What is an inverse of a function?

Reflexive :for (a, b) R (a, b)

-> ab– ab = 0 is divisible by 5.

So (a, b) R (a, b) " a, b Î Z

R is reflexive

Symmetric:

For (a, b) R (c, d)

If ad – bc is divisible by 5.

Then bc – ad is also divisible by 5.

-> (c, d) R (a, b) "a, b, c, dÎZ

R is symmetric

Transitive:

If (a, b) R (c, d) ->ad –bc divisi

$x={\displaystyle \sum _{n=0}^{\infty }co{s}^{2n}\theta =co{s}^0\theta +co{s}^2\theta +co{s}^4\theta +.....=1+co{s}^2\theta +co{s}^4\theta +.....}$

a = 1, r = cos2

$x=S_{\infty }=\frac{a}{1-r}=\frac{1}{1-co{s}^2\theta }=\frac{1}{si{n}^2\theta }$

Similarly, y = $\frac{1}{co{s}^2\theta }\Rightarrow \frac{1}{y}=co{s}^2\theta$ $\frac{}{{}^{}}\frac{}{}{}^{}$

$z=\frac{xy}{xy-1}\Rightarrow xyz-z=xy...........\left(i\right)$

Also, $\frac{1}{x}+\frac{1}{y}=1\Rightarrow x+y=xy..........\left(ii\right)$ $\frac{}{}\frac{}{}$

(i) & (ii) xyz = xy + z (x + y) z = xy + z

Total number of possible relation = $2^{n^2}=2^4=16$ ${}^{{}^{}}{}^{}$

Favourable relations = $?,\left\{\left(x,x\right)\right\},\left\{\left(y,y\right)\right\}$ $$

$\left\{\left(x,x\right),\left(y,y\right)\right\}$

$\left\{\left(x,x\right),\left(y,y\right),\left(x,y\right),\left(y,x\right)\right\}$

Probability = $\frac{5}{16}$

A = {1, 2, 3, 4}. As (4, 4) $\notin$  R so not reflexive

Also not transitive

$3-2^x-2^{1-x}\ge 0put{2}^x=t$

$3-t-\frac{2}{t}\ge 0\Rightarrow t^2-3t+2\le 0$

$\left(t-1\right)\left(t-2\right)\le 0\Rightarrow t\in \left[1,2\right]$