The function f : [−1/2, 1/2, 1/2]→[−π/2, π/2] , defined by

$f(x)=\sin ^{-1}\left(3 x-4 x^{3}\right)$

$\Rightarrow f(x)=3 \sin ^{-1} x$

Injectivity:

Let $x$ and $y$ be two elements in the domain $\left[\frac{-1}{2}, \frac{1}{2}\right]$, such that

$f(x)=f(y)$

$\Rightarrow 3 \sin ^{-1} x=3 \sin ^{-1} y$

$\Rightarrow \sin ^{-1} x=\sin ^{-1} y$

$\Rightarrow x=y$

So. $f$ is one-one.

Surjectivity:
Let y be any element in the co-domain, such that

$f(x)=y$

$\Rightarrow 3 \sin ^{-1}(x)=y$

$\Rightarrow \sin ^{-1}(x)=\frac{y}{3}$

$\Rightarrow x=\sin \frac{y}{3} \in\left[\frac{-1}{2}, \frac{1}{2}\right]$

$\Rightarrow f$ is onto.

$\Rightarrow f$ is a bijection.

So, the answer is (a).