Maths

  $z+\alpha \left|z-1\right|+2i=0$

Let z = x + iy

$\Rightarrow x+i\left(y+2\right)=-\alpha \sqrt[]{{\left(x-1\right)}^2+y^2}$            

 for $\alpha \in Ry+2=0\Rightarrow y=-2$  

$x^2={\alpha }^2\left[{\left(x-1\right)}^2+4\right]$      = 0

$\frac{1}{{\alpha }^2}\ge \frac{4}{5}\Rightarrow {\alpha }^2\le \frac{5}{4}\Rightarrow \frac{-\sqrt[]{5}}{2}\le \alpha \le \frac{\sqrt[]{5}}{2}$

$4\left(p^2+q^2\right)=4\left(\frac{5}{4}+\frac{5}{4}\right)=10$

  ${\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{1}{1+r+r^2}={\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{\left(r+1\right)}{1+r\left(r+1\right)}}}$

$=ta{n}^{-1}2-ta{n}^{-1}+......+ta{n}^{-1}\left(n+1\right)-ta{n}^{-1}n$            

For n  $\infty$ value = $\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$  

$\therefore tan\left(\frac{\pi }{4}\right)=1$           

$\frac{4}{sinx}+\frac{1}{1-sinx}=a\forall x\in \left(0,\frac{\pi }{2}\right)$  

 Let sin x = t, t $\in$ (0, 1)

$g\left(t\right)=\frac{4}{t}+\frac{1}{1-t}$

$g\text{'}\left(t\right)=0\Rightarrow t=\frac{2}{3}$

$g\text{'}\text{'}\left(\frac{2}{3}\right)>0$

$\therefore g{\left(t\right)}_{minimum}=\frac{4}{2/3}+\frac{1}{1-2/3}=9$  

 Minimum value of a for which solution exist = 9

$M=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill c_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right]$    

$M^{T}M=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill b_1\hfill & \hfill c_1\hfill \\ \hfill a_2\hfill & \hfill b_2\hfill & \hfill c_2\hfill \\ \hfill a_3\hfill & \hfill b_3\hfill & \hfill c_3\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill b_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right]$

$Tr\left(M^{T}M\right)=a_1^2+b_1^2+c_1^2+a_2^2+b_2^2+c_2^2+a_3^2+b_3^2+c_3^2=7$           

all   $a_i,b_i,c_i\in \left\{0,1,2\right\}for$ i = 1, 2, 3

Case 1 7 one's and two zeroes which can occur in ways

Case 2 One 2 three 1's five zeroes =

total such matrices = 504 + 36 = 540

  $z+\alpha \left|z-1\right|+2i=0$

Let z = x + iy

$\Rightarrow x+i\left(y+2\right)=-\alpha \sqrt[]{{\left(x-1\right)}^2+y^2}$            

 for $\alpha \in Ry+2=0\Rightarrow y=-2$  

$x^2={\alpha }^2\left[{\left(x-1\right)}^2+4\right]$      = 0

$\frac{1}{{\alpha }^2}\ge \frac{4}{5}\Rightarrow {\alpha }^2\le \frac{5}{4}\Rightarrow \frac{-\sqrt[]{5}}{2}\le \alpha \le \frac{\sqrt[]{5}}{2}$

$4\left(p^2+q^2\right)=4\left(\frac{5}{4}+\frac{5}{4}\right)=10$

  ${\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{1}{1+r+r^2}={\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{\left(r+1\right)}{1+r\left(r+1\right)}}}$

$=ta{n}^{-1}2-ta{n}^{-1}+......+ta{n}^{-1}\left(n+1\right)-ta{n}^{-1}n$            

For n  $\infty$ value = $\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$  

$\therefore tan\left(\frac{\pi }{4}\right)=1$           

Equation of normal is 4x – 3y + 1 = 0

Equation of tangent is 3x + 4y – 43 = 0

Area of triangle = $\frac{1}{2}\left(\frac{43}{3}+\frac{1}{4}\right)*7$  

$=\frac{1}{2}*\left(\frac{172+3}{12}\right)*7=\frac{1225}{24}$          

  $\therefore 24A=1225$        

${\sigma }^2=\frac{\sum x^2}{n}-{\left(\frac{\sum x}{n}\right)}^2=\frac{9+k^2}{10}-{\left(\frac{9+k}{10}\right)}^2<10$

$\Rightarrow k<\frac{10\sqrt[]{10}}{3}+1\Rightarrow k\le 11$        

$\therefore$ maximum value of k is 11

$af\left(x\right)+\alpha f\left(\frac{1}{x}\right)=bx+\frac{\beta }{x}............\left(i\right)$  

Replace x by   $\frac{1}{x}$

$af\left(\frac{1}{x}\right)+\alpha f\left(x\right)=\frac{b}{x}+\beta x..............\left(ii\right)$  

$a\left(f\left(x\right)+f\left(\frac{1}{x}\right)\right)+\alpha \left(f\left(x\right)+f\left(\frac{1}{x}\right)\right)=b\left(x+\frac{1}{x}\right)+\beta \left(x+\frac{1}{x}\right)$           

$\therefore \frac{f\left(x\right)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}}=\frac{b+\beta }{a+\alpha }=\frac{2}{1}=2$