Maths

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

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

$\therefore Y=\alpha l+\beta X+\gamma X^2=\left[\begin{array}{ccc}\hfill \alpha \hfill & \hfill \beta \hfill & \hfill \gamma \hfill \\ \hfill 0\hfill & \hfill \alpha \hfill & \hfill \beta \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \alpha \hfill \end{array}\right]$

$\begin{array}{l}\therefore \alpha =5,\beta =10,y=15\\ \therefore {\left(\alpha -\beta +y\right)}^2=100\end{array}$

$z^2+z+1=0\Rightarrow z=w,w^2$

$\left|{\displaystyle \sum _{n=1}^{15}{\left(z^n+{\left(-1\right)}^n\frac{1}{z^n}\right)}^2}\right|=\left|{\displaystyle \sum _{n=1}^{15}\left(z^{2n}+\frac{1}{z^{2n}}+2{\left(-1\right)}^n\right)}\right|=\left|{\displaystyle \sum _{n=1}^{15}{w}^{2n}+\frac{1}{w^{2n}}+2{\left(-1\right)}^n}\right|$

$=\left|\frac{w^2\left(1-1\right)}{1-w^2}+\frac{\frac{1}{w^2}\left(1-1\right)}{1-\frac{1}{w^2}}-2\right|=2$

  $?p+q=3$ $$ ….(i)

and $p^4+q^4$ ${}^{}{}^{}$ = 369 ….(ii)

${\left\{{\left(p+q\right)}^2-2pq\right\}}^2-2p^2{q}^2=369$

or ${\left(9-2pq\right)}^2-2{\left(pq\right)}^2=369$ ${}^{}{}^{}$

or ${\left(pq\right)}^2-18pq-144=0$ ${}^{}$

$\therefore pq=-6or24$

But pq 24 is not possible

$\therefore pq=-6$

$Hence,{\left(\frac{1}{p}+\frac{1}{q}\right)}^{-2}={\left(\frac{pq}{p+q}\right)}^2={\left(-2\right)}^2=4$

$f\left(x-y\right)=2^{x}f\left(y\right)+4^{y}f\left(x\right)$ ……. (i)

$$ $f\left(y+x\right)=2^{y}f\left(x\right)+4^{x}f\left(y\right)$  ……. (ii)

(i) – (ii)

$\therefore \frac{f\text{'}\left(4\right)}{f\text{'}\left(2\right)}=\frac{k\left(16\mathcal{l}n2-256\mathcal{l}n4\right)}{k\left(4\mathcal{l}n2-16\mathcal{l}n4\right)}$

$=\frac{\left(16-512\right)\mathcal{l}n2}{\left(4-32\right)\mathcal{l}n2}=\frac{496}{28}$

$14\frac{f\text{'}\left(4\right)}{f\text{'}\left(2\right)}=248$

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.

$co{s}^{-1}\left(\frac{3}{10}cos\left(ta{n}^{-1}\left(\frac{4}{3}\right)\right)+\frac{2}{5}sin\left(ta{n}^{-1}\left(\frac{4}{3}\right)\right)\right)$

$=co{s}^{-1}\left(\frac{3}{10}.\frac{3}{5}+\frac{2}{5}.\frac{4}{5}\right)$

$=co{s}^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$

$16sin20°.sin40°.sin80°$

$=4sin60°\left\{?4sin\theta .sin\left(60°-\theta \right).sin\left(60°+\theta \right)=sin3\theta \right\}=2\sqrt[]{3}$

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

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

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

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 :

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