Problem 64 - A Complete Resource Book in Mathematics for JEE Main

Question

The tangent to the curve \(y=f(x)\) at the point with abscissa \(x=1\) form an angle of \(\pi / 6\) and at the point \(x=2\) an angle of \(\pi / 3\) and at the point \(x=3\) an angle of \(\pi / 4\). If \(f^{n}(x)\) is continuous, then the value of \(\int_{1}^{3} f^{\prime \prime}(x) f^{\prime}(x) d x+\int_{2}^{3} f^{\prime \prime}(x) d x\) is (A) \(\frac{4 \sqrt{3}-1}{3 \sqrt{3}}\) (B) \(\frac{3 \sqrt{3}-1}{2}\) (C) \(\frac{4-3 \sqrt{3}}{3}\) (D) None of these

Step-by-Step Solution

Answer

Option D: None of these.

Step 1: Understand Tangent and Angles

When a tangent to the curve at a point forms an angle with the x-axis, the slope of the tangent (i.e., the derivative at that point) can be given by \( \tan(\theta) \), where \( \theta \) is the angle the tangent makes with the x-axis. Given angles \( \frac{\pi}{6} \), \( \frac{\pi}{3} \), and \( \frac{\pi}{4} \) at points \( x = 1, 2, 3 \) respectively, the slopes of the tangents are \( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \), \( \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \), and \( \tan\left(\frac{\pi}{4}\right) = 1 \).

Step 2: Derive Relations from Slopes

The relations we deduce from the slopes are \( f'(1) = \frac{1}{\sqrt{3}} \), \( f'(2) = \sqrt{3} \), and \( f'(3) = 1 \).

Step 3: Analyze the Integral Expression

We need to evaluate \[ \int_{1}^{3} f''(x) f'(x) \, dx + \int_{2}^{3} f''(x) \, dx \]. Notice the first integral involves the product of derivatives, suggesting possible simplification via substitution.

Step 4: Evaluate \( \int f''(x) f'(x) \, dx \)

Set \( u = f'(x) \), then \( du = f''(x) \, dx \), converting the first integral to \( \int_{u(1)}^{u(3)} u \, du = \left[ \frac{u^2}{2} \right]_{u(1)}^{u(3)} \). Plugging in \( u(1) = \frac{1}{\sqrt{3}} \) and \( u(3) = 1 \), we find \( \frac{1^2}{2} - \frac{\left(\frac{1}{\sqrt{3}}\right)^2}{2} = \frac{1}{2} - \frac{1}{6} = \frac{1}{3} \).

Step 5: Evaluate \( \int_{2}^{3} f''(x) \, dx \)

This integral represents the change in \( f'(x) \) from \( x = 2 \) to \( x = 3 \), which is \( f'(3) - f'(2) = 1 - \sqrt{3} \).

Step 6: Sum Results from Steps 4 and 5

Add the results from Steps 4 and 5: \( \frac{1}{3} + (1 - \sqrt{3}) = 1 + \frac{1}{3} - \sqrt{3} = \frac{4}{3} - \sqrt{3} \). This matches none of the provided options, therefore the answer is (D) None of these.

Key Concepts

Derivative of a Function

The derivative of a function is a fundamental concept in calculus, representing the rate at which a function changes as its input changes. It can be thought of as the "instantaneous rate of change" or the "slope of the tangent line" at any given point along the curve of the function. When we talk about the slope of the tangent to a curve at a certain point, we're essentially discussing the derivative of the function at that point. For example, in the problem, the slopes of the tangents at points with abscissas 1, 2, and 3 are given as \( \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} \), \( \tan(\frac{\pi}{3}) = \sqrt{3} \), and \( \tan(\frac{\pi}{4}) = 1 \) respectively.

These values imply that the derivative of the function, denoted as \( f'(x) \), evaluates to these tangent values at the respective x positions: \( f'(1) = \frac{1}{\sqrt{3}} \), \( f'(2) = \sqrt{3} \), and \( f'(3) = 1 \). Understanding derivatives is crucial for analyzing how curves behave, enabling us to determine which way they are turning, how steep they are going, and where they have any inflection points.

Integral Calculus

Integral calculus, often seen as the inverse process to differentiation, deals with accumulation and area under curves. It is essential for solving problems where we need to find the total size, area, or volume given a certain rate of change. In this problem, integrals are used to evaluate expressions involving derivatives of a function. Specifically, the aim is to find the values of \( \int_{1}^{3} f''(x) f'(x) \, dx \) and \( \int_{2}^{3} f''(x) \, dx \).

These integrals require knowledge of how integration can transform the product or the sum of the derivatives and what this represents in terms of change over an interval. The expression involving \( f''(x) f'(x) \) highlights a method where substitution can simplify complex integrals; by letting \( u = f'(x) \), the process can simplify the integration of products and reveal valuable insights into how the function behaves implicitly across different intervals.

This manipulation through substitution is one of the powerful tools in integral calculus, allowing one to compute areas or changes in quantities even when dealing with products of derivatives.

Angle of Tangent

The angle that a tangent makes with the x-axis is a geometrically intuitive way to understand the slope of a function at a particular point. This angle, denoted as \( \theta \), impacts how we perceive the steepness or flatness of a curve at that point and is directly related to the slope of the tangent line. In mathematical terms, the tangent of the angle \( \theta \) is equal to the slope of the tangent line, or \( \tan(\theta) = m \), where \( m \) is the derivative at that point.

For example, a curve with a tangent line forming an angle of \( \pi / 6 \) with the x-axis has a derivative equal to \( \tan(\pi / 6) = \frac{1}{\sqrt{3}} \). As angles change, so do these calculated slopes, thus playing a crucial role in determining and analyzing how curves behave across different sections. Understanding these angles helps one not only solve problems conceptually but also visually comprehend how curves might rise or fall relative to the horizontal axis.