If A = [1, 2, 3}, B = {5, 6, 7} and f: A → B is a function such that f(x) = x + 4 then what type of function is f?

The correct option is one-one onto.

Let's analyze the given function step-by-step to understand why it is a one-one onto function.

1. Identify the Domain, Codomain, and Function Rule:

The domain of the function is set A:

$A=\{1,2,3\}$

The codomain of the function is set B:

$B=\{5,6,7\}$

The function rule is given by:

$f\left(x\right)=x+4$

where the input variable satisfies:

$x\in A$

2. Calculate the Images of Elements in Domain A:
We substitute each element of the domain A into the function rule:

For the element
$x=1$
we calculate the image as:
$f\left(1\right)=1+4=5$

For the element
$x=2$
we calculate the image as:
$f\left(2\right)=2+4=6$

For the element
$x=3$
we calculate the image as:
$f\left(3\right)=3+4=7$

3. Determine if the Function is One-One (Injective):
A function is defined as one-one if distinct elements in the domain map to distinct elements in the codomain. Mathematically, if:
$x_1\ne x_2$
then their images must also satisfy:
$f\left(x_1\right)\ne f\left(x_2\right)$
In this case, we have distinct elements mapping to distinct outputs (1 to 5, 2 to 6, and 3 to 7). Since no two different elements in set A map to the same element in set B, the function is one-one.

4. Determine if the Function is Onto (Surjective):
A function is defined as onto if every element in the codomain B has at least one corresponding pre-image in the domain A. This means the range of the function must equal the codomain.
From our calculations, the range (set of all actual outputs) is:
$\text{Range}=\{5,6,7\}$
Since the Range is exactly equal to the codomain B, which is:
$\{5,6,7\}$
every element in B has a pre-image. Therefore, the function is onto.

Conclusion:
Since the function
$f$
is both one-one and onto, it is classified as a one-one onto function (also known as a bijection).