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Question: The function \(f:\left\lbrack - \frac{1}{2},\frac{1}{2} \right\rbrack \rightarrow \left\lbrack \frac...

The function f:[−12,12]→[π2,π2]f:\left\lbrack - \frac{1}{2},\frac{1}{2} \right\rbrack \rightarrow \left\lbrack \frac{\pi}{2},\frac{\pi}{2} \right\rbrack defined by

f(x)=sin⁡−1(3x−4x3)f(x) = \sin^{- 1}\left( 3x - 4x^{3} \right) is

A

Both one-one and onto

B

Neither one-one nor onto

C

Onto but not one- one

D

One- one but not onto

Answer

Onto but not one- one

Explanation

Solution

Since sin⁡−1(3x−4x3)=3sin⁡−1x∈[−π2,π2]\sin^{- 1}\left( 3x - 4x^{3} \right) = 3\sin^{- 1}x \in \left\lbrack - \frac{\pi}{2},\frac{\pi}{2} \right\rbracki.e.

sin⁡01x∈[−π6,π6]\sin^{01}x \in \left\lbrack - \frac{\pi}{6},\frac{\pi}{6} \right\rbrackor x∈[−12,12]x \in \left\lbrack - \frac{1}{2},\frac{1}{2} \right\rbrackso ffis onto. More ever the function y=sin⁡xy = \sin xis one-one on [−π2,π2]\left\lbrack - \frac{\pi}{2},\frac{\pi}{2} \right\rbrackso if f(x1)=f(x2)f\left( x_{1} \right) = f\left( x_{2} \right) then sin⁡−1(3x1−4x13)=sin⁡−1(3x2−4x23)\sin^{- 1}\left( 3x_{1} - 4x_{1}^{3} \right) = \sin^{- 1}\left( 3x_{2} - 4x_{2}^{3} \right) which implies that 3x1−4x13=3x2−4x13=sin⁡−1(3x2−4x23)3x_{1} - 4x_{1}^{3} = 3x_{2} - 4x_{1}^{3} = \sin^{- 1}\left( 3x_{2} - 4x_{2}^{3} \right) which implies that 3x1−4x13=3x2−4x23.3x_{1} - 4x_{1}^{3} = 3x_{2} - 4x_{2}^{3}. The real solution of the last equation is given by x1=x2x_{1} = x_{2}. Hence ffis one-one.