Given function, $f(x)=\cos x, \forall x \in R$

Now, $$f\left(\frac{\pi}{2}\right)=\cos \frac{\pi}{2}=0$$

$$\begin{array}{lc} \Rightarrow & f\left(\frac{-\pi}{2}\right)=\cos \frac{\pi}{2}=0 \\ \Rightarrow & f\left(\frac{\pi}{2}\right)=f\left(\frac{-\pi}{2}\right) \\ \text { But } & \frac{\pi}{2} \neq \frac{-\pi}{2} \end{array}$$

So, $f(x)$ is not one-one.

Now, $f(x)=\cos x, \forall x \in R$ is not onto as there is no pre-image for any real number. Which does not belonging to the intervals $[-1,1]$, the range of $\cos x$.

Given that,

$$\begin{aligned} X & =\{1,2,3\} \text { and } Y=\{4,5\} \\ X \times Y & =\{(1,4),(1,5),(2,4),(2,5),(3,4),(3,5)\} \end{aligned}$$

(i) $f=\{(1,4),(1,5),(2,4),(3,5)\}$

$f$ is not a function because $f$ has not unique image.

(ii) $g=\{(1,4),(2,4),(3,4)\}$

Since, $g$ is a function as each element of the domain has unique image.

(iii) $h=\{(1,4),(2,5),(3,5)\}$

It is clear that $h$ is a function.

(iv) $k=\{(1,4),(2,5)\}$

$k$ is not a function as 3 has not any image under the mapping.

Given that,

$f: A \rightarrow B$ and $g: B \rightarrow A$ satisfy $g \circ f=I_A$

$$\begin{aligned} \because\quad g \circ f & =I_A \\ \Rightarrow\quad\operatorname{g\circ f}\left\{f\left(x_1\right)\right\} & =g \circ f\left\{f\left(x_2\right)\right\} \\ \Rightarrow\quad g\left(x_1\right) & =g\left(x_2\right) \quad \left[\because g \circ f=I_A\right]\\ \therefore\quad x_1 & =x_2 \end{aligned}$$

Hence, f is one-one and g is onto.

$$\begin{aligned} \text{Given function,}\quad & f(x)=\frac{1}{2-\cos x}, \forall x \in R \\ \text{Let,}\quad & y=\frac{1}{2-\cos x} \\ \Rightarrow\quad & 2 y-y \cos x=1 \\ \Rightarrow\quad & y \cos x=2 y-1 \\ \Rightarrow\quad & \cos x=\frac{2 y-1}{y}=2-\frac{1}{y} \Rightarrow \cos x=2-\frac{1}{y} \\ \Rightarrow\quad & -1 \leq \cos x \leq 1 \quad \Rightarrow \quad-1 \leq 2-\frac{1}{y} \leq 1 \\ \Rightarrow\quad & -3 \leq-\frac{1}{y} \leq-1 \quad \Rightarrow \quad 1 \leq \frac{1}{y} \leq 3 \\ \Rightarrow\quad & \frac{1}{3} \leq \frac{1}{y} \leq 1 \end{aligned}$$

So, $y$ range is $\left[\frac{1}{3}, 1\right]$.

Given that, $\forall a, b \in Z, a R b$ if and only if $a-b$ is divisible by $n$.

Now,

I. Reflexive

$a R a \Rightarrow(a-a)$ is divisible by $n$, which is true for any integer $a$ as ' $O$ ' is divisible by $n$. Hence, $R$ is reflexive.

II. Symmetric

$$\begin{array}{ll} & a R b \\ \Rightarrow & a-b \text { is divisible by } n . \\ \Rightarrow & -b+a \text { is divisible by } n \\ \Rightarrow & -(b-a) \text { is divisible by } n \\ \Rightarrow & (b-a) \text { is divisible by } n . \\ \Rightarrow & b R a \end{array}$$

Hence, R is symmetric.

III. Transitive

Let $a R b$ and $b R c$

$\Rightarrow\quad (a-b)$ is divisible by $n$ and $(b-c)$ is divisible by $n$

$\Rightarrow\quad(a-b)+(b-c)$ is divisibly by $n$

$\Rightarrow\quad(a-c)$ is divisible by $n$

$\Rightarrow\quad aRc$

Hence, R is transitive.

So, R is an equivalence relation.