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Range of f(x) = arcsin(x) - arccos(x):
A[-π/2, π/2]
B[-3π/2, π/2]
C[-π, 0]
D[-π/2, π/2] (from -π/2 - π = -3π/2? no — let me compute: at x = -1: arcsin(-1) - arccos(-1) = -π/2 - π = -3π/2; at x = 1: π/2 - 0 = π/2)
Answer & Solution
Correct answer: B. [-3π/2, π/2]
f(x) = arcsin(x) - arccos(x) = arcsin(x) - (π/2 - arcsin(x)) = 2 arcsin(x) - π/2. Range of 2 arcsin(x): [-π, π]. So f range: [-π - π/2, π - π/2] = [-3π/2, π/2].
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