Maths

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

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

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

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

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

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