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

Question

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

Step-by-Step Solution

Answer

I: No discontinuity, II: Infinite, III: Oscillating, IV: Removable.

Step 1: Determine discontinuity of function I at x=0

The function is given by \(f(x) = |2 \sin 2x| + 2\). At \(x=0\), calculate \(f(0)\): \(|2 \sin 0| + 2 = 2\). Since the sine function is continuous everywhere, \(f(x)\) is continuous at \(x = 0\). Hence, there is no discontinuity.

Step 2: Analyze function II at x and its points of discontinuity

The function is defined as a piecewise function:\[f(x) = \begin{cases} \tan \frac{\pi x}{2}, & x < 1 \ x - 1, & 1 \leq x < 2 \end{cases}\]- For \(x < 1\), as \(x\) approaches 1 from the left, \(\tan\left(\frac{\pi x}{2}\right)\) approaches infinity. Hence, there's an infinite discontinuity at \(x=1\).- For \(1 \leq x < 2\), there is no actual expression for \(x=1\), so the function is undefined here, supporting the infinite discontinuity.

Step 3: Determine type of discontinuity in function III at x=0

The function is given by:\[f(x) = \begin{cases} \sin \left(\frac{1}{x}\right), & x eq 0 \ 0, & x = 0 \end{cases}\]At \(x=0\), the function value is \(0\). However, as \(x\) approaches 0 from either side, \(\sin\left(\frac{1}{x}\right)\) oscillates between \(-1\) and \(1\). This creates an oscillating discontinuity at \(x=0\).

Step 4: Analyze discontinuity type of function IV at x=-2

The function is \(f(x) = \frac{|x+2|}{\tan^{-1}(x+2)}\). Evaluating at \(x=-2\), we substitute into the function:\[f(-2) = \frac{|0|}{\tan^{-1}(0)} = \frac{0}{0},\]an indeterminate form. Simplifying with a limit approach, since both the numerator and denominator approach 0 as \(x\to -2\), this suggests a potential removable discontinuity. This function's limit at \(x = -2\) must be simplified to determine a defined value. In this form, because the limit can be resolved, this indicates a removable discontinuity.

Key Concepts

Piecewise Functions

Piecewise functions are a type of function that have different expressions based on the input value. They are especially useful when a function has different behaviors in different parts of its domain.
For example, consider function II from the exercise, which is a piecewise function:

• For values of \(x < 1\), the function follows the expression \(\tan \frac{\pi x}{2}\).
• For values where \(1 \leq x < 2\), it follows the expression \(x - 1\).

Because these functions exhibit different rules in different intervals, analyzing their behavior at the boundaries of these intervals can be particularly interesting. Discontinuities often occur at these boundaries where the rule changes, as shown with the infinite discontinuity at \(x=1\). When dealing with piecewise functions, understanding each section's definition separately is crucial for exploring continuity or discontinuity.

Types of Discontinuity

Discontinuity is an important concept in understanding how functions behave. There are several types of discontinuity, and they describe how and why a function might "break" or behave unexpectedly.
Each type tells a different story:

• Infinite Discontinuity: Occurs when the function approaches infinity at a certain point. For example, in problem II at \(x=1\), the function \(\tan \frac{\pi x}{2}\) becomes undefined. As x approaches 1, the tangent function grows infinitely large, leading to an infinite discontinuity.
• Oscillating Discontinuity: This occurs when the function doesn't settle near any number as it approaches a point. It continuously oscillates between values, like function III at \(x=0\), where \(\sin \left(\frac{1}{x}\right)\) rapidly flips between values as \(x\) approaches 0.
• Removable Discontinuity: When a function is undefined at a point but can be "fixed" by assigning a particular value, such as in problem IV at \(x=-2\).
• Jump Discontinuity: Characterized by abrupt changes in value as the independent variable goes over a point, although this kind wasn't explicitly outlined in the exercise solution.

Removable Discontinuity

Removable discontinuities occur in a function where there is a hole at a certain point. However, this hole can be "filled" by redefining the function at that particular point.
In the case presented in problem IV with the function \(f(x) = \frac{|x+2|}{\tan^{-1}(x+2)}\) at \(x=-2\), an indeterminate form \(\frac{0}{0}\) occurs. Nevertheless, it is possible to evaluate this with limits:

• Approach \(-2\) from both sides using limits techniques.
• If these limits reveal a consistent value, the "hole" can be filled by assigning this value, making the discontinuity removable.

Thus, removable discontinuities are special cases where "fixing" the function by defining it exactly at the problematic point can ensure a smoother function overall. They often appear in rational functions.

Infinite Discontinuity

Infinite discontinuity can happen when a function diverges towards infinity as it approaches a certain point. This results in gaps that cannot be fixed by merely assigning a single value at that point.
In the exercise provided, function II illustrates infinite discontinuity at \(x=1\), because:

• The function \(\tan \frac{\pi x}{2}\) rapidly increases or decreases without bounds as \(x\) approaches 1.
• The mathematical expression doesn't allow for a single quantifiable value at the point of discontinuity, which means the function essentially "blows up" at this point.

Infinite discontinuities are prevalent in functions that involve asymptotes, where inputs drive the function value to extreme and often unmanageable numbers. Recognizing these are particularly important in calculus, where these behaviors affect limits and integral evaluations.