Relations and Functions

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

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

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

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

${\mathrm{l}\mathrm{i}\mathrm{m}}_{x\to 9^{-}}?\left. [x^2+1\right]+\left. [x^2-1\right]+\left\{x\right\}$

$\Rightarrow \mathrm{}{\mathrm{l}\mathrm{i}\mathrm{m}}_{x\to 9^{-}}?2\left. [x^2\right]+\left\{x\right\}\Rightarrow {\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?2\left. [(9-x{)}^2\right]+\{9-h\}=160+1=161$

  $\left|x\right|+\left|y\right|\le \frac{1}{2}$            

 If x = y,   $2\left|x\right|\le \frac{1}{2},-\frac{1}{4}\le x\le \frac{1}{4}$

  $\therefore \left(x,x\right)\notin Rvx\in realno$          

    $\therefore$         R is not reflexive

$If\left|x\right|+\left|y\right|\le \frac{1}{2}\Rightarrow \left|y\right|+\left|x\right|\le \frac{1}{2}$            

  $\therefore \left(x,y\right)\in R\Rightarrow \left(y,x\right)\in R$          

$\therefore$ R is symmetri

? P? = 6!/2! = 360

for reflexive (x, x)
x³ - 3x²x + 3x³ = 0
. reflexive
For symmetric
(x, y)? R
x³ - 3x²y - xy² + 3y³ = 0
? (x - 3y) (x² - y²) = 0
For (y, x)
(y - 3x) (y² - x²) = 0
? (3x - y) (x² - y²) = 0
Not symmetric

Given f (1) = a = 3, and assuming the function form is f (x) = a?
So f (x) = 3?
∑? f (i) = 363
⇒ 3 + 3² + . + 3? = 363
This is a geometric progression. The sum is S? = a (r? -1)/ (r-1).
3 (3? -1)/ (3-1) = 363
3 (3? -1)/2 = 363
3? - 1 = 242
3? = 243
3? = 3? ⇒ n = 5