Class 12th

Total number of numbers from given condition = n (s) = 26

Every required number is of the form

${}^{}$ $A=7.\left(10^{a_1}+10^{a_2}+10^{a_3}+....\right)$  + 111111

Here 111111 is always divisible by 21.

$$  $\therefore$ Required probability = $\frac{22}{2^5}=p$ $\frac{}{{}^{}}$ $\frac{}{}$ $\therefore \frac{11}{32}$ = p $$ $\therefore$ 96p = 33

Equation of L₁ = is

$\frac{xsec\theta }{4}-\frac{ytan\theta }{2}=1$ ….(i)

Equation of line L₂ is

$\frac{xtan\theta }{2}+\frac{ysec\theta }{4}=0$ ….(ii)

$?$ Required point of intersection of L₁ and L₂ is (x₁, y₁) then

$\frac{x_{1}sec\theta }{4}-\frac{y_{1}tan\theta }{2}-1=0$ ….(iii)

$and\frac{y_{1}sec\theta }{4}+\frac{x_{1}tan\theta }{2}=0$ ……(iv)

From equations (iii) and (iv)

$sec\theta =\frac{4x_1}{x_1^2+y_1^2}andtan\theta =\frac{-2y_1}{x_1^2+y_1^2}$        

$\therefore$ Required locus of (x₁, y₁) is

${\left(x^2+y^2\right)}^2=16x^2-4y^2$   

$\begin{array}{l}\therefore \alpha =16,\beta =-4\\ \therefore \alpha =\beta =12\end{array}$     

Equation of perpendicular bisector of AB is

$y-\frac{3}{2}=-\frac{1}{5}\left(x-\frac{5}{2}\right)\Rightarrow x+5y=10$

Solving it with equation of given circle,

${\left(x-5\right)}^2+{\left(\frac{10-x}{5}-1\right)}^2=\frac{13}{2}$

$\Rightarrow x-5=\pm \frac{5}{2}\Rightarrow x=\frac{5}{2}or\frac{15}{2}$

But

$x\ne \frac{5}{2}$

because AB is not the diameter.

So, centre will be

$\left(\frac{15}{2},\frac{1}{2}\right)$

Now,

$r^2={\left(\frac{15}{2}-2\right)}^2+{\left(\frac{1}{2}+1\right)}^2=\frac{65}{2}$

Clearly r must be equal to $\sim$ $$ p

$?\sim p\vee \sim p=\sim p$

$and\left(p\wedge q\right)\vee \sim p=p$

$$ $\therefore \sim p\Rightarrow p=$  tautology.

$\overline{x}=15,\sigma =2\Rightarrow {\sigma }^2=4$

$\therefore x_1+x_2+....+x_{50}=15*50=750$

$4=\frac{x_1^2+x_2^2+.....+x_{50}^2}{50}-225$

Let a be the correct observation and b is the incorrect observation then a + b = 70 and

$16=\frac{75-b+a}{50}$

$=\frac{50*229+60^2-10^2}{50}-256=43$

$\overrightarrow{v}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}$

$\overrightarrow{v}=\lambda \left(1,1,2\right)+\mu \left(2,-3,1\right)$

$\overrightarrow{v}.\widehat{j}=7\overrightarrow{v}.\frac{\overrightarrow{c}}{\left|\overrightarrow{c}\right|}=\frac{2}{\sqrt[]{3}}$

$\lambda +2\mu -\lambda +3\mu +2\lambda +\mu =2$

$\lambda -3\mu =7$

$\begin{array}{l}2\lambda =8\\ \lambda =4\end{array}$ $\begin{array}{l}\mu =-1\\ \overrightarrow{v}=\left(2,7,7\right)\end{array}$

The given two lines are coplanar

$\therefore \left|\begin{array}{ccc}\hfill 0\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill -3\hfill \\ \hfill 1-\alpha \hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|=0\Rightarrow \alpha =\frac{5}{3}$

Now, $\overrightarrow{n}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill -3\hfill \end{array}\right|=\widehat{i}\left(-9\right)-\widehat{j}\left(2\right)+\widehat{k}\left(-6\right)=\left(9,2,6\right)$

Equation of plane :

$=\left|\frac{\left(9.\frac{5}{3}+0+0-13\right)}{\sqrt[]{81+36+4}}\right|=\frac{2}{\sqrt[]{121}}=\frac{2}{11}$

  $P_1:2x-y-52=0,P_2:3x-y+4z-7=0\text{?}\text{?}P$

Equation of plane passing through the line of intersection between planes p₁ = 0 & p₂ = 0 is

    $P:P_1+\lambda P_2$            

$\therefore P:8x-y+32-14=0$      

It passes through the point (1, 0, 2)

Given hyperbola :

$\frac{x^2}{a^2}-\frac{y^2}{9}=1$

$$ $?$  it passes through

$\left(8,3\sqrt[]{3}\right)$

$?\frac{64}{a^2}-\frac{27}{9}=1\Rightarrow a^2=16$

Now, equation of normal to hyperbola

$\frac{16x}{8}+\frac{9y}{3\sqrt[]{3}}=16+9$

$\text{exception 25:}$ $\left(-1,9\sqrt[]{3}\right)$  satisfied

  $E:\frac{x^2}{4}+\frac{y^2}{2}=1$

any pt on it is P  $\left(2cos\theta ,\sqrt[]{2}sin\theta \right)$

$\therefore$ M (h, k) be mid point of P & A (4, 3)

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

Required locus (x – 2)² +   $\frac{{\left(y-\frac{3}{2}\right)}^2}{\frac{1}{2}}=1$

$\therefore e=\sqrt[]{1-\frac{1}{2}}=\frac{1}{\sqrt[]{2}}$