Maths

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

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

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

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

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

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

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

$If{\widehat{n}}_1$ is a vector normal to the plane determined by $\widehat{i}and\widehat{i}+\widehat{j}then{\widehat{n}}_1=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right|=\widehat{k}$

$If{\widehat{n}}_2$ is a vector normal to the pane determined by $\widehat{i}-\widehat{j},$ and $\widehat{i}+\widehat{k}$ then ${\widehat{n}}_2$ = $\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|=\widehat{i}-\widehat{j}+\widehat{k}$

Vector $\widehat{a}$ is parallel to ${\widehat{n}}_1*{\widehat{n}}_2$ i.e. $\widehat{a}$ is parallel to $\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right|=\widehat{i}-\widehat{j}$

Given $\overrightarrow{b}=\widehat{i}-2\widehat{j}+2\widehat{k}$

consine of acute angle between $\widehat{a}and\widehat{b}$ = $\left|\frac{\widehat{a}.\widehat{b}}{\left|\widehat{a}\right|.\left|\widehat{b}\right|}\right|=\frac{1}{\sqrt[]{2}}$

obtuse angle between $\widehat{a}and\widehat{b}=\frac{3\pi }{4}$

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

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

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

Given hyperbola : $\frac{k{x}^2}{6}-\frac{y^2}{6}=1$ so eccentricity e = $\sqrt[]{1+k}$ and directrices $x=\pm \frac{a}{e}$

$\Rightarrow x=\pm \frac{\sqrt[]{6}}{\sqrt[]{k}\sqrt[]{k+1}}\Rightarrow \frac{\sqrt[]{6}}{\sqrt[]{k}\sqrt[]{k+1}}=1$

k = 2 therefore equation of hyperbola is $\frac{x^2}{3}-\frac{y^2}{6}=1$

hence it passes through the point $\left(\sqrt[]{5},-2\right)$