Respuesta :
For those of you with differently ordered answers, the correct answers according to Apex are as followed: y=sec x and y=tan x
Answer:
Option 2 and 4
Step-by-step explanation:
Given : Function's graph has asymptotes located at the values [tex]x=\frac{\pi}{2}\pm n\pi[/tex]
To find : Which function graph has asymptote given?
Solution :
An asymptote is a line or curve that approaches a given curve.
To find vertical asymptote the limit has to go to either ∞ or − ∞ , which happens when the denominator becomes zero.
The sine are defined for all real x, y is defined for all real x, so there are no vertical asymptotes.
So, Option 1 is not true.
In option 3,
[tex]y=\cot x=\frac{\cos x}{\sin x}[/tex]
When we put Denominator = 0
[tex]\sin x=0[/tex]
Value of x lie between [tex](0,n\pi)[/tex]
So, Option 3 is not true.
In Option 2,
[tex]y=\sec x=\frac{1}{\cos x}[/tex]
When we put Denominator = 0
[tex]\cos x=0[/tex]
When cos x=0 the values of x is
[tex]x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\pm\frac{5\pi}{2},.....[/tex]
[tex]x=\frac{\pi}{2}\pm n\pi[/tex]
Therefore, The graph of y= sec x has asymptote located at the values [tex]x=\frac{\pi}{2}\pm n\pi[/tex]
So, Option 2 is correct.
In Option 4,
[tex]y=\tan x=\frac{sin x}{\cos x}[/tex]
When we put Denominator = 0
[tex]\cos x=0[/tex]
When cos x=0 the values of x is
[tex]x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\pm\frac{5\pi}{2},.....[/tex]
[tex]x=\frac{\pi}{2}\pm n\pi[/tex]
Therefore, The graph of y= tan x has asymptote located at the values [tex]x=\frac{\pi}{2}\pm n\pi[/tex]
So, Option 4 is correct.
Therefore, Option 2 and 4 are correct.
Hence Option A is correct - 2 only
