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Sagot :
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]
[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]
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