Problem 41 - A Complete Resource Book in Mathematics for JEE Main

Question

For \(x \in R\), let \([x]\) denotes the greatest integer \(\leq x\), then the value of \(\left(-\frac{1}{3}\right)+\left(-\frac{1}{3}-\frac{1}{100}\right)+\left(-\frac{1}{3}-\frac{2}{100}\right)\) \(+, \ldots,+\left(-\frac{1}{3}-\frac{99}{100}\right)\) is (A) \(-100\) (B) \(-123\) (C) \(-135\) (D) \(-153\)

Step-by-Step Solution

Answer

The total sum is \(-202\).

Step 1: Understanding the Expression

The expression involves a sum of terms of the form \(-\frac{1}{3} - \frac{k}{100}\), where \(k\) ranges from 0 to 99. For each \(k\), compute the greatest integer less than or equal to this expression, which is represented as \([x]\).

Step 2: Compute Each Term Statement

For each term \(-\frac{1}{3} - \frac{k}{100}\), identify the range of integers that include or floor that value. Initially, \(-\frac{1}{3}\approx -0.3333\) is already less than a whole number.

Step 3: Apply Greatest Integer Function

When evaluating each term \(-\frac{1}{3} - \frac{k}{100}\), calculate the value, floor it to the nearest lower integer, using the negative ceiling method as it always rounds towards zero. For instance, starting at \(-\frac{1}{3} - \frac{0}{100} = -\frac{1}{3}\), the greatest integer function floors value to \(-1\), similarly for \(-\frac{1}{3} - \frac{k}{100}\).

Step 4: Identify Pattern in the Series

Observe the values calculated by \([x]\):- For \(k=0\) to \(k=32\), the terms are \([-1]\) because their floor values are \(-0.33\geq x>-1\).- For \(k=33\) to \(k=65\), the terms are \([-2]\) and so on, until \(k=99\), which reaches another interval.

Step 5: Sum the Values

Sum the values based on identified sequences: 33 terms of \(-1\), 33 terms of \(-2\), 33 terms of \(-3\), and 1 term of \(-4\):- Sum = \(33(-1) + 33(-2) + 33(-3) + 1(-4) = -33 - 66 - 99 - 4 = -202\).

Key Concepts

Floor Function

The floor function, also known as the greatest integer function, is a mathematical operation that maps a real number to the largest integer less than or equal to it. This function is denoted by square brackets containing the number, like \([x]\). For example, if we have \([3.7]\), the floor function would round this down to 3, the largest integer less than or equal to 3.7.

The floor function is particularly useful when dealing with real numbers and matching them to integer values. It finds applications in various mathematical problems, including sequences and sums where we need exact integer results. In our exercise, the function helps break down complex expressions into manageable parts by determining the integer values for each term in the sequence.

• Symbol: [x]
• Purpose: Rounds down to the nearest integer
• Example: \([5.9] = 5\)

By understanding this function, solving intricate numerical problems where approximation might lead to errors becomes easier.

Flooring Negative Values

When flooring negative values, it's important to remember that the floor function continues to work by moving downward to the next integer. For positive numbers, we usually see this function operate as expected—rounding downwards. However, with negative numbers, rounding still occurs downwards, which takes into account the element of moving more negative.

In the example from the exercise, evaluating \(-\frac{1}{3} - \frac{k}{100}\) for different values of \(k\) results in observing how the floor function rounds these negative numbers. Let's break it down:

• Start with \(-\frac{1}{3} = -0.3333\) which floors to \(-1\).
• Continue adding fractions like \(-\frac{1}{100}\) does not change the floor because \(-0.3433\) still floors to \(-1\).
• As we layer in more fractions \((-\frac{k}{100})\), at certain points (for instance, around \(k=34\)), the sum dips below another integer level, prompting a floor to \(-2\), then beyond that eventually to \(-3\) and so forth.

This rounding downwards function allows for calculation management where tracking precise lowering of values is crucial, providing exact integer intervals in sequences.

Mathematical Pattern Recognition

Identifying patterns in sequences is a key component in solving mathematical exercises. In the context of our problem, recognizing how the sequence of terms transforms and groups them based on their floored integer results leads directly to obtaining the sum.

Each group in the sequence of terms shows a predictable mathematical pattern. Examine how \(-\frac{1}{3}-\frac{k}{100}\) displays its behavior as the following pattern emerges:

• For \(k\) from 0 to 32, each term floors to \([-1]\).
• A shift then occurs from 33 to 65, where each term now floors to \([-2]\).
• This pattern continues, with terms \(66\) to \(98\) flooring to \([-3]\), and finally, the last term, \(k=99\), reaches \([-4]\).

By observing these intervals, we can efficiently compute sums over large sequences—something common in calculus and integer-based problems. Because these intervals appear consistently, they offer solutions not only specific to the floor function application but broadly in contexts involving repetitive numerical behavior.