Q. The function $f:X \rightarrow Y$ defined by $f(x) = \sin x$ is one-one but not onto, if $X$ and $Y$ are respectively equal to
A) $R$ and $R$
B) $\lbrack 0,\ \pi\rbrack$ and $\lbrack 0,\ 1\rbrack$
C) $\left\lbrack 0,\ \frac{\pi}{2} \right\rbrack$ and $\lbrack - 1,\ 1\rbrack$
D) $\left\lbrack \frac{- \pi}{2},\frac{\pi}{2} \right\rbrack$ and $\lbrack - 1,\ 1\rbrack$
Explanation
(c)
Since, $f:X \rightarrow Y$ and $f(x) = \sin x$
Now, take option (c).
Domain$\ = \left\lbrack 0,\ \frac{\pi}{2} \right\rbrack$, Range$\ = \lbrack - 1,\ 1\rbrack$
For every value of $x$, we get unique value of $y$. But the value of $y$ in $\lbrack - 1,\ 0)$ does not have any preimage.
$\therefore$ Function is one-one but not onto.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.