Problem 146 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(x>0\) and \(\int_{0}^{x}[x] d x=[x]\left(\frac{1}{2} A+B\right)\), where \([.]\) denotes the greatest integer function, then (A) \(A=[x]-1\) (B) \(B=x-[x]\) (C) \(A=[x]+1\) (D) \(B=x+[x]\)

Step-by-Step Solution

Answer

(A) and (B) are the correct answers: \(A = [x] - 1\) and \(B = x - [x]\).

Step 1: Understanding the Problem

The problem involves solving an integral within certain limits and expressing the result using the greatest integer (floor) function, denoted by \([x]\). We need to relate the integral from 0 to \(x\) to expressions involving \([x]\).

Step 2: Understanding the Greatest Integer Function

The greatest integer function, \([x]\), outputs the greatest integer less than or equal to \(x\). For example, \([3.7] = 3\) and \([4] = 4\). We will use this property in our calculations.

Step 3: Setting Up the Integral

The definite integral given in the problem is \( \int_{0}^{x} [u] \, du \). We can break the integral into segments where \([u]\) is constant. For example, if \( [x] = n \), then across each integer interval, \([u]\) remains the same.

Step 4: Evaluating the Integral

Consider the interval from 0 to \(x\), covering integer segments \([0, 1), [1, 2), \, \ldots , [n, x]\) where each segment contributes to the integral. The integral becomes a summation: \[ \int_{0}^{x} [u] \, du = \sum_{k=0}^{n-1} k + n(x - n) \].

Step 5: Simplifying the Summation

The sum \( \sum_{k=0}^{n-1} k = \frac{n(n-1)}{2} \). Thus, the integral becomes: \[ \frac{n(n-1)}{2} + n(x-n) \].

Step 6: Applying Constraints

According to the problem, the integral equals \([x](\frac{1}{2}A + B)\). Substituting our integral expression and equating gives: \( \frac{n(n-1)}{2} + n(x-n) = n\left(\frac{1}{2} A + B\right) \).

Step 7: Solving for A and B

Set the expressions for the integral equal and match coefficients: 1. \( \frac{n(n-1)}{2} = \frac{n^2}{2} - \frac{n}{2} \) contributes to \( \frac{n}{2} A \), so \( A = n = [x] - 1 \). 2. The term \( n(x-n) \) matches \( nB \), so \( B = x-n = x-[x] \).

Step 8: Verifying the Answer

Compare the computed values of \(A\) and \(B\) with the options provided in the question. Confirm that \(A = [x] - 1\) and \(B = x - [x]\) match options (A) and (B).

Step 9: Conclusion

The values of \(A\) and \(B\) that satisfy the equation and problem constraints confirm the selection of options (A) and (B).

Key Concepts

Greatest Integer Function

The greatest integer function, often represented as \([x]\), is a step function that assigns each real number \(x\) to the largest integer less than or equal to \(x\). This function is also commonly known as the floor function.
It provides a way to work with decimal numbers by converting them into integers in a very controlled manner.
For instance:

• If \(x = 4.75\), then \([x] = 4\).
• If \(x = -2.3\), then \([x] = -3\).
• If \(x = 7\), then \([x] = 7\).

This function is particularly useful because it helps simplify expressions during integral evaluation especially when dealing with definite integrals that have stepwise constant properties. By breaking intervals at integer boundaries, calculations become straightforward and manageable.

Definite Integral Evaluation

Definite integrals are essential in calculus and represent the area under a curve between two bounds. In the problem given, it involves evaluating an integral with a function that changes at certain points, specifically integers.
This is where the greatest integer function adds complexity to the evaluation of the integral.
The key step in solving this problem is breaking the integral over the interval \( [0, x] \) into segments where the greatest integer function remains constant.Consider the structure:

• The function stays constant in each subinterval.
• For instance, from 0 to 1, \([u]=0\); from 1 to 2, \([u]=1\).
• This pattern continues until the integral's upper limit, \([n, x]\) where \([x] = n\).

The evaluation becomes feasible because for each subinterval, the integral simplifies to a summation. The solution involves understanding these segment contributions and assessing them through convenient algebraic substitutions and simplifications.

Integral Properties

Integral properties are the rules and formulas that govern how integrals are calculated and manipulated. These properties were instrumental in solving the problem at hand. Let's take a closer look at some of the important ones:

• Linearity: Integrals can be split into parts and constants can be pulled out, such as \( \int (cf(x) + g(x)) dx = c \int f(x) dx + \int g(x) dx \).
• Summation: Integrals over segmented intervals, especially when dealing with step functions, can be expressed as sums over those intervals.
• Property of Definite Integrals: If a function changes over specific boundaries, then the integral can be computed by considering each interval separately, just as done with the greatest integer function.

In solving the problem, these properties were used to simplify and recognize patterns within the definite integral to match the form given in the equation. By recomposing the definite integral as a sum of simpler parts, it allowed for direct comparison with the required function expressions, ultimately leading to the correct identification of the coefficients \(A\) and \(B\).