Question

what are the vertical asymptotes of y = cot x for -3π ≤ x ≤ 3π? only ±π, ±2π, ±3π; 0, ±π, ±2π, ±3π; ±π/2, ±3π/2, ±5π/2

Explanation:

Step1: Recall cotangent definition

Cotangent is defined as \(\cot x=\frac{\cos x}{\sin x}\). Vertical asymptotes occur where \(\sin x = 0\) (since division by zero is undefined).

Step2: Solve \(\sin x = 0\)

The solutions to \(\sin x=0\) are \(x = n\pi\), where \(n\) is an integer. But wait, no—wait, \(\cot x=\frac{\cos x}{\sin x}\), so when \(\sin x = 0\), but also, wait, no: actually, \(\sin x = 0\) at \(x = n\pi\), but let's check the domain \(-3\pi\leq x\leq3\pi\). Wait, no, wait a second—wait, \(\cot x\) has vertical asymptotes where \(\sin x = 0\), which is at \(x = n\pi\)? Wait, no, no: \(\tan x=\frac{\sin x}{\cos x}\) has asymptotes at \(\cos x = 0\) (i.e., \(x=\frac{\pi}{2}+n\pi\)), and \(\cot x=\frac{\cos x}{\sin x}\) has asymptotes where \(\sin x = 0\), which is \(x = n\pi\)? Wait, no, that's a mistake. Wait, \(\sin x = 0\) when \(x = n\pi\), \(n\in\mathbb{Z}\). But let's check the options. Wait, the third option has \(\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\pm\frac{5\pi}{2}\). Wait, no, I must have messed up. Wait, no: \(\cot x=\frac{\cos x}{\sin x}\), so vertical asymptotes when \(\sin x = 0\), so \(x = n\pi\). But the options: first option: \(\pm\pi,\pm2\pi,\pm3\pi\) (and maybe 0? Wait, first option says "Only \(\pm\pi,\pm2\pi,\pm3\pi\)"—no, wait the first option's text: "Only \(\pm\pi,\pm2\pi,\pm3\pi\)"? Wait, no, the second option has \(0,\pm\pi,\pm2\pi,\pm3\pi\). Wait, but let's list \(x = n\pi\) for \(n\) such that \(-3\pi\leq x\leq3\pi\). So \(n\) can be \(-3,-2,-1,0,1,2,3\), so \(x=-3\pi,-2\pi,-\pi,0,\pi,2\pi,3\pi\). But wait, the third option has \(\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\pm\frac{5\pi}{2}\). Wait, that's for \(\tan x\), which has asymptotes at \(\cos x = 0\) (i.e., \(x=\frac{\pi}{2}+n\pi\)). Wait, I see my mistake! Oh no, I confused \(\tan\) and \(\cot\). \(\cot x=\frac{\cos x}{\sin x}\), so vertical asymptotes when \(\sin x = 0\) (denominator zero), so \(x = n\pi\). But \(\tan x=\frac{\sin x}{\cos x}\) has asymptotes when \(\cos x = 0\) (i.e., \(x=\frac{\pi}{2}+n\pi\)). Wait, so I had it backwards. So \(\cot x\) asymptotes at \(x = n\pi\), \(n\in\mathbb{Z}\). Now, let's list \(x = n\pi\) in \(-3\pi\leq x\leq3\pi\):

For \(n=-3\): \(x=-3\pi\)

\(n=-2\): \(x=-2\pi\)

\(n=-1\): \(x=-\pi\)

\(n=0\): \(x=0\)

\(n=1\): \(x=\pi\)

\(n=2\): \(x=2\pi\)

\(n=3\): \(x=3\pi\)

Now check the options:

First option: "Only \(\pm\pi,\pm2\pi,\pm3\pi\)"—missing 0, so wrong.

Second option: "0,\pm\pi,\pm2\pi,\pm3\pi"—these are the \(x = n\pi\) for \(n=-3,-2,-1,0,1,2,3\) (since \(-3\pi\leq x\leq3\pi\), \(n=-3\) gives \(-3\pi\), \(n=-2\) gives \(-2\pi\), \(n=-1\) gives \(-\pi\), \(n=0\) gives 0, \(n=1\) gives \(\pi\), \(n=2\) gives \(2\pi\), \(n=3\) gives \(3\pi\)). Wait, but the second option is "0,\pm\pi,\pm2\pi,\pm3\pi"—wait, but \(-3\pi\) and \(3\pi\) are included? Wait, the second option says "0,\pm\pi,\pm2\pi,\pm3\pi"—so \(x = -3\pi,-2\pi,-\pi,0,\pi,2\pi,3\pi\)? Wait, no, the second option's text is "0, ±π, ±2π, ±3π"—so that's 0, -π, π, -2π, 2π, -3π, 3π? Wait, but the first option says "Only ±π, ±2π, ±3π"—so missing 0. The third option has \(\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\pm\frac{5\pi}{2}\), which are the asymptotes for \(\tan x\) (since \(\tan x\) has asymptotes at \(\frac{\pi}{2}+n\pi\)). So I must have made a mistake earlier. Wait, no: \(\cot x=\frac{\cos x}{\sin x}\), so vertical asymptotes when \(\sin x = 0\), i.e., \(x = n\pi\). So in the interval \(-3\pi\leq x\leq3\pi\), the values of \(x\) where \(\sin x = 0\) are \(x=-3\pi,-2\pi,-\pi,0,\pi,2\pi,3\pi\). Now check the options:

First option: "Only ±π, ±2π, ±3π"—missing -3π, 0, 3π? Wait, no, the first option's text is "Only ±π, ±2π, ±3π"—so that's -π, π, -2π, 2π, -3π, 3π? But missing 0. Second option: "0, ±π, ±2π, ±3π"—so that's 0, -π, π, -2π, 2π, -3π, 3π? Wait, but \(-3\pi\) is included? Wait, the domain is \(-3\pi\leq x\leq3\pi\), so \(x=-3\pi\) is included. Wait, but the second option's text is "0, ±π, ±2π, ±3π"—so that's 0, -π, π, -2π, 2π, -3π, 3π. But the first option says "Only ±π, ±2π, ±3π"—so missing 0. But wait, let's check the definition again. Wait, \(\cot x\) is undefined at \(x = n\pi\), \(n\in\mathbb{Z}\). So in the interval \(-3\pi\leq x\leq3\pi\), the vertical asymptotes are at \(x=-3\pi,-2\pi,-\pi,0,\pi,2\pi,3\pi\). Now look at the options:

First option: "Only ±π, ±2π, ±3π"—so x = -π, π, -2π, 2π, -3π, 3π (missing 0).

Second option: "0, ±π, ±2π, ±3π"—so x = 0, -π, π, -2π, 2π, -3π, 3π (includes 0 and the others).

Third option: asymptotes for \(\tan x\) (since \(\tan x\) has asymptotes at \(\frac{\pi}{2}+n\pi\), which are \(\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\pm\frac{5\pi}{2}\) in the interval? Wait, \(\frac{\pi}{2}+n\pi\) for \(n=-2,-1,0,1,2\): \(n=-2\): \(-\frac{3\pi}{2}\), \(n=-1\): \(-\frac{\pi}{2}\), \(n=0\): \(\frac{\pi}{2}\), \(n=1\): \(\frac{3\pi}{2}\), \(n=2\): \(\frac{5\pi}{2}\), and \(n=-3\): \(-\frac{5\pi}{2}\), \(n=3\): \(\frac{7\pi}{2}\) (but \(\frac{7\pi}{2}>3\pi\) since \(3\pi=\frac{6\pi}{2}\), so \(\frac{7\pi}{2}\) is outside). So in \(-3\pi\leq x\leq3\pi\) (i.e., \(-\frac{6\pi}{2}\leq x\leq\frac{6\pi}{2}\)), the asymptotes for \(\tan x\) are \(-\frac{5\pi}{2},-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2}\), which is the third option. But we're dealing with \(\cot x\), which has asymptotes at \(x = n\pi\). So where is the mistake?

Wait, no! I confused \(\cot x\) and \(\tan x\). \(\tan x=\frac{\sin x}{\cos x}\) has vertical asymptotes where \(\cos x = 0\) (i.e., \(x=\frac{\pi}{2}+n\pi\)), and \(\cot x=\frac{\cos x}{\sin x}\) has vertical asymptotes where \(\sin x = 0\) (i.e., \(x = n\pi\)). So the third option is for \(\tan x\), the second option is for \(\cot x\) (since \(\cot x\) has asymptotes at \(x = n\pi\)). Wait, but let's check the interval \(-3\pi\leq x\leq3\pi\). For \(x = n\pi\), \(n\) can be -3, -2, -1, 0, 1, 2, 3. So \(x=-3\pi,-2\pi,-\pi,0,\pi,2\pi,3\pi\). The second option is "0, ±π, ±2π, ±3π"—which includes 0, -π, π, -2π, 2π, -3π, 3π (since ±3π means -3π and 3π, ±2π means -2π and 2π, ±π means -π and π, plus 0). The first option is "Only ±π, ±2π, ±3π"—so missing 0. So the correct option should be the second one? Wait, no, wait the third option: let's check the values. \(\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\pm\frac{5\pi}{2}\): \(-\frac{5\pi}{2},-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2}\). These are the asymptotes for \(\tan x\), because \(\tan x\) has asymptotes at \(x=\frac{\pi}{2}+n\pi\), so for \(n=-2\): \(-\frac{3\pi}{2}\), \(n=-1\): \(-\frac{\pi}{2}\), \(n=0\): \(\frac{\pi}{2}\), \(n=1\): \(\frac{3\pi}{2}\), \(n=2\): \(\frac{5\pi}{2}\), \(n=-3\): \(-\frac{5\pi}{2}\) (since \(-3\pi\leq x\leq3\pi\), \(-\frac{5\pi}{2}\geq -3\pi\) (because \(-3\pi=-\frac{6\pi}{2}\), so \(-\frac{5\pi}{2}>-\frac{6\pi}{2}\)) and \(\frac{5\pi}{2}\leq3\pi\) (since \(3\pi=\frac{6\pi}{2}\), so \(\frac{5\pi}{2}<\frac{6\pi}{2}\)). So the third option is for \(\tan x\), but the question is about \(\cot x\). So \(\cot x\) has asymptotes at \(x = n\pi\), so in \(-3\pi\leq x\leq3\pi\), that's \(x=-3\pi,-2\pi,-\pi,0,\pi,2\pi,3\pi\). The second option is "0, ±π, ±2π, ±3π"—which is 0, -π, π, -2π, 2π, -3π, 3π. So that's correct. Wait, but the first option says "Only ±π, ±2π, ±3π"—so missing 0. So the correct answer is the second option.

Answer:

B. \(0, \pm\pi, \pm2\pi, \pm3\pi\)