Maths

Let a_n be the side of square A_n

$a_n=\sqrt[]{2}{a}_{n+1}$       

a₁ = 12

a_n = 12 $*{\left(\frac{1}{\sqrt[]{2}}\right)}^{n-1}$  

${\left(a_n\right)}^2<1$

$\frac{144}{2^{\left(n-1\right)}}<1$

$\Rightarrow 2^{\left(n-1\right)}>144$   

$\Rightarrow n-1\ge 8$

$\Rightarrow n\ge 9$        

$1+{\alpha }^2=1\Rightarrow \alpha =0$

$\&{\alpha }^2+{\beta }^2=1\Rightarrow {\beta }^2=\pm 1$

$\&\alpha -\alpha \beta =0\Rightarrow \alpha \left(1-\beta \right)=0\Rightarrow \alpha =0or\beta =1$

= 0 & = 1 or = 0 & = 1

4 + 4 = 1

$A=\left[\begin{array}{ccc}\hfill x\hfill & \hfill y\hfill & \hfill z\hfill \\ \hfill y\hfill & \hfill z\hfill & \hfill x\hfill \\ \hfill z\hfill & \hfill x\hfill & \hfill y\hfill \end{array}\right],\left|A\right|=3xyz-\left(x^3+y^3+z^3\right)=-\left(x+y+z\right)\left[{\left(x+y+z\right)}^2-3\left(xy+yz+zx\right)\right]$

A² = l

A. A' = l    (as A = A')

$\therefore x^2+y^2+z^2=1andxy+yz+zx=0$         

$x^3+y^3+z^3=3*2+1*\left(1-0\right)=7$             

Slope = 1

$\frac{dy}{dx}=x=1$

$P\left(1,\frac{1}{2}\right)$

Equation of tangent at P : y - $\frac{1}{2}$ = 1 (x – 1)

$\Rightarrow y=x-\frac{1}{2}$

dist= $\frac{\frac{1}{2}}{\sqrt[]{2}}=\frac{1}{2\sqrt[]{2}}$  

$\overrightarrow{a}=\widehat{i}+2\widehat{j}-\widehat{k},\overrightarrow{b}=\widehat{i}-\widehat{j},\overrightarrow{c}=\widehat{i}-\widehat{j}-\widehat{k}$      

-> $\overrightarrow{r}*\overrightarrow{a}=\overrightarrow{c}*\overrightarrow{a}$

$\overrightarrow{r}=\overrightarrow{c}+\lambda \overrightarrow{a}$

Now, $0=\overrightarrow{b}.\overrightarrow{c}+\lambda \overrightarrow{a}.\overrightarrow{b}as\overrightarrow{r}.\overrightarrow{b}=0$  

$\lambda =\frac{-\overrightarrow{b}.\overrightarrow{c}}{\overrightarrow{a}.\overrightarrow{b}}=2$        

$\therefore \overrightarrow{r}.\overrightarrow{a}=\overrightarrow{a}.\overrightarrow{c}+2a^2=12$           

$\left[\begin{array}{cc}\hfill 0\hfill & \hfill -tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 0\hfill \end{array}\right],I_2+A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill -tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 1\hfill \end{array}\right],I_2-A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill -tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 1\hfill \end{array}\right]$           

  $\left(I_2+A\right){\left(I_2-A\right)}^{-1}=\left[\begin{array}{cc}\hfill a\hfill & \hfill -b\hfill \\ \hfill b\hfill & \hfill a\hfill \end{array}\right]$          

  $a^2+b^2=\left|\left(I_2+A\right){\left(I_2-A\right)}^{-1}\right|=se{c}^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.*co{s}^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.=1$          

$\therefore 13\left(a^2+b^2\right)=13*1=13$

Critical point of function are x = $\frac{-1}{2},$ 1 and -2 but x = -2 is making zero .

$\therefore nondifferentiableatx=\frac{-1}{2},1$      

$A={\displaystyle \underset{\pi /4}{\overset{5\pi /4}{\int }}\left(sinx-cosx\right)dx={\left[-cosx-sinx\right]}_{\pi /4}^{5\pi /4}}=\left(\frac{1}{\sqrt[]{2}}+\frac{1}{\sqrt[]{2}}\right)-\left(-\frac{1}{\sqrt[]{2}}-\frac{1}{\sqrt[]{2}}\right)$

A =    $2\sqrt[]{2}$

A⁴ = 64

 $\underset{n\to \infty }{lim}{\displaystyle \sum _{r=0}^{n-1}\frac{n}{{\left(n+r\right)}^2}}$

$\underset{n\to \infty }{lim}{\displaystyle \sum _{r=0}^{n-1}\frac{1}{n}.\frac{1}{{\left(1+\frac{r}{n}\right)}^2}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{{\left(1+x\right)}^2}={\left[-\frac{1}{1+x}\right]}_0^1}=-\frac{1}{2}-\left(-1\right)=\frac{1}{2}}$