Maths

$\frac{\left(28-25\right)}{\left(45-28\right)}=\frac{3}{17}$

Rahul can occupy any one of the two corner positions and therefore, Rahul can occupy a position in two ways.

The other four people can seat themselves in 4! ways, that is, 24 ways.

Total ways = 2 * 24 = 48 ways.

f : R -> R.

$f\left(x\right)=[\begin{array}{l}\frac{x^3}{{\left(1-cos2x\right)}^2}log\left(\frac{1+2x{e}^{-2x}}{{\left(1-x{e}^{-x}\right)}^2}\right)x\ne 0\\ \alpha x=0\end{array}$                

As f is continuous at x = 0

$\therefore \alpha =\underset{x\to 0}{Lim}f\left(x\right)$              

$=\underset{x\to 0}{Lim}\frac{1}{2}\left[e^{-2x}+e^{-x}\right]=\frac{1}{2}*2=1$            

$\therefore \alpha =1$              

As per the problem, we have Length of the first train = 120 m

Length of the second train = 180 m

As per the problem,

$\frac{(120+180)}{(8+s)}=5$

8 + s = 60

s = 52 m/sec

6X + 5Y = 218 . (1)

5X - 3Y = 24 . (2)

Solve to get X = 18, Y = 22.

  $S_{10}=530\Rightarrow 5\left[2a+9d\right]=530$

2a + 9d = 106 . (i)

$S_5=140\Rightarrow \frac{5}{2}\left[2a+4d\right]=140$              

a + 2d = 28 . (ii)

Solving (i) & (ii) a = 88 d = 10

$\therefore S_{20}-S_6=14a+175d=1862$            

$\overrightarrow{a}=\alpha \overrightarrow{b}+\beta \overrightarrow{c}=\left(2\alpha +\beta \right)i+\left(\alpha -\beta \right)J+\left(\alpha +\beta \right)k$

is perpendicular to   $\overrightarrow{d}=3\widehat{i}+2\widehat{j}+6\widehat{k}$

$\Rightarrow 3\left(2\alpha +\beta \right)+2\left(\alpha +\beta \right)+6\left(\alpha +\beta \right)=0$              

->14a + 7b = 0 Þ b = -2a .(i)

$\Rightarrow \overrightarrow{a}=3\alpha j-\alpha k$

$\left|\overrightarrow{a}\right|=\sqrt[]{9{\alpha }^2+{\alpha }^2}=\sqrt[]{10}\Rightarrow \left|\alpha \right|=1\Rightarrow \alpha \pm 1,for\alpha =1,\overrightarrow{a}=3\widehat{j}-\widehat{k}$       

for $\alpha =-1,\overrightarrow{a}=-3\widehat{j}+\widehat{k}$  

$\overrightarrow{d}*\overrightarrow{a}=\pm \left|\begin{array}{ccc}\hfill i\hfill & \hfill j\hfill & \hfill k\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill -1\hfill \end{array}\right|$

$=\pm 42$              

Let x be the required percent.

Using the successive percentage change rule, we have

150 + x + $\frac{150x}{100}$ $\frac{}{}$ = 0

$\frac{}{}$ $\frac{250x}{100}$  = – 150

x = – 60%

so 60%

Required time t =

$\frac{1760}{2\pi r}=\frac{1760}{\left[2*\left(\frac{22}{7}\right)*\frac{3.5}{2}\right]}=160min.$

  $f\left(x\right)=[\begin{array}{l}-\frac{4}{3}{x}^3+2x^2+3x,x>0\\ 3x{e}^x,x\le 0\end{array}$

  $f\text{'}\left(x\right)>0forx<0e^x\left(x+1\right)>0\Rightarrow X>-1\Rightarrow x\in \left(-1,0\right)$            

for  x > 0  $-4x^2+4x+3>0$   

$\Rightarrow x\in \left(0,\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$  

Again    at  x = 0  f'(0) = 3 > 0

$\therefore f\left(x\right)isincreasingin\left(-1,\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$