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Question
The function f : [-1/2, 1/2, 1/2] → [-π /2,π/2], defined by f (x) = `sin^-1` (3x - `4x^3`), is
Options
bijection
injection but not a surjection
surjection but not an injection
neither an injection nor a surjection
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Solution
\[f\left( x \right) = \sin^{- 1} \left( 3x - 4 x^3 \right)\]
\[ \Rightarrow f\left( x \right) = 3 \sin^{- 1} x\]
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
\[f\left( x \right) = y\]
\[ \Rightarrow 3 \sin^{- 1} \left( x \right) = y\]
\[ \Rightarrow \sin^{- 1} \left( x \right) = \frac{y}{3}\]
\[ \Rightarrow x = \sin\frac{y}{3} \in \left[ \frac{- 1}{2}, \frac{1}{2} \right]\]
\[\Rightarrow\] f is onto.
\[\Rightarrow\] f is a bijection.
So, the answer is (a).
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