Class 12th

$\begin{array}{ccc}\hfill 3R\hfill & \hfill 2R\hfill & \hfill 4B\hfill \\ \hfill 5B\hfill & \hfill 3W\hfill & \hfill 2W\hfill \end{array}$  

I             II

Let

E₁ : a red ball is transferred from I to II

E₂ : a black is transferred from I to II

E₃ :a white transferred from I to II

E : a black ball is drawn from 2^nd bag after a ball from I to II was transferred.

$P\left(\frac{E_1}{E}\right)=\frac{P\left(E_1\cap E\right)}{P\left(E\right)}$  

$P\left(E\right)=P\left(E_1\cap E\right)+P\left(E_2\cap E\right)+P\left(E_3\cap E\right)$  

=  $P\left(E_1\right)\cdot P\left(\frac{E}{E_1}\right)+....+.....$  

=   $\frac{3}{10}\cdot \frac{5}{10}+\frac{4}{10}\cdot \frac{6}{10}+\frac{3}{10}\cdot \frac{5}{10}=\frac{54}{100}$

$P\left(E_1/E\right)=\frac{15/100}{54/100}=\frac{5}{18}$

Final range = 3 Initial Range

$\Rightarrow d_f=3d_i$ $\begin{array}{l}\Rightarrow \sqrt[]{2h\text{'}R}=3\sqrt[]{2hR}\\ \Rightarrow h\text{'}=9h=900m\end{array}$

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