Question

Which function's graph has asymptotes located at the values x=pi/2+-npi?

1. y = sin x
2. y = sec x
3. y = cot x
4. y = tan x 

A) 2 only
B) 1 only
C) 1 and 3 only
D) 2 and 4 only

Answer 1

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 2

Option 2 and 4

Explanation:

Given : Function's graph has asymptotes located at the values
`x=(\pi)/(2)\pm n\pi`

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,

`y=\cot x=(\cos x)/(\sin x)`

When we put Denominator = 0

`\sin x=0`

Value of x lie between
`(0,n\pi)`

So, Option 3 is not true.

In Option 2,

`y=\sec x=(1)/(\cos x)`

When we put Denominator = 0

`\cos x=0`

When cos x=0 the values of x is

`x=\pm(\pi)/(2),\pm(3\pi)/(2),\pm(5\pi)/(2),.....`

`x=(\pi)/(2)\pm n\pi`

Therefore, The graph of y= sec x has asymptote located at the values
`x=(\pi)/(2)\pm n\pi`

So, Option 2 is correct.

In Option 4,

`y=\tan x=(sin x)/(\cos x)`

When we put Denominator = 0

`\cos x=0`

When cos x=0 the values of x is

`x=\pm(\pi)/(2),\pm(3\pi)/(2),\pm(5\pi)/(2),.....`

`x=(\pi)/(2)\pm n\pi`

Therefore, The graph of y= tan x has asymptote located at the values
`x=(\pi)/(2)\pm n\pi`

So, Option 4 is correct.

Therefore, Option 2 and 4 are correct.

Hence Option A is correct - 2 only