Class 11th

Kindly consider the following figure

K $=\frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}$  

K =    $\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$

$\therefore \frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}=\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$   

$\frac{x^2}{16}-\frac{y^2}{48}=1$

48 = 16 (e² – 1)

$e=\sqrt[]{4}=2$      

n = 1 mole

$W_{AB}=nRTln\frac{v_2}{v_1}=1*RTln\frac{2v_1}{v_1}=RTln2$

$W_{BC}=0$

$W_{CA}=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}=\frac{\frac{P_1}{4}*2V_1-P_1*V_1}{\gamma }$

$\Rightarrow W_{CA}=\frac{-RT}{2\left(\gamma -1\right)}$

$\therefore W_{total}=RT\left[\mathcal{l}n2-\frac{1}{2\left(\gamma -1\right)}\right]$

$f\left(x\right)=\frac{5^x}{5^x+5}$

$f\left(2-x\right)=\frac{5^{2-x}}{5^{2-x}+5}=\frac{25/5^x}{\frac{25}{5^x}+5}=\frac{25}{25+5^{x+1}}=\frac{5}{5+5^x}$

$f\left(x\right)+f\left(2-x\right)=1$

$S={\displaystyle \sum _{k=1}^{39}f\left(\frac{k}{20}\right).....\left(i\right)}$

$S={\displaystyle \sum _{k=1}^{39}f}\left(2-\frac{k}{20}\right)..........\left(ii\right)$

(i) + (ii) 2S = ${\displaystyle \sum _{k=1}^{39}\left(f\left(\frac{k}{20}\right)+f\left(2-\frac{k}{20}\right)\right)}$

$S=\frac{39}{2}$

Let a_n be the side of square A_n

$a_n=\sqrt[]{2}{a}_{n+1}$       

a₁ = 12

a_n = 12 $*{\left(\frac{1}{\sqrt[]{2}}\right)}^{n-1}$  

${\left(a_n\right)}^2<1$

$\frac{144}{2^{\left(n-1\right)}}<1$

$\Rightarrow 2^{\left(n-1\right)}>144$   

$\Rightarrow n-1\ge 8$

$\Rightarrow n\ge 9$        

Let x = A sin $\omega t$ $$ & v = $A\omega cos\omega t$ $$

$\therefore v=\omega \sqrt[]{A^2-x^2}$

$v^2={\omega }^2{x}^2-{\omega }^2{A}^2$

$\frac{v^2}{{\omega }^2{A}^2}+\frac{x^2}{A^2}=1$

Slope = 1

$\frac{dy}{dx}=x=1$

$P\left(1,\frac{1}{2}\right)$

Equation of tangent at P : y - $\frac{1}{2}$ = 1 (x – 1)

$\Rightarrow y=x-\frac{1}{2}$

dist= $\frac{\frac{1}{2}}{\sqrt[]{2}}=\frac{1}{2\sqrt[]{2}}$  

Here, $K_{eq}=2k$ ${}_{}$

$T=2\pi \sqrt[]{\frac{m}{2k}}$

$\therefore f=\frac{1}{2\pi }\sqrt[]{\frac{2k}{m}}$

$A={\displaystyle \underset{\pi /4}{\overset{5\pi /4}{\int }}\left(sinx-cosx\right)dx={\left[-cosx-sinx\right]}_{\pi /4}^{5\pi /4}}=\left(\frac{1}{\sqrt[]{2}}+\frac{1}{\sqrt[]{2}}\right)-\left(-\frac{1}{\sqrt[]{2}}-\frac{1}{\sqrt[]{2}}\right)$

A =    $2\sqrt[]{2}$

A⁴ = 64

 $\underset{n\to \infty }{lim}{\displaystyle \sum _{r=0}^{n-1}\frac{n}{{\left(n+r\right)}^2}}$

$\underset{n\to \infty }{lim}{\displaystyle \sum _{r=0}^{n-1}\frac{1}{n}.\frac{1}{{\left(1+\frac{r}{n}\right)}^2}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{{\left(1+x\right)}^2}={\left[-\frac{1}{1+x}\right]}_0^1}=-\frac{1}{2}-\left(-1\right)=\frac{1}{2}}$