Class 12th

$f\left(x\right)=\frac{2e^{2x}}{e^{2x}+e^x}andf\left(1-x\right)=\frac{2e^{2-2x}}{e^{2-2x}+e^{1-x}}$

$\therefore \frac{f\left(x\right)+f\left(1-x\right)}{2}=1$

i.e. f (x) + f (1 – x) = 2

$\therefore f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+....+f\left(\frac{99}{100}\right)$

${\displaystyle \sum _{x=1}^{49}f\left(\frac{x}{100}\right)+f\left(1-\frac{x}{100}\right)+f\left(\frac{1}{2}\right)}$

= 49 * 2 + 1 = 99

Consider the below image

  $si{n}^{-1}\left(\frac{\sqrt[]{3}}{2}\right)+co{s}^{-1}\left(\frac{-\sqrt[]{3}}{2}\right)+ta{n}^{-1}\left(-1\right)$

$=\frac{\pi }{3}+\frac{5\pi }{6}-\frac{\pi }{4}$

$=\frac{4\pi +10\pi -3\pi }{12}$ =   $\frac{11\pi }{12}$

$\frac{2\pi }{7}+cos\frac{4\pi }{7}+cos\frac{6\pi }{7}=\frac{sin3\left(\frac{\pi }{7}\right)}{sin\frac{\pi }{7}}cos\frac{\left(\frac{2\pi }{7}+\frac{6\pi }{7}\right)}{2}$

$=\frac{sin\left(\frac{3\pi }{7}\right).cos\left(\frac{4\pi }{7}\right)}{sin\left(\frac{\pi }{7}\right)}=\frac{2sin\frac{4\pi }{7}cos\frac{4\pi }{7}}{2sin\frac{\pi }{7}}=\frac{sin\left(\frac{8\pi }{7}\right)}{2sin\frac{\pi }{7}}=\frac{-sin\frac{\pi }{7}}{2sin\frac{\pi }{7}}=\frac{-1}{2}$

Given P (X = 3) = 5P (X = 4) and n = 7

${\Rightarrow }^7{C}_3{p}^3{q}^4=5^7{C}_4{p}^4{q}^3$

-> q = 5p and also p + q = 1

$\Rightarrow p=\frac{1}{6}andq=\frac{5}{6}$  

Mean = $\frac{7}{6}$  and variance =   $\frac{35}{36}$

Mean + Variance

$=\frac{7}{6}+\frac{35}{36}+\frac{77}{36}$

Total case = ¹⁸C₅

Favourable cases

(Select x₁) (Select x₃)       (Select x₅)

$\overrightarrow{a}=\widehat{i}+\widehat{j}-\widehat{k}$

$\overrightarrow{c}=2\widehat{i}-3\widehat{j}+2\widehat{k}$                

Now,

$\overrightarrow{b}*\overrightarrow{c}=\overrightarrow{a}$      

$\Rightarrow \left(\widehat{i}+\widehat{j}-\widehat{k}\right)\left(2\widehat{i}-3\widehat{j}+2\widehat{k}\right)=0$            

 = 2 – 3 – 2 = 0

->-3 = 0 (Not possible)

->No possible value of

$\overrightarrow{b}$  is possible.

 l + m – n = 0 Þ n = l + m

$3l^2+m^2+cnl=0$  

$3l^2+m^2+cl\left(l+m\right)=0$    

$=\left(3+c\right){\left(\frac{l}{m}\right)}^2+c\left(\frac{l}{m}\right)+1=0$    

$?$    Lines are parallel

D = 0

$c=4$     (as c > 0)

$\frac{dy}{dx}+\frac{2^{x-y}\left(2^y-1\right)}{2^x-1}$ = 0

x, y > 0, y(1) = 1

$\frac{dy}{dx}=-\frac{2^x\left(2^y-1\right)}{2^y\left(2^x-1\right)}$         

${\displaystyle \int \frac{2^y}{2^y-1}dy=-{\displaystyle \int \frac{2^x}{2^x-1}dx}}$           

$=\frac{lo{g}_e\left(2^y-1\right)}{lo{g}_{e}2}=-\frac{lo{g}_e\left(2^x-1\right)}{lo{g}_{e}2}+\frac{lo{g}_{e}c}{lo{g}_{e}2}$  

Taking log of base 2.

$\therefore$  y = 2 – log₂