Maths

$2si{n}^2\theta =cos2\theta =1-2si{n}^2\theta$

$\Rightarrow 4si{n}^2\theta =1\Rightarrow sin\theta =\pm \frac{1}{2}$

$\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$

$Sum\frac{7\pi }{6}+\frac{11\pi }{6}=3\pi \Rightarrow k=3$

 ${\displaystyle \sum _{k=1}^{10}\frac{k}{k^4+k^2+1}}$

$=\frac{1}{2}{\displaystyle \sum _{k=1}^{10}\left[\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right]}$

$=\frac{1}{2}\left[1-\frac{1}{111}\right]=\frac{110}{222}=\frac{55}{111}=\frac{m}{n}$

$\therefore m+n=166$

Last two digit must be in form

$\begin{array}{l}23,4,516\\ 324\\ 32\\ 52\end{array}\}3*4=12$

Total number of required number = 12 + 18 = 30

 $0\le \frac{2+3p}{6}\le 1\Rightarrow p\in \left[-\frac{2}{3},\frac{4}{3}\right],0\le \frac{2-p}{8}\le 1\Rightarrow p\in \left[-6,2\right]and0\le \frac{1-p}{2}\le 1\Rightarrow p\in \left[-1,1\right]$

$0<P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)\le 10\le \frac{13}{12}-\frac{p}{8}\le 1\Rightarrow p\in \left[\frac{2}{3},\frac{26}{3}\right]$ Taking intersection to all $p\in \left[\frac{2}{3},1\right]$

$\therefore p_1+p_2=\frac{5}{3}$

The normal vector to the plane is $\overline{n_1}*\overline{n_2}=\left|\begin{array}{ccc}\hfill i\hfill & \hfill 3\hfill & \hfill k\hfill \\ \hfill 1\hfill & \hfill a\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill -a\hfill \end{array}\right|=\left(1-a\right)\widehat{i}+\widehat{j}+\widehat{k}$

$\therefore equationofplaneis\left(1-a\right)\left(x-1\right)+\left(y-1\right)+z=0$

(1 – a)x + y + z = 2 – a …… (i)

Now distance from (2, 1, 4) = $\sqrt[]{3}$

$\Rightarrow \sqrt[]{3}=\left|\frac{2\left(1-a\right)+1+4-\left(2-a\right)}{\sqrt[]{{\left(1-a\right)}^2+1+1}}\right|$

$\Rightarrow a^2+2a-8=0\Rightarrow a=-4,2$ the largest value of a = 2.

Given, mean = np = . and variance = npq = $\frac{\alpha }{3}\Rightarrow q=\frac{1}{3}andp=\frac{2}{3}$

$P\left(X=1\right)=n{p}^1{q}^{n-1}=\frac{4}{243}\Rightarrow n{\left(\frac{2}{3}\right)}^1{\left(\frac{1}{3}\right)}^{n-1}=\frac{4}{243}\Rightarrow n=6$

$P\left(X=4or5\right){=}^6{C}_4{\left(\frac{2}{3}\right)}^4{\left(\frac{1}{3}\right)}^2{+}^6{C}_5{\left(\frac{2}{3}\right)}^5{\left(\frac{1}{3}\right)}^1=\frac{16}{27}$

 $C\subseteq AandC\cap B=\phi$

If C is formed only by {1, 2, 4, 5} total number of subsets of A = 27.

Total number of subsets of {1, 2, 4, 5} = 24

$\therefore$ Number of subsets where $C\cap B\ne \phi$

= 27 – 24 = 112

Given : $\overrightarrow{a}=\left(\alpha ,1,-1\right)and\overrightarrow{b}=\left(2,1,-\alpha \right)\overrightarrow{c}=\overrightarrow{a}*\overrightarrow{b}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill \alpha \hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill -\alpha \hfill \end{array}\right|$

$=\left(-\alpha +1\right)\widehat{i}+\left({\alpha }^2-2\right)\widehat{j}+\left(\alpha -2\right)\widehat{k}$ Projection of $\overrightarrow{c}$ on $\overrightarrow{d}=-\widehat{i}+2\widehat{j}-2\widehat{k}=\left|\overrightarrow{c}.\frac{\overrightarrow{d}}{\left|\overrightarrow{d}\right|}\right|=30\left\{Given\right\}$

$\Rightarrow \left|\frac{\alpha -1-4+2{\alpha }^2-2\alpha +4}{\sqrt[]{1+4+4}}\right|=30$

On solving $\alpha =\frac{-13}{2}$ (Rejected as > 0) and = 7

 $Area=2{\displaystyle \underset{0}{\overset{\sqrt[]{2}}{\int }}\left(1-\left|x^2-1\right|\right)dx=2\left[{\displaystyle \underset{0}{\overset{1}{\int }}\left(1-\left(1-x^2\right)\right)dx+{\displaystyle \underset{1}{\overset{\sqrt[]{2}}{\int }}\left(2-x^2\right)dx}}\right]}=\frac{8}{3}\left(\sqrt[]{2}-1\right)$