Question

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

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

Answer:

Option 2 and 4

Step-by-step explanation:

Given : Function's graph has asymptotes located at the values $x=\frac{\pi}{2}\pm n\pi$

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,

$y=\cot x=\frac{\cos x}{\sin x}$

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,

$y=\sec x=\frac{1}{\cos x}$

When we put Denominator = 0

$\cos x=0$

When cos x=0 the values of x is

$x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\pm\frac{5\pi}{2},.....$

$x=\frac{\pi}{2}\pm n\pi$

Therefore, The graph of y= sec x has asymptote located at the values $x=\frac{\pi}{2}\pm n\pi$

So, Option 2 is correct.

In Option 4,

$y=\tan x=\frac{sin x}{\cos x}$

When we put Denominator = 0

$\cos x=0$

When cos x=0 the values of x is

$x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\pm\frac{5\pi}{2},.....$

$x=\frac{\pi}{2}\pm n\pi$

Therefore, The graph of y= tan x has asymptote located at the values $x=\frac{\pi}{2}\pm n\pi$

So, Option 4 is correct.

Therefore, Option 2 and 4 are correct.

Hence Option A is correct - 2 only