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Let f: (−1, 1) → B be given by f(x) = tan⁻¹((−1 + 2x)/(1 − x²)). For f to be both injective and surjective, which interval must B be?

1. (0, π/2)
2. [0, π/2)
3. [−π/2, π/2]
4. (−π/2, π/2)

Correct answer: (−π/2, π/2)

Solution

The function f(x) = tan⁻¹((−1 + 2x)/(1 − x²)) is defined for x in the interval (−1, 1) and maps to the interval (−π/2, π/2) because the arctangent function outputs values in this range. For f to be both injective (one-to-one) and surjective (onto), B must cover the entire range of the function, which is (−π/2, π/2).

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