Problem 101 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f\) be a real-valued function defined on an interval \(I\). If \(f\) be derivable at a point \(x_{0} \in I\), then it is continuous at \(x_{0}\). The converse of the above statement does not hold. That is, a function may be continuous at a point but may fail to be derivable at that point. Thus, derivability is a more restrictive property than continuity. In fact, there are functions which are continuous everywhere but differentiable nowhere. If \(R f^{\prime}\left(x_{0}\right)\) and \(L f^{\prime}\left(x_{0}\right)\) are finite (they may or may not be equal), then \(f(x)\) is continuous at \(x=x_{0}\). Let \(f(x)=\left\\{\begin{array}{cc}\int_{0}^{x}(5+|1-t|) d t, & x>2 \\ \qquad 5 x+1, & x \leq 2\end{array}\right.\), then at \(x=2\) (A) \(f(x)\) is continuous (B) \(f(x)\) is not continuous (C) \(f(x)\) is differentiable (D) \(f(x)\) is not differentiable

Step-by-Step Solution

Answer

(B) and (D) - Function is not continuous and not differentiable at \(x=2\).

Step 1: Calculate the Value of the Function at x=2

We need to find the value of the function \( f(x) \) at \( x = 2 \). By definition, where \( x \leq 2 \), \( f(x) = 5x + 1 \). Hence, \( f(2) = 5(2) + 1 = 11 \).

Step 2: Check the Left-Hand Limit as x Approaches 2

Since \( x = 2 \) falls under the case \( x \leq 2 \), the left-hand limit of \( f(x) \) as \( x \) approaches 2 is the same as the function value: \( \lim_{{x \to 2^-}} f(x) = 11 \).

Step 3: Check the Right-Hand Limit as x Approaches 2

Evaluate the integral \( \int_{0}^{2}(5 + |1 - t|) \, dt \) for the right-hand limit: Evaluate separately: For \( 0 \leq t < 1 \), \( |1-t| = 1-t \); For \( 1 \leq t \leq 2 \), \( |1-t| = t-1 \).\( \int_{0}^{1}(5 + 1 - t)\, dt = \int_{0}^{1}(6-t) \, dt = [6t - \frac{1}{2}t^2]_{0}^{1} = 6 - \frac{1}{2} = 5.5 \)\( \int_{1}^{2}(5 + t - 1) \, dt = \int_{1}^{2}(4+t) \, dt = [4t + \frac{1}{2}t^2]_{1}^{2} = 8 + 2 - 4.5 - 0.5 = 5 \)Summing both parts: \( 5.5 + 5 = 10.5 \).So, \( \lim_{{x \to 2^+}} f(x) = 10.5 \).

Step 4: Determine Continuity at x=2

A function \( f(x) \) is continuous at \( x = 2 \) if \( \lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^+}} f(x) = f(2) \). From our calculations: - \( \lim_{{x \to 2^-}} f(x) = 11 \)- \( \lim_{{x \to 2^+}} f(x) = 10.5 \)- \( f(2) = 11 \)Since \( \lim_{{x \to 2^-}} f(x) eq \lim_{{x \to 2^+}} f(x) \), \( f(x) \) is not continuous at \( x = 2 \). Thus, option (B) is correct.

Step 5: Determine Differentiability at x=2

A function is differentiable at a point only if it is continuous at that point. Since \( f(x) \) is not continuous at \( x = 2 \), \( f(x) \) cannot be differentiable at \( x = 2 \). Thus, option (D) is correct.

Key Concepts

Continuity

Let's begin by understanding the concept of continuity, a fundamental idea in calculus. A function is continuous at a point if you can draw the graph of the function at that point without lifting your pencil. Mathematically, a function \( f(x) \) is continuous at a point \( x_0 \) if the following three conditions are met:

• \( f(x_0) \) is defined.
• The left-hand limit \( \lim_{{x \to x_0^-}} f(x) \) exists.
• The right-hand limit \( \lim_{{x \to x_0^+}} f(x) \) exists.

Moreover, these limits must equal the value of the function at \( x_0 \). Therefore, \( \lim_{{x \to x_0}} f(x) = f(x_0) \).
If any of these conditions fails, the function is not continuous at that point.

In the original exercise, we evaluated both limits as \( x \) approaches 2 from the left and the right. Unfortunately, these did not match, leading to the conclusion that the function \( f(x) \) is not continuous at \( x = 2 \). This discrepancy illustrates the fairly common scenario where a function is defined everywhere but encounters discontinuity at a specific point.

Real-Valued Function

Now, let's explore real-valued functions. These are functions that yield real numbers as their output for every input from their domain. The domain typically comprises numbers within an interval on the real number line, hence the term "real-valued."

Understanding how a real-valued function behaves is crucial for analyzing its continuity and differentiability. For instance, in our problem, we dealt with a piecewise real-valued function \( f(x) \) defined differently over two intervals: \( x > 2 \) and \( x \leq 2 \). The function involved an integral component for values greater than 2 and a linear algebraic expression when \( x \leq 2 \).

• This dual nature can often cause challenges in continuity or differentiability at points where the definition changes, such as \( x = 2 \) in this function.
• Handling piecewise functions requires careful examination of each segment to ensure that they "match" smoothly at their boundaries.

This type of evaluation is critical for any real-valued functions that might switch formulae based on input value ranges.

Integral Calculus

Integral calculus provides tools for finding volumes, areas, central points, among other things representing accumulation or aggregation. The integral part of our function nomination employs definite integrals to sum up effects over an interval.

In the specific exercise, the given function's integral portion expressed through \( \int_{0}^{x}(5+|1-t|) dt \) coupled integral calculus with absolute values. Evaluating this type of integral requires us to address different cases for the absolute value expression:

• For this situation, we resolved it piecewise over subintervals depending on the behavior of \(|1-t|\).
• Absolute values necessitate splitting the integral evaluation into segments over which the expression within the absolute value maintains consistent behavior.

Understanding these principles of integral calculus helps in crafting accurate solutions and seamless transitions between mathematical ideas.