The function f : [-$\frac{1}{2}$, $\frac{1}{2}$, $\frac{1}{2}$] $\to$ [-$\pi$ /2,$\pi$/2], defined by f (x) = sin^-1 (3x - 4x^3), is

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f\left( x \right) = \sin^{- 1} \left( 3x - 4 x^3 \right) \Rightarrow f\left( x \right) = 3 \sin^{- 1} x Injectivity: Let x and y be two elements in the domain \left[ - $\frac{1}{2}$, $\frac{1}{2}$ \right] , such that f\left( x \right) = f\left( y \right) \Rightarrow 3 \sin^{- 1} x = 3 \sin^{- 1} y \Rightarrow \sin^{- 1} x = \sin^{- 1} y \Rightarrow x = y So, f is one-one. Surjectivity: Let y be any element in the co-domain, such that f\left( x \right) = y \Rightarrow 3 \sin^{- 1} \left( x \right) = y \Rightarrow \sin^{- 1} \left( x \right) = y/3 \Rightarrow x = \siny/3 \in \left[ - $\frac{1}{2}$, $\frac{1}{2}$ \right] \Rightarrow f is onto. \Rightarrow f is a bijection. So, the answer is (a).

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