We have,

f (x) = cosx, ∀ x ∈ R

In order to prove that f is one-one, it is sufficient to prove that f(x₁)=f(x₂) ⇒ x₁=x₂∀ x₁, x₂ ∈ A .

Let x₁ = 0 and x₂ = 2π are two different elements in R.

Now,

f(x₁) = f(0) = cos0 = 1

f(x₂) = f(2π) = cos2π = 1

we observe that f(x₁)=f(x₂) but x₁ ≠ x_2.

This shows that different element in R may have same image.

Thus, f(x) is not one-one.

We know that cosx lies between -1 and 1.

So, the range of f is [-1,1] which is not equal to its co-domain.

i.e., range of f ≠ R (co-domain)

In other words, range of f is less than co-domain, i.e there are elements in co-domain which does not have any pre-image in domain.

so, f is not onto.

Hence, f is neither one-one nor onto.