Problem 108 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Function Character of discontinuity I. \(f(x)=|2 \sin 2 x|+2\) at \(x=0\) (A) Oscillating discontinuity II. \(f(x)=\left\\{\begin{array}{cc}\tan \frac{\pi x}{2}, & x<1 \\ x-1, & 1 \leq x<2\end{array}\right.\) at \(x=1\) (B) Infinite discontinuity III. \(f(x)=\left\\{\begin{array}{cc}\sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{array}\right.\) at \(x=0\) (C) Removable discontinuity IV. \(f(x)=\frac{|x+2|}{\tan ^{-1}(x+2)}\) at \(x=-2\) (D) Jump discontinuity

Step-by-Step Solution

Answer

I: None; II: B (Infinite); III: A (Oscillating); IV: C (Removable)

Step 1: Analyze Function I

The function \( f(x) = |2 \sin 2x| + 2 \) at \( x = 0 \) is continuous since \( \sin 2x \) approaches zero at \( x = 0 \), leading to \( f(0) = 2 \). Hence, there is no discontinuity at \( x = 0 \). This rules out an oscillating discontinuity.

Step 2: Analyze Function II

For \( f(x) = \tan \frac{\pi x}{2} \) when \( x < 1 \), the limit as \( x \to 1^- \) is infinite since the tangent function approaches infinity as its argument approaches \( \frac{\pi}{2} \). For the piece \( x-1 \) when \( 1 \leq x < 2 \), \( f(x) = 0 \) as \( x \to 1^+ \). Hence, there is an infinite discontinuity at \( x = 1 \).

Step 3: Analyze Function III

For \( f(x) = \sin \frac{1}{x} \) where \( x eq 0 \), the function oscillates between -1 and 1 as \( x \to 0 \). At \( x = 0 \), the function \( f(0) = 0 \). Since the limit does not exist due to oscillations, the discontinuity is oscillating at \( x = 0 \).

Step 4: Analyze Function IV

For \( f(x) = \frac{|x+2|}{\tan^{-1}(x+2)} \) at \( x = -2 \), direct substitution gives an indeterminate form \( \frac{0}{0} \). Taking the limit as \( x \to -2 \) using L'Hôpital's rule or algebraic manipulation shows that the limit exists and equals 1. Thus, it is a removable discontinuity at \( x = -2 \).

Key Concepts

Oscillating Discontinuity

An oscillating discontinuity occurs when a function behaves irregularly near a specific point. This irregular behavior involves rapid changes, causing the function not to settle at any particular value as it approaches the point of discontinuity. Let’s illustrate this with \( f(x) = \sin \frac{1}{x} \) at \( x = 0 \). Here, the output of \( \sin \frac{1}{x} \) changes erratically between -1 and 1 as \( x \) nears zero. This rapid oscillation means that the limit does not exist because the function does not approach any fixed value.
Moreover, the lack of a defined limit results in an oscillating discontinuity at \( x = 0 \). For students tackling this concept, it's helpful to visualize how the graph zigzags infinitely, never approaching a single value exactly.

Infinite Discontinuity

Infinite discontinuity is identified at a point where a function’s value grows uncontrollably without bound. A classic example is the behavior of the tangent function when it nears the vertical asymptotes. In the given problem, consider the function \( f(x) = \tan \frac{\pi x}{2} \) for \( x < 1 \). As \( x \) approaches 1 from the left, the value of \( \tan \frac{\pi x}{2} \) tends towards infinity. This is because tangent functions increase indefinitely near their vertical asymptotes.
At \( x = 1 \), the results exhibit an infinite jump, hence an infinite discontinuity. Such a discontinuity causes an undefined behavior at the point where the function leaps to infinity in a finite interval.

Removable Discontinuity

Removable discontinuities present themselves when a function has a gap at some point that can be "fixed" by appropriately defining or modifying a single point. Imagine examining the function \( f(x) = \frac{|x+2|}{\tan^{-1}(x+2)} \) at \( x = -2 \). Initially, substituting \( x = -2 \) leads to the indeterminate form \( \frac{0}{0} \). However, by simplifying or using calculus techniques like L'Hôpital's Rule, one finds that the limit indeed exists and equals 1 as \( x \to -2 \).
This discovery indicates that placing the value 1 at \( x = -2 \) makes the function continuous at that point. Thus, it's a removable discontinuity, solvable by redefining the function at \( x = -2 \). Recognizing this type of discontinuity helps in mathematical analysis and in understanding how functions might only *appear* broken, yet hold predictable behavior.