Class 11th

$M=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill c_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right]$    

$M^{T}M=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill b_1\hfill & \hfill c_1\hfill \\ \hfill a_2\hfill & \hfill b_2\hfill & \hfill c_2\hfill \\ \hfill a_3\hfill & \hfill b_3\hfill & \hfill c_3\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill b_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right]$

$Tr\left(M^{T}M\right)=a_1^2+b_1^2+c_1^2+a_2^2+b_2^2+c_2^2+a_3^2+b_3^2+c_3^2=7$           

all   $a_i,b_i,c_i\in \left\{0,1,2\right\}for$ i = 1, 2, 3

Case 1 7 one's and two zeroes which can occur in ways

Case 2 One 2 three 1's five zeroes =

total such matrices = 504 + 36 = 540

  ${\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{1}{1+r+r^2}={\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{\left(r+1\right)}{1+r\left(r+1\right)}}}$

$=ta{n}^{-1}2-ta{n}^{-1}+......+ta{n}^{-1}\left(n+1\right)-ta{n}^{-1}n$            

For n  $\infty$ value = $\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$  

$\therefore tan\left(\frac{\pi }{4}\right)=1$           

Equation of normal is 4x – 3y + 1 = 0

Equation of tangent is 3x + 4y – 43 = 0

Area of triangle = $\frac{1}{2}\left(\frac{43}{3}+\frac{1}{4}\right)*7$  

$=\frac{1}{2}*\left(\frac{172+3}{12}\right)*7=\frac{1225}{24}$          

  $\therefore 24A=1225$        

${\sigma }^2=\frac{\sum x^2}{n}-{\left(\frac{\sum x}{n}\right)}^2=\frac{9+k^2}{10}-{\left(\frac{9+k}{10}\right)}^2<10$

$\Rightarrow k<\frac{10\sqrt[]{10}}{3}+1\Rightarrow k\le 11$        

$\therefore$ maximum value of k is 11

Let a, ar, ar², ar³ are in G.P.

a + ar + ar² + ar³ = $\frac{65}{12}$  

$\frac{1}{a}+\frac{1}{ar}+\frac{1}{a{r}^2}+\frac{1}{a{r}^3}=\frac{65}{18}$        

${\left(ar\right)}^{2}r=\frac{3}{2}\Rightarrow a=\frac{2}{3},r=\frac{3}{2}$          

  $\therefore$   third term = ar² = $\frac{2}{3}*\frac{9}{4}=\frac{3}{2}$  

  $\therefore \alpha =\frac{3}{2}\Rightarrow 2\alpha =3$          

Let P (h, k)

$\sqrt[]{{\left(h-5\right)}^2+k^2}=3\sqrt[]{{\left(h+5\right)}^2+k^2}$           

$h^2-10h+25+k^2=9h^2+90h+225+9k^2$           

  $8h^2+8k^2+100h+200=0$

$x^2+y^2+\frac{25}{2}x+25=0$

$r^2=\frac{625}{16}-25=\frac{625-400}{16}=\frac{225}{16}$

$4r^2=\frac{225}{4}=56.25$         

${\left(x+1\right)}^2+\left|x-5\right|=\frac{27}{4}$

Case l        x < 5

$x^2+2x+1-x+5=\frac{27}{4}$       

$\left(2x+3\right)\left(2x-1\right)=0\Rightarrow x=-\frac{3}{2},\frac{1}{2}$     

Case II   $\ge$   x 5

$x^2+2x+1+x-5=\frac{27}{4}$       

$x=\frac{-12\pm \sqrt[]{144+43*16}}{2*4}=\frac{-12\pm \sqrt[]{832}}{8}\left(rejected\right)$           

           because x > 5

${\displaystyle \sum _{i=0}^k\left(\begin{array}{l}10\\ i\end{array}\right)}\left(\begin{array}{l}15\\ K-i\end{array}\right)+{\displaystyle \sum _{i=0}^{k+1}\left(\begin{array}{l}12\\ i\end{array}\right)\left(\begin{array}{l}13\\ k+1-i\end{array}\right)}$  

= Equating the coefficient of $x^{k}in{\left(1+x\right)}^{25}{=}^{25}{C}_k$ .(i)

= Equating the coefficient of   $x^{k+1}in{\left(1+x\right)}^{12}{\left(x+1\right)}^{13}$

From (i) and (ii)

Hence   $26\ge k+1\Rightarrow k\le 25$

But maximum value of k is not defined bonus

W = 4.75 g

$n=\frac{4.75}{26}$ T = 323 K, R = 0.0826

  $P=\frac{740}{760}atm$          

 V = $\frac{nRT}{P}=\frac{4.75*0.0826*323}{26\left(\frac{740}{760}\right)}=5litre$