Class 11th

Clearly PM. PN $={\mathrm{O}\mathrm{P}}^2$ =OP2

i.e. $PM\cdot PN=16$ PM⋅PN=16

$\Delta APC,AC=AP=12$  

$\Delta ABP,tan60°=\frac{12}{AB}$           

$AB=\frac{12}{\sqrt[]{3}}=4\sqrt[]{3}$            

$Area=4\sqrt[]{3}.4\sqrt[]{6}=48\sqrt[]{2}$            

Using F = MA = $m\frac{V}{T}$ $\frac{}{}$

m = FTV¹

Load of mass will be equally distributed among the four colours so force on each columns will be 125 * 10³ N.

Cross section area of the column = $\pi \left[{\left(1\right)}^2-{\left(0.5\right)}^2\right]=2.355m^2$

Using young's modulus : $\epsilon =\frac{\sigma }{Y}=\frac{F}{AY}=\frac{125*10^3}{2.355*2*10^{11}}=2.65*10^{-7}$  

$S_1=1+2+3\dots \dots \dots .n$

$S_2=1+3+5\dots \dots \dots .n$

$S_3=1+4+7\dots \dots \dots .$

$\begin{array}{ll}\left(S_1+S_3\right)=\lambda S_2& S_1=\frac{n}{2}[2a+(n-1\left)1\right]\\ \frac{n}{2}[2a+(n-1\left)1\right]+\frac{n}{2}[2a+(n-1\left)3\right]& S_2=\frac{n}{2}[2a+(n-1\left)2\right]\\ =\lambda \frac{n}{2}[2a+(n-1\left)2\right]& S_3=\frac{n}{2}[2a+(n-1\left)3\right]\end{array}$

$\lambda =2$

$\sim (\sim p\vee \sim q)\to (\sim q\vee r)\Rightarrow p\wedge q\to \sim q\vee r$

$\mathrm{}\mathrm{T}\mathrm{F}\mathrm{T}\Rightarrow \mathrm{F}\to \mathrm{T}\to \mathrm{T}$

$\mathrm{F}\mathrm{T}\mathrm{T}\Rightarrow \mathrm{F}\to \mathrm{T}\to \mathrm{T}$

$\mathrm{T}\mathrm{T}\mathrm{T}\Rightarrow \mathrm{T}\to \mathrm{T}\to \mathrm{T}$

$\mathrm{T}\mathrm{T}\mathrm{F}\Rightarrow \mathrm{T}\to \mathrm{F}\to \mathrm{F}$

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

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

Let $z=x+yi\mathrm{}z-2\bar{z}=1\Rightarrow x=-1\mathrm{}y=0$

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

 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 symmetric

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

$\cancel{\Rightarrow }\left|x\right|+\left|z\right|\le \frac{1}{2}$            

$\therefore$ R is not transitive.