Class 12th

$\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{1}{4},\frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{1}{12}$

divide both $\frac{P\left(B\right)}{P\left(A\right)}=\frac{1}{3}$

$P\left(A\right)=\frac{1}{11},P\left(B\right)=\frac{1}{33}$

$\frac{dy}{dx}=3a{x}^2-2bx$

${\frac{dy}{dx}|}_{x=1}=3a-2b=3$

$a=\frac{2b+3}{3}\in \left[1,2\right]$

$2b+3\in \left[3,6\right]$

$2b\in \left[0,3\right]$

$b\in \left[0,\frac{3}{2}\right]$

${\displaystyle \int e^{sinx}}\left(\frac{5+co{s}^{2}x}{\left(2-sinx\right)\sqrt[]{3+co{s}^{2}x}}\right)cosxdx$

put sin x = t

${\displaystyle {}^{}}\frac{{}^{}}{\text{exception 25:}}$  ${\displaystyle \int e^t}\left(\frac{6-t^2}{\left(2-t\right)\sqrt[]{4-t^2}}\right)$ dt

${\displaystyle \int e^t}\left(\sqrt[]{\frac{2+t}{2-t}}+\frac{2}{{\left(2-t\right)}^{3/2}{\left(2+t\right)}^{1/2}}\right)dt$

If $g\left(t\right)=\sqrt[]{\frac{2+t}{2-t}},g\text{'}\left(t\right)=\frac{2}{{\left(2-t\right)}^{3/2}{\left(2+t^2\right)}^{1/2}}$ $\text{exception 25:}\frac{}{{}^{}{{}^{}}^{}}$

$e^t\sqrt[]{\frac{2+t}{2-t}}+c$

$f\left(\frac{\pi }{2}\right)=\sqrt[]{3}e$

When the ring rotates about its axis with a uniform frequency fHz, the current flowing in the ring is

I=q/T=qf

Magnetic field at the centre of the ring is

Line L is   $\frac{x-1}{1}=\frac{y-2}{2\sqrt[]{2}}=\frac{z}{0}$

this line L makes an angle of 45° with the plane  $\sqrt[]{2}x+y-z=1$

$\therefore$ Required distance PQ is perpendicular distance of plane from P is i.e.,

$PQ=\frac{\left|\sqrt[]{2}+7-\sqrt[]{2}-1\right|}{\sqrt[]{2+1+1}}=3$

$\left|A-xI\right|=\left|\begin{array}{cc}\hfill 1/2-x\hfill & \hfill 1/2\hfill \\ \hfill a\hfill & \hfill b-x\hfill \end{array}\right|=x^2-\left(b+\frac{1}{2}\right)x+\frac{b-a}{2}$

from Cayley – Hamilton Theorem

$A^2-\left(b+\frac{1}{2}\right)A+\left(\frac{b-a}{2}\right)I=0$

$\Rightarrow A^3=\left({\left(b+\frac{1}{2}\right)}^2-\left(\frac{b-a}{2}\right)\right)A-\left(b+\frac{1}{2}\right)\left(\frac{b-a}{2}\right)I$

$\therefore {\left(b+\frac{1}{2}\right)}^2-\left(\frac{b-a}{2}\right)=1\&\left(b+\frac{1}{2}\right)\left(\frac{b-a}{2}\right)=0$

$\therefore \left(a,b\right)=\left(\frac{1}{2},\frac{1}{2}\right),\left(-\frac{3}{2},-\frac{3}{2}\right),\left(\frac{3}{2},-\frac{1}{2}\right)$

$xdy+ydx=\frac{xydx}{\sqrt[]{1-x^2}}$

$\frac{d\left(xy\right)}{xy}=\frac{dx}{\sqrt[]{1-x^2}}$

Integrate both sides we get

logexy = sin1 x + C

$y^2=x,a=\frac{1}{4}$

$x^2=-y,b=\frac{1}{4}$

$A=\frac{16\left|ab\right|}{3}=\frac{1}{3}$

$\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=0$

$\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 6\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill \alpha +1\hfill & \hfill \beta -1\hfill & \hfill 1\hfill \end{array}\right|=0$

$8\beta -24=0\Rightarrow \beta =3$

$\left|\overrightarrow{c}\right|=\sqrt[]{6}$

${\left(\alpha +1\right)}^2+{\left(\beta -1\right)}^2+1=6\Rightarrow {\left(\alpha +1\right)}^2=1$

a = 1 = 1, 1 $\Rightarrow \alpha =-2,0$ $$

+ = 1, 3

$f\left(x\right)=[\begin{array}{l}-\frac{1}{x}x\in \left(-\infty ,-1\right)\\ a{x}^2+bx\in \left(-1,1\right)\\ \frac{1}{x}x\in \left[1,\infty \right)\end{array}$

cont at x = 1, a + b = 1

diff at x = 1, 2a = 1 $\Rightarrow a=-\frac{1}{2}\Rightarrow b=\frac{3}{2}$