Problem 20 - A Complete Resource Book in Mathematics for JEE Main

Question

If the area bounded by the curve \(y=f(x), x\)-axis and the ordinates \(x=1\) and \(x=b\) is \((b-1) \sin (3 b+4)\), then (A) \(f(x)=\cos (3 x+4)+3(x-1) \sin (3 x+4)\) (B) \(f(x)=\sin (3 x+4)+3(x-1) \cos (3 x+4)\) (C) \(f(x)=\sin (3 x+4)-3(x-1) \cos (3 x+4)\) (D) None of these

Step-by-Step Solution

Answer

Option (B) is correct.

Step 1: Understand Area Bounded by a Curve

The exercise mentions finding the area bounded by the curve \( y = f(x) \), the x-axis, and the ordinates \( x = 1 \) and \( x = b \). Mathematically, this area can be expressed using the integral: \( \int_{1}^{b} f(x) \, dx = (b-1) \sin(3b+4) \). We need to find \( f(x) \) so that this equality holds.

Step 2: Derive the Expression for Area

Given that \( \int_{1}^{b} f(x) \, dx = (b - 1) \sin(3b + 4) \), we differentiate both sides with respect to \( b \) to find \( f(x) \). This yields \( f(b) = \left(\sin(3b + 4) + (b-1) \cdot 3 \cos(3b + 4) \right) \). This step uses the Fundamental Theorem of Calculus and the product rule for differentiation.

Step 3: Compare with Given Options

From differentiation, we find \( f(b) = \sin(3b + 4) + 3(b-1) \cos(3b + 4) \). We compare this result with the given options. Option (B) states \( f(x) = \sin (3x + 4) + 3(x-1) \cos (3x + 4) \), which matches the derived expression.

Step 4: Select the Correct Option

Since Option (B) matches the expression we derived for \( f(x) = \sin(3x + 4) + 3(x-1) \cos(3x + 4) \), this is the correct choice. Therefore, \( f(x) \) is as stated in Option (B).

Key Concepts

Integrals in Calculus

Integrals in calculus are a fundamental concept used to compute the area under curves. This concept is central to understanding how to evaluate the total "accumulated" area from a function's graph as it passes through an interval. When you see the symbol \( \int \), it represents the integral, which in its basic form means you are finding this accumulated value. For instance, the integral from 1 to \( b \) of \( f(x) \) is written as \( \int_{1}^{b} f(x) \, dx \), where \( dx \) indicates the variable of integration.
Integrals can be thought of as an "anti-derivative," as they essentially reverse the process of differentiation. They help in calculating areas bounded by curves, like in the given exercise, providing a visual approximation of the amount of space under a specific graph.

• Definite integrals give the exact area between the curve, x-axis, and vertical lines at the limits of integration.
• Indefinite integrals produce a general form including a constant of integration "C" because there are infinite possible solutions differing by a constant value.

Integrals have wide applications not only in pure mathematics but also in physics, engineering, and beyond, illustrating their significant role in not just academia but also real-world problem-solving.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus serves as a bridge connecting differentiation and integration, two main branches of calculus. This theorem is pivotal because it links the concept of the derivative of a function to the concept of an integral.
The theorem has two key parts:

• The first part states that if a function \( f \) is continuous on the interval \( [a, b] \) and \( F \) is the indefinite integral of \( f \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). This indicates how integration and differentiation can reverse each other under certain conditions.
• The second part asserts that if \( F \) is an antiderivative of \( f \), then the derivative of the integral of \( f \) from \( a \) to \( x \) is \( f(x) \). Mathematically, it can be expressed as \( \frac{d}{dx}\Big(\int_{a}^{x} f(t) \, dt\Big) = f(x) \).

This theorem simplifies the calculation of exact areas and provides a deeper insight into understanding the accumulation of quantities over an interval.

Differentiation

Differentiation is another foundational concept in calculus that focuses on determining the rate of change of a quantity. It produces the derivative of a function, which tells us how a function value changes as its input changes.
Understanding differentiation involves visually imagining how steep a curve is at a specific point. You can think of it as calculating the slope of the tangent line to a function's graph at a given point.

• The derivative is a powerful tool for finding local maximum and minimum values, helping to optimize functions in various contexts.
• It also assists in assessing the behavior of functions around certain points, such as when they are speeding up or slowing down.

In solving the exercise, differentiating the expression derived from the integral was crucial. By employing techniques like the product rule, the correct function \( f(x) \) was determined, showcasing how differentiation is used to solve problems involving rates and areas.
Differentiation extends into many fields, serving as an underpinning for theories and practical techniques in science, engineering, and economics.