Class 12th

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

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:}$

  ${\displaystyle \int \frac{\left(1-\frac{1}{\sqrt[]{3}}\right)\left(cosx-sinx\right)}{\left(1+\frac{2}{\sqrt[]{3}}sin2x\right)}}dx={\displaystyle \int \frac{\left(\frac{\sqrt[]{3}-1}{\sqrt[]{3}}\right)\sqrt[]{2}sin\left(\frac{\pi }{4}-x\right)}{\left(\frac{2}{\sqrt[]{3}}\right)\left(sin\frac{\pi }{3}+sin2x\right)}dx}$

$={\displaystyle \int \frac{\left(\frac{\sqrt[]{3}-1}{\sqrt[]{2}}\right)sin\left(\frac{\pi }{4}-x\right)}{\left(sin\frac{\pi }{3}+sin2x\right)}}dx={\displaystyle \int \frac{\left(\frac{\sqrt[]{3}-1}{2\sqrt[]{2}}\right)sin\left(\frac{\pi }{4}-x\right)}{sin\left(\frac{\pi }{6}+x\right)cos\left(\frac{\pi }{6}-x\right)}dx}$

$=\frac{1}{2}\left[lo{g}_e\left|tan\left(\frac{x}{2}+\frac{\pi }{12}\right)\right|-lo{g}_e\left|tan\left(\frac{x}{2}+\frac{\pi }{6}\right)\right|\right]+C=\frac{1}{2}lo{g}_e\left|\frac{tan\left(\frac{x}{2}+\frac{\pi }{12}\right)}{tan\left(\frac{x}{2}+\frac{\pi }{6}\right)}\right|+C$

$v=f\lambda$

$f=\frac{v}{\lambda }=\frac{3*10^8}{10^3*10^{-9}}=3*10^{14}{H}_z$

Channels =  $\frac{2\mathrm{\%}of3*10^{14}{H}_z}{8*10^3}$

$\frac{2\mathrm{\%}of3*10^{14}{H}_z}{8*10^3}$

$\frac{2\mathrm{\%}of3*10^{14}{H}_z}{8*10^3}$

$\frac{=\frac{2x}{100}3*10^{14}}{8*10^3}=75*10^7$

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.

Mean = 4 = $\mu$ = np

Variance = ${\sigma }^2=np\left(1-p\right)=\frac{4}{3}\Rightarrow 4\left(1-p\right)=\frac{4}{3}\Rightarrow p=\frac{2}{3}\Rightarrow n=6$

${=}^6{C}_0{\left(\frac{1}{3}\right)}^6{+}^6{C}_1{\left(\frac{2}{3}\right)}^1{\left(\frac{1}{3}\right)}^6{+}^6{C}_2{\left(\frac{2}{3}\right)}^2{\left(\frac{1}{3}\right)}^4=\frac{146}{27}$

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

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

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

Let point P : (h, k)

Therefore according to question,  ${\left(h-1\right)}^2+{\left(k-2\right)}^2+{\left(h+2\right)}^2+{\left(k-1\right)}^2=14$

$\therefore$ locus of P (h, k) is $x^2+y^2+x-3y-2=0$

Now intersection with x – axis are $x^2+x-2=0\Rightarrow x=-2,1$

Now intersection with y – axis are $y^2-3y-2=0\Rightarrow y=\frac{3\pm \sqrt[]{17}}{2}$

Therefore are of the quadrilateral ABCD is = $\frac{1}{2}\left(\left|x_1\right|+\left|x_2\right|\right)\left(\left|y_1\right|+\left|y_2\right|\right)=\frac{1}{2}*3*\sqrt[]{17}=\frac{3\sqrt[]{17}}{2}$

 $Let\frac{si{n}^{-1}x}{\alpha }=\frac{co{s}^{-1}x}{\beta }=k\Rightarrow si{n}^{-1}x+co{s}^{-1}x=k\left(\alpha +\beta \right)\Rightarrow \alpha +\beta =\frac{\pi }{2k}$

Now $\frac{2\pi \alpha }{\alpha +\beta }=\frac{2\pi \alpha }{\frac{\pi }{2k}}\Rightarrow 4k\alpha =4si{n}^{-1}x.Heresin\left(\frac{2\pi \alpha }{\alpha +\beta }\right)=sin\left(4si{n}^{-1}x\right)$

$\Rightarrow sin4\theta =4x\sqrt[]{1-x^2}\left(1-2x^2\right)$

${\beta }_1=\frac{{\lambda }_1{D}_1}{d_1}$

${\beta }_2=\frac{{\lambda }_2{D}_2}{d_2}$

$\frac{{\beta }_1}{{\beta }_2}=\frac{{\lambda }_1}{d_1}*\frac{d_2}{{\lambda }_2}$

$\frac{0.5}{{\beta }_2}=\frac{5000*d_1}{d_1*600}$

${\beta }_2=\frac{0.5*6}{10}$

2 = 0.3mm