Mark (√) against the correct answer in the following: $\mathrm{f}:\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \rightarrow[-1,1]: \mathrm{f}(\mathrm{x})=\sin \mathrm{x}$ is A. one - one and into B. one - one and onto C. many - one and into D. many - one and onto

Home » Mark (√) against the correct answer in the following: is A. one – one and into B. one – one and onto C. many – one and into D. many – one and onto

Mark (√) against the correct answer in the following: $\mathrm{f}:\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \rightarrow[-1,1]: \mathrm{f}(\mathrm{x})=\sin \mathrm{x}$ is
A. one – one and into
B. one – one and onto
C. many – one and into
D. many – one and onto

CBSE | Class 12 | Functions | Maths | RS Aggarwal

Mark (√) against the correct answer in the following: $\mathrm{f}:\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \rightarrow[-1,1]: \mathrm{f}(\mathrm{x})=\sin \mathrm{x}$ is
A. one – one and into
B. one – one and onto
C. many – one and into
D. many – one and onto

Solution:

Graph
Option (B) is correct.
One-one Function

According to the graph for $\sin (\mathrm{x})$ , for given range of $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right], \mathrm{f}(\mathrm{x})$ is not repeating its value.
Thus, its one-one.
Onto function
Range of the function $\mathrm{f}(\mathrm{x})$ is also the co-domain of the function, Therefore it is onto.
Hence, $f:\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \rightarrow[-1,1]: \mathrm{f}(\mathrm{x})=\sin (\mathrm{x})$ is one-one onto.