Class 12th

f' (x) = x (1+y)
dy/ (y+1) = xdx
ln (y+1) = x²/2 + c
(0,0)
c = 0
y+1 = e^ (x²/2) ⇒ e^ (x²/2) = 2
x²/2 = k = (ln2 > 0)
x = ±√2k

${\stackrel{?}{B}}_1=\frac{E_1}{c}\mathrm{c}\mathrm{o}\mathrm{s}?(kz-\omega t){\stackrel{?}{B}}_1$    where   ${\stackrel{?}{B}}_1={\hat{c}}_1*{\stackrel{?}{E}}_1=\hat{k}*\hat{i}={\stackrel{?}{B}}_1=\hat{j}$

${\stackrel{?}{B}}_2=\frac{E_2}{c}\mathrm{c}\mathrm{o}\mathrm{s}?(kz+\omega t){\stackrel{?}{B}}_2$ where ${\stackrel{?}{B}}_2={\hat{c}}_2*{\stackrel{?}{E}}_2=-\hat{k}*\hat{j}={\stackrel{?}{B}}_2=\hat{i}$

$\therefore \mathrm{}\stackrel{?}{B}={\stackrel{?}{B}}_1+{\stackrel{?}{B}}_2$

$\stackrel{?}{B}=\frac{E_1}{c}\hat{j}\mathrm{c}\mathrm{o}\mathrm{s}?(kz-\omega t)+\frac{E_2}{c}\hat{i}\mathrm{c}\mathrm{o}\mathrm{s}?(kz+\omega t)$

f (x) = x? /20 - x? /12 + 5
f' (x) = x? /4 - x³/3 = x³ (x/4 - 1/3)
Local maxima at 0, Local minima at 4/3
f' (x) = x³ - x² = x² (x-1)
x = 1 point of inflection

dy/√ (1-y²) = dx/x²
sin? ¹ (y) = -1/x + c ⇒ c = π/2
sin? ¹ (y) = -π/3 + π/2 = π/6
y = 1/2

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

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

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

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

$V_1=\frac{k{Q}_3}{3R}+\frac{k{Q}_2}{2R}+\frac{k{Q}_1}{R}=30$

$V_2=\frac{k{Q}_3}{3R}+\frac{k{Q}_2}{2R}+\frac{k{Q}_1}{2R}=15$

$V_3=\frac{k{Q}_3}{3R}+\frac{k{Q}_2}{3R}+\frac{k{Q}_1}{3R}=10$

Eq. (1) $\&$  (2) $\to \frac{k{Q}_1}{R}=30$  

Eq. (2) $\&$  (3) $\to \frac{k{Q}_2}{6R}+\frac{k{Q}_1}{6R}=5\Rightarrow \frac{k{Q}_2}{R}=-\frac{k{Q}_1}{R}+30=0$

By (3) $\to \frac{k{Q}_3}{R}=30-\frac{k{Q}_2}{R}-\frac{k{Q}_1}{R}=0$

Potential of innermost shell (earthed) $=\frac{k{Q}_1^{\mathrm{\text{'}}}}{R}+\frac{k{Q}_2}{2R}+\frac{k{Q}_3}{3R}=0$

$\Rightarrow \mathrm{}\frac{k{Q}_1^{\mathrm{\text{'}}}}{R}=0\mathrm{}\therefore \mathrm{}{Q}_1^{\mathrm{\text{'}}}=0$

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

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

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

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

$k\ne 0\Rightarrow k=4$

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

DNA and RNA both are the types of nucleic acids.

• DNA (Deoxyribonucleic acid) store the genetic  information.

• RNA (Ribonucleic acid) helps in protein synthesis. It carry and translates the genetic codes.

  $\overrightarrow{c}=\alpha \overrightarrow{a}+\beta \overrightarrow{b}.....\left(i\right)$

  $\overrightarrow{a}.\overrightarrow{c}=7\overrightarrow{b}.\overrightarrow{c}=0$         

$\overrightarrow{a}=-\widehat{i}+\widehat{j}+\widehat{k}\Rightarrow \left|\overrightarrow{a}\right|=\sqrt[]{3}$           

$\overrightarrow{b}=2\widehat{i}+\widehat{k}\Rightarrow \left|\overrightarrow{b}\right|=\sqrt[]{5}\overrightarrow{a}.\overrightarrow{b}=-2+1=-1$           

From   $\left(i\right)\overrightarrow{a}.\overrightarrow{c}=\alpha {\left|\overrightarrow{a}\right|}^2-\beta$

$3\alpha -\beta =7..........\left(ii\right)$           

$\overline{b}.\overline{c}=\alpha \overline{b}.\overline{a}+\beta {\left|\overrightarrow{b}\right|}^2\Rightarrow -\alpha +5\beta =0.......\left(iii\right)$           

Solving $\alpha =\frac{5}{2}and\beta =\frac{1}{2}$

$\Rightarrow 2{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|}^2=75$            

f (x) = λ (x-2)²
⇒ 12 = λ (2)² ⇒ λ = 3
f (x) = 3 (x-2)² f (6) = 3 * 4² = 48