Problem 62 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The value of \(f(0)\) so that the function \(f(x)=\frac{\cos ^{-1}\left(1-\\{x\\}^{2}\right) \sin ^{-1}(1-\\{x\\})}{\\{x\\}-\\{x\\}^{3}}, x \neq 0(\\{x\\}\) denotes fractional part of \(x\) ) becomes continuous at \(x=0\) is (A) \(\frac{\pi}{\sqrt{2}}\) (B) \(\frac{\pi}{4}\) (C) \(\frac{\pi}{2}\) (D) None of these

Step-by-Step Solution

Answer

None of these.

Step 1: Identify the Problem

We need to find the value of \( f(0) \) such that the function \( f(x) = \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\} - \{x\}^3} \) is continuous at \( x = 0 \), where \( \{ x \} \) represents the fractional part of \( x \).

Step 2: Understanding the Fractional Part

The fractional part of \( x \), denoted \( \{ x \} \), is defined as \( x - \lfloor x \rfloor \), where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \). For any integer \( x \), \( \{ x \} = 0 \). For \( x \) close to zero, \( \{ x \} \approx x \).

Step 3: Simplifying the Expression

Near \( x = 0 \), \( \{x\} \approx x \) and the function simplifies to \( f(x) = \frac{\cos^{-1}(1-x^2) \sin^{-1}(1-x)}{x - x^3} \). When \( x \rightarrow 0 \), this becomes an indeterminate form \( \frac{0}{0} \). We need to find \( \lim_{x \to 0} f(x) \).

Step 4: Applying L'Hôpital's Rule

Use L'Hôpital's Rule because we have the form \( \frac{0}{0} \). Differentiate the numerator and denominator. Differentiating, the numerator becomes: \(-2x \sin \cos^{-1}(1-x^2) + \cos \sin^{-1}(1-x)\), and the denominator becomes \(1 - 3x^2\).

Step 5: Evaluate the Limit

Evaluating the limit as \( x \to 0 \), the numerator approaches \(-2x(0) + \frac{1}{\sqrt{1-0}}\), which simplifies to 1. The denominator approaches 1. Thus, \( \lim_{x \to 0} f(x) = \frac{1}{1} = 1 \).

Step 6: Ensure Continuity

Since the limit as \( x \to 0 \) is 1, set \( f(0) = 1 \) for the function to be continuous at \( x = 0 \). Check the options. None match \( f(0) = 1 \). This implies the correct answer is 'None of these'.

Key Concepts

Fractional Part Function

The fractional part of a function is an interesting aspect of continuity problems. It is represented as \( \{ x \} \), which is defined as \( x - \lfloor x \rfloor \), where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \).
For instance, if \( x \) is 3.7, the fractional part, \( \{ x \} \), is 0.7 because the greatest integer less than 3.7 is 3. For integer values of \( x \), the fractional part \( \{ x \} \) always equals zero.
This concept is crucial when evaluating the function's behavior as \( x \) approaches specific values, such as zero. It's particularly significant in scenarios like the one in the original problem, where calculating the fractional part assists in simplifying complex expressions.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool used in calculus for evaluating limits of indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule requires differentiating the numerator and the denominator separately.
When faced with an indeterminate limit, instead of directly evaluating the limit, you calculate the derivatives. If the limit exists after distinguishing derivatives, it is considered the limit of the original function.
In the provided solution, L'Hôpital's Rule was applied to transform an indeterminate limit into a solvable form. Here's a quick rundown of how the rule was applied:

• The function transformed from \( \frac{0}{0} \) into one where differentiability clarified the numerator and denominator separately.
• This simplification helped evaluate the limit at \( x = 0 \), resulting in a coherent solution.

L'Hôpital's Rule is a fundamental technique, often bridging complex problems and their solutions.

Limit Evaluation

Limit evaluation defines how functions behave as input values near a specific point. In this context, evaluating \( \lim_{x \to 0} f(x) \) is essential for ensuring function continuity at a particular point.
The original problem required calculating the limit when \( x \rightarrow 0 \), which presented an indeterminate form, necessitating tools like L'Hôpital’s Rule. The process is about:

• Breaking down the expression to manageable parts that simplify the function around the limiting behavior.
• Ensuring both the numerator and denominator approach non-zero forms, allowing further simplification.
• Finally arriving at a finite value that confirms the continuity of the function around the desired point.

Through thorough evaluation, limits give insights into the behavior and continuity standards necessary for smooth function analysis.

Trigonometric Functions

Trigonometric functions are mathematical functions of angles central to periodic phenomena. They include sine, cosine, and their respective inverse forms, \( \sin^{-1} \) and \( \cos^{-1} \). In mathematical problems like the original, trigonometric functions were used within the context of limit evaluations.
These functions are pivotal because:

• They exhibit symmetry and periodicity, which define their behavior over specific intervals.
• Inverse trigonometric functions, such as \( \sin^{-1} \) and \( \cos^{-1} \), allow calculation of angles given a ratio, which is crucial for evaluating functions where direct solutions are not possible.
• Their properties simplify complex expressions when determining continuity and limits.

Understanding trigonometric functions deepens comprehension about their application in evaluating limits and ensuring function continuity.