Maths

$x^2+9y^2-4x+3=0,x,y\in R.........\left(i\right)$

$x^2-4x+9y^2+3=0$        

Since $x\in R,D\ge 0i.e.16-4\left(9y^2+3\right)\ge 0$

or $9y^2-1\le 0$

$\Rightarrow y\in \left[-\frac{1}{3},\frac{1}{3}\right]$

$\left(i\right)\Rightarrow 9y^2=-x^2+4x-3$

Since L.H.S. $\ge 0soR.H.S.\ge 0i.e.-x^2+4x-3\ge 0$

or $x^2-4x+3\le 0$

$\left(x-3\right)\left(x-1\right)\le 0$

$\Rightarrow x\in \left[1,3\right]$

Let $z_1=x_1+i{y}_1,z_2=x_2+i{y}_2\&z=x+iythen$  

and $\left(z_1-z_2\right)=\frac{\pi }{4}gives\frac{y_1-y_2}{x_1-x_2}=1or{y}_1-y_2=x_1-x_2............\left(i\right)$  

y² – 6x + 9 = 0 .(ii)

as z₁ & z₂ lies on (ii) so ${y_1}^2-6x_1+9=0$    .(iii)

& $y_{2}2-6x_2+9=0.........\left(iv\right)$  

(iii) & (iv) $\left(y_1+y_2\right)\left(y_1-y_2\right)-6\left(x_1-x_2\right)=0$

$\left(x_1-x_2\right)\left(y_1+y_2-6\right)=0from\left(i\right)$

-> $y_1+y_2-6=0i.e.,y_1+y_2=6$  

Let $t=\frac{3}{2}{x}^2+\frac{5}{3}{x}^3+\frac{7}{4}{x}^4+......$

$=\left(2-\frac{1}{2}\right)x^2+\left(2-\frac{1}{3}\right)x^3+\left(2-\frac{1}{4}\right)x^4+......$   

= $2\left(x^2+x^3+x^4+.....\right)-\left(\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+.....\right)$

$=2\frac{x^2}{1-x}+ln\left(1-x\right)+x=\frac{x^2+x}{1-x}+ln\left(1-x\right)=\frac{x\left(1+x\right)}{1-x}+ln\left(1-x\right)$              

$=2\frac{x^2}{1-x}+ln\left(1-x\right)+x=\frac{x^2+x}{1-x}+ln\left(1-x\right)=\frac{x\left(1+x\right)}{1-x}+ln\left(1-x\right)$

$\sum P\left(x\right)=1\Rightarrow k+2k+2k+3k+k=1sok=\frac{1}{9}$

 Now, $P\left(\frac{1<x<4}{x\angle 3}\right)=P\frac{\left(x=2\right)}{P\left(x\angle 3\right)}=\frac{\frac{2k}{9k}}{\frac{k}{9k}+\frac{2k}{9k}}=\frac{2}{3}$  

$\Rightarrow P=\frac{2}{3}$          

So, $5P=\lambda kgives\frac{10}{3}=\lambda *\frac{1}{9}\Rightarrow \lambda =30$  

Let $l={\displaystyle \underset{6}{\overset{16}{\int }}\frac{lo{g}_e{x}^2}{lo{g}_e{x}^2+lo{g}_e\left(x^2-44x+484\right)}dx..........\left(i\right)}$

By property ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}}$

$\left(i\right)\Rightarrow l={\displaystyle \underset{6}{\overset{16}{\int }}\frac{lo{g}_e{\left(22-x\right)}^2}{lo{g}_e{\left(22-x\right)}^2+lo{g}_e{x}^2}dx}..........\left(ii\right)$      

(i) + (ii) 2l = ${\displaystyle \underset{6}{\overset{16}{\int }}1dx=10}$

l = 5

 Normal vector to the given plane be

$2\widehat{i}-\widehat{j}+3\widehat{k}so$                   

Equation of line QS :

$\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=\lambda$

So let P $\left(2\lambda +1,-\lambda +3,\lambda +4\right)$  

Now P lies on given plane so

$4\lambda +2+\lambda -3+8\lambda +4+3=0$  

So, S (-3, 5, 2)

also given R lies on given plane so

6 – 5 + $\gamma$ + 3 = 0 so   $\gamma$ = -4

So, R (3, 5, -4)

SR² = 72

Kindly consider the following figure

$3*7^{22}+2*10^{22}-44=3*{\left(1+6\right)}^{22}+2{\left(1+9\right)}^{22}-44$

$=3\left[1{+}^{22}{C}_1*6{+}^{22}{C}_2*6^2{+}^{22}{C}_3*6^3+....{.}^{22}{C}_{22}{6}^{22}\right]$                

= -39 on division by 18

= (-54 + 15) on division by 18 = 15

$A=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]then{A}^2=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2k\hfill & \hfill -2\hfill \\ \hfill -k\hfill & \hfill 2k+1\hfill \end{array}\right]$

$A^3=\left[\begin{array}{cc}\hfill 2k\hfill & \hfill -2\hfill \\ \hfill -k\hfill & \hfill 2k+1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill -2k\hfill & \hfill 4k+2\hfill \\ \hfill 2k^2+k\hfill & \hfill -4k-1\hfill \end{array}\right]$              

Now, A(A³ + 3l) = 2l gives A³ + 3l = 2A^-1

$\Rightarrow -2k+3=\frac{1}{k}\Rightarrow 2k^2-3k+1=0$    

  $k=\frac{1}{2},1$            

& $4k+2=\frac{2}{k}or2k+1-\frac{1}{k}soor2k^2+k-1=0$

=> $k=\frac{1}{2}$

 Slope of C₁C₂ = $\frac{3}{4}=tan\theta$  

By parametric form C₁ (1 + 5 cosθ, 2 + 5sinθ)

& C₂ (1 – 5 cos θ, 2 – 5 sinθ)

C_{1 $\left(1+5*\frac{4}{5}.2+5*\frac{3}{5}\right)\&C_2\left(1-5*\frac{4}{5}.2-5*\frac{3}{5}\right)$}

So, $\left|\left(\alpha +\beta \right)\left(r+\delta \right)\right|=40$