Class 12th

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

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

Equation of any tangent to   $\frac{x^2}{16}+\frac{y^2}{9}=1$

is y = mx + $\sqrt[]{16m^2+9}$  if this line is also tangent to x² + y² = 12

then  $\sqrt[]{12}$ =   $\left|\frac{\sqrt[]{16m^2+9}}{\sqrt[]{1+m^2}}\right|$

^{$\Rightarrow 12+12m^2=16m^2+9$}

$\Rightarrow 12m^2=9$        

RM = $\left|\frac{3+7-5}{\sqrt[]{2}}\right|=\frac{5}{\sqrt[]{2}}$

$\mathcal{l}sin60°=\frac{5}{\sqrt[]{2}}\Rightarrow \mathcal{l}=5\sqrt[]{\frac{2}{3}}$

$\therefore Areaof\Delta PQR=\frac{\text{exception 25:}}{4}{\mathcal{l}}^2=$=25/2√3

Put x = cos 2θ

dx = -2 sin 2θ . dθ

$={\displaystyle \int \frac{1}{cos2\theta }tan\theta \left(-4sin\theta .cos\theta \right)d\theta }$

$g\left(\frac{1}{2}\right)=\mathcal{l}n\left|2-\sqrt[]{3}\right|+\frac{\pi }{3}$

= $\mathcal{l}n\left|\frac{\sqrt[]{3}-1}{\sqrt[]{3}+1}\right|+\frac{\pi }{3}$