Answer :

We have,

f (x) = cosx, ∀ x ∈ R

In order to prove that f is one-one, it is sufficient to prove that f(x_{1})=f(x_{2}) ⇒ x_{1}=x_{2}∀ x_{1}, x_{2} ∈ A .

Let x_{1} = 0 and x_{2} = 2π are two different elements in R.

Now,

f(x_{1}) = f(0) = cos0 = 1

f(x_{2}) = f(2π) = cos2π = 1

we observe that f(x_{1})=f(x_{2}) but x_{1} ≠ 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.

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