Problem 151 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

In the following questions an Assertion (A) is given, followed by a Reason (R). Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason(R) is a correct explanation for Assertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True \mathrm{\\{} A s s e r t i o n : ~ I f ~ \(f^{\prime}(x)=\frac{1}{1+x^{2}}\) for all \(x\) and \(f(0)=0\), then \(0.4

Step-by-Step Solution

Answer

B

Step 1: Understand the Assertion

The assertion states that if the derivative of a function \( f(x) \) is given by \( f^{\prime}(x) = \frac{1}{1+x^{2}} \) and \( f(0) = 0 \), then the value of \( f(2) \) lies between 0.4 and 2. We need to verify this statement by calculating \( f(2) \).

Step 2: Calculate the function \( f(x) \)

To find \( f(x) \), integrate \( f^{\prime}(x) = \frac{1}{1+x^{2}} \). The integral of \( \frac{1}{1+x^{2}} \) is \( \arctan(x) + C \). Since \( f(0) = 0 \), substitute \( x = 0 \) into \( f(x) = \arctan(x) + C \), resulting in \( 0 = \arctan(0) + C \), thus \( C = 0 \). Therefore, \( f(x) = \arctan(x) \).

Step 3: Evaluate \( f(2) \)

With \( f(x) = \arctan(x) \), compute \( f(2) = \arctan(2) \). The value of \( \arctan(2) \) is approximately 1.107 (using a calculator). This value lies within the range 0.4 to 2 specified in the assertion.

Step 4: Analyze the Reason

The reason states that by the Mean Value Theorem, there exists a \( c \) in \( (0, 2) \) such that \( f^{\prime}(c) = \frac{f(2) - f(0)}{2 - 0} \). According to the theorem, this is a correct statement because it formalizes the existence of such a point \( c \) where the tangent slope equals the average slope between \( f(0) \) and \( f(2) \). This doesn't directly explain why \( 0.4 < f(2) < 2 \).

Step 5: Determine the Relationship between Assertion and Reason

Both the assertion and the reason are true statements, but the reason provided does not directly explain the specific range \( 0.4 < f(2) < 2 \). Instead, it discusses the application of the Mean Value Theorem which is related to the general behavior of \( f(x) \) but not the specific bounds given in the assertion.

Key Concepts

Integral Calculus

Integral calculus is a branch of calculus that helps us find the area under curves, among other things. In the context of this problem, we have to integrate the derivative of a function to get the function itself. The derivative given is \( f'(x) = \frac{1}{1+x^2} \). By integrating this expression, we can find the original function \( f(x) \).
To perform the integration, we note that the integral of \( \frac{1}{1+x^2} \) is a well-known result: it is the inverse tangent function, denoted as \( \arctan(x) \). So the integral calculus process gives us \( f(x) = \arctan(x) + C \).
Since we know \( f(0) = 0 \), we can find the constant \( C \) by setting \( x = 0 \). This gives \( 0 = \arctan(0) + C \), leading to \( C = 0 \). Thus, the function is \( f(x) = \arctan(x) \). This step is crucial as it transforms the derivative information into the full function we need for further calculations.

Derivative of a function

The derivative of a function represents how the function changes as its input changes. For this problem, the derivative provided is \( f'(x) = \frac{1}{1+x^2} \). This derivative tells us two things:

• First, it describes the slope of the tangent to the curve of \( f(x) \) at any point \( x \).
• Second, it is specifically the rate of change for the inverse tangent, or arctan, function, which indicates how the angle whose tangent is \( x \) changes.

Derivatives are essential tools for understanding the behavior of functions. They help locate maxima, minima, and points of inflection, which describe the points where functions take unique shapes or scales.
Here, knowing the derivative allows us to go backwards and identify the original function through integration. Once we recover \( f(x) \) by integrating the derivative, it provides the values we need in subsequent computations, like finding specific values such as \( f(2) \).
This reverse process from derivative back to function is a core aspect of calculus that shows the interconnectedness of differentiation and integration.

Arctan function

The arctan function, also known as the inverse tangent function, is symbolized as \( \arctan(x) \). Unlike the standard tangent function, which takes an angle and gives the ratio of the opposite to the adjacent side, the arctan function does the reverse.
The arctan function gives the angle whose tangent is \( x \). This is critical in trigonometry and calculus because it converts a ratio back into an angle. In this exercise, we ultimately find \( f(x) = \arctan(x) \) after integration.
Evaluation of \( \arctan(2) \), as seen in this problem, yields an approximate value of 1.107. This value lies between 0.4 and 2, verifying the assertion made for \( f(2) \). The mean value theorem contributes by showing that some values in \( (0,2) \) will have a slope equal to the average slope between \( 0 \) and \( 2 \).

• Arctan is widely used in computations involving angles, such as projecting into different dimensions.
• It plays a pivotal role in expressing angles in various branches of physics and engineering.

Understanding the properties and behaviors of the arctan function is key in fields that deal with circular and periodic functions, where inverse trigonometric functions are often applied.