Problem 112 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(f\) be function defined on set of non-negative integers and taking values in the same set. Given that:- i. \(x-f(x)=19\left[\frac{x}{19}\right]-90\left[\frac{f(x)}{90}\right]\) for all non-negative integers; ii. \(1900

Step-by-Step Solution

Answer

The possible value of \(f(1990)\) is 1911.

Step 1: Analyze the given functional equation

We are given the following functional equation for all non-negative integers \(x\): \( x - f(x) = 19\left[\frac{x}{19}\right] - 90\left[\frac{f(x)}{90}\right] \) We know that: - \([\frac{x}{19}]\) is the greatest integer function, which represents the largest integer less than or equal to \(\frac{x}{19}\). - \([\frac{f(x)}{90}]\) is the greatest integer function, which represents the largest integer less than or equal to \(\frac{f(x)}{90}\). Let's put a constraint on x: If \(x \leq 18\), then \(\frac{x}{19}<1 \Rightarrow \left[\frac{x}{19}\right]=0\). So, the equation becomes: \( x - f(x) = -90\left[\frac{f(x)}{90}\right] \) Now notice that for all such x values, both \(x\) & \(f(x)\) are non-negative: \( x \geq 90\left[\frac{f(x)}{90}\right] \Rightarrow f(x) \leq x \)

Step 2: Determine bounds on \(f(1990)\)

From the given condition, \(1900 < f(1990) < 2000\). As the function \(f\) takes values in non-negative integers, we can rewrite the given condition as: 1901 ≤ f(1990) ≤ 1999 Now, we have to find all possible integer values of \(f(1990)\) under these constraints. Consider \(x = 1990\) in the given functional equation: \( 1990 - f(1990) = 19\left[\frac{1990}{19}\right] - 90\left[\frac{f(1990)}{90}\right] \) We know that \(\frac{1990}{19} = 105\), so the equation becomes: \( 1990 - f(1990) = 19(105) - 90\left[\frac{f(1990)}{90}\right] \)

Step 3: Find possible values for \(f(1990)\)

Since \(1901 \le f(1990) \le 1999\), we have \(0 \le 1990 - f(1990) \le 89 \) Now, we need to find all possible values of \(\left[\frac{f(1990)}{90}\right]\) to satisfy the equation: \( 1990 - f(1990) = 19(105) - 90\left[\frac{f(1990)}{90}\right] \) Since \(\left[\frac{f(1990)}{90}\right]\) is an integer and \(1901 \le f(1990) \le 1999\), we can divide this interval into the following cases: 1) \(1901 \le f(1990) \le 1990\): In this case, we have \(\left[\frac{f(1990)}{90}\right] = 21\). So the equation becomes: \( 1990 - f(1990) = 19(105) - 90(21) \) Solving this equation gives \(f(1990) = 1990 - 19(105) + 90(21) = 1911\) . 2) \(1991 \le f(1990) \le 1999\): In this case, we have \(\left[\frac{f(1990)}{90}\right] = 22\). So the equation becomes: \( 1990 - f(1990) = 19(105) - 90(22) \) Solving this equation gives \(f(1990) = 1990 - 19(105) + 90(22) = 1891\) . However, this value does not satisfy the given constraint that \(1901 \le f(1990) \le 1999\). So, we discard this value.

Step 4: Conclusion

We have found that the only possible value for \(f(1990)\) that satisfies all the given conditions is \(f(1990) = 1911\).

Key Concepts

Greatest Integer Function

The Greatest Integer Function, also known as the floor function, is denoted by \( [x] \) and is a crucial concept in mathematics, often tested in competitive exams like IIT JEE. The function effectively 'rounds down' to the nearest integer. For instance, \( [3.7] = 3 \) and \( [-1.2] = -2 \) represent how the function works for positive and negative values, respectively.

Understanding this function helps in breaking complex problems into smaller, more manageable parts, particularly in functional equations where variables are manipulated based on their integer parts. It's essential to grasp this concept because it introduces the idea of partitioning the real number line into discrete segments, each associated with an integer value. In the given exercise, the greatest integer function helps to establish bounds on the function \( f(x) \) based on the input \( x \) and to exploit the properties of the function to narrow down potential solutions.

Non-negative Integers

Non-negative integers are whole numbers that include zero and all positive integers. They form a fundamental part of number theory and are often the domain or range for functions in IIT JEE Mathematics problems. When we consider functions defined on non-negative integers, we operate within a discrete and bounded spectrum of values.

Having a function defined on non-negative integers simplifies the problem to some extent because the function values are restricted to whole numbers. This property is advantageous when determining the possible values of the function or solving equations, as it confines the solution set and makes the process more straightforward. In our given problem, knowing the function takes on non-negative integer values allows us to use properties of divisibility and modular arithmetic to find the specific value of \( f(1990) \) under the constraints given.

Bounds on Functions

Establishing bounds on functions involves finding intervals within which the function's value must lie. In the context of the textbook problem, constraining \( f(1990) \) within an interval of \( 1900 < f(1990) < 2000 \) narrows down the possible integer values it can take. Knowing the bounds is critical because it eliminates many potential solutions and allows for a more targeted approach when solving for \( f(x) \).

To effectively apply bounds, one must consider the properties of the function and the constraints set by the problem. By examining how the greatest integer function behaves with division and multiplication, we can leverage the bounds of the function to pinpoint the exact values that satisfy the functional equation within the given range.

IIT JEE Mathematics

The IIT Joint Entrance Exam (IIT JEE) is one of the most competitive engineering entrance exams worldwide, and it places a strong emphasis on Mathematics. Mastery in topics like functional equations, the greatest integer function, and number theory is essential for success in this exam. Problems are designed not only to test conceptual knowledge but also the application of concepts to novel situations.

Candidates are often required to connect different areas of mathematics to arrive at a solution, much like in the problem provided, which combines the understanding of the greatest integer function and bounds on functions to solve a functional equation. As a part of the IIT JEE Mathematics syllabus, students must practice such problems extensively to develop problem-solving speed and accuracy under pressure, which is crucial for a high score in the exam.