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 = $\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$  

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

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   $\frac{1}{x}$

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

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

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

Let a, ar, ar², ar³ are in G.P.

a + ar + ar² + ar³ = $\frac{65}{12}$  

$\frac{1}{a}+\frac{1}{ar}+\frac{1}{a{r}^2}+\frac{1}{a{r}^3}=\frac{65}{18}$        

${\left(ar\right)}^{2}r=\frac{3}{2}\Rightarrow a=\frac{2}{3},r=\frac{3}{2}$          

  $\therefore$   third term = ar² = $\frac{2}{3}*\frac{9}{4}=\frac{3}{2}$  

  $\therefore \alpha =\frac{3}{2}\Rightarrow 2\alpha =3$          

Let P (h, k)

$\sqrt[]{{\left(h-5\right)}^2+k^2}=3\sqrt[]{{\left(h+5\right)}^2+k^2}$           

$h^2-10h+25+k^2=9h^2+90h+225+9k^2$           

  $8h^2+8k^2+100h+200=0$

$x^2+y^2+\frac{25}{2}x+25=0$

$r^2=\frac{625}{16}-25=\frac{625-400}{16}=\frac{225}{16}$

$4r^2=\frac{225}{4}=56.25$         

${\left(x+1\right)}^2+\left|x-5\right|=\frac{27}{4}$

Case l        x < 5

$x^2+2x+1-x+5=\frac{27}{4}$       

$\left(2x+3\right)\left(2x-1\right)=0\Rightarrow x=-\frac{3}{2},\frac{1}{2}$     

Case II   $\ge$   x 5

$x^2+2x+1+x-5=\frac{27}{4}$       

$x=\frac{-12\pm \sqrt[]{144+43*16}}{2*4}=\frac{-12\pm \sqrt[]{832}}{8}\left(rejected\right)$           

           because x > 5

  $\frac{2^{21}{\left(\frac{-1+i\sqrt[]{3}}{2}\right)}^{21}}{{\left(\sqrt[]{2}\right)}^{24}{\left(\frac{1-i}{\sqrt[]{2}}\right)}^{24}}+\frac{2^{21}{\left(\frac{1+\sqrt[]{3}i}{2}\right)}^{21}}{{\left(\sqrt[]{2}\right)}^{24}{\left(\frac{1+i}{\sqrt[]{2}}\right)}^{24}}=k$

=   ${\displaystyle \sum _{j=0}^5\left(j+5\right)\left(j+5-1\right)={\displaystyle \sum _{j=0}^5\left(j+5\right)\left(j+4\right)}}$

= ${\displaystyle \sum _{j=0}^5\left(j^2+9\widehat{j}+20\right)=\frac{5*6*11}{6}+\frac{9*5*6}{2}+20*6}$  

55 + 135 + 120 = 310