Given that f : [–1/2, 1/2] → [π/2, π/2] where f(x) = sin^–1 (3x – 4x³)

Put x = sinθ in f(x) = sin^–1 (3x – 4x³)

⇒ f(x=sinθ) = sin^–1 (3sinθ – 4sinθ³)

⇒ f(x) = sin^–1 (sin3θ)

⇒ f(x) = 3θ

⇒ f(x) = 3 sin^–1x

If f(x) = f(y)

Then

3 sin^–1x = 3 sin^–1y

⇒ x = y

So, f is one-one.

y = 3 sin^–1x

∵ x ϵ R also y ϵ R so f is onto.

Hence, f is bijection.