Maths

$2sin\left(\frac{\pi }{22}\right)sin\left(\frac{3\pi }{22}\right)sin\left(\frac{5\pi }{22}\right)sin\left(\frac{7\pi }{22}\right)sin\left(\frac{9\pi }{22}\right)$

$=\frac{2sin\frac{32\pi }{11}}{2^{5}sin\frac{\pi }{11}}=\frac{1}{16}$

Median = $\frac{2k+12}{2}=k+6$

Mean deviation = $\sum \frac{\left|x_i-M\right|}{n}=6$

$\Rightarrow \frac{\left(k+3\right)+\left(k+1\right)+\left(k-1\right)+\left(6-k\right)+\left(6-k\right)+\left(10-k\right)+\left(15-k\right)+\left(18-k\right)}{8}$

$\therefore \frac{58-2k}{8}=6$

k = 5

$Median=\frac{2*5+12}{2}=11$

$\overrightarrow{a}=\widehat{i}-\widehat{j}+2\widehat{k}$

$\overrightarrow{a}*\overrightarrow{b}=2\widehat{i}-\widehat{k}$

$\overrightarrow{a}.\overrightarrow{b}=3$

${\left|\overrightarrow{a}*\overrightarrow{b}\right|}^2+{\left|\overrightarrow{a}.\overrightarrow{b}\right|}^2={\left|\overrightarrow{a}\right|}^2.{\left|\overrightarrow{b}\right|}^2$$$

$=\frac{\overrightarrow{b}.\overrightarrow{a}-{\left|\overrightarrow{b}\right|}^2}{\left|\overrightarrow{a}-\overrightarrow{b}\right|}=\frac{3-\frac{7}{3}}{\sqrt[]{\frac{7}{3}}}=\frac{2}{\sqrt[]{21}}$

 $A\left(-7\widehat{i}+6\widehat{j}\right)C\left(7\widehat{i}+2\widehat{j}+6\widehat{k}\right)$

$\overrightarrow{b}=-6\widehat{i}+7\widehat{j}+\widehat{k}$ $\overrightarrow{d}=-2\widehat{i}+\widehat{j}+\widehat{k}$

$\overrightarrow{b}*\overrightarrow{d}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill -6\hfill & \hfill 7\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right|=3\widehat{i}+2\widehat{j}+4\widehat{k}$

the shortest distance between the lines

$=$$\left|\frac{42-8+24}{\text{exception 25:}}\right|$$=$$\frac{58}{\text{exception 25:}}$$=$$2$

First plane, P₁ = 2x – 2y + z = 0,

normal vector $\equiv n_1=\left(2,-2,1\right)$

Second plane, P₂ = x – y + 2z = 4,

normal vector $\equiv n_2=\left(1,-1,2\right)$

Plane perpendicular to P₁ and P₂ will have normal vector n₃ where n₃ = (n₁ * n₂)

Hence,

n₃ = (3, 0)

Distance PQ

$=\sqrt[]{21}\Rightarrow {\left(PQ\right)}^2=21$

Ellipse : $\frac{x^2}{16}+\frac{y^2}{7}=1$

Eccentricity = $\sqrt[]{1-\frac{7}{10}}=\frac{3}{4}$

Foci $\equiv \left(\pm ae,0\right)$ $\equiv \left(\pm 3,0\right)$

Hyperbola : $\frac{x^2}{\left(\frac{144}{25}\right)}-\frac{y^2}{\left(\frac{\alpha }{25}\right)}=1$

Eccentricity = $\sqrt[]{1+\frac{\alpha }{144}}=\frac{1}{12}\sqrt[]{144+\alpha }$

$=2.\frac{\frac{81}{25}}{\frac{12}{5}}=\frac{27}{10}$

$y^2-2x-2y=1$

$\Rightarrow {\left(y-1\right)}^2=2\left(x+1\right)$ …. (i)

Equation of tangent at A is 2x – y – 5 = 0 ………. (ii)

D is mid point of AB solving (ii) with y = 1 P (3, 1)

$\therefore PD=4,AD=2$

$Areaof\Delta APD=\frac{1}{2}\left(PD\right)\left(AD\right)=4$

$\therefore Areaof\Delta APB=8sq.units$

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

(B) $$ $q\vee \left(-r\vee p\right)$

(C)  $\left(\sim p\right)\vee \left(q\vee r\right)$

(D) $$ $\left(\sim p\vee q\right)\vee r$

Using Venn diagram we get B as the correct option.

$-1\le \frac{x^2-3x+2}{x^2+2x+7}\le 1$                

$\Rightarrow 0\le \frac{2x^2-x+9}{x^2+2x+7}\&\frac{-5x-5}{x^2+2x+7}\le 0$                       

$x\in R\&-1\le x<\infty$

Given : $\widehat{a}\cdot \widehat{b}=\widehat{b}\cdot \widehat{c}=\widehat{c}\cdot \widehat{a}=cos\theta \left(say\right)$  

$\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|=14$

$\left(\overrightarrow{a}*\overrightarrow{b}\right)\cdot \left(\overrightarrow{b}*\overrightarrow{c}\right)$            

  $=\overrightarrow{a}\cdot \left[\left(\overrightarrow{b}\cdot \overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{b}\cdot \overrightarrow{b}\right)\overrightarrow{c}\right]$             

$=\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)\left(\overrightarrow{b}\cdot \overrightarrow{c}\right)-{\left|\overrightarrow{b}\right|}^2\overrightarrow{a}\cdot \overrightarrow{c}$          

  $=\left|\overrightarrow{a}\right|{\left|\overrightarrow{b}\right|}^2\left|\overrightarrow{c}\right|\left(co{s}^2\theta -cos\theta \right)=14\left|\overrightarrow{b}\right|\left(co{s}^2\theta -cos\theta \right)$            

Similarly,  $\left(\overrightarrow{b}*\overrightarrow{c}\right)\cdot \left(\overrightarrow{c}*\overrightarrow{a}\right)$

=  $\left|\overrightarrow{b}\right|{\left|\overrightarrow{c}\right|}^2\left|\overrightarrow{a}\right|\left(co{s}^2\theta -sin\theta \right)=14\left|\overrightarrow{c}\right|\left(cos2\theta -sin\theta \right)\&\left(\overrightarrow{c}*\overrightarrow{a}\right)\cdot \left(\overrightarrow{a}*\overrightarrow{b}\right)$  

$=\left|\overrightarrow{c}\right|{\left|\overrightarrow{a}\right|}^2\left|\overrightarrow{b}\right|\left(co{s}^2\theta -sin\theta \right)=14\left|\overrightarrow{a}\right|\left(co{s}^2\theta -sin\theta \right)$

Given : 14  $\left(co{s}^2\theta -cos\theta \right)\left(\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|+\left|\overrightarrow{c}\right|\right)=168$  

$\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|+\left|\overrightarrow{c}\right|=\frac{12}{co{s}^2\theta -cos\theta }=\frac{12}{\frac{1}{4}-\left(-\frac{1}{2}\right)}=\frac{12}{\frac{3}{4}}=16$

Given :  $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$  are coplanar & pair wise equal angle.