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Let f: R -> R be defined by f(x) = 2x + sin x for all real x. Then f is:

1. one to one and onto
2. one to one but not onto
3. onto but not one to one
4. neither one to one nor onto

Correct answer: one to one and onto

Solution

f'(x) = 2 + cos x. Since cos x is in [-1, 1], f'(x) is in [1, 3] > 0 for all x, so f is strictly increasing and hence one-to-one (injective). As x -> +infinity, f -> +infinity, and as x -> -infinity, f -> -infinity; being continuous and strictly increasing, the range is all of R, so f is onto. Therefore f is both one-to-one and onto (a bijection).

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