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Chapter 3: Problem 229

If \(\mathrm{f}(\mathrm{x})=\mathrm{x}-[\mathrm{x}], \mathrm{x} \in \mathrm{R}-\\{0\\}\) where \([\mathrm{x}]=\) the greatest integer not greater than \(\mathrm{x}\), than number of solution \(\mathrm{f}(\mathrm{x})+\mathrm{f}(1 / \mathrm{x})=1 \ldots \ldots \ldots \ldots \ldots\) (a) 0 (b) 1 (c) 2 (d) infinite

Short Answer

Expert verified
The function \(f(x) = x - [x]\) represents the fractional part of x, which is a value between 0 and 1 (including 0 and excluding 1). The function \(f(1/x)\) follows the same properties. Hence, their sum should lie between 0 and 2 (excluding 2). Taking into consideration the right side of the equation, which is an integer value, the equation holds true when this integer is 1. After solving, we find that there are two values of x that satisfy the equation: x= 1 and x= -2. Therefore, there are 2 solutions for the given equation. The correct answer is (c) 2.

Step by step solution

01

Identify the properties of the function f(x)

The function \(f(x) = x - [x]\) represents the fractional part of x. For any given value of \(x\), \(f(x)\) returns the value between 0 and 1 (including 0 and excluding 1).
02

Rewrite the given equation in terms of f(x)

Let's rewrite the given equation in terms of \(f(x)\): \(f(x) + f\left(\frac{1}{x}\right) = 1\)
03

Manipulate the equation

Since we know that \(f(x)\) returns the fractional part of x, let's write the equation separately for \(x\) and \(\frac{1}{x}\) and subtract the greatest integer not greater than x: \(x - [x] + \frac{1}{x} - \left[ \frac{1}{x} \right] = 1\)
04

Rearrange the terms

To arrange the given equation, we will group the fractional parts and integer parts together: \((x - [x]) + \left(\frac{1}{x} - \left[ \frac{1}{x} \right] \right) = ([x] + \left[ \frac{1}{x} \right]) - 1\) Now, we can write this equation as: \(f(x) + f\left(\frac{1}{x}\right) = [x] + \left[ \frac{1}{x} \right] - 1\)
05

Analyze the properties of the equation

We have the following properties of the equation: 1. \(f(x)\) is a value between 0 and 1 (including 0 and excluding 1). 2. \(f\left(\frac{1}{x}\right)\) is also a value between 0 and 1 (including 0 and excluding 1). Hence, the sum of \(f(x)\) and \(f\left(\frac{1}{x}\right)\) lies between 0 and 2 (excluding 2). Now, let's look closely at the right side of the equation: \([x] + \left[ \frac{1}{x} \right] - 1\) Since \([x]\) and \(\left[ \frac{1}{x} \right]\) are both integers, their sum minus 1 must also be an integer. We can deduce that for the equation to hold true, the right side must be equal to 1, as the sum of \(f(x)\) and \(f\left(\frac{1}{x}\right)\) lies between 0 and 2 (excluding 2).
06

Solve for x

Since the right side of the equation must equal 1, we can write: \([x] + \left[ \frac{1}{x} \right] - 1 = 1\) Adding 1 to both sides, we have: \([x] + \left[ \frac{1}{x} \right] = 2\) Now, we can test different values of x, such as x = 1, x = -1, x = 2, and x = -2. Upon testing, we find out that there are only two values of x that satisfy the equation: x = 1 and x = -2. Thus, our answer is:
07

Answer: Number of solutions

There are 2 solutions for the given equation, so the correct answer is (c) 2.

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Most popular questions from this chapter

If \(0 \leq \mathrm{x} \leq \pi\) and \(16^{(\sin ) 2 \mathrm{x}}+16^{(\cos ) 2 \mathrm{x}}=10\) then \(\mathrm{x}=\) (a) \((\pi / 3)\) (b) \((\pi / 2)\) (c) \((\pi / 4)\) (d) \((3 \pi / 4)\)

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