Class 12th

$\Delta -\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 2\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 0\hfill & \hfill -8\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|,\left[R_2-R_1-2R_3\right]$

= -8 (-3 + k)

For inconsistent $\Delta =0\Rightarrow k=3$  

${\Delta }_x=\left|\begin{array}{ccc}\hfill 10\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 6\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 5m\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=32-40m\ne 0\Rightarrow m\ne \frac{4}{5}$              

$A=A_0{e}^{-\lambda t_1}$ [Radio active decay law]

$\frac{A}{5}=A_0{e}^{-\lambda \left(t_2-t_1\right)}$

$\Rightarrow ln5=\lambda \left(t_2-t_1\right)$

$\Rightarrow Averagelife=\frac{1}{\lambda }=\frac{\left(t_2-t_1\right)}{\mathcal{l}n5}$

y = x³

${\frac{dy}{dx}=3x^2\Rightarrow \frac{dy}{dx}|}_{\left(t,t^3\right)}=3t^2$        

Equation of tangent y – t³ = 3t² (x – t)  

Let again meet the curve at $Q\left(t_1,t_1^3\right)$  

$\Rightarrow t_1^3-t^3=3t^2\left(t_1-t\right)$           

  $t_1^2+t{t}_1+t^2=3t^2\left[?t_1\ne t\right]$          

$t_1^2+t{t}_1-2t^2=0$            

->t₁ = -2t

Required ordinate =    $\frac{2t^3+t_1^3}{3}=\frac{2t^3-8t^3}{3}=-2t^3$

$\frac{dp}{dt}=0.5p-450andP\left(0\right)=850$        

$\Rightarrow \frac{dp}{P-900}=0.5dt$

${\displaystyle \underset{850}{\overset{0}{\int }}\frac{dp}{P-900}={\displaystyle \underset{0}{\overset{T}{\int }}0.5dt}}$           

  $\Rightarrow ln\left(P-900\right){|}_{805}^0=0.5T$          

  $\frac{T}{2}=ln\left|\frac{900}{50}\right|=ln18$

T = 2 ln 18

Lyman Series n_f = 1, ni = 2, 3.

Paschen series n_f = 3, ni = 4, 5, 6 -

$\frac{1}{{\lambda }_1}=R{Z}^2\left(\frac{1}{1^2}-\frac{1}{4^2}\right)$

$\frac{1}{{\lambda }_2}=R{Z}^2\left(\frac{1}{3^2}-\frac{1}{4^2}\right)$

$\frac{1/{\lambda }_1}{1/{\lambda }_2}=\frac{1-\frac{1}{16}}{\frac{1}{9}-\frac{1}{16}}\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=\frac{\frac{15}{16}}{\frac{7}{9*16}}=\frac{15*9}{7}\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\frac{7}{135}$

$l=l_{1}sin\omega t+l_{2}cos\omega t=\sqrt[]{l_1^2+l_2^2}\left[\left(\frac{l_1}{\sqrt[]{l_1^2+l_2^2}}\right)sin\omega t+\frac{l_2}{\sqrt[]{l_1^2+l_2^2}}cos\omega t\right]$

$=\sqrt[]{l_1^2+l_2^2}sin\left(\omega t+\alpha \right)=l_{0}sin(\omega t+\alpha$

$l_{rms}=\frac{l_0}{\sqrt[]{2}}=\frac{\sqrt[]{l_1^2+l_2^2}}{\sqrt[]{2}}$

Let total number of throws = n

Probability of getting 2 times = Probability of getting an even number 3 times.

[as probability of getting odd number = probability of getting even number = $\frac{1}{2}$ ]

Probability of getting an odd number for odd number of times =

  ${\displaystyle \underset{-a}{\overset{a}{\int }}\left(\left|x\right|+\left|x-2\right|\right)dx=22,a>2}$

${\displaystyle \underset{-a}{\overset{0}{\int }}\left(-2x+2\right)dx+{\displaystyle \underset{0}{\overset{2}{\int }}\left(x-x+2\right)dx+{\displaystyle \underset{2}{\overset{a}{\int }}\left(2x-2\right)dx=22}}}$

$\Rightarrow 2a^2+2=20\Rightarrow a^2=9\Rightarrow a=3$

$\therefore {\displaystyle \underset{3}{\overset{-3}{\int }}\left(x+\left[x\right]\right)dx=-{\displaystyle \underset{-3}{\overset{3}{\int }}\left(2x-\left\{x\right\}\right)dx=}}-{\displaystyle \underset{-3}{\overset{3}{\int }}2xdx+6}{{\displaystyle \underset{0}{\overset{1}{\int }}xdx=6.\frac{x^2}{2}}|}_0^1=3$           

$P=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill \alpha \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 0\hfill \end{array}\right]andQ=\left[q_{ij}\right]\Rightarrow PQ=k{l}_3$           

$q_{23}=-\frac{k}{8}and\left|Q\right|=\frac{k^2}{2}$            

  $PQ=k{l}_3\Rightarrow P^{-1}=\frac{Q}{k}={\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill \alpha \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 0\hfill \end{array}\right)}^{-1}$

$\left|p\right|\left|Q\right|=\left(k{l}_3\right)\Rightarrow 8.\frac{k^2}{2}=k^3$

$k\ne 0\Rightarrow k=4$

$\therefore {\alpha }^2+k^2=1+16=17.$