Problem 21 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(f:[1,3] \rightarrow R\) be a function satisfying \(\frac{x}{[x]} \leq f(x) \leq \sqrt{6-x}\), for all \(x \neq 2\) and \(f(2)=1\), where \(R\) is the set of all real numbers and \([x]\) denotes the largest integer less than or equal to \(x\). Statement 1: \(f(x)\) exists. [Online May 19, 2012] \(x \rightarrow 2\) Statement 2: \(f\) is continuous at \(x=2\). (a) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1 . (b) Statement 1 is false, Statement 2 is true. (c) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1 . (d) Statement 1 is true, Statement 2 is false.

Step-by-Step Solution

Answer

Statement 1 is true and Statement 2 is false. (d)

Step 1: Check if f(x) exists as x approaches 2

To determine if the limit of \( f(x) \) exists as \( x \rightarrow 2 \), we need to evaluate both the left-hand limit and right-hand limit of \( f(x) \) at \( x=2 \), which gives \( \lim_{{x \to 2^-}} f(x) \) and \( \lim_{{x \to 2^+}} f(x) \).

Step 2: Evaluate the left-hand limit

For \( x \) approaching 2 from the left, \( [x] = 1 \). Thus, \( \frac{x}{[x]} = x \). Since \( f(x) \leq \sqrt{6-x} \), as \( x \rightarrow 2^- \), \( \sqrt{6-x} \rightarrow \sqrt{4} = 2 \). Therefore, \[ \lim_{{x \to 2^-}} f(x) \leq 2. \] Because \( \frac{x}{1} = x \), as \( x \rightarrow 2^- \), \( \lim_{{x \to 2^-}} f(x) \geq 2 \). Thus, \( \lim_{{x \to 2^-}} f(x) = 2 \).

Step 3: Evaluate the right-hand limit

For \( x \) approaching 2 from the right, \( [x] = 2 \). Hence, \( \frac{x}{[x]} = \frac{x}{2} \). As \( x \rightarrow 2^+ \), \( \frac{x}{2} \rightarrow 1 \). If \( f(x) \leq \sqrt{6-x} \), as \( x \rightarrow 2^+ \), \( \sqrt{6-x} \rightarrow 2 \), therefore, \[ 1 \leq \lim_{{x \to 2^+}} f(x) \leq 2. \] So, \( \lim_{{x \to 2^+}} f(x) = 2 \).

Step 4: Conclude the existence of the limit

Since both the left-hand limit and right-hand limit as \( x \rightarrow 2 \) are equal and equal to 2, \( \lim_{{x \to 2}} f(x) = 2 \). Thus, Statement 1 is true.

Step 5: Check continuity at x=2

For \( f(x) \) to be continuous at \( x = 2 \), \( f(2) = \lim_{{x \to 2}} f(x) \). Since \( f(2) = 1 \) and \( \lim_{{x \to 2}} f(x) = 2 \), these do not match, indicating \( f(x) \) is not continuous at \( x=2 \). Thus, Statement 2 is false.

Step 6: Analyze the given statements

Statement 1 is confirmed true because the limit exists. Statement 2 is false because \( f \) is not continuous at \( x=2 \). Therefore, the correct choice is that Statement 1 is true, Statement 2 is false (option d).

Key Concepts

Limits of Functions

The concept of the limit of a function is key in understanding many aspects of calculus and real analysis. When we talk about the limit of a function, we are looking at the value that a function approaches as the input approaches a particular point. In the exercise above, they are focused on evaluating the limit of the function \( f(x) \) as \( x \) approaches 2.

To determine if this limit exists, we need to check the left-hand limit (as \( x \) approaches 2 from the left) and the right-hand limit (as \( x \) approaches 2 from the right). For the limit to exist, these two limits must be equal.

**Key Points About Limits:**

• To establish that a limit exists at a point, it’s crucial to calculate both the left-hand limit \( \lim_{{x \to 2^-}} f(x) \) and the right-hand limit \( \lim_{{x \to 2^+}} f(x) \).
• If these limits are the same, the common value is the limit of the function at that point.
• In this problem, both the left and the right-hand limits are calculated to be 2, hence \( \lim_{{x \to 2}} f(x) = 2 \).

Piecewise Functions

Piecewise functions can behave differently on different parts of their domain. Each section of the function can be defined by a particular expression based on the interval of the domain. In the given exercise, \( f(x) \) is constrained by a set of inequalities, depending on the segment of \( x \) considered.

**Understanding with Constraints:**

• The exercise specifies that \( \frac{x}{[x]} \leq f(x) \leq \sqrt{6-x} \). This indicates that the behavior and output values of \( f \) are restricted by these functions over different segments.
• This kind of problem often requires analyzing each piece separately, calculating the relevant limits or values.
• For \( x eq 2 \), the constraints \( \frac{x}{[x]} \) and \( \sqrt{6-x} \) help determine possible range behaviors of the piecewise function.
• When \( x=2 \), \( f(2)=1 \) is set explicitly, showing another typical piecewise definition aspect, where values can be assigned regardless of the general function rule.

Continuity

Continuity of functions is an essential concept that helps understand how functions behave when plotted on a graph. A function is continuous if, roughly speaking, you can draw it without lifting your pencil from the paper. Mathematically, a function \( f(x) \) is continuous at a point \( x = c \) if:

**Criteria for Continuity:**

• \( f(c) \) is defined.
• The limit of \( f(x) \) as \( x \to c \) exists.
• \( \lim_{{x \to c}} f(x) = f(c) \).

In the given exercise, the lack of continuity at \( x=2 \) is due to the fact that \( \lim_{{x \to 2}} f(x) = 2 \), while \( f(2) = 1 \). These values do not match, thus breaking the continuity condition.

Looking at this within the context of piecewise functions helps us understand why such discontinuities might occur. Often, piecewise functions may have defined values that create breaks in continuity, particularly at points where different pieces meet.