Maths NCERT Exemplar Solutions Class 12th Chapter One

 $tan\left(2ta{n}^{-1}\frac{1}{5}+se{c}^{-1}\frac{\sqrt[]{5}}{2}+2ta{n}^{-1}\frac{1}{8}\right)tan\left(2ta{n}^{-1}\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5}*\frac{1}{8}}+se{c}^{-1}\frac{\sqrt[]{5}}{2}\right)$

$=tan\left(ta{n}^{-1}\frac{3}{4}+ta{n}^{-1}\frac{1}{2}\right)=tan\left(ta{n}^{-1}\frac{\frac{3}{4}+\frac{1}{2}}{1-\frac{3}{8}}\right)$

$=tan\left(ta{n}^{-1}\frac{\frac{5}{4}}{\frac{5}{8}}\right)=2$

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

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

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

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

$=4-2\left(4\right)=-4$

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

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

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

Given, mean = np = . 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

he 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 $\stackrel{}{}$$\overline{QN}\left(\frac{1}{2},2,\frac{-5}{2}\right)and\overline{\left|QN\right|}=\text{exception 25:}$

Any tangent to y² = 24x at (α, β) is βy = 12 (x + α) therefore Slope = $\frac{12}{\beta }$

and perpendicular to 2x + 2y = 5 =>12 =β and α= 6 Hence hyperbola is $\frac{x^2}{6^2}-\frac{y^2}{12^2}$ = 1 and normal is drawn at (10, 16)

therefore equation of normal $\frac{36x}{10}+\frac{144y}{16}=36+144\Rightarrow \frac{x}{50}+\frac{y}{20}=1$ This does not pass through (15, 13) out of given option.

$p\iff \left(q\Rightarrow p\right)\sim \left(p\iff \left(q\iff p\right)=p\iff \sim \left(q\Rightarrow p\right)\right)$

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

$=\left(\sim p\wedge \sim q\right)$