1.

Find the domain and range of the function `f(x)=1/(2-sin3x)`.

Answer» `f(x) = 1/(2-sin3x)`
Here, `2-sin3x` can never be zero as `sin3x` will always less than `2`.
`:.` Domain of `f(x)` will be `x in R.`
Now, `f(x)` will be maximum when `2-sin3x` is minimum.
`2-sin3x` will be minimum when `sin3x = 1.`
`:. f(x)_max = 1/(2-1) = 1`
`f(x)` will be minimum when `2-sin3x` is maximum.
`2-sin3x` will be maximum when `sin3x = -1`.
`:. f(x)_min = 1/(2-(-1) = 1/3`
So, range of `f(x)` will be `[1/3,1].`


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