Class 12th

A - B has the lowest potential energy and it means it has stronger bond.

Candidates seeking admission to the GNM programme can enrol for admission with Class 12 marks. Candidates must complete Class 12 to enrol for the GNM course. Aspirants from recognised State and National Open Schools or those who have completed a vocational ANM course and are registered ANMs are also eligible to apply. Aspirants must have completed Class 12 in English and obtained a minimum of 40% aggregate in the qualifying examination, with English passed individually from any recognised board.

 $\overrightarrow{a}=2\widehat{i}+\widehat{j}+3\widehat{k}$

$\begin{array}{l}\overrightarrow{b}=3\widehat{i}+3\widehat{j}+\widehat{k}\\ \overrightarrow{c}=c_1\widehat{i}+c_2\widehat{j}+c_3\widehat{k}\end{array}$

$Coplnanar\Rightarrow \left|\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right|=0$

$\begin{array}{l}\Rightarrow -8c_1+7c_2+12c_3=0........(i)\\ \overrightarrow{a}.\overrightarrow{c}=5\Rightarrow 2c_1+c_2+3c_3=5........(ii)\\ \overrightarrow{b}.\overrightarrow{c}=0\Rightarrow 3c_1+3c_2+c_3=0........(iii)\end{array}$

Solving (i), (ii), (iii)

$C_1=\frac{10}{122},c_2=\frac{-85}{122},c_3=\frac{225}{122}$

$\therefore 122\left(c_1+c_2+c_3\right)=150$

 $Mean\frac{{\displaystyle \sum _{i=1}^{15}{x}_i}}{15}=8$

$\Rightarrow {\displaystyle \sum _{i=1}^{15}{x}_i}=120.......\left(i\right)$

S.D = 3

$\Rightarrow {\displaystyle \sum _{i=1}^{15}{x_i}^2}=15\left(9+64\right)=15*73..........\left(ii\right)$

Given 20 has misread as 5

$\therefore$ in new case

${\displaystyle \sum _{q=1}^{15}{x}_{i^2}=15*73-25+400=1470}$

mean in new case

$\overline{x}=\frac{120-5+20}{15}$

$\therefore v$ariance in new case

${\sigma }_{new}^2=\frac{1}{15}\left(1470\right)-81=17$

$e^{4x}+4e^{3x}-58e^{2x}+4e^x+1=0$

$\left(e^{2x}+\frac{1}{e^{2x}}\right)+4\left(e^x+\frac{1}{e^x}\right)-58=0$

$\Rightarrow {\left(e^x+\frac{1}{e^x}+2\right)}^2=64$

$e^x$$=$$\frac{6\pm \text{exception 25:}}{2}$$=$$3$$\pm$$2$

A = {1, 2, 3, ….50}

R₁ = (2, 1), (2, 2), (2, 4)…. (2, 32)

(3, 1) (3, 3) (3, 9) (3, 27)

(13, 1) (13, 13)   etc.

R₂ = { (2, 1), (2, 2), (3, 1), (3, 3), (5, 1), (5, 5), (7, 1), (7, 7), (1, 1), (11, 11)…}

$\therefore R_1-R_2$ contains only 9 elements.

Consider the following image

${\mathrm{O}}^{2-}>{\mathrm{F}}^{-}>{\mathrm{N}\mathrm{a}}^{+}>{\mathrm{M}\mathrm{g}}^{2+}$ . Greater the Zeff, smaller is size.

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