Maths

Kindly go through the solution

= 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

$\begin{array}{l}\left(i\right){\left|a{z}_1?b{z}_2\right|}^2+{\left|b{z}_1+a{z}_2\right|}^2\\ \text{?}\text{?}\text{?}\text{?}\text{?}\text{?}={\left|a{z}_1\right|}^2+{\left|b{z}_2\right|}^2?2Re\left(a{z}_1.b{\overline{z}}_2\right)+{\left|b{z}_1\right|}^2+{\left|a{z}_2\right|}^2+2Re\left(a{z}_1.b{\overline{z}}_2\right)\\ \text{?}\text{?}\text{?}\text{?}\text{?}\text{?}={\left|a{z}_1\right|}^2+{\left|b{z}_2\right|}^2+{\left|b{z}_1\right|}^2+{\left|a{z}_2\right|}^2\\ \text{?}\text{?}\text{?}\text{?}\text{?}\text{?}=\left(a^2+b^2\right)\left({\left|z_1\right|}^2+{\left|z_2\right|}^2\right)\\ \text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}Hence,\text{?}\text{?}the\text{?}\text{?}value\text{?}\text{?}of\text{?}\text{?}the\text{?}\text{?}filler\text{?}\text{?}is\text{?}\text{?}\left(a^2+b^2\right)\left({\left|z_1\right|}^2+{\left|z_2\right|}^2\right).\\ \left(ii\right)\sqrt[]{?25}*\sqrt[]{?9}=\sqrt[]{?1}.\sqrt[]{25}*\sqrt[]{?1}\sqrt[]{9}\\ \text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}=5i*3i=15i^2=?15\\ \text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}Hence,\text{?}\text{?}the\text{?}\text{?}value\text{?}\text{?}of\text{?}\text{?}the\text{?}\text{?}filler\text{?}\text{?}is\text{?}\text{?}?15.\end{array}$

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

$\alpha ,\beta$ 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

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

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

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

$=98$             

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

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

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

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.