f: R+→R defined by f (x)=2x, x∈ (0,1), f (x) = 3x, x∈ [1,∞) is one -one, onto neither one-one nor onto  one-one, not onto onto

The function f: R+→R defined by f (x)=2x, x∈ (0,1), f (x) = 3x, x∈ [1,∞) is one-one but not onto.

Here's why:

1. One-one (Injective): A function is said to be one-one (or injective), if the images of distinct elements of the domain are distinct, i.e., for every x1, x2 in the domain, if x1 ≠ x2, then f(x1) ≠ f(x2). In this case, for any two distinct x in the intervals (0,1) and [1,∞), their images under f are distinct. Hence, the function is one-one.

2. Onto (Surjective): A function is said to be onto (or surjective), if every element of the co-domain has a pre-image in the domain, i.e., for every y in the co-domain, there exists an x in the domain such that f(x) = y. In this case, for y < 2 or y > 3, there doesn't exist an x in the domain such that f(x) = y. Hence, the function is not onto.