Let N denote the set of all non-negative integers and Z denote the set of all integers. The function f : Z → N given by f(x) = |x| is:
Let N denote the set of all non-negative integers and Z denote the set of all integers. The function f : Z → N given by f(x) = |x| is: Correct Answer Onto but not one-one
Concept:
One–One Function or Injective Function:
A function f: A → B is said to be a one–one function, if different elements in A have different images or associated with different elements in B i.e if
f (x1) = f (x2) ⇒ x1 = x2, ∀ x1, x2 ∈ A.
Into Function:
Any function f: A → B is said to be into function if there exist at least one element in B which does not has a pre-image in A, then the function f is said to be into function.
If Range of function f ⊂ Co-domain of function f, then f is into
Many-one Function:
Any function f: A → B is said to be many-one, if two (or more than two) distinct elements in A have same images in B.
Onto Function / Surjective Function :
Any function f: A → B is said to be onto if every element in B has atleast one pre-image in A.
If Range of function f = Co-domain of function f, then f is onto
Calculation:
Here, f(x) = |x|
f(x) = -x, when x < 0 and f(x) = x, when x > 0
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It cuts the line parallel to the x-axis at two points so, it's not one-one function.
Given function is linear and it is defined Z → N
Hence the range of the function is a natural number.
Therefore co-domain = range
So, the given function is onto.
Hence, option (2) is correct.