Class 12th

Diborane $\left({\mathrm{B}}_2{\mathrm{H}}_6\right)$  is used to reduced carboxylic acid to alcohol.

According to Hardy-Schulze Rule coagulation value $\alpha \frac{1}{\text{Coagulation Power}}$

$=\frac{128}{3}$

$\mathrm{}3+2(\stackrel{?}{a}\cdot \stackrel{?}{b}+\stackrel{?}{b}\cdot \stackrel{?}{c}+\stackrel{?}{c}\cdot \stackrel{?}{a})=0$

$\begin{array}{rr}& \stackrel{?}{a}\cdot \stackrel{?}{b}+\stackrel{?}{b}\cdot \stackrel{?}{c}+\stackrel{?}{c}\cdot \stackrel{?}{a}=-\frac{3}{2}\\ & \stackrel{?}{d}=\stackrel{?}{a}*\stackrel{?}{b}+\stackrel{?}{b}*\stackrel{?}{c}+\stackrel{?}{c}*\stackrel{?}{a}\\ & \stackrel{?}{d}=3(\stackrel{?}{a}*\stackrel{?}{b})\end{array}$

$x=1$

$4x^2=8x+12$

Area $x=-\mathrm{1,3}$

$={\int }_{-1}^3?\left. [(8x+12)-4x^2\right]dx$

$={\left. [\frac{8x^2}{2}+12x-\frac{4x^3}{3}\right]}_{-1}^3$

Kindly consider the following Image

At a fixed temperature, $\mathrm{X}$ $$ having more vapour pressure as compared to $$ $\mathrm{Y}$ . So, intermolecular interaction is lower as compared to $\mathrm{Y}$ $$ .

Except ${\mathrm{O}}_2$ , remarginal ${\mathrm{C}\mathrm{O}}_2,{\mathrm{O}}_3,{\mathrm{H}}_2\mathrm{O},\mathrm{C}\mathrm{F}\mathrm{C}\mathrm{s}$  are green house gases

$\mathrm{}\left. [\mathrm{M}{\mathrm{a}}_3{\mathrm{b}}_3\right]$  - type complex compound show fac-and mer-isomer.

$\therefore \left. [\mathrm{C}\mathrm{o}{\left({\mathrm{N}\mathrm{H}}_3\right)}_3{\left({\mathrm{N}\mathrm{O}}_2\right)}_3\right]$

$\mathrm{}{f}^{\mathrm{\text{'}}}\left(x\right)=a(x+1)(x-1)x^2$

$f^{\mathrm{\text{'}}}\left(x\right)=a{x}^4-x^{2}f$

$f\left(x\right)=\frac{a{x}^5}{5}-\frac{a{x}^3}{5}+C$
$?f\left(0\right)=0\Rightarrow c=0$

${\mathrm{l}\mathrm{i}\mathrm{m}}_{x\to 0}?\frac{f\left(x\right)}{x^3}=2$

$\Rightarrow \mathrm{a}=-6$

Minima at $\therefore f^{\mathrm{\text{'}}}\left(x\right)=-6\left(x^2-1\right)\left(x^2\right)$

Maxima at $x=-1$

Differentiating

y.  $\frac{-2x}{2\sqrt[]{1-x^2}}+\sqrt[]{1-x^2}\cdot y^{\mathrm{\text{'}}}=\frac{x\cdot 2y{y}^{\mathrm{\text{'}}}}{2\sqrt[]{1-y^2}}-\sqrt[]{1-y^2}$

Put $x=\frac{1}{2},y=-\frac{1}{4}$ and $x,y=-\frac{1}{8}$

$y^{\mathrm{\text{'}}}=\frac{-\sqrt[]{5}}{2}$