Maths

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

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

(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

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

The bag has 25-paise, 50-paise and 1 -rupee coins in the ratio 8 : 2 : 7. Now this is the ratio of the number of coins while the total value of the coins has been given. So we first need to convert coins into value.

Let the coins be 8x, 2x and 7x respectively.

8x coins of 1/4 rupee each = value is Rs. 2x.

2x coins of 1/2 rupee each = value is Rs. x.

7x coins of 1 rupee each = value is Rs. 7x.

Now

2x + x + 7x = 770

10x = 770 x = 77

Number of 25 paise and 50 coins = 10 x = 770 coins

$E_1:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

  $e_1^2=1-\frac{b^2}{a^2}$ -(i)

Let $E_2:\frac{x^2}{a^2}+\frac{y^2}{B^2}=1$  

B > a

 $e_2^2=1-\frac{a^2}{B^2}andgivenB{e}_2=b$

$e^2=\frac{3-\sqrt[]{5}}{2}={\left(\frac{\sqrt[]{5}-1}{2}\right)}^2$

$\Rightarrow e=\frac{\sqrt[]{5}-1}{2}$

Let daily working hours & wages per hour before increase be H and W respectively

So his Salary/day = HW

Working hours after increase = 0.82H, wages per hour = 1.45W

So his daily salary = $\frac{82}{100}H*\frac{29}{20}W=\frac{1189}{1000}HW$ $\frac{}{}\frac{}{}\frac{}{}$ . Thus an Increase of 18.9%

A + B + C = 18500…. (1)

A = B + 25% of   $B=\frac{5B}{4}$ $\frac{}{}$ …. (2)

B = C + 20% of $C=\frac{6C}{5}$ $\frac{}{}$ …. (3)

Equation (1) gives,

$\frac{5B}{4}+\frac{6C}{5}+C=18500$

$\Rightarrow \frac{3C}{2}+\frac{6C}{5}+C=18500$

37C = 185000 C = 5000

B = 6000, A = 7500.

A + B = 13500

System of equations can be written as

$\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 5\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill \lambda \hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}6\\ 26\\ \mu \end{array}\right)$                          

$R{\text{'}}_3=R_3-R_1,R{\text{'}}_2=R_2-5R_1\Rightarrow$                

$\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill \lambda -1\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}6\\ -4\\ \mu -6\end{array}\right)$                              

Again $R\text{'}{\text{'}}_3=R{\text{'}}_3-R{\text{'}}_1\Rightarrow \left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \lambda -2\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}4\\ -4\\ \mu -10\end{array}\right)$  

The system will have no solution for

$\lambda =2and\mu \ne 10.$  

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

For domain $0\le x^2-x+1\le 1$

$\&x^2-x+1\ge 0\forall x\in R$

$x^2-x\le 0\&x\left(x-1\right)\le 0\Rightarrow x\in \left[0,1\right]...........\left(i\right)$              .

  $\frac{1}{2}\le x\le \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.................\left(ii\right)$             

$\left(i\right)\cap \left(ii\right)x\in \left(\frac{1}{2},1\right]\equiv \left(\alpha ,\beta \right]$

then α + β = $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$