Question

Let $f$ be a function from a set $X$ to a set $Y$. Consider the following statements:
$P:$ For each x $\in$ X t,here exists uniqueX $\in$ Ysuch that $f$ (x)
$Q:$ For each y $\in$ Y ,there exists x $\in$ X such that $f$ (x) = y .
$R:$ There exist x${}_1$,x${}_2$ $\in$ X such that x${}_1$ $\ne$ x${}_2$ and $f$ (x${}_1$) = $f$ (x${}_2$).
The negation of the statement "$f$ is one-to-one and onto " is

• A. P or not R
• B. R or not P
• C. R or not Q
• D. P and not R
• E. R and not Q

Answer 1

Answer: (C)

We know that,
(i) A function $f:X\to Y$ is said to be one-one, if distinct elements of $X$ have distinct images in $Y$.
$f$ is one-one when $f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow x_1=x_2$
(ii) A function $f:X\to Y$ is said to be onto, if every element in $Y$ has atleast one pre-image in $X$. Thus, if $f$ is onto, then for each $y\in X\exists$ atleast one element $x\in X$ such that $y=f(x)$.
$\therefore$ Negative of the statement " $f$ is one-one and onto" iś
" $f$ is not one-to-one and onto".
which hold the logical statement, " $R$ or not $Q$ "