F(x)=|x| is a decreasing function that is not one-one. For x=-1 and x =1 ,f(x)=1 . So it is decreasing function but it is not one-one(since two domain values have same value in codomain)
Incorrect!$f(x) = | x| $ is not a decreasing function with domain $A$ and co-domain $B$ being subsets of $R$
proof
To be decreasing function it has to satisfy $$\forall x\in A ~~\forall y \in A ( x < y \to f(x) \geq f(y))$$
But for $x=1$ and $y=2$, $f(x)=|1|=1$ and $f(y) = |2|=2$
Here $x<y$ but $f(x)\ngeq f(y)$
$\therefore$ By proof by counter example, f(x)=|x| from $R \to R$ is not a decreasing function