Maths

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

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

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

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

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

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

  $a_0{C}_0-a_1{C}_1+a_2{C}_2-a_3{C}_3+.....+a_{2012}{c}_{2012}$

$=a_0\left(C_0-C_1+C_2+....+C_{2012}\right)$            

  $-d\left(C_1-2C_2+3C_3-4C_4+....{-}^{2012}{C}_{2012}\right)$           

  $=a_0\left(0\right)-d\left(C_1-2C_2+3C_3{-}^4{C}_4\right)$           

+ …….-  

 =   $a_0\left(0\right)-d\left(0\right)$

= 0

$(x+y{)}^n={}^n{C}_r{x}^{n-r}{y}^r$

${\left(\frac{3}{2}{x}^2+\frac{2}{x}\right)}^{12}?{}^{12}{C}_r{\left(\frac{3}{2}{x}^2\right)}^{12-r}{\left(\frac{2}{x}\right)}^r$

For independent of X .

$x^{24-2r}{x}^{-r}=x^0;24=3r\Rightarrow r=8$

${}^{12}{C}_8{\left(\frac{3}{2}\right)}^{12-8}(2{)}^8;\frac{\lfloor 12}{\overline{12}44}*\frac{3^4}{2^4}*2^8\Rightarrow \lambda ;\frac{\lambda }{81}\Rightarrow 7920$

$Inx\in \left(-2,2\right)$  

 |x|- 1| is not differentiable at x = -1, 0, 1

|cospx| is not differentiable at x =  $\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2}$ -  

$\int {\mathrm{c}\mathrm{o}\mathrm{t}}^{-1}?\left(\frac{1}{1+x}\right)dx$

$=\int {\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)dx$

$=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\int \frac{x}{1+(1+x{)}^2}dx;=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\int \frac{x}{x^2+2x+2}dx$

$=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\int \frac{\frac{1}{2}(2x+2)-1}{x^2+2x+2}dx=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\frac{1}{2}\mathrm{l}\mathrm{o}\mathrm{g}?\left(x^2+2x+2\right)+\int \frac{dx}{(x+1{)}^2+1}$

$=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\frac{1}{2}\mathrm{l}\mathrm{o}\mathrm{g}?\left(x^2+2x+2\right)+{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(x+1)+c$

$A^2{\left|A\right|}^2-{\left|B\right|}^{3.2}B$

$=25A^2-\left|B\right|.B$

$\begin{array}{l}=25A^2+\left|B\right|A^2\\ =0\end{array}$

Put x³/² = t
√xdx = (2/3)dt
⇒ (2/3) ∫ dt/√ (1-t²) = (2/3)sin? ¹t + c
= (2/3)sin? ¹x³/² + c
⇒ g (x) = sin? ¹x
g (0) = 0

? P? = 6!/2! = 360