Let P={10,20,30} and Q={5,10,15,20}. Which one of the following functions is one – one and not onto?

The correct option is:
f = {(10, 5), (20, 10), (30, 15)}

Let us understand why this function satisfies the criteria of being one-to-one (injective) and not onto (surjective) from set $P=\{10,20,30\}$ to set $Q=\{5,10,15,20\}$.

1. Definition of a Function:
For a relation from $P$ to $Q$ to be a function, every element in the domain $P$ must map to exactly one unique element in the codomain $Q$.
In the option $f=\left\{\right(10,5),(20,10),(30,15\left)\right\}$:
- The element 10 maps to 5.
- The element 20 maps to 10.
- The element 30 maps to 15.
Since every element in the domain $P=\{10,20,30\}$ has exactly one image in $Q$, $f$ is a valid function.

2. Checking if the Function is One-to-One (Injective):
A function is one-to-one if distinct elements in the domain map to distinct elements in the codomain. That is, if $a\ne$b , then $f\left(a\right)\ne f\left(b\right)$.
Looking at the mappings of $f$:
- $f\left(10\right)=5$
- $f\left(20\right)=10$
- $f\left(30\right)=15$
All the outputs (5, 10, and 15) are distinct. Therefore, the function is one-to-one.

3. Checking if the Function is Onto (Surjective):
A function is onto if the range of the function is equal to its codomain $Q$. In other words, every element in $Q$ must have at least one pre-image in $P$.
The codomain is $Q=\{5,10,15,20\}$.
The range (set of actual outputs) is $\{5,10,15\}$.
Notice that the element 20 in the codomain $Q$ does not have any corresponding pre-image in the domain $P$ (no element in $P$ maps to 20).
Since the range is not equal to the codomain, the function is not onto.

Conclusion:
Since the function $f=\left\{\right(10,5),(20,10),(30,15\left)\right\}$ is both one-to-one and not onto, it perfectly satisfies all criteria of the question.