Problem 8 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(f:[-1,3] \rightarrow \mathrm{R}\) be defined as \(f(x)= \begin{cases}|x|+[x], & -1 \leq x<1 \\ x+|x|, & 1 \leq x<2 \\ x+[x], & 2 \leq x \leq 3\end{cases}\) where \([t]\) denotes the greatest integer less than or equal to \(t\). Then, \(f\) is discontinuous at: \(\quad\) (a) only one point (b) only two points (c) only three points (d) four or more points

Step-by-Step Solution

Answer

The function is discontinuous at two points: \(x = 1\) and \(x = -1\) (domain boundary). Choice (b).

Step 1: Understand the function definition

Look at the piecewise function definition for \( f(x) \). It consists of three parts:- \( |x| + [x] \) applies when \(-1 \leq x < 1\).- \( x + |x| \) applies when \(1 \leq x < 2\).- \( x + [x] \) applies when \(2 \leq x \leq 3\).

Step 2: Identify potential points of discontinuity

Discontinuities in piecewise functions can occur at the points where the function definition changes. In this function, the change points are at \( x = 1 \) and \( x = 2 \), as well as the boundary points \( x = -1 \) and \( x = 3 \).

Step 3: Check continuity at \( x = -1 \)

At \( x = -1 \), the function value \( f(-1) = |-1| + [-1] = 1 - 1 = 0 \). This is the boundary of the domain, so check within the piecewise function. As there's no left-side domain, continuity at \( x = -1 \) doesn't need testing from the left.

Step 4: Check continuity at \( x = 1 \)

For \( x = 1 \), check the limit from the left: \( \lim_{{x \to 1^-}} f(x) = |1| + [1] = 1 + 0 = 1 \).The value at \( x = 1 \) is given by the next piece, \( f(1) = 1 + |1| = 2 \).The steps show that there is a discontinuity because the left-hand limit and the value at \( x = 1 \) differ: \( \lim_{{x \to 1^{+}}} f(x) = 1 eq 2 \).

Step 5: Check continuity at \( x = 2 \)

For \( x = 2 \), check the limit from the left: \( \lim_{{x \to 2^-}} f(x) = 2 + |2| = 4 \).The value at \( x = 2 \) is from the next part, \( f(2) = 2 + [2] = 2 + 2 = 4 \).As the left-hand limit and the value match, \( \lim_{{x \to 2^-}} f(x) = 4 \), there's no discontinuity at \( x = 2 \).

Step 6: Check continuity at \( x = 3 \)

At \( x = 3 \), it's on the boundary of the function's domain again so only consider from the left: \( \lim_{{x \to 3^-}} f(x) = 3 + [3] = 6 \).The value at \( x = 3 \) confirms: \( f(3) = 3 + 3 = 6 \). There's no discontinuity here.

Key Concepts

Piecewise Function

Piecewise functions are like mathematical multitools. They allow us to define a function using different expressions for different parts of its domain. This is especially useful when a function's behavior changes at specific points. In the given problem, the function \( f(x) \) is a piecewise function, meaning it has multiple 'pieces' based on the value of \( x \):

• \( |x| + [x] \) when \(-1 \leq x < 1\)
• \( x + |x| \) when \(1 \leq x < 2\)
• \( x + [x] \) when \(2 \leq x \leq 3\)

Each of these expressions gives a specific rule for calculating the output of the function in its particular range. Piecewise functions like this are quite common in mathematics, and they help us model situations where a rule might change based on conditions. When working with piecewise functions, it is essential always to consider each piece and what it represents for different intervals of \( x \).

Discontinuity

A discontinuity in a function occurs when there's a break, jump, or gap at a certain point in its graph. For piecewise functions, discontinuities often occur where the pieces of the function meet or where the domain changes. In our exercise, the function \( f(x) \) potentially has discontinuities at the points where its formula changes, which are at \( x = 1 \) and \( x = 2 \). Additionally, discontinuities might appear at the domain boundaries, \( x = -1 \) and \( x = 3 \), making these important points to test. In mathematical terms, a discontinuity happens if the left-hand limit \( \lim_{{x \to c^-}} f(x) \), the right-hand limit \( \lim_{{x \to c^+}} f(x) \), and the function value \( f(c) \) are not all equal.

Greatest Integer Function

The greatest integer function, denoted as \( [x] \), returns the largest integer less than or equal to \( x \). It's a step function that 'jumps' at each integer value. For example, \([3.7] = 3\), and \([-2.3] = -3\). In the exercise, the greatest integer function appears in the expression \([x]\) for the pieces of the piecewise function. It plays a significant role, especially in determining discontinuity points because of its inherent step-like nature. When working with this function, remember:

• It is not continuous at integer points.
• The value remains constant between two consecutive integers, then jumps at the integer.

This behavior can contribute to discontinuities if part of a wider function change as well.

Continuity Testing

Testing a function for continuity involves checking whether it has an uninterrupted output in its domain. For piecewise functions, the critical test points are often at the boundaries of the pieces. Here's how to test for continuity:1. **Consider Necessary Limits:** - Calculate the left-hand limit and right-hand limit for each boundary point between pieces. - Check if these limits and the actual function value at key points are equal.2. **Check Domain Boundaries:** - Confirm if the domains include appropriate inequalities. - Ensure there aren't unexpected gaps in the domain.In the provided exercise, continuity checks occurred at points \( x = -1, 1, 2, \) and \( 3 \). Applying these steps:

• At \( x = 1 \), the function showed discontinuity because \( \lim_{{x \to 1^-}} eq f(1) \).
• No discontinuity at \( x = 2 \) and \( x = 3 \) was observed since limits matched the function values.

This systematic approach helps to precisely understand where and why discontinuities occur within complex functions.