Question

The asymptotes of q(x) = 1/2 csc 2x are

a. Πn/2 + Πn/4 radians

b. Πn/2 - Πn/4 radians

c. Πn/4 radians

d. Πn/2 radians


https://courseplayer.avalearning.com/NeutralLMS/coursefiles/Questions/23819/L15%2016.png

I think it's d but I'd rather have a second opinion.

Answer 1

is that csc squared or csc of 2*x?

\[q(x)=\frac{ 1 }{ 2 }*\csc(2x)\]
\[q(x)=\frac{ 1 }{ 2 }*\csc^2(x)\]

Answer 2

its csc(2x)

Answer 3

ok, the csc(x) is the same as 1/sin(x), so the function becomes
\[\large q(x)=\frac{ 1 }{ 2 }*\frac{ 1 }{ \sin(2x) }\]

Answer 4

The vertical asymptotes are where the function is undefined. if you get a zero in the denominator, then q(x) is undefined..

Answer 5

so where does sin(2x) = 0 ?

recall the double angle thing from trig, sin(2x)=2*sin(x)*cos(x)

\[q(x)=\frac{ 1 }{ 2 }*\frac{ 1 }{ 2\sin(x)\cos(x) }\]

Answer 6

So q(x) will be undefined whenever either sin(x) or cos(x) equals zero.

sin(x)=0
or
cos(x)=0

Answer 7

@e28195 you are right, it is d

Answer 8

sin(x)=0, for x=0 and x=pi

cos(x)=0, for x=pi/2, and x=3pi/2

so combining those, you get every half pie increase starting at 0.

Answer 9

\[\large \frac{ \pi }{ 2 }*n~~~~~~for,~~~n=0,1,2,3...\]

Answer 10

@DanJS ahh ty so much

Answer 11

welcome