Question

find range and domain of the function?
f(x)=1/(2-sin3x)

Answer 1

For domain:

Denominator must not be equal to 0,

So, 2 - sin3x not=0

sin3x not=2, which is never possible as sin function varies from -1 to 1 only.

So, it can take any value of x.

Therefore, domain is x€R(x belongs to R).

Now, range:

There are two methods for range, one is shortcut and other is long including concepts.

First, long method:

Write y=1/2-sin3x

So, 2y-ysin3x=1

(2y-1)/ y=sin3x

Therefore, x=1/3sin^-1(2–1/y)

Now,just calculating domain of it will give u range of f(x).

So, -1 <= 2–1/y <=1

-3<=-1/y<=-1

1<=1/y<=3

1/3<=y<=1.

Therefore, range is y€[1/3,1].

Shortcut:

As, we know greaest and least value of sin is 1 & -1. So, put them in function , you will get extereme values of ur range