Maths

First we arrange 5 red cubes in a row and assume  number of blue cubes between them

$Here,x_1+x_2+x_3+x_4+x_5+x_6=11$            

and    $x_2,x_3,x_4,x_5\ge 2$

so $x_1+x_2+x_3+x_4+x_5+x_6=3$  

No. of solutions = ⁸C₅ = 56

$\left|adj\left(adj\left(A\right)\right)\right|={\left|A\right|}^{2^2}={\left|A\right|}^4$

$\therefore {\left|A\right|}^4=\left|\begin{array}{ccc}\hfill 14\hfill & \hfill 28\hfill & \hfill -14\hfill \\ \hfill -14\hfill & \hfill 14\hfill & \hfill 28\hfill \\ \hfill 25\hfill & \hfill -14\hfill & \hfill 14\hfill \end{array}\right|$

$={\left(14\right)}^3\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right|$

$={\left(14\right)}^3\left(3-2\left(-5\right)-1\left(-1\right)\right)$

${\left|A\right|}^4={\left(14\right)}^4\Rightarrow \left|A\right|=14$

Let e^x = t then equation reduces to

$t^2-11t-\frac{45}{t}+\frac{81}{2}=0$                

$\Rightarrow 2t^3-22t^2+81t-45=0$                     ……. (i)

If roots of

$e^{2x}-11e^x-45e^{-x}+\frac{81}{2}=0$            

$\Rightarrow {\alpha }_1+{\alpha }_2+{\alpha }_3=\mathcal{l}n45\Rightarrow p=45$                

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

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

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

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

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

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

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

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

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.