Maths

The average age of n men is X year. Total age is given by N years

$\begin{array}{l}Newaverageage =\frac{NX+X-1+X+1+X+2}{N+3}\\ =\frac{NX+3X+2}{N+3}\\ =\frac{X[N+3]}{N+3}+\frac{2}{N+3}\\ \Rightarrow X+\frac{2}{N+3}\end{array}$

1^st position can be filled in 4 ways as zero cannot appear in 1^st position.

2^nd position can be filled in 4 ways and so on.

Total cases = 4 * 4 * 3 * 2 * 1 = 96

$A={\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)}_{3*3}$

 B is a matrix of same order with entries from {1,2,3,4,5}. and satisfying AB = BA.

$LetB=\left(\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right)$           

$\therefore \left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$

$a_{21}=a_{12},a_{22}=a_{11},a_{23}=a_{13},a_{31}=a_{32}$

$\therefore$ there exist only 5 distinct entries in the matrix B so that possible case = 5⁵ = 3125

Total time = 182s

2 + 4 + 6 + 8 +……………+ n terms = 182

i.e. n=13

Total distance covered = 2* [1² + 2 ²+ 3² +……. …………. + 13²]

= 2 * 13 * 14 * $\frac{27}{6}$  = 9 * 13 * 14m

Average speed = 9 * 13 * $\frac{4}{182}$ = 9m/s

$x^2=-2\left(y-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$  

y² = -4 (x – 1)

Required area = $2{\displaystyle {\int }_{-1}^0\left[2\sqrt[]{x+1}-\frac{1-x^2}{2}\right]dx}$  

$=2{\left[\frac{4}{3}{\left(x+1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{1}{2}\left(x-\frac{x^3}{3}\right)\right]}_{-1}^0$          

= 2

  $11^n-9^n>10^n$             

$\Rightarrow {\left(10+1\right)}^n-{\left(10-1\right)}^n>10^n$           

->For  $n\ge 5$ the inequality Satisfies.

For n= 4.             2 [4 * 10³ + 4 * 10] > 10⁴

->8 * 1010 > 10⁴ which is a contradiction

$\therefore n\ge 5$ all values satisfies. Hence possible values of n is 96

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

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

(1) -> $v+\left(x+2\right)\frac{dv}{dx}=e^v+v.$  

$\Rightarrow \frac{dv}{e^v}=\frac{dx}{x+2}$ Integrating both side

Let $e^{e-2/3}=a\Rightarrow 0<\left|\frac{x+2}{3}\right|<a$  

$\therefore a=-3a-2\&\beta =3a-2$              

$\Rightarrow \left|\alpha +\beta \right|=4$               

$f\left(x\right)=[\begin{array}{l}3\left(1-\frac{\left|x\right|}{2}\right);\left|x\right|\le 2\\ 0;\left|x\right|>2\end{array}$

$\therefore g\left(x\right)=f\left(x+2\right)-f\left(x-2\right)[\begin{array}{l}0x<-4\\ 3\left(1-\frac{\left|x-2\right|}{2}\right)-4\le x\le 0\\ -3\left(1-\frac{\left|x-2\right|}{2}\right)0<x\le 4\\ 0x>4.\end{array}$

g(x) is continuous every where but not differentiable

at x = -4, -2, 2 and 4

$\therefore n=0\&m=4\Rightarrow n+m=4$