Answer

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**Hint:**To solve this question, we have to remember that a function $f:A \to B$ is a bijection if it is one-one as well as onto (surjection).

(i) one-one i.e. $f\left( x \right) = f\left( y \right) \Rightarrow x = y\forall x,y \in A$

(ii) onto i.e. for all $y \in B$, there exists $X \in A$ such that $f\left( x \right) = y$

**Complete step-by-step answer:**

Given that,

$f:R \to R,f\left( x \right) = \cos x$

We have to show that the given function is neither one-one nor onto (surjection).

So,

First, let us check for one-one.

Let x and y be two arbitrary elements of R (domain of f) such that $f\left( x \right) = f\left( y \right)$

Then,

$ \Rightarrow \cos x = \cos y$

We know that, $f\left( 0 \right) = \cos 0 = 1$ and $f\left( {2\pi } \right) = \cos 2\pi = 1$,

i.e. $f\left( 0 \right) = f\left( {2\pi } \right)$, but $0 \ne 2\pi $

hence, different elements in R may have the same image. So, it is not a one-one function.

Now, for surjection (onto).

Let y be an arbitrary element of R.

Then,

$ \Rightarrow f\left( x \right) = y$

Since, the value of cos x lies between -1 and 1, it follows that the range of $f\left( x \right)$ is not equal to its co-domain.

So, f is not a surjection.

Hence, we can say that the mapping $f:R \to R,f\left( x \right) = \cos x$ is neither one-one nor surjective.

**Note:**In this type of questions, we have to observe the properties of the given function such as its domain, co-domain and range. Using these properties, we can easily identify that either the given function is an injection, surjective or bijection. Through this, we will get the answer.

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