Problem 119 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \([\cdot]\) denotes the greatest integer function, then \(\int_{0}^{2}[x+[x+[x]]] d x=\) (A) 1 (B) 2 (C) 3 (D) 0

Step-by-Step Solution

Answer

C) 3

Step 1: Understand the Problem

We are asked to evaluate the integral \[ \int_{0}^{2}[x+[x+[x]]] dx, \]where \([\cdot]\) is the greatest integer function, also known as the floor function. This function returns the largest integer less than or equal to a given number.

Step 2: Break Down the Function

The function \([x+[x+[x]]]\) involves several nested greatest integer functions. First, evaluate \([x]\), which changes as \(x\) varies. Next, recognize that \([x + [x]]\) slices the intervals further, and finally, calculate \([x + [x + [x]]]\).

Step 3: Identify Function Behavior

We need to understand how \([x + [x + [x]]]\) behaves over the interval \([0, 2]\). Notice that for \(x \in [0, 1)\), \([x] = 0\). Therefore,\[ [x + [x + [x]]] = [x + 0 + 0] = [x]. \] Similarly, for \(x \in [1, 2)\), \([x] = 1\), so,\[ [x + [x + [x]]] = [x + 1 + [x]] = [x + 1]. \]

Step 4: Evaluate the Integral

We split the integral at the critical point (1) where the function changes:\[ \int_{0}^{2}[x+[x+[x]]] dx = \int_{0}^{1} 0 \: dx + \int_{1}^{2} 2 \: dx. \]Calculate each part: the first integral evaluates to 0, whereas the second evaluates to 2.

Step 5: Calculate Final Result

Perform the integration:\[ \int_{0}^{1} [x] \, dx = 0, \quad \int_{1}^{2} [x + 1] \, dx = \int_{1}^{2} 2 \, dx. \] Finally, \[ \int_{0}^{2}[x+[x+[x]]] \, dx = \int_{1}^{2} 2 \, dx = 2(2 - 1) = 2. \]

Key Concepts

Understanding Definite Integrals

The concept of a definite integral is essential in calculus. It allows us to calculate the area under a curve between two specific points on a graph. For a function \( f(x) \), the definite integral from \( a \) to \( b \) is denoted by \( \int_{a}^{b} f(x) \, dx \). This integral gives us a number representing the total area under the curve of \( f(x) \) from \( x = a \) to \( x = b \).

• The limits \( a \) and \( b \) are called the bounds of integration.
• The result of a definite integral is a real number, not another function.
• It is crucial for applications in physics, engineering, economics, and probability.

By understanding how to break down complex functions in these problems, you can accurately find areas or sums. This involves recognizing when to split an integral at points where the behavior of the function changes. Such techniques show how powerful integration is in handling functions that vary over different intervals.

Exploring the Floor Function

The floor function, denoted as \( \lfloor x \rfloor \), is a mathematical function that rounds a real number down to the nearest integer less than or equal to that number. This function is called the greatest integer function.

• For example, \( \lfloor 3.7 \rfloor = 3 \) and \( \lfloor -2.3 \rfloor = -3 \).
• This function is essential when dealing with discrete steps or divisions.
• It operates piecewise, meaning its structure and behavior vary over different intervals.

In contexts like the original exercise, understanding how the floor function works helps solve integrals involving nested expressions. By analyzing the behavior of the numbers inside the brackets, you can determine the values the floor function returns at specific intervals.

Concept of Piecewise Functions

Piecewise functions are functions defined by different expressions depending on the value of the input variable. These functions are characterized by having "pieces" where each piece applies to a certain interval of the domain.

• Each section may have a distinct rule or formula.
• They are widely used for representing situations where decisions or occurrences change at specific points.
• Optimal for modeling real-world systems like tax brackets or shipping costs.

In the context of definite integrals, understanding piecewise functions is critical because it allows you to break down complex behavior into simpler integrals over different intervals. The behavior within each interval is much easier to integrate and then added together to find the total effect across the span of input values.