Answered

Which of the following is an asymptote of [tex]$y=\sec (x)$[/tex]?

A. [tex]$x=-2 \pi$[/tex]
B. [tex]$x=-\frac{\pi}{6}$[/tex]
C. [tex][tex]$x=\pi$[/tex][/tex]
D. [tex]$x=\frac{3 \pi}{2}$[/tex]



Answer :

To determine which of the given [tex]$x$[/tex] values is an asymptote of the function [tex]\( y = \sec(x) \)[/tex], we first need to understand the nature of the secant function. The function [tex]\( y = \sec(x) \)[/tex] is defined as the reciprocal of the cosine function:

[tex]\[ y = \sec(x) = \frac{1}{\cos(x)} \][/tex]

A vertical asymptote for [tex]\( \sec(x) \)[/tex] occurs where the cosine function is zero because the secant function will then approach infinity. The cosine function [tex]\( \cos(x) \)[/tex] equals zero at the points:

[tex]\[ x = \frac{(2k+1)\pi}{2} \][/tex]

where [tex]\( k \)[/tex] is any integer. These points are where vertical asymptotes of [tex]\( \sec(x) \)[/tex] occur.

Next, we will check each of the given [tex]\( x \)[/tex] values to see if it matches this form.

1. [tex]\( x = -2\pi \)[/tex]

[tex]\[ -2\pi = \frac{(2k+1)\pi}{2} \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ -2 = \frac{2k+1}{2} \][/tex]
[tex]\[ -4 = 2k + 1 \][/tex]
[tex]\[ 2k = -5 \][/tex]
[tex]\[ k = -\frac{5}{2} \][/tex]
Since [tex]\( k \)[/tex] must be an integer, [tex]\( x = -2\pi \)[/tex] is not a vertical asymptote.

2. [tex]\( x = -\frac{\pi}{6} \)[/tex]

[tex]\[ -\frac{\pi}{6} = \frac{(2k+1)\pi}{2} \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ -\frac{1}{6} = \frac{2k+1}{2} \][/tex]
[tex]\[ -\frac{1}{6} = 2k + 1 \][/tex]
[tex]\[ -\frac{1}{3} = 2k + 1 \][/tex]
[tex]\[ 2k = -\frac{1}{3} - 1 \][/tex]
[tex]\[ 2k = -\frac{4}{3} \][/tex]
[tex]\[ k = -\frac{2}{3} \][/tex]
Since [tex]\( k \)[/tex] must be an integer, [tex]\( x = -\frac{\pi}{6} \)[/tex] is not a vertical asymptote.

3. [tex]\( x = \pi \)[/tex]

[tex]\[ \pi = \frac{(2k+1)\pi}{2} \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ 1 = \frac{2k+1}{2} \][/tex]
[tex]\[ 2 = 2k + 1 \][/tex]
[tex]\[ 2k = 1 \][/tex]
[tex]\[ k = \frac{1}{2} \][/tex]
Since [tex]\( k \)[/tex] must be an integer, [tex]\( x = \pi \)[/tex] is not a vertical asymptote.

4. [tex]\( x = \frac{3\pi}{2} \)[/tex]

[tex]\[ \frac{3\pi}{2} = \frac{(2k+1)\pi}{2} \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ 3 = 2k + 1 \][/tex]
[tex]\[ 2k = 2 \][/tex]
[tex]\[ k = 1 \][/tex]
Since [tex]\( k \)[/tex] is an integer, [tex]\( x = \frac{3\pi}{2} \)[/tex] is a vertical asymptote.

Therefore, the asymptote of [tex]\( y = \sec(x) \)[/tex] among the given options is:

[tex]\[ x = \frac{3\pi}{2} \][/tex]