Class 12th

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

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

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

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

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

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 $x,y=-\frac{1}{8}$

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

$\mathrm{}4\alpha {\int }_{-1}^0?e^{\alpha x}dx+{\int }_0^2?e^{-\alpha x}dx=5$

$\Rightarrow 4\alpha \left(\frac{1-{\mathrm{e}}^{-\alpha }}{\alpha }-\frac{{\mathrm{e}}^{-2\alpha }-1}{-\alpha }\right)=5$

Let ${\mathrm{e}}^{-\alpha }=\mathrm{t}$

$\therefore 4{\mathrm{t}}^2+4\mathrm{t}-3=0$

$\Rightarrow \mathrm{t}=\frac{1}{2}$

$\alpha =\mathrm{l}\mathrm{n}?2$

Using LMVT

$f^{\mathrm{\text{'}}}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$

$3c^2-8c+8=\frac{16-11}{1-0}$

$3c^2-8\mathrm{c}+3=0$

$c=\frac{4-\sqrt[]{7}}{3}\in \left(\mathrm{0,1}\right)$

$\mathrm{}\left|B\right|=\left|\begin{array}{lll}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right|=\left|\begin{array}{lll}{3}^0{a}_{11}& 3^1{a}_{21}& 3^2{a}_{31}\\ 3^1{a}_{12}& 3^2{a}_{22}& 3^3{a}_{32}\\ 3^2{a}_{13}& 3^3{a}_{23}& 3^4{a}_{33}\end{array}\right|$

$81=3^3\cdot 3^3\cdot 3^2\left|\mathrm{A}\right|$

$\Rightarrow \left|A\right|=\frac{1}{9}$

The dotted ones are ${\mathrm{s}\mathrm{p}}^2$  carbons

$30\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}$ of $\mathrm{H}\mathrm{C}\mathrm{l}$

Then in $250\mathrm{m}\mathrm{L}\mathrm{H}\mathrm{C}\mathrm{l}$ will be $\frac{30}{4}\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}$

${\mathrm{H}}_2{\mathrm{S}\mathrm{O}}_4$ will be $\frac{1}{2}*\frac{30}{4}\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}=98*\frac{1}{2}*\frac{30}{4}*{16}^{-3}\mathrm{g}=0.3675\mathrm{g}$

$\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}+{\mathrm{K}}_2{\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7\stackrel{{\mathrm{H}}^{+}}{?}{\mathrm{C}\mathrm{r}\mathrm{O}}_2{\mathrm{C}\mathrm{l}}_2\stackrel{\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}}{?}{\mathrm{N}\mathrm{a}}_2{\mathrm{C}\mathrm{r}\mathrm{O}}_4\underset{{\mathrm{H}}^{+}}{\stackrel{{\mathrm{H}}_2{\mathrm{O}}_2}{?}\underset{\left(\mathrm{C}\right)}{{\mathrm{C}\mathrm{r}\mathrm{O}}_5}}$

 ${\mathrm{C}\mathrm{r}\mathrm{O}}_2{\mathrm{C}\mathrm{l}}_2\to 5$

${\mathrm{N}\mathrm{a}}_2{\mathrm{C}\mathrm{r}\mathrm{O}}_4\to 7$

${\mathrm{C}\mathrm{r}\mathrm{O}}_3\to 6$

Vapour pressure over solvent is greater than that over solution.