Problem 131
Question
If the period of the function \(f(x)=\sin (\sqrt{[n]} x)\), where \([n]\) denotes
the greatest integer less than or equal to \(n\), is \(2 \pi\), then
(A) \(1 \leq n<2\)
(B) \(1
Step-by-Step Solution
Verified Answer
(A) \( 1 \leq n < 2 \)
1Step 1: Understand the Given Function
The function is given as \( f(x) = \sin(\sqrt{[n]} \cdot x) \). The period of a function of the form \( \sin(kx) \) is \( \frac{2\pi}{k} \). Since the period is given as \( 2\pi \), we should equate and solve for \( k \) such that \( \frac{2\pi}{k} = 2\pi \).
2Step 2: Determine the Value of k
From \( \frac{2\pi}{k} = 2\pi \), solving gives \( k = 1 \). For the function \( \sin(\sqrt{[n]} \cdot x) \), \( k = \sqrt{[n]} \). Therefore, \( \sqrt{[n]} \) must equal 1.
3Step 3: Solve for [n]
Since \( \sqrt{[n]} = 1 \), squaring both sides gives \( [n] = 1^2 = 1 \). The greatest integer less than or equal to \( n \) is 1. Thus, \( n \) must be within the range such that \( [n] = 1 \).
4Step 4: Translate the Condition Into an Interval for n
The condition \( [n] = 1 \) means the greatest integer less than or equal to \( n \) is 1. This implies \( 1 \leq n < 2 \). This is because \( [n] = 1 \) includes values from 1 up to (but not including) 2.
Key Concepts
Greatest Integer FunctionSine FunctionInterval Notation
Greatest Integer Function
The greatest integer function, denoted as \([n]\), is also known as the floor function. It serves a unique purpose by finding the largest integer that is less than or equal to a given number \(n\). This means \([n]\) outputs a value where \(n\) falls between two integers, it "rounds down" to the nearest whole number.
- If \({n} = 1.9\), then the greatest integer function \([n] = 1\).
- If \({n} = 2\), then \([n] = 2\) because \({2}\) is an integer.
Sine Function
The sine function is one of the primary trigonometric functions and is periodic in nature. Typically represented as \(\sin (x)\), this function outputs a value between \(-1\) and \(1\) for any given \(x\). The standard period for \(\sin(x)\) is \(2\pi\), meaning it repeats its values over every interval of \(2\pi\). In our exercise, \(f(x) = \sin(\sqrt{[n]} \cdot x)\), we introduce a factor \(\sqrt{[n]}\) multiplying \(x\). The periodicity formula for \(\sin(kx)\) is given by \(\frac{2\pi}{k}\). Here, by setting the period to \(2\pi\), we solve the equation \(\frac{2\pi}{k} = 2\pi\) which gives \(k = 1\). Thus, \(\sqrt{[n]}\) must equal 1. Understanding the sine function's periodic nature helps solve problems involving periodicity where transformations or manipulated terms alter its standard period. This knowledge is essential for mathematical studies involving oscillations, waves, or any phenomena with cyclical patterns.
Interval Notation
Interval notation is a concise way of writing subsets of the real number line. It's particularly useful for expressing solutions to inequalities and ranges of values. An interval is denoted by brackets, with each bracket's type signaling inclusion or exclusion of endpoints:
- Square brackets \([\text{ and } ]\)mean the number is included.
- Round parentheses \( (\text{ and } )\) signal that the endpoint is not included.
Other exercises in this chapter
Problem 129
Let \(f(x)=\frac{9^{x}}{9^{x}+3}\). Then, (A) \(f(x)+f(1-x)=1\) (B) \(f(x)+f(1-x)=-1\) (C) \(f\left(\frac{1}{1996}\right)+f\left(\frac{2}{1996}\right)\) \(+f\le
View solution Problem 130
Let \(n\) be a positive integer with \(f(n)=1 !+2 !+3 !+\) \(\ldots+n !\) and \(P(x)\) and \(Q(x)\) be polynomials in \(x\) such that \(f(n+2)=P(n) f(n+1)+Q(n)
View solution Problem 134
The period of the function \(f(x)=3 x+3-[3 x+3]+\) \(\sin \frac{\pi x}{2}\), where \([x]\) denotes the greatest integer \(\leq x\), is (A) 4 (B) 1 (C) 2 (D) Non
View solution Problem 135
\(\pi\) is the period of the function (A) \(|\sin x|+|\cos x|\) (B) \(\sin ^{4} x+\cos ^{4} x\) (C) \(\sin (\sin x)+\sin (\cos x)\) (D) \(\frac{1+2 \cos x}{\sin
View solution