x) \), is another crucial inverse trigonometric function. This function calculates the angle whose tangent is \( x \). Essentially, if you know the tangent of an angle, the arctangent function helps you retrieve the actual angle.

The output or range of the arctangent function is restricted to \((-\frac{\pi}{2}, \frac{\pi}{2})\). This restriction ensures that each possible output corresponds to a single, unique angle in this interval, thus making it a one-to-one function.

• The primary use of the arctangent function is in applications requiring angle determination from its tangent value.
• It's critical in both mathematics and engineering fields, used in scenarios involving angles and their relationships.

Range of Functions

Understanding the range of functions is vital to working with trigonometric functions and their inverses. The range of a function is the set of all possible values it can output, given its input. Knowing the range helps in predicting possible outputs and making informed calculations.

For instance, in inverse trigonometric functions like \( \cot^{-1}(z) \), identifying the range helps determine the limits of outputs, crucial for solving equations and understanding behavior. When \( f(x) = \frac{\pi}{2} - \tan^{-1}(z) \), you adjust the range of the arctangent function \((-\frac{\pi}{2}, \frac{\pi}{2})\) to find the new range \((0, \pi)\) by applying transformations via subtraction.

• Assessing the range provides a boundary to the results, guiding expectations of the function's outputs.
• This becomes even more critical when transitioning between functions and their inverses, as discrepancies in range affect function composition.