Class 12th

$S=\left\{\theta \in \left[0,2\pi \right]:8^{2si{n}^{2}x}+8^{2co{s}^{2}x}=16\right\}$

Now apply AM $\ge GM$ for $\frac{8^{2si{n}^{2}x}+8^{2co{s}^{2}x}}{2}\le {\left(8^{2si{n}^{2}x+2co{s}^{2}x}\right)}^{\frac{1}{2}}\Rightarrow 8^{2si{n}^{2}x}=8^{2co{s}^{2}x}$  

Þ $si{n}^2\theta =co{s}^2\theta$               $\therefore \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$

$=4+\left[cosec\left(\frac{\pi }{2}+\pi \right)+cosec\left(\frac{\pi }{2}+3\pi \right)+cosec\left(\frac{\pi }{2}+5\pi \right)+cosec\left(\frac{\pi }{2}+7\pi \right)\right]$

$=4-2\left(4\right)=-4$  

$\frac{sin{\theta }_C}{V_B}=\frac{sin90°}{V_A}\Rightarrow sin{\theta }_C=\frac{V_B}{V_A}=\frac{1.5*10^{10}}{2.0*10^{10}}=\frac{3}{4}$                    

According to question, we can write

$\theta >\theta =si{n}^{-1}\left(\frac{3}{4}\right)$

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

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

$\therefore p_1+p_2=\frac{5}{3}$  

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

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

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

According to question, we can write

^{$F_{CA}=F_{CB}=\frac{KqQ}{x^2+{\left(\frac{d}{2}\right)}^2}\Rightarrow F=2F_{CA}cos\theta =\frac{2KqQx}{{\left[x^2+{\left(\frac{d}{2}\right)}^2\right]}^{\frac{3}{2}}}$}                         

For maxima of force  $\frac{dF}{dx}=0,so$  

$x=\frac{d}{2\sqrt[]{2}}$

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

The line x + y – z = 0 = x – 2y + 3z – 5 is parallel to the vector

  $\overrightarrow{b}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 3\hfill \end{array}\right|=\left(1,4,-3\right)$ Equation of line through P(1, 2, 4) and parallel to $\overrightarrow{b}\frac{x-1}{1}=\frac{y-2}{-4}=\frac{z-4}{-3}$  

Let  $N\equiv \left(\lambda +1,-4\lambda +2,-3\lambda +4\right)\overline{QN}=\left(\lambda ,-4\lambda +4,-3\lambda -1\right)$  

$\overline{QN}$ is perpendicular to $\overrightarrow{b}\Rightarrow \left(\lambda ,-4\lambda +4,-3\lambda -1\right).\left(1,4,-3\right)=0\Rightarrow \lambda =\frac{1}{2}.$  

Hence  $\overline{QN}\left(\frac{1}{2},2,\frac{-5}{2}\right)and\overline{\left|QN\right|}=\sqrt[]{\frac{21}{2}}$  

$\frac{dy}{dx}+2ytanx=sinx,\therefore I.F.e^{{\displaystyle \int 2tanxdx}}=se{c}^{2}x$  

= cos x – 2 cos² x = $-2{\left(cosx-\frac{1}{4}\right)}^2+\frac{1}{8}\therefore y_{max}=\frac{1}{8}$

$a=\underset{n\to \infty }{lim}{\displaystyle \sum _{k=1}^n\frac{2n}{n^2+k^2}=\underset{n\to \infty }{lim}\frac{1}{n}}{\displaystyle \sum _{k=1}^n\frac{2}{1+{\left(\frac{k}{n}\right)}^2}}$

 $\therefore a={\displaystyle {\int }_0^1\frac{2}{1+x^2}dx=2ta{n}^{-1}x}]{}_0^1=\frac{\pi }{2}$

⇒ $f\text{'}\left(\frac{a}{2}\right)=\sqrt[]{2}f\left(\frac{a}{2}\right)$