Q62 / 180 · COMEDK 2016 · Mathematics
The function $f: X \rightarrow Y$ defined by $f(x)=\sin x$ is one-one but not onto if $X$ and $Y$ are respectively, is equal to
- A$\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ and $[-1,1]$
- Correct answerB$\left[0, \frac{\pi}{2}\right]$ and $[-, 1,1]$
- C$[0, \pi]$ and $[0,1]$
- D$R$ and $R$
Solution
We have,
$f: X \rightarrow Y$
$X:$ domain and $Y$ : codomain
In one-one function,
$x_{1} \neq x_{2}$ $f\left(x_{1}\right) \neq f\left(x_{2}\right)$
In onto function,
Range $=$ domain .
$f(x)=\sin x$
So, if $X=\left[0, \frac{\pi}{2}\right]$ and $Y=[-1,1]$
For each value of $x \in X$ and range $=[0,1]$, there exists a unique value for $y \in Y$
Hence, function is one-one but not onto.
Hence, the correct answer is option: 'B'.