Class 12th

Let daily working hours & wages per hour before increase be H and W respectively

So his Salary/day = HW

Working hours after increase = 0.82H, wages per hour = 1.45W

So his daily salary = $\frac{82}{100}H*\frac{29}{20}W=\frac{1189}{1000}HW$ $\frac{}{}\frac{}{}\frac{}{}$ . Thus an Increase of 18.9%

A + B + C = 18500…. (1)

A = B + 25% of   $B=\frac{5B}{4}$ $\frac{}{}$ …. (2)

B = C + 20% of $C=\frac{6C}{5}$ $\frac{}{}$ …. (3)

Equation (1) gives,

$\frac{5B}{4}+\frac{6C}{5}+C=18500$

$\Rightarrow \frac{3C}{2}+\frac{6C}{5}+C=18500$

37C = 185000 C = 5000

B = 6000, A = 7500.

A + B = 13500

System of equations can be written as

$\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 5\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill \lambda \hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}6\\ 26\\ \mu \end{array}\right)$                          

$R{\text{'}}_3=R_3-R_1,R{\text{'}}_2=R_2-5R_1\Rightarrow$                

$\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill \lambda -1\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}6\\ -4\\ \mu -6\end{array}\right)$                              

Again $R\text{'}{\text{'}}_3=R{\text{'}}_3-R{\text{'}}_1\Rightarrow \left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \lambda -2\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}4\\ -4\\ \mu -10\end{array}\right)$  

The system will have no solution for

$\lambda =2and\mu \ne 10.$  

At minimum deviation :

i = i'

r = r'

$r+r\text{'}=A\Rightarrow r=\frac{A}{2}$

Snell's law :

1 sin I = $\sqrt[]{3}sinr=\sqrt[]{3}sin\frac{i}{2}$ $\text{exception 25:}\text{exception 25:}\frac{}{}$

$\Rightarrow 2cos\frac{i}{2}=\sqrt[]{3}$

$\Rightarrow i=60°$

$\begin{array}{l}i=2r\Rightarrow r=30°\\ \Rightarrow A=60°\end{array}$

$f\left(x\right)=\frac{co{s}^{-1}\sqrt[]{x^2-x+1}}{\sqrt[]{si{n}^{-1}\left(\frac{2x-1}{2}\right)}}$

For domain $0\le x^2-x+1\le 1$

$\&x^2-x+1\ge 0\forall x\in R$

$x^2-x\le 0\&x\left(x-1\right)\le 0\Rightarrow x\in \left[0,1\right]...........\left(i\right)$              .

  $\frac{1}{2}\le x\le \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.................\left(ii\right)$             

$\left(i\right)\cap \left(ii\right)x\in \left(\frac{1}{2},1\right]\equiv \left(\alpha ,\beta \right]$

then α + β = $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$  

Using gauss law in differential form

$\overrightarrow{\nabla }.\overrightarrow{E}=\frac{\rho }{{\epsilon }_0}$

$\overrightarrow{\nabla }.\left({\epsilon }_0\overrightarrow{E}\right)=\rho$

$\Rightarrow \overrightarrow{\nabla }.\overrightarrow{D}=\rho$

$\Rightarrow \left(\frac{\partial }{\partial x}\widehat{i}+\frac{\partial }{\partial y}\widehat{j}+\frac{\partial }{\partial z}\widehat{k}\right).\overrightarrow{D}=\rho$

$\begin{array}{l}\Rightarrow \rho =-e^{-x}siny+e^{-x}siny+2=2\\ \Rightarrow Q=2\left(Volume\right)=4*10^{-9}c=4nc\end{array}$

$F=\left[\frac{2k\lambda }{r_1}-\frac{2k\lambda }{r_2}\right]q=2k\lambda q\left[\frac{1}{r_1}-\frac{1}{r_2}\right]$

$2*9*10^9*3*10^{-6}\left[\frac{1000}{10}-\frac{1000}{12}\right]q$

$\begin{array}{l}\Rightarrow 4=9*10^{5}q\\ \Rightarrow q=4.44\mu C\end{array}$

  $A={\left[a_{ij}\right]}_{3*3}=\left[\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right]$             

$a_{i1}+a_{i2}+a_{i3}=1;i=1,2,3$            

$LetX=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]them$  

given $\left[\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right]$ $\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]$  

->AX = X .(i)

replace x by A x we have

A (AX) = AX

->A²X = AX = X .(ii)

Again replace X by AX

A³X = AX = X.

As  $X=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right],$ Sum of all entries in A³ = sum of entries in X = 1 +1 + 1 = 3

$R_1={?}_1\frac{\mathcal{l}}{A}{R}_2={?}_2\frac{\mathcal{l}}{A}$

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{A}{\mathcal{l}}\left[\frac{1}{{?}_1}+\frac{1}{{?}_2}\right]$

$=\frac{3*10^{?6}}{\frac{1}{4}}\left[\frac{1}{1.7*10^{?8}}+\frac{1}{2.6*10^{?8}}\right]$

$R=0.858*10^{?3}?$

$\overrightarrow{r}.\left(\widehat{i}-\widehat{j}+2\widehat{k}\right)=2$

&  $\overrightarrow{r}.\left(2\widehat{i}+\widehat{j}-\widehat{k}\right)=2$ are two planes the direction ratio of the line of intersection of then is collinear to   

$\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill -1\hfill \end{array}\right|=-\widehat{i}+5\widehat{j}+3\widehat{k}$

Any point on the line in given by x – y = 2

& 2x + y = 2

$\Rightarrow x=\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.,y=\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.,z=0$

$\therefore equationoflineL:\frac{x-\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{-1}=\frac{y+\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{5}=\frac{z}{3}=r$

$\therefore pointP\left(\frac{33}{35},\frac{45}{35},\frac{41}{35}\right)\equiv \left(\alpha ,\beta ,\gamma \right)$

$\therefore 35\left(\alpha +\beta +\gamma \right)=119$