Class 11th

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

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

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

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

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

(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

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

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

Number divisible by 3;

(a) sum of the digit must be divisible by 3

(i)  $\boxed{1}\boxed{2}\boxed{3}$  3! = 6

(ii) 1, 3, 5 -> 3! = 6

(iii) 2, 3, 4 -> 3! = 6

(iv) 3, 4, 5 = 3! = 6

Total = 24

(b) Divisible by

$\boxed{}\boxed{}\boxed{5}$  4 * 3 = 12

(c) Now common divisible by both

$\boxed{1}\boxed{3}\boxed{5}$ 2! = 2

For 3, 4  $\boxed{}\boxed{}\boxed{5}$  2! = 2

$\therefore$ Total ways = 24 + 12 – 4 = 32

$F=-\alpha x^2\Rightarrow \frac{mvdv}{dx}=-\alpha x^2$

$\Rightarrow {\displaystyle \underset{v_0}{\overset{0}{\int }}vdv=-\frac{\alpha }{m}{\displaystyle \underset{0}{\overset{x}{\int }}{x}^{2}dx\Rightarrow \frac{0^2}{2}-\frac{v_0^2}{2}=\frac{-\alpha }{3m}{x}^3}}$

$\Rightarrow \frac{-v_0^2}{2}=-\frac{\alpha }{3m}{x}^3\Rightarrow x={\left(\frac{3v_0^{2}m}{2\alpha }\right)}^{1/3}$

$\frac{a^{a^x}+a^{1-a^x}}{2}\ge \sqrt[]{a^{a^x}.a^{1-a^x}}$ $\left(AM\ge GM\right)$

$\Rightarrow a^{a^x}+a^{1-a^x}\ge 2\sqrt[]{a}$