Class 11th

According to Bohr's theory,

$K.E.=13.6\frac{z^2}{n^2}$ frequency $\left(\upsilon \right)$ of revolution of electron $\propto \frac{z^2}{n^3}$

  $v=2.18*10^6\frac{z}{n}$          

$F=\frac{k{q}_1{q}_2}{r^2}=\frac{Kz{e}^2}{r^2}\left(r\propto \frac{n^2}{z}\right)$       

$F\propto \frac{z^3}{n^4}$         

$\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)-1+\lambda \left(\overrightarrow{r}.\left(\widehat{i}-2\widehat{j}\right)+2\right)=0$

$\therefore point\left(1,0,2\right)=\widehat{i}+2\widehat{k}$

$\therefore \overrightarrow{r}\left(\frac{\widehat{i}}{3}+\frac{7\widehat{j}}{3}+\widehat{k}\right)-\frac{7}{3}=0$

$\overrightarrow{r}.\left(\widehat{i}+7\widehat{j}+3\widehat{k}\right)=7$

${=}^{n+1}{C}_2+2{\displaystyle \sum _{r=1}^{n-1}\frac{\left(r+1\right)!}{\left(r-1\right)!2!}}{=}^{n+1}{C}_2+{\displaystyle \sum _{r=1}^{n-1}\left(r+1\right)r}$

  =   $\frac{\left(n+1\right)!}{2!\left(n+1\right)!}+\frac{\left(n-1\right)n\left(2\left(n-1\right)+1\right)}{6}+\frac{\left(n-1\right)n}{2}=\frac{n\left(n+1\right)}{2}+\frac{n\left(n-1\right)\left(2n-1\right)}{6}+\frac{n\left(n-1\right)}{2}$

=   $\frac{n\left(6n+2n^2-3n+1\right)}{6}=\frac{\left(2n^2+3n+1\right)}{6}=\frac{n\left(2n+1\right)\left(n+1\right)}{6}$

$x+\sqrt[]{3}y=2\sqrt[]{3},andm=-\frac{1}{\sqrt[]{3}},C=2$          not possible

 (2) , $\left(1\right)\frac{x^2}{\frac{9}{2}}-\frac{y^2}{\frac{1}{2}}=1\Rightarrow c=\sqrt[]{a^2{m}^2-b^2}=\sqrt[]{\frac{9}{2}*\frac{1}{3}-\frac{1}{2}}=1,$ not possible

(3) $c=a\sqrt[]{1+m^2}=\sqrt[]{7}*2=2\sqrt[]{7},$ not possible

(4) $\frac{x^2}{9}+\frac{y^2}{1}=1,C=\sqrt[]{9m^2+1}=\sqrt[]{9*\frac{1}{3}+1}=2$  

Hence

CH₃ – CH₂ – CHO can't be formed.

In case of unsymmetrical alkynes, more positive charge stabilizing carbon attacked by H₂O and finally converted into carbonyl.

MgO has more lattice enthalpy than NaCl

 $x=y^4,xy=k$

$\frac{dy}{dx}=\frac{1}{4y^3},\frac{dy}{dx}=\frac{-k}{x^2}$

P (x₁, y₁)

where $x_1=y_1^4\&x_1{y}_1=k$

$\Rightarrow y_1=k^{1/5},x_1{y}_1=k$

$m_1{m}_2=-1$

$\Rightarrow \frac{-1}{4.k^{6/5}}=-1\Rightarrow k^{6/5}=\frac{1}{4}\Rightarrow k^6=\frac{1}{4^5}=\frac{1}{2024}$

${\left(4k\right)}^6=2^{12}.\frac{1}{2^{10}}=2^2=4$

Hydrogen is most abundant element in earth crust and the lightest element. But in troposphere nitrogen gas in most abundant element

Kindly consider the following figure

let f : R -> R

$f\text{'}\left(x\right)=\{\begin{array}{l}-55,x<-5\\ 6x^2-6x-120,-5<x<4\\ 6x^2-6x-36,x>4\end{array}$            

f' (x) increasing in x $\in \left(-5,-4\right)\cup \left(4,\infty \right)$