Class 12th

Resolving power

$\frac{h-cos\theta }{2}=\frac{k-sin\theta }{3}=\frac{w-0}{1}$

$=\frac{-1\left(2cos\theta +3sin\theta -6\right)}{14}\Rightarrow h=\frac{cos-2\left(2cos\theta +3sin\theta -6\right)}{14}$

$k=\frac{5sin\theta -6cos\theta +18}{14}$

${\left(5h+6k-12\right)}^2+4{\left(3h+5k-9\right)}^2=1$

  $S_1=\left\{z,\in c:\lfloor z_1-3\rfloor =\frac{1}{2}\right\}and{S}_2=\left\{z_2\in c:\left|z_2-\left|z_2+1\right|\right|=\left|z_2+\left|z_2-1\right|\right|\right\}$

Then, for $z_1\in S_1$  and $z_2\in S_2,$  the least value of |z₂ – z₁|

${\left|z_2+\left|z_2-1\right|\right|}^2={\left|z_2-\left|z_2+1\right|\right|}^2$      

$\Rightarrow \left(z_2+{\overline{z}}_2\right)\left(\left|z_2-1\right|+\left|z_2+1\right|-2\right)=0$      

$\therefore z_2+{\overline{z}}_2=0or\left|z_2-1\right|+\left|z_2+1\right|-2=0$          

$\therefore$ z₂ lies on imaginary axis or on real axis with in [-1, 1]

also $\left|z_1-3\right|=\frac{1}{2}$  lie on circle having centre 3 and radius   $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Clearly $\left|z_1-z_2\right|min=\frac{5}{2}-1=\frac{3}{2}$

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

$=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$

$a\leftrightarrow$ R2

$=-\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$

$R_2\leftrightarrow R_3$

$=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ =1 

$\begin{array}{l}{B}_0=A^{49}+2A^{98}=A+2I\\ B_n=Adj\left(B_n-1\right)\\ B_4=Adj\left(Adj\left(Adj\left(Adj{B}_0\right)\right)\right)\end{array}$

$={\left|B_0\right|}^{{\left(n-1\right)}^4}={\left|B_0\right|}^{16}$

$B_0=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 2\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill \end{array}\right]$

$=2\left(4-0\right)-1\left(0-1\right)=9$

$B_4{\left(9\right)}^{16}=3^{32}$

$\frac{dy}{dx}+\left(\frac{2x^2+11x+13}{x^3+6x^2+11x+6}\right)y=\frac{x+3}{x+1},x>-1$

IF = $e^{{\displaystyle \int pdx}}=\frac{{\left(x+1\right)}^2\left(x+2\right)}{x+3}$

${\displaystyle \int Pdx={\displaystyle \int \frac{2x^2+11x+13}{x^3+6x^2+11x+6}dn={\displaystyle \int \left(\frac{2}{x+1}+\frac{1}{x+2}-\frac{1}{x+3}\right)dx}}}$

$=\mathcal{l}n\left({\left(x+1\right)}^2\cdot \left(x+2\right)/\left(x+3\right)\right)$

$\frac{2x^2+11x+13}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}$

$2x^2+11x+13=A\left(x+2\right)\left(x+3\right)+B\left(x+1\right)\left(x+3\right)+C\left(x+1\right)\left(x+2\right)$

x = -1

->4 = 2A Þ A = 2

x = -2

-> -1 = -B Þ B = 1

x₂ – 3 Þ -2 = 2c

c = -1

$y\cdot \frac{{\left(x+1\right)}^2\left(x+2\right)}{x+3}={\displaystyle \int \frac{x+3}{x+1}\cdot \frac{{\left(x+1\right)}^2\left(x+2\right)}{x+3}dx}$

= ${\displaystyle \int \left(x+1\right)\left(x+2\right)dx}$

= $\frac{x^3}{3}+\frac{3x^2}{2}+2x+c$

$\left(0,1\right)\Rightarrow 1\cdot \frac{2}{3}=c$

x = 1  $y\cdot \left(3\right)=\frac{1}{3}+\frac{3}{2}+2+\frac{2}{3}=\frac{3}{2}+3=\frac{9}{2}$

y =  $\frac{3}{2}$

$E_y=900sin\omega \left(t-\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$c$}\right.\right)$

$\frac{F_E}{F_B}=\frac{qE}{\left|q\overline{V}*\overline{B}\right|}=\frac{C}{V}=\frac{3*10^8}{3*10^7}=10$

$\Rightarrow \frac{F_E}{F_B}=\frac{10}{1}$

$7^{2022}+3^{2022}$

$={\left(49\right)}^{1011}+{\left(9\right)}^{1011}$

$={\left(50-1\right)}^{1011}+{\left(10-1\right)}^{1011}$

$=5\lambda -1+5k-1$

= 5m – 2

Remainder = 5 – 2 = 3

In DC :-                In AC:-

              i₁ = 4A                 i₂ = 4 sin   $\omega t$

              R₁ = 3   $\Omega$             R₂ =   $2\Omega$

              In same time internal of 't' - $\frac{H_1}{H_2}=\frac{i_1^2{R}_{1}t}{{\left({\mathcal{l}}_{2^{-rms}}\right)}^2{R}_{2}t}=\frac{16*3t}{{\left(\frac{4}{\sqrt[]{2}}\right)}^2*2*t}=\frac{48}{\left(\frac{32}{2}\right)}=3:1$

If y = y(x), $x\in \left(0,\frac{\pi }{2}\right)$ $\frac{}{}$ be the solution curve of the different equation

$\frac{}{}$ $\therefore \frac{dy}{dx}+\left(8+4cot2x\right)y=\frac{2e^{-4x}}{si{n}^{2}2x}\left(2sinx+cos2x\right)$  which is a linear different equation

I.F = $e^{{\displaystyle \int \left(8+4cot2x\right)dx}}=e^{8x+2cos\left(sin2x\right)}=e^{8x}.si{n}^{2}2x$ ${}^{{\displaystyle }}{}^{}{}^{}{}^{}$

Given, $y\left(\frac{\pi }{4}\right)=e^{-\pi }\Rightarrow c=0$ $\frac{}{}{}^{}$

$\therefore y=\frac{e^{-4x}}{sin2x}$

$\therefore y\left(\frac{\pi }{6}\right)=\frac{e^{-4\frac{\pi }{6}}}{sin\left(2.\frac{\pi }{6}\right)}=\frac{2}{\sqrt[]{3}}{e}^{\frac{2\pi }{3}}$

  $\frac{dy}{dx}=\frac{x+y-2}{x-y}$  

Let x – 1 = X, y – 1 = Y

then DE: $\frac{dY}{dX}=\frac{X+Y}{X-Y}=\frac{1+\frac{Y}{X}}{1-\frac{Y}{X}}$  

Put y = vx

then  $\frac{dY}{dX}=V+X\frac{dV}{dX}$  

V + $X\frac{dV}{dX}=\frac{1+V}{1-V}$

$X\frac{dV}{dX}=\frac{1+V}{1-V}-V$

$=\frac{1+V^2}{1-V}$

$\frac{1-V}{1+V^2}dV=\frac{dX}{X}$

$\frac{V-1}{V^2+1}dV+\frac{dX}{X}=0$

$\frac{1}{2}\mathcal{l}n\left|V^2+1\right|-ta{n}^{-1}V+\mathcal{l}n\left|X\right|=c$

$\mathcal{l}n\left(\sqrt[]{1+{\left(\frac{Y-1}{X-}\right)}^2}\cdot \left|X-1\right|\right)ta{n}^{-1}\frac{Y-1}{X-1}=c$

$\left(2,1\right)\Rightarrow \mathcal{l}n\left(\sqrt[]{1+0}\cdot 1\right)-0=c$

c = 0  

$\mathcal{l}n\sqrt[]{{\left(X-1\right)}^2+{\left(Y-1\right)}^2}=ta{n}^{-1}\frac{Y-1}{X-1}$

point (k + 1, 2) $\mathcal{l}n\sqrt[]{k^2+1}=ta{n}^{-1}\frac{1}{k}$

$\Rightarrow \frac{1}{2}\mathcal{l}n\left(k^2+1\right)=ta{n}^{-1}\frac{1}{k}$