Question

The  function f : [-1/2, 1/2, 1/2] → [-π /2,π/2], defined by f (x) = `sin^-1` (3x - `4x^3`), is

 

पर्याय

• bijection

• injection but not a surjection

• surjection but not an injection

• neither an injection nor a surjection

Answer 1

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

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).