Problem 104 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f\) be a function defined on an interval \(\mathrm{I}\). If \(f\) be discontinuous at a point \(p \in I\), then we say that (i) \(f\) has a removable discontinuity at \(p\) if \(\lim _{x \rightarrow p} f(x)\) exists but is not equal to \(f(p)\). (ii) \(f\) has a discontinuity of the first kind at \(p\) if \(\lim _{x \rightarrow p-0} f(x)\) and \(\lim _{x \rightarrow p+0} f(x)\) exist but are unequal. (iii) \(f\) has a discontinuity of the second kind at \(p\) if neither of \(\lim _{x \rightarrow p-0} f(x)\) and \(\lim _{x \rightarrow p+0} f(x)\) exists. The function \(f(x)=t^{3}\), where \(t=\left\\{\begin{array}{cl}x-1, & x \leq 0 \\ x+1, & 0

Step-by-Step Solution

Answer

(A) and (B) are correct: first kind at \(x=0\) and removable at \(x=1\).

Step 1: Identify the form of the function around the points of interest

The function is defined piecewise:- For \(x \leq 0\), the function is \(f(x) = (x-1)^3\).- For \(0 < x < 1\), the function is \(f(x) = (x+1)^3\).- At \(x=1\), the function is \(f(x) = 1^3 = 1\).- For \(1 < x\), the function is \(f(x) = (3-x)^3\).We need to examine the behavior of the function at the points \(x=0\) and \(x=1\).

Step 2: Check discontinuity at \(x=0\)

Evaluate the left-hand limit: \(\lim_{x \to 0^-} f(x) = ((-1)^3) = -1 \)Evaluate the right-hand limit:\(\lim_{x \to 0^+} f(x) = (1^3) = 1 \)Since the left-hand and right-hand limits exist but are unequal, there is a first kind of discontinuity at \(x=0\).

Step 3: Check discontinuity at \(x=1\)

Evaluate the left-hand limit:\(\lim_{x \to 1^-} f(x) = (2^3) = 8 \)Evaluate the right-hand limit:\(\lim_{x \to 1^+} f(x) = (2^3) = 8 \)Both limits from the left and right exist and are equal to 8. The value \(f(1)\) is 1, which is different from the limits (8). Therefore, there is a removable discontinuity at \(x=1\).

Step 4: Determine the correct answer

Reviewing the results, the function has: - A first kind of discontinuity at \(x=0\).- A removable discontinuity at \(x=1\).The options are:- (A) discontinuity of first kind at \(x=0\)- (B) removable discontinuity at \(x=1\) Thus, options (A) and (B) are correct.

Key Concepts

Removable Discontinuity

In mathematics, a removable discontinuity in a function occurs at a point where the limit of the function exists, but it is not equal to the function's value at that point. This means that while there's a disruption in the graph of the function, it can be "fixed" by redefining the function at the point of discontinuity to match the limit.
The classic example of a removable discontinuity is the hole in a graph. Imagine a scenario where the plot of the function has a gap at a particular point. By filling in this gap, the graph becomes continuous.
A tangible instance can be seen with the function defined for this exercise, where at point \(x=1\), the left-hand and right-hand limits are equal and exist, both being equal to 8, but the function value is 1, not 8.
You could "remove" this discontinuity by redefining \(f(1)=8\), making the function continuous at this point.

First Kind Discontinuity

The discontinuity of the first kind, also sometimes referred to as a jump discontinuity, occurs in a function when its left-hand and right-hand limits both exist at a point but are not equal.
This results in a "jump" in the graph of the function at that point. Imagine a route that suddenly has a step up or down, leading to an abrupt transition between two distinct segments.
In the given exercise, the function demonstrates a first kind of discontinuity at \(x=0\). The left-hand limit evaluates to \(-1\) while the right-hand limit equals \(1\).
Such unequal limits indicate that while the behavior of the function from either direction settles to a steady number, these numbers don't coincide, slicing the graph into two disjointed parts.

Piecewise Functions

Piecewise functions are functions that use different expressions for different intervals of their domain. They are defined by multiple sub-functions, each applying to a certain interval of the main function's domain. This gives them the flexibility to model complex behaviors by stitching simpler functions together.
In the exercise, the function \(f\) is defined piecewise for various intervals:

• For \(x \leq 0\), \(f(x) = (x-1)^3\).
• For \(0 < x < 1\), \(f(x) = (x+1)^3\).
• At \(x=1\), \(f(x) = 1^3\).
• For \(1 < x\), \(f(x) = (3-x)^3\).

Each of these intervals might be dictated by certain real-world conditions or theoretical constructs, making piecewise functions very practical in illustrating phenomena that cannot be neatly encapsulated by a single formula across the entire domain.
They often lead to different kinds of discontinuities because the transition between the sub-functions might not be smooth. As seen in this exercise, transitions at certain points (like \(x=0\) and \(x=1\)) cause discontinuities.