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Let f: (-1, 1) -> B be defined by f(x) = arctan(2x/(1 - x²)). For f to be both one-one (injective) and onto (surjective), what must the interval B be?

1. [-pi/2, pi/2]
2. (-pi/2, pi/2)
3. (0, pi/2)
4. [0, pi/2)

Correct answer: (-pi/2, pi/2)

Solution

Using the identity arctan(2x/(1 - x²)) = 2 arctan x for |x| < 1. As x ranges over (-1, 1), arctan x ranges over (-pi/4, pi/4), so 2 arctan x ranges over (-pi/2, pi/2). The function is continuous and strictly increasing (one-one), so it is onto exactly when B = (-pi/2, pi/2).

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