Saved Bookmarks
| 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].` |
|