Maths

$\left(a-1\right)x+0y+z=\alpha$        ……(1)

$x+\left(b-1\right)y+0z=\beta$                       ……(2)

$0x+y+\left(c-1\right)z=\gamma$                       ……(3)

$\left|\begin{array}{ccc}\hfill \left(a-1\right)\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill \left(b-1\right)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill \left(c-1\right)\hfill \end{array}\right|=0$           For no unique solution D = 0

$\left(a-1\right)\left(b-1\right)\left(c-1\right)+1=0$            

$\therefore a=2;b=2;c=0$            

Hence, |a + b + c| = 4

  $A\left(a\right)=2{\displaystyle \underset{0}{\overset{\sqrt[]{1-a}}{\int }}\left(\left(1-x^2\right)-a\right)dx=\frac{4}{3}{\left(1-a\right)}^{3/2}}$  

  $\therefore A\left(0\right)=\frac{4}{3}$            

and    ^{$A\left(\frac{1}{2}\right)=\frac{4}{3}{\left(\frac{1}{2}\right)}^{\frac{3}{2}}\Rightarrow \frac{A\left(0\right)}{A\left(\frac{1}{2}\right)}=2\sqrt[]{2}$}

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

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

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

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

$\lambda =2$

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

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

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

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

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

$\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}\right|=\sum {\left|\overrightarrow{a}\right|}^2+2\sum \overrightarrow{a}.\overrightarrow{b}$  

$=4+2\overrightarrow{d}.\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)$  (a, b, c are mutually ^ r)

Let   $\overrightarrow{d}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}+v\overrightarrow{c}$

Also   ${\lambda }^2+{\mu }^2+v^2=1=3co{s}^2\theta orcos\theta =\frac{1}{\sqrt[]{3}}$

$\therefore {\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}\right|}^2=4\pm 2.\frac{3}{\sqrt[]{3}}$        

$=4\pm 2\frac{3}{\sqrt[]{3}}$       

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

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

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

$y(ydx+xdy)=x^5\left(\frac{xdy-ydx}{x^2}\right)$

$\Rightarrow d\left(xy\right)=x^5{y}^{-1}\left(\frac{xdy-ydx}{x^2}\right)\Rightarrow d\left(xy\right)=(xy{)}^{\alpha }{\left(\frac{y}{x}\right)}^{\beta }d\left(\frac{y}{x}\right)$

$\alpha -\beta =5\mathrm{}\alpha +\beta =-1$

$2\alpha =4\mathrm{}\alpha =2\mathrm{}\beta =-3$

$\Rightarrow \mathrm{}\left(xy{)}^{-2}d\right(xy)={\left(\frac{y}{x}\right)}^{-3}d\left(\frac{y}{x}\right)\Rightarrow -(xy{)}^{-1}=-\frac{1}{2}{\left(\frac{y}{x}\right)}^{-2}+c$

$\Rightarrow \mathrm{}-(xy{)}^{-1}=-\frac{1}{2}{\left(\frac{y}{x}\right)}^{-2}+c\mathrm{}$ Passing $\left(\mathrm{1,1}\right)$

$\Rightarrow \mathrm{}-1=-\frac{1}{2}+c\Rightarrow c=-\frac{1}{2};-(xy{)}^{-1}=-\frac{1}{2}{\left(\frac{y}{x}\right)}^{-2}-\frac{1}{2}$

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

 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.

Clearly mean will increase by 5 units and variance will be un changed so the sum would be $7+8+5=20$ .7+8+5=20