Let f:[12,1]→[1,1] is given by f(x)=4x3−3x then f−1(x) is given by

f(x)=4x3-3x x=cosθ⇒θ=cos-1x f(x)=y⇒x=f-1(x) cos3θ=y⇒θ=13cos-1y⇒x=cos(13cos-1y) f-1(x)=cos(13cos-1x)

Let f:[12,1]→[1,1] is given by f(x)=4x3−3x then f−1(x) is given by

f(x)=4x3-3x x=cosθ⇒θ=cos-1x f(x)=y⇒x=f-1(x) cos3θ=y⇒θ=13cos-1y⇒x=cos(13cos-1y) f-1(x)=cos(13cos-1x)

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Let $f : [\frac{1}{2}, 1] \to [1, 1]$ is given by $f (x) = 4 x^{3} - 3 x$ then $f^{- 1} (x)$ is given by

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$\cos [\frac{1}{3} \cos^{- 1} x]$

$3 \cos [\sin^{- 1} x]$

$3 \sin^{- 1} [\cos x]$

$\sin [\frac{1}{3} \cos^{- 1} x]$

answer is A.

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Detailed Solution

$f (x) = 4 x^{3} - 3 x x = \cos θ \Rightarrow θ = \cos^{- 1} x f (x) = y \Rightarrow x = f^{- 1} (x) \cos 3 θ = y \Rightarrow θ = \frac{1}{3} \cos^{- 1} y \Rightarrow x = \cos (\frac{1}{3} \cos^{- 1} y) f^{- 1} (x) = \cos (\frac{1}{3} \cos^{- 1} x)$