Class 12th

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 $\Rightarrow a=-\frac{1}{2}\Rightarrow b=\frac{3}{2}$

${\displaystyle \underset{0}{\overset{\pi /2}{\int }}\left(\frac{1}{\left(1+ta{n}^{2}x\right)\left(1+2^x\right)}+\frac{1}{\left(1+ta{n}^{2}x\right)\left(1+2^{-x}\right)}\right)dx}$

${\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{dx}{\left(1+ta{n}^{2}x\right)}={\displaystyle \underset{0}{\overset{\pi /2}{\int }}co{s}^{2}xdx=\frac{\pi }{4}}}$

Kindly consider the following figure

Eq. = no. of faraday

$\frac{10*0.2*10^{-2}*19.7}{197}*3=\frac{0.193*t}{96500}$

$\sqrt[]{2}a=\frac{258.36}{\sqrt[]{2}}$

$\alpha =258.36=2r$