Maths

  $T_r=\frac{r}{{\left(2r^2\right)}^2+1}$

$=\frac{r}{{\left(2r^2+1\right)}^2-{\left(2r\right)}^2}$

$=\frac{1}{4}\frac{4r}{\left(2r^2+2r+1\right)\left(2r^2-2r+1\right)}$

$S_{10}=\frac{1}{4}{\displaystyle \sum _{r=1}^{10}\left(\frac{1}{\left(2r^2-2r+1\right)}-\frac{1}{\left(2r^2+2r+1\right)}\right)}$

$\Rightarrow S_{10}=\frac{1}{4}.\frac{220}{221}=\frac{55}{221}=\frac{m}{n}$

$\therefore m+m=276$

  $T_{r+1}{=}^{60}{C}_r{\left(x^{\frac{1}{2}}\right)}^{60-r}{\left(x^{-\frac{1}{3}}\right)}^r{\left(5^{\frac{-1}{4}}\right)}^{60-r}{\left(5^{\frac{1}{2}}\right)}^r$

for

$x^{10}\frac{60-r}{2}-\frac{r}{3}=10$    

$\Rightarrow 180-3r-2r=60$             

->r = 24

k = 3 + exponent of 5 in 

= $3+\left(\left[\frac{60}{5}\right]+\left[\frac{60}{5^2}\right]-\left[\frac{24}{5}\right]-\left[\frac{24}{5^2}\right]-\left[\frac{36}{5}\right]-\left[\frac{36}{5^2}\right]\right)$  

= 3 + (12 + 2 – 4 – 0 – 7 – 1)

= 3 + 2 = 5

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

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

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

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

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

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