Problem 119 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right] .\) If \(f(x)\) is continuous in \(\left[0, \frac{\pi}{2}\right]\), then \(f\left[\frac{\pi}{4}\right]\) is \(\quad\) [2004] (A) 1 (B) \(\frac{1}{2}\) (C) \(-\frac{1}{2}\) (D) \(-1\)

Step-by-Step Solution

Answer

(C) \(-\frac{1}{2}\)

Step 1: Identify the Problem

We need to determine the value of \( f\left(\frac{\pi}{4}\right) \) under the condition that \( f(x) \) is continuous over the interval \( \left[0, \frac{\pi}{2}\right] \). This involves determining the limit of \( f(x) \) as \( x \to \frac{\pi}{4} \).

Step 2: Write the Function and Identify Issue

The function given is \( f(x) = \frac{1 - \tan x}{4x - \pi} \). There is a problem of direct substitution at \( x = \frac{\pi}{4} \) since \( \tan \frac{\pi}{4} = 1 \) and results in a 0/0 indeterminate form.

Step 3: Apply L'Hôpital's Rule

Since the form is 0/0, we can apply L'Hôpital's Rule which allows us to find the limit by differentiating the numerator and the denominator separately. Differentiate \( 1 - \tan x \) to get \( -\sec^2 x \), and differentiate \( 4x - \pi \) to get 4.

Step 4: Evaluate the Limit

Now, evaluate the limit: \[ \lim_{x \to \frac{\pi}{4}} \frac{-\sec^2 x}{4} = \frac{-\sec^2 \frac{\pi}{4}}{4} = \frac{-2}{4} = -\frac{1}{2}. \]

Step 5: Conclude Continuity Condition

To ensure \( f(x) \) is continuous over \( \left[0, \frac{\pi}{2}\right] \), \( f\left(\frac{\pi}{4}\right) \) must equal this limit. Thus, the value of \( f\left(\frac{\pi}{4}\right) \) is \(-\frac{1}{2}\).

Key Concepts

L'Hôpital's Rule

When we encounter a mathematical limit problem that results in an indeterminate form like \(\frac{0}{0}\), L'Hôpital's Rule comes to the rescue. This rule provides a technique for evaluating limits of indeterminate forms by differentiating the numerator and the denominator separately. The main condition for applying L'Hôpital’s Rule is that both the numerator and the denominator must be differentiable at the point of interest.**Steps to Apply L'Hôpital's Rule**:

• First, confirm the limit is an indeterminate form such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
• Differentiate the numerator and the denominator independently.
• Calculate the limit of the new function obtained after differentiation.

This method can be repeated if the first application still results in an indeterminate form. In many cases, this rule simplifies the function significantly, making it easier to evaluate the limit.

Limits and Continuity

Continuity and limits are fundamental concepts in calculus that allow us to analyze functions and their behavior. A function is considered continuous at a point if the limit of the function as it approaches that point is equal to the function's value at that point. This seamlessness, or lack of "holes" or "jumps" in the graph, is key for continuity.**Understanding Continuity**:

• A function \(f(x)\) is continuous at point \(c\) if \(\lim_{x \to c} f(x) = f(c)\).
• For a function to be continuous over an interval, this condition must hold true for every point within that range.

To check if a function satisfies this condition, we often need to find and evaluate limits. If a function is continuous on an interval, a break or jump is not expected unless explicitly accounted for, such as in piecewise functions.

Indeterminate Forms

Indeterminate forms are scenarios that occur when evaluating limits, where the direct substitution of the limit point into a function results in ambiguous expressions like \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), or others such as \(0 \times \infty\) or \(\infty - \infty\). These forms do not immediately suggest the limit's behavior, which is why additional methods like L'Hôpital's Rule are needed.**Common Indeterminate Forms Include**:

• \(\frac{0}{0}\), which often arises when both the numerator and denominator approach zero.
• \(\frac{\infty}{\infty}\), occurring when both the numerator and denominator tend to infinity.
• \(0^0, \ 1^\infty,\) and \(\infty^0\), which can arise in exponential forms.

These forms highlight the complexity underlying calculus problems. Each necessitates special attention and techniques to resolve, often requiring algebraic manipulation or the application of calculus theorems.