Class 11th

 $\frac{x^2}{a^2}+\frac{y^2}{4}=1$

$\therefore \Delta =\frac{1}{2}*a\left(1+cos\theta \right).4sin\theta$

$\therefore {\Delta }_{max}^{\Delta }=6\sqrt[]{3}\Rightarrow a=4$

$\therefore e=\frac{\sqrt[]{3}}{2}$

Equation of tangent at P (x, y) is Y = $\frac{dy}{dx}\left(X-x\right)$

$A/q,2x-y\frac{dx}{dy}=0\Rightarrow 2\frac{dy}{y}=\frac{dx}{x}$

$\begin{array}{l}\Rightarrow 2\mathcal{l}ny=\mathcal{l}nx+\mathcal{l}nc\\ y^2=xc\end{array}$

It passes through (3, 3), c = 3

$\therefore y^2=3x\therefore$ Length of latus rectum = 3

 $x_{1}y>0and{x}^3{y}^2=2^{15}$

$AM\ge GM$

$\frac{3x+2y}{5}\ge {\left(x^3{y}^2\right)}^{\frac{1}{5}}$

$\Rightarrow 3x+2y\ge 40$

$\left(x^2+1\right)*1=x*2$

$\Rightarrow {\left(x^2+1\right)}^2+1=x^2+8\Rightarrow x^2=2$

$2si{n}^{-1}\left(\frac{x^4+x^2-2}{x^4+x^2+2}\right)=2si{n}^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$

 $\frac{dT}{dt}=-k\left(T-T_0\right)$

$\frac{dT}{\left(T-T_0\right)}=-kdt$

$\Rightarrow k=-\frac{1}{6}\mathcal{l}n\left(\frac{2}{3}\right)$ - (1)

Now $\frac{dT}{dt}=-k\left(T-T_0\right)$

t = 10.257

t₂ = 6 + 10.257 = 16.257 minutes

16.25 and or 16

$Q=\Delta \upsilon +w\Rightarrow nC\Delta T=n{C}_v\Delta T=\frac{nC\Delta T}{4}=\frac{3}{4}nC\Delta T=n{C}_v\Delta T$

$C=\frac{4}{3}{C}_v=\frac{4}{3}*\frac{3}{2}R=2R$

$y=\left(10cos\pi xsin\frac{2\pi t}{T}\right)cm$

$atx=\frac{4}{3}$

$y=10cos\left(\frac{4\pi }{3}\right)sin\left(\frac{2\pi t}{T}\right)cm$

$=-5sin\left(\frac{2\pi t}{T}\right)cm$

Then Amplitude will be 5 cm.

after 2 sec, v = 20 m/sec

$\begin{array}{l}\overrightarrow{v}=ucos45°\widehat{i}+\left(usin45°-gt\right)\widehat{j}\\ v=20=\sqrt[]{{\left(ucos45°\right)}^2+{\left(usin45°-20\right)}^2}\end{array}$

$\Rightarrow 400=u^2+400-40usin45°$

u = 40 sin 45°

$H_{max}=\frac{u^{2}si{n}^{2}45}{2g}=\frac{40^{2}si{n}^{2}45°*si{n}^{2}45}{2g}$

$=\frac{40*40}{2*10}*\frac{1}{2}*\frac{1}{2}=20m$

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$\tilde{z}=i{z}^2+z^2-z$

$z+\overline{Z}=z^2\left(i+1\right)$

$z+\overline{Z}=z^2\left(i+1\right)$ Let z be equal to (x + iy)

(x + iy) + (x – iy) = (x + iy)² (i + 1)

$2x=\left(x^2-y^2+2ixy\right)\left(i+1\right)$               

Equating the real & in eg part.

$\left(x^2-y^2+2ixy\right)=0.........\left(i\right)$

$\left(x^2-y^2-2xy\right)=\left(2x\right)..............\left(ii\right)$               

(i) & (ii)

 4xy = -2x Þ x = 0 or y = $\left(\frac{-1}{2}\right)$  

(for x = 0, y = 0)

For y = $\frac{-1}{2}$  

x²   $-\frac{1}{4}+2\left(\frac{-1}{2}\right)x=0$

x =   $\frac{4\pm \sqrt[]{16+16}}{2.4}$

=  $\left(\frac{1+\sqrt[]{2}}{2}\right)or\left(\frac{1-\sqrt[]{2}}{2}\right)$  

$\therefore sum$ of   ${\left|z\right|}^2={\left(\frac{1+\sqrt[]{2}}{2}\right)}^2+\frac{1}{4}+{\left(\frac{1-\sqrt[]{2}}{2}\right)}^2+\frac{1}{4}+0^2+O^2$

=   $\frac{3}{4}+\frac{\sqrt[]{2}}{2}+\frac{1}{4}+\frac{3}{4}-\frac{\sqrt[]{2}}{2}+\frac{1}{4}=\frac{3}{2}+\frac{1}{2}=2$