Class 12th

Slope of any point P (x, y) to y = f (x) is $\frac{dy}{dx}=-k\frac{y}{x}$

$\Rightarrow \frac{dy}{y}+k\frac{dx}{x}=0$

Solving the equation the curve is xky = c

It passes (1, 2) c = 2 xky = 2 again it passes (8, 1) 8k = 2 k = $\frac{1}{3}$

$\therefore$ the equation of curve is $x^{1/3}$ y = 2 …… (i)

$\therefore \left|y\left(\frac{1}{8}\right)\right|=\left|\frac{2}{{\left(\frac{1}{8}\right)}^{1/3}}\right|=4$

$A=\frac{A_0}{16}=\frac{A_0}{2^{30/t_{1/2}}}\Rightarrow 2^4=2^{30/t_{1/2}}$

$\Rightarrow \frac{A_0}{16}=\frac{A_0}{2^{30/t_{1/2}}}\Rightarrow 2^4=2^{30/t_{1/2}}$

$\Rightarrow t_{1/2}=\frac{30}{4}=7.5$ years

$f\left(x\right)=e^x.{\displaystyle \underset{0}{\overset{x}{\int }}\frac{f\text{'}\left(t\right)}{e^t}dt}$

$f\text{'}\left(x\right)=e^x.{\displaystyle \underset{0}{\overset{x}{\int }}\frac{f\text{'}\left(t\right)}{e^t}dt+e^x.\frac{f\text{'}\left(x\right)}{e^x}-\left[\left(2x-1\right).e^x+\left(x^2-x+1\right).e^x\right]}$

$\begin{array}{l}f\left(x\right)=\left(2x+1\right).e^x-2e^x+c\\ f\left(x\right)=e^x\left(2x-1\right)c=0\end{array}$

$f\left(-\frac{1}{2}\right)=\frac{-2}{\sqrt[]{e}}$

Given that   $\overrightarrow{v}=v_{0}iand\overrightarrow{E}=-E_0\widehat{i}$

  $\overrightarrow{v}\left(t\right)=v_0\widehat{i}+\frac{eEt}{m}\widehat{i}$              

So, de Broglie wave length,   $\lambda =\frac{h}{mv}$

$\Rightarrow \lambda =\frac{h}{m{v}_0\left(1+\frac{q{E}_0}{m{v}_0}t\right)}=\frac{{\lambda }_0}{\left(1+\frac{q{E}_0}{m{v}_0}t\right)}$

Case I

= = 0 then

f (x) = yx

$g\left(x\right)=\frac{x}{y}$

$\frac{1}{x}{\displaystyle \sum _{i=1}^{n}f\left(a_i\right)\Rightarrow \frac{y}{x}\left(a_1+a_2+....+a_x\right)=0}$

$\Rightarrow f\left(g\left(0\right)\right)\Rightarrow f\left(0\right)$

=0

$\frac{5x^2}{2}+\frac{\alpha }{2x^5}+\frac{\alpha }{2x^5}\ge 7{\left(\frac{{\alpha }^2}{2^7}\right)}^{\frac{1}{7}}$

$\frac{7.{\left(\alpha \right)}^{2/7}}{2}=14$

${\left({\alpha }^2\right)}^{\frac{1}{7}}=2^2\Rightarrow \alpha ={\left(2^2\right)}^{\frac{7}{2}}=2^7$

=128

Resolving power

$\frac{h-cos\theta }{2}=\frac{k-sin\theta }{3}=\frac{w-0}{1}$

$=\frac{-1\left(2cos\theta +3sin\theta -6\right)}{14}\Rightarrow h=\frac{cos-2\left(2cos\theta +3sin\theta -6\right)}{14}$

$k=\frac{5sin\theta -6cos\theta +18}{14}$

${\left(5h+6k-12\right)}^2+4{\left(3h+5k-9\right)}^2=1$

  $S_1=\left\{z,\in c:\lfloor z_1-3\rfloor =\frac{1}{2}\right\}and{S}_2=\left\{z_2\in c:\left|z_2-\left|z_2+1\right|\right|=\left|z_2+\left|z_2-1\right|\right|\right\}$

Then, for $z_1\in S_1$  and $z_2\in S_2,$  the least value of |z₂ – z₁|

${\left|z_2+\left|z_2-1\right|\right|}^2={\left|z_2-\left|z_2+1\right|\right|}^2$      

$\Rightarrow \left(z_2+{\overline{z}}_2\right)\left(\left|z_2-1\right|+\left|z_2+1\right|-2\right)=0$      

$\therefore z_2+{\overline{z}}_2=0or\left|z_2-1\right|+\left|z_2+1\right|-2=0$          

$\therefore$ z₂ lies on imaginary axis or on real axis with in [-1, 1]

also $\left|z_1-3\right|=\frac{1}{2}$  lie on circle having centre 3 and radius   $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Clearly $\left|z_1-z_2\right|min=\frac{5}{2}-1=\frac{3}{2}$