Q. 205.0( 1 Vote )

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Answer :

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.

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