cmi-entrance 2024 Q18

cmi-entrance · India · ugmath 1 marks Not Maths
Suppose $f$ is a function whose domain is $X$ and codomain is $Y$. It is given that $|X|>1$ and $|Y|>1$. No other information is known about $X$, $Y$ and $f$. Instruction: Write the number of a single correct option for the given statement S.
$\mathrm{S} =$ "For each $y$ in $Y$, there exists $x$ in $X$ such that $f(x) = y$." [1 point]
Options:
  1. S is always true.
  2. S is always false.
  3. S is true if and only if $f$ is one-to-one.
  4. If S is true then $f$ is one-to-one but the converse is false.
  5. If $f$ is one-to-one then S is true but the converse is false.
  6. S is true if and only if $f$ is onto.
  7. If S is true then $f$ is onto but the converse is false.
  8. If $f$ is onto then S is true but the converse is false.
  9. S is true if and only if $f$ is a constant function.
  10. If S is true then $f$ is a constant function but the converse is false.
  11. If $f$ is a constant function then S is true but the converse is false.
  12. None of the above.
Suppose $f$ is a function whose domain is $X$ and codomain is $Y$. It is given that $|X|>1$ and $|Y|>1$. No other information is known about $X$, $Y$ and $f$. Instruction: Write the number of a single correct option for the given statement S.

$\mathrm{S} =$ "For each $y$ in $Y$, there exists $x$ in $X$ such that $f(x) = y$." [1 point]

Options:
\begin{enumerate}
  \item S is always true.
  \item S is always false.
  \item S is true if and only if $f$ is one-to-one.
  \item If S is true then $f$ is one-to-one but the converse is false.
  \item If $f$ is one-to-one then S is true but the converse is false.
  \item S is true if and only if $f$ is onto.
  \item If S is true then $f$ is onto but the converse is false.
  \item If $f$ is onto then S is true but the converse is false.
  \item S is true if and only if $f$ is a constant function.
  \item If S is true then $f$ is a constant function but the converse is false.
  \item If $f$ is a constant function then S is true but the converse is false.
  \item None of the above.
\end{enumerate}