Problem 43 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f^{\prime}(x)=\frac{1}{1+x^{2}}\) for all \(x\) and \(f(0)=0\), then (A) \(f(2)<0.4\) (B) \(f(2)>2\) (C) \(0.4

Step-by-Step Solution

Answer

Option (C): \(0.4 < f(2) < 2\).

Step 1: Understand the Problem

We are given the derivative of a function, \( f'(x) = \frac{1}{1+x^2} \), and an initial condition \( f(0) = 0 \). We need to find \( f(2) \) and then determine which of the given options is true based on this result.

Step 2: Set Up the Integral

To find the function \( f(x) \) from its derivative \( f'(x) \), we need to integrate \( f'(x) \). Since \( f'(x) = \frac{1}{1+x^2} \), the integral of \( f'(x) \) will give us \( f(x) \). Thus, set up the integral: \( f(x) = \int_0^x \frac{1}{1+t^2} \, dt \).

Step 3: Recognize the Integral Formula

The integral \( \int \frac{1}{1+x^2} \, dx \) is a standard result that equals \( \arctan(x) + C \), where \( C \) is the constant of integration. We can apply this knowledge to integrate \( f'(x) \).

Step 4: Find the Function f(x)

From step 3, we have \( f(x) = \arctan(x) + C \). Using the initial condition \( f(0) = 0 \), we substitute to find \( C \): \( f(0) = \arctan(0) + C = 0 \). Since \( \arctan(0) = 0 \), we get \( C = 0 \). Therefore, \( f(x) = \arctan(x) \).

Step 5: Evaluate f(2)

Now that we have \( f(x) = \arctan(x) \), we can find \( f(2) \). So, \( f(2) = \arctan(2) \). Use a calculator or a known value to approximate \( \arctan(2) \), which is approximately 1.107 radians.

Step 6: Compare and Conclude

With \( f(2) \approx 1.107 \), which option does it fall into? Compare this to the given options: \( f(2) = 1.107 \) satisfies the condition \( 0.4 < f(2) < 2 \). Therefore, option (C) is correct.

Key Concepts

Derivative of a function

The derivative of a function, denoted as \( f'(x) \), provides the rate at which the function \( f(x) \) changes with respect to \( x \). It is a fundamental concept in calculus that helps us understand how functions behave. In the given exercise, the derivative is presented as \( f'(x) = \frac{1}{1+x^2} \). This expression tells us how \( f(x) \) is changing at any point \( x \). The role of derivatives is critical in finding the slope of the tangent line to the curve and in optimizing functions, among other applications.
To find the original function \( f(x) \), we need to "reverse" the derivative process by integrating \( f'(x) \). This process will give us the antiderivative, or the integral of the function, which includes an arbitrary constant since differentiation of a constant is zero.

Arctangent function

The arctangent function, denoted as \( \arctan(x) \), is the inverse of the tangent function. This function returns the angle whose tangent is \( x \). The integral of \( \frac{1}{1+x^2} \) is a standard result in calculus that equals \( \arctan(x) + C \), where \( C \) is a constant of integration. Understanding how these inverse trigonometric functions work is crucial because they often appear in engineering and physics when dealing with angles and periodic functions.
In this exercise, obtaining \( \arctan(x) \) as part of the solution illustrates the inverse operation of differentiation over a standard form. Calculating \( \arctan(2) \) gives us approximately 1.107 radians, which helps us evaluate the function \( f(x) \) at specific points.

Constant of integration

When integrating a function, we always add a constant of integration, denoted as \( C \). This is because differentiation of a constant is zero, and there are infinitely many antiderivatives for a given function. In our exercise, the integral of \( \frac{1}{1+x^2} \) leads to \( \arctan(x) + C \). Initially, \( C \) is unknown, but we can determine it using given conditions.
In this context, the initial condition \( f(0) = 0 \) helps us find \( C \). By substituting \( x = 0 \) into \( f(x) = \arctan(x) + C \) and knowing \( \arctan(0) = 0 \), it's clear that \( C = 0 \). Thus, the function simplifies to \( f(x) = \arctan(x) \), eliminating any ambiguity about the constant. These steps show how essential initial conditions are in solving integration problems.

Initial conditions

Initial conditions are specific values provided for a function or its derivative at a particular point. They help determine the exact solution of differential equations or integrals. In this problem, we have \( f(0) = 0 \), which is an initial condition.
Initial conditions are crucial because they allow us to solve for any unknown constants, such as the constant of integration. By applying \( f(0) = 0 \), we replace \( x \) in the integrated result \( \arctan(x) + C \) to determine \( C \). This application results in determining that \( C = 0 \), giving a more precise function \( f(x) = \arctan(x) \).

• Initial conditions specify the particular solution among the family of solutions given by an indefinite integral.
• They provide crucial information that shapes the function's graph and behavior.

Understanding initial conditions helps us see the connection between the abstract mathematical solution and its real-world applications.