Question

Show that the Modulus Function f: R → R given by, is neither one-one nor onto, where is x, if x is positive or 0 andis − x, if x is negative.

Answer 1

Show that the Modulus Function f : R → R, given by f (x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is – x, if x is negative.

solution :- $\text{\bf{one - one function}}$ A function f : A --> B is said to be a one - one function or injective mapping, if different elements of A have different f images in B.

$\text{\bf{onto function}}$ If the function f :A → B is such that each element in B (codomain) is the f image of atleast one element in A. then we say that f is a function of A onto B.

given f : R → R, given by f (x) = | x|
We can see that f(-1) = |-1| = 1, f(1) = |1| = 1
⇒ f(-1) = f(1), but -1 ≠ 1.
therefore, f is not one-one.

Now, we consider -1 ∈ R
We know that f(x) = |x| is always positive.
Therefore, there doesn't exist any element x in domain R such that f(x) = |x| = -1
e.g., Range ∈ R⁺
hence, codomain ≠ range
therefore, f is not onto.

Therefore, modulus function is neither one-one nor onto.