Maths

  $ta{n}^{-1}\frac{1}{2r^2}=ta{n}^{-1}\frac{2}{1+\left(4r^2-1\right)}$             

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

${\displaystyle \sum _{r=1}^{50}\left[ta{n}^{-1}\left(2r+1\right)-ta{n}^{-1}\left(2r-1\right)\right]}$

$=ta{n}^{-1}101-ta{n}^{-1}1=ta{n}^{-1}\frac{100}{102}$

$\Rightarrow tanP=\frac{100}{102}=\frac{50}{51}$             

$\frac{dy}{dx}+\frac{e^y-2x}{2x^2}=0$  

$e^{-y}\frac{dy}{dx}-\frac{e^{-y}}{x}=-\frac{-1}{2x^2}$              

Put $e^{-y}=t\Rightarrow -e^{-y}\frac{dy}{dx}=\frac{dt}{dx}$  

$\frac{dt}{dx}+\frac{t}{x}=\frac{1}{2x^2}$              

$\frac{dt}{dx}+\frac{t}{x}=\frac{1}{2x^2}$

 I. F. $=e^{{\displaystyle \int \frac{dx}{x}}}=e^{lnx}=x$  

Soln. tx = ${\displaystyle \int \frac{x}{2x^2}dx=\frac{1}{2}lnx+c}$  

$x=1\Rightarrow e^{-y}=\frac{1}{2}\Rightarrow y=ln2$              

$P\left(-2\sqrt[]{6},\sqrt[]{3}\right)lieson\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$              

$\Rightarrow \frac{24}{a^2}-\frac{3}{b^2}=1.........\left(i\right)$              

$b^2=\frac{a^2}{4}.........\left(ii\right)$

solving (i) & (ii) a² = 12 Þ b² = 3 hyperbola $\frac{x^2}{12}-\frac{y^2}{3}=1$  

Cuts conjugate axis at $R\left(0,R\sqrt[]{3}\right)$     

$\therefore QR=6\sqrt[]{3}$        

Let height of the wall = h than A(0, 0, 0) G (10, 10, h)

$\overrightarrow{AG}=10\widehat{i}+10\widehat{j}+h\widehat{k}$              

B(10, 0, 0) A (0, 10, h)

$\overrightarrow{BH}=-10\widehat{i}+10\widehat{j}+h\widehat{k}$              

  $cos\theta =\frac{1}{5}=\frac{\overrightarrow{BH}.\overrightarrow{AG}}{\left|\overrightarrow{BH}\right|\left|\overrightarrow{AG}\right|}$             

$=\frac{-100+100+h^2}{\left(200+h^2\right)}$              

$5h^2=200+h^2\Rightarrow h^2=50$              

$h=5\sqrt[]{2}$

$n\left(n+1\right)x^2+2\left(2n+1\right)x+4$

$=n\left(n+1\right)x^2+\left\{\left(2n+2\right)+2n\right\}x+4$

${\displaystyle \sum _{n=1}^9\frac{x}{\left(nx+2\right)\left(\left(n+1\right)x+2\right)}}$

$={\displaystyle \sum _{n=1}^9\left[\frac{1}{nx+2}-\frac{1}{\left(n+1\right)x+2}\right]}$

$\left[\left(\frac{1}{x+2}-\frac{1}{2x+2}\right)+\left(\frac{1}{2x+2}-\frac{1}{3x+2}\right)+....+\left(\frac{1}{9x+2}-\frac{1}{10x+2}\right)\right]$

$\therefore \underset{x\to 2}{Lt}\frac{9x}{\left(10x+2\right)\left(x+2\right)}=\frac{9}{44}$              

${\left(\sqrt[]{3}+i\right)}^{100}=2^{100}{\left(\frac{\sqrt[]{3}}{2}+i\frac{1}{2}\right)}^{100}$

$=2^{100}{\left(cos\frac{\pi }{6}+isin\frac{\pi }{6}\right)}^{100}$

$=2^{99}\left(-1+i\sqrt[]{3}\right)=2^{99}\left(p+iq\right)$

$\therefore p=-1\&q=\sqrt[]{3}$              

$\therefore$ Required equation $x^2-\left(\sqrt[]{3}-1\right)x-\sqrt[]{3}=0$

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left[\frac{1+si{n}^{2}x}{1+{\pi }^{sinx}}+\frac{1+si{n}^{2}x}{1+{\pi }^{-sinx}}\right]dx}$

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1+si{n}^{2}x\right)dx=\frac{\pi }{2}+\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1-cos2x\right)dx}}$

$=\frac{\pi }{2}+\frac{1}{2}{\left[x-\frac{sin2x}{2}\right]}_0^{\frac{\pi }{2}}$
$=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}$

$A^2=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$

$A^3=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$

$A^{2025}-A^{2020}$

$=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2024\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)-\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2019\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 5\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=A^6-A$

Equation of required circle

$C:{\left(x-2\right)}^2+{\left(y-1\right)}^2+\lambda \left(2y-x\right)=0..........\left(i\right)$              

Intersect C₁x² + y² + 2y – 5 = 0 .(ii)

$\therefore$ Equation of radical axis

    $-4x-4y+10+\lambda \left(2y-x\right)=0..........\left(ii\right)$          

Centre of C₁ (0, -1) lies on .(ii)

$\Rightarrow 4+10-2\lambda =0\Rightarrow \lambda =7$        

Equation of circle C is

$x^2+y^2-11x+12y+5=0$       

Diameter = $\sqrt[]{245}=7\sqrt[]{5}$  

Given limit
$={\displaystyle \underset{0}{\overset{2}{\int }}\frac{dx}{1+4x^2}}$

$={\left[\frac{1}{2}ta{n}^{-1}2x\right]}_0^2=\frac{1}{2}ta{n}^{-1}4$