Class 12th

$L_1:4x+3y+2=0$  

$L_2:3x-4y-11=0$               

Since circle C touches the line L2 at Q intersection point L1 and L2 is (1, -2)

P lies of L₁

$\therefore P\left(x,-\frac{1}{3}\left(2+4x\right)\right)$               

Now,

PQ = 5 ? (x – 1)² + ${\left(\frac{4x+2}{3}-2\right)}^2=25$  

$\Rightarrow x=4,-2$                     

$?$ The circle lies below the axis

y = -6

p (4, -6)

Now distance of P from 5x – 12 y + 51 = 0

$=\left|\frac{20+72+51}{13}\right|=\frac{143}{13}=11$                

$x^2-3x+p=0$

$\alpha ,\beta ,\gamma ,\delta$ in G.P.

$\alpha +\alpha r=3$

$x^2-6x+q=0$

$\alpha r^2+\alpha r^3=6$

$\left(2\right)÷\left(1\right)\Rightarrow r^2=2$

So,  $\frac{2q+p}{2q-p}=\frac{2r^5+r}{2r^5-r}=\frac{2r^4+1}{2r^4-1}=\frac{9}{7}$

$\left(1-x^2\right)dy=\left(xy+\left(x^3+2\right)\sqrt[]{1-x^2}\right)dx$

$\therefore \frac{dy}{dx}-\frac{x}{1-x^2}y=\frac{x^3+3}{\sqrt[]{1-x^2}}$

$\therefore l.F.=e^{{\displaystyle \int \frac{X}{1-x^2}dx}}=\sqrt[]{1-x^2}$

$\therefore y\left(x\right)=\frac{x^4+12x}{4\sqrt[]{1-x^2}}$

$\therefore {\displaystyle \underset{\frac{-1}{2}}{\overset{\frac{1}{2}}{\int }}\sqrt[]{1-x^2}y\left(x\right)dx={\displaystyle \underset{\frac{-1}{2}}{\overset{\frac{1}{2}}{\int }}\left(\frac{x^4+12x}{4}\right)}dx}$

$\therefore k=\frac{1}{320}$

$\therefore k^{-1}=320$

Area of shaded region

$={\displaystyle {\int }_{-\left(\frac{1}{2}\right)}^0\frac{3}{2}\left({\left(1-x^{\frac{2}{3}}\right)}^{\frac{3}{2}}+x\right)dx+{\displaystyle {\int }_0^1{\left(1-x^{\frac{2}{3}}\right)}^{\frac{3}{2}}dx}}$

$x=si{n}^3\theta$

$\therefore dx=3si{n}^2\theta cos\theta d\theta$

$={\displaystyle {\int }_{-\frac{\pi }{4}}^{\frac{\pi }{2}}3si{n}^2\theta co{s}^4\theta d\theta +\left(0-\frac{1}{16}\right)}$

$=\frac{9\pi }{64}+\frac{1}{16}-\frac{1}{16}=\frac{36\pi }{256}=A$

$\therefore \frac{256A}{\pi }=36$

$?y\left(x\right)={\left(x^x\right)}^x$

$\therefore y=x^{x^2}$

$\therefore \frac{dy}{dx}=x^2.x^{x^2-1}-x^{x^2}\mathcal{l}nx.2x$

$\therefore \frac{dx}{dy}=\frac{1}{x^{x^2+1}\left(1+2lnx\right)}$ ….(i)

$\frac{d^{2}x}{dx}=\frac{d}{dx}\left({\left(x^{x^2+1}\left(1+2lnx\right)\right)}^{-1}\right).\frac{dx}{dy}$

$\begin{array}{l}{\left(\frac{d^{2}x}{d{y}^2}\right)}_{x=1}=-4\\ {\left(\frac{d^{2}x}{d{y}^2}\right)}_{x=1}+20=16\end{array}$

 $f\left(x\right)=\left[1+x\right]+\frac{{\alpha }^{2\left|x\right|}+\left\{x\right\}+\left[x\right]-1}{2\left[x\right]+\left\{x\right\}}$

$li{m}_{x\to 0}-f\left(x\right)=\alpha -\frac{4}{3}$

$\Rightarrow li{m}_{h\to 0}1-1+\frac{{\alpha }^{-h-1}-1-1}{-h-1}=\alpha -\frac{4}{3}$

$\therefore \frac{{\alpha }^{-1}-2}{-1}\alpha -\frac{4}{3}$

32 - 10 + 3 = 0

$\therefore \alpha =3or1/3$

$?\alpha$ in integer, hence = 3

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

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