Class 12th

= 939

$\therefore r=4$

$?7n-nr-5r=0$

And r = 4 then

$n>\frac{20}{3}$

And r should not be 5

$\therefore n<\frac{25}{2}$

$\therefore$ possible values of n are 7, 8, 9, 10, 11, 12

$\therefore$ Sum of integral value of n = 57

Kindly go through the solution

Sum of all entries of matrix A must be prime p such that 2 < p < 8 then sum of entries may be 3, 5 or 7

If sum is 3 then possible entries are

(0, 5), (0, 1, 4), (0, 2, 3), (0, 1, 3)

(0, 1, 2, 2) and (1, 2)

$\therefore$ Total number of matrices = 4 + 12 + 12 + 12 + 12 + 4 = 56

If sum is 7 then possible entries are

(0, 2, 25), (0, 03, 4), (0, 1, 5), (0, 3, 1), (0, 2, 3), (1, 4), (1, 2, 2), (1, 2, 3) and (0, 1, 2, 4)

Total number of matrices with sum 7 = 104

$\therefore$ total number of required matrices

= 20 + 56 = 104

= 108

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.