Maths

$x=1$

$4x^2=8x+12$

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

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

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

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

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

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

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

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

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

Maxima at $x=-1$

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

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

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

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

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

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

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

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

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

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

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

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

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

$3+4+8+9+13+14+\dots .$ . upto 40 terms

$\Rightarrow 7+17+27+\dots \dots ..20$ terms

$\mathrm{S}=\frac{20}{2}[2*7+19*10]$

$=102*20=102\mathrm{m}$

$\therefore \mathrm{m}=20$

$y=-\frac{3x}{4}+3\sqrt[]{2}$ line is tangent to ellipse

$\therefore {\mathrm{c}}^2={\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2$

$18=\frac{9{\mathrm{a}}^2}{16}+9$

${\mathrm{a}}^2=16$

$\therefore {\mathrm{e}}^2=1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}$

$\mathrm{e}=\frac{\sqrt[]{7}}{4}$

Distance between focii $=2\mathrm{a}\mathrm{e}$

$=2\sqrt[]{7}$

$\mathrm{}6\cdot {}^{35}{\mathrm{C}}_{\mathrm{r}}=\left({\mathrm{k}}^2-3\right)\cdot \frac{36}{\mathrm{r}+1}{}^{35}{\mathrm{C}}_{\mathrm{r}}$

$\Rightarrow {\mathrm{k}}^2-3=\frac{\mathrm{r}+1}{6},{\mathrm{k}}^2-3>0$

(i) $k=\pm 2$ gives $r=5$

(ii) $k=\pm 3$ gives $r=35$

4 ordered pairs