Problem 53 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)=x-[x], x(\neq 0) \in R\), where \([x]\) is the greatest integer less than or equal to \(x\), then the number of solutions of \(f(x)+f\left(\frac{1}{x}\right)=1\) are (A) 0 (B) 1 (C) infinite (D) 2

Step-by-Step Solution

Answer

The number of solutions is 2.

Step 1: Understanding the Function

Given the function \( f(x) = x - [x] \), where \([x]\) is the greatest integer less than or equal to \( x \). This means \( f(x) \) represents the fractional part of \( x \). For example, if \( x = 2.3 \), then \([x] = 2\) and \( f(x) = 0.3 \).

Step 2: Formulating the Equation

We need to find the number of solutions for the equation \( f(x) + f\left(\frac{1}{x}\right) = 1 \). This translates to finding \( x \) such that the sum of its fractional part and the fractional part of its reciprocal equals 1.

Step 3: Analyzing the Fractional Parts

Since \( 0 \leq f(x) < 1 \) and \( 0 \leq f\left(\frac{1}{x}\right) < 1 \), the sum \( f(x) + f\left(\frac{1}{x}\right) = 1 \) implies both fractional parts must add up exactly to 1. Since each fractional part is strictly less than 1, neither can be zero.

Step 4: Finding Possible Solutions

Let \( f(x) = a \) and \( f\left(\frac{1}{x}\right) = b \), where \( a+b=1 \). Thus, \( a = f(x) = x - [x] \) and \( b = f\left(\frac{1}{x}\right) = \frac{1}{x} - \left[ \frac{1}{x} \right] \). Solving, we obtain \( a = 1 - b \).

Step 5: Solving the Equations Together

Assume \( a = 1 - b \), then \( f\left( \frac{1}{x} \right) = 1 - f(x) \). Plug in values to find: considering the equality \( x + \frac{1}{x} = [x] + \left[ \frac{1}{x} \right] + 1 \), we deduce form solutions when \( x = 1 \) or \( x = -1 \).

Step 6: Verifying the Solutions

Substitute \( x = 1 \) and \( x = -1 \) into the original equation: \( f(1) + f\left(\frac{1}{1}\right) = 0 + 0 = 0 \) (no solution). However, substituting non-trivial fractional \( x \) gives two solutions \(x \approx \phi,~\frac{1}{\phi}\approx - \phi \), providing that there are two solutions.

Key Concepts

Greatest Integer Function

The greatest integer function, also known as the floor function, is a special type of function that rounds down any given real number to the nearest integer that is less than or equal to that number. For instance, if you have the number 3.7, the greatest integer function will convert it to 3.
Similarly, if it is -1.2, the result would be -2, because -2 is the greatest integer less than or equal to -1.2.
This function is often denoted using brackets:

• For a number x, \([x]\) represents the greatest integer less than or equal to x.

It's commonly used in mathematical problems dealing with real numbers and their integer parts, such as breaking down a number into its integer part and fractional part.In the context of this exercise, understanding the greatest integer function helps us identify the fractional part of any real number, which is crucial for solving the given equation.

Reciprocal Function

The reciprocal function is another fundamental concept in mathematics. It involves taking any non-zero number and flipping it to get its reciprocal, or multiplicative inverse.
For example, the reciprocal of 4 is \(\frac{1}{4}\).
Similarly, if you have \(\frac{1}{5}\), its reciprocal is 5.

• Note that the reciprocal of a number multiplied by the number itself always equals 1.
• In mathematical terms, the reciprocal of x is given by \(\frac{1}{x}\).

In this exercise, the reciprocal function is important because it involves analyzing the fractional part of the reciprocal of x, \(f\left(\frac{1}{x}\right)\).Together with \(f(x)\), these functions are used to find a set of solutions where their combined fractional parts sum up to 1.

Solutions of Equations

Finding the solutions to equations often demands finding numbers that satisfy a given mathematical expression or condition.
In this exercise, we are tasked with solving the equation \(f(x) + f\left(\frac{1}{x}\right) = 1\).

• The goal is to discern what values of x will make the equation hold true.
• This involves analyzing both the fractional part of x and its reciprocal, as their sum must equate to 1.

By exploring the properties of the fractional parts and using logical reasoning, we determined that there are two specific non-trivial solutions to this equation.
Thus, understanding the individual components and how they interact is essential to solve such problems efficiently.