Class 12th

eq of ${\mathrm{H}}_2{\mathrm{O}}_2=$ eq of ${\mathrm{K}\mathrm{M}\mathrm{n}\mathrm{O}}_4$

$\begin{array}{rr}& =\frac{0.316}{158}*5=0.01\\ & =\frac{0.01*17}{0.2}*100=85\mathrm{\%}\end{array}$

$\alpha ,\beta$ are roots of x² - $4\lambda x+5=0$  

$\therefore \alpha +\beta =4\lambda and\alpha \beta =5$               

Also $\alpha ,\gamma$  are roots of

$x^2-\left(3\sqrt[]{2}+2\sqrt[]{3}\right)x+7+3\sqrt[]{3}\lambda =0,\lambda >0$

$\therefore \alpha +\gamma =3\sqrt[]{2}+2\sqrt[]{3},\alpha \gamma =7+3\sqrt[]{2\lambda }$               

$?\alpha$ is common root

$\therefore {\alpha }^2-4\lambda \alpha +5=0$     …….(i)

And

${\alpha }^2-\left(3\sqrt[]{2}+2\sqrt[]{3}\right)\alpha +7+3\sqrt[]{3}\lambda =0$      ….(ii)

From (i) – (ii) ; we get

$\alpha =\frac{2+3\sqrt[]{3}\lambda }{3\sqrt[]{2}+2\sqrt[]{3}-4\lambda }$               

$?\beta +\gamma =3\sqrt[]{2}$               

$\therefore {\left(\alpha +2\beta +\gamma \right)}^2={\left(\alpha +\beta +\beta +\gamma \right)}^2$               

$=98$             

Kindly go through the solution

${\mathrm{N}}_2+3{\mathrm{H}}_2?2{\mathrm{N}\mathrm{H}}_3$

$\begin{array}{ll}\frac{2.8*{10}^3}{28}& \frac{1*{10}^3}{2}\\ \frac{0.1*{10}^3\mathrm{m}\mathrm{o}\mathrm{l}}{\mathrm{L}\mathrm{R}}& 0.5*{10}^3\mathrm{m}\mathrm{o}\mathrm{l}\end{array}$

$\text{Mass of}{\mathrm{N}\mathrm{H}}_3\text{produced}=0.2*{10}^3*17=3.4\mathrm{k}\mathrm{g}=3400\mathrm{g}$

. $?f\left(n\right)=\{\begin{array}{c}\hfill 2n,n=1,2,3,4,5\hfill \end{array}$

$\begin{array}{c}\hfill \mathrm{ \; \; \; \; \; \; \; \; \; \; 2}n-11,n=6,7,8,9,10\hfill \end{array}$

$\therefore f\left(1\right)=2,f\left(2\right)=4,....,f\left(5\right)=10$

And f(6) = 1, f(7) = 3, f(8) = 5, …., f(10) = 9

Now,

$f\left(g\left(n\right)\right)=\{\begin{array}{cc}\hfill n+1\hfill & \hfill ifnisodd\hfill \\ \hfill n-1,\hfill & \hfill ifniseven\hfill \end{array}$

$\therefore f\left(g\left(10\right)\right)=9\Rightarrow g\left(1\right)=1$

$\therefore g\left(10\right)\left(g\left(1\right)+g\left(2\right)+g\left(3\right)+g\left(4\right)+g\left(5\right)\right)=190$

 $=\frac{4+5+6+6+7+8+x+y}{8}=6$

$\therefore x+y=12$ …. (i)

And variance

$=\frac{2^2+1^2+0^2+0^2+1^2+2^2+{\left(x-6\right)}^2+{\left(y-6\right)}^2}{8}$

$=\frac{9}{4}$

$\therefore {\left(x-6\right)}^2+{\left(y-6\right)}^2=8$ ……. (ii)

From (i) and (ii)

x = 4 and y = 8

$\therefore x^4+y^2=320$

 $?\left(\left(\sim q\right)\wedge p\right)\vee \left(pv\left(\sim p\right)\right)$

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

$\equiv t$

$\therefore$ option (C) is correct.

$\alpha =sin36°=x\left(say\right)$

$\therefore x=\frac{\sqrt[]{10-2\sqrt[]{5}}}{4}$

$\Rightarrow 16x^2=10-2\sqrt[]{5}$

$\Rightarrow 16x^4-80x^2+20=0$

$\therefore 4x^4-20x^2+5=0$

$cot\left({\displaystyle {\sum }_{n=1}^{50}ta{n}^{-1}}\left(\frac{1}{1+n+n^2}\right)\right)$

$=cot\left(ta{n}^{-1}51-ta{n}^{-1}1\right)$

$=cot\left(co{t}^{-1}\left(\frac{52}{50}\right)\right)$

$=\frac{26}{25}$

The required probability

$=\frac{AreaofRegionPQCAP}{AreaofRegionABCA}$

$=\frac{\frac{1}{2}*8*6-\frac{1}{2}*2*4}{\frac{1}{2}*8*6}$

$=\frac{5}{6}$