Class 12th

${\int }_0^3?g\left(x\right)-f\left(x\right)={\int }_0^3?\left|\right|x-2|-2|dx-{\int }_0^3?|x-2|dx$

$=\left(\frac{1}{2}*2*2+1+\frac{1}{2}*1*1\right)-\left(\frac{1}{2}*2*2+\frac{1}{2}*1*1\right)$

$=\left(2+1+\frac{1}{2}\right)-\left(2+\frac{1}{2}\right)=1$

P (E_n) = n/36 for n = 1, 2, 3, …., 8

$P\left(A\right)=\frac{Anypossiblesumof\left(1,2,3,.........,8\right)\left(=\alpha say\right)}{36}$

$\frac{\alpha }{36}\ge \frac{4}{5}$

$\therefore a\ge 29$

If one of the number from {1, 2, ….8} is left then total $\ge$ 29 by 3 ways

Similarly by leaving terms more 2 or 3 we get 16 more combinations

$\therefore$ Total number of different set a possible is 16 + 3

= 19

The direction of ratios of the lines,   $\frac{x-3}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}\&\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ , are $-3,2k,2and3k,1,-5$ respectively.

It is known that two lines with direction ratios,   $a_1, b_1, c_1 and a_2, b_2,c_2$ , are perpendicular, if  $a_1{a}_2 + b_1{b}_2 + c_1{c}_2 =0$

$\begin{array}{l}\therefore -3(3k)+2k*1+2(-5)=0\\ \Rightarrow -9k+2k-10=0\\ \Rightarrow 7k=-10\\ \Rightarrow k=\frac{-10}{7}\end{array}$

Therefore, for k= -10/7, the given lines are perpendicular to each other.

$L_1:4x+3y+2=0$  

$L_2:3x-4y-11=0$               

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