Maths

  $A={\left[a_{ij}\right]}_{3*3}=\left[\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right]$             

$a_{i1}+a_{i2}+a_{i3}=1;i=1,2,3$            

$LetX=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]them$  

given $\left[\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right]$ $\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]$  

->AX = X .(i)

replace x by A x we have

A (AX) = AX

->A²X = AX = X .(ii)

Again replace X by AX

A³X = AX = X.

As  $X=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right],$ Sum of all entries in A³ = sum of entries in X = 1 +1 + 1 = 3

$\overrightarrow{r}.\left(\widehat{i}-\widehat{j}+2\widehat{k}\right)=2$

&  $\overrightarrow{r}.\left(2\widehat{i}+\widehat{j}-\widehat{k}\right)=2$ are two planes the direction ratio of the line of intersection of then is collinear to   

$\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill -1\hfill \end{array}\right|=-\widehat{i}+5\widehat{j}+3\widehat{k}$

Any point on the line in given by x – y = 2

& 2x + y = 2

$\Rightarrow x=\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.,y=\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.,z=0$

$\therefore equationoflineL:\frac{x-\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{-1}=\frac{y+\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{5}=\frac{z}{3}=r$

$\therefore pointP\left(\frac{33}{35},\frac{45}{35},\frac{41}{35}\right)\equiv \left(\alpha ,\beta ,\gamma \right)$

$\therefore 35\left(\alpha +\beta +\gamma \right)=119$                                           

$\overrightarrow{a}*\overrightarrow{b}=\overrightarrow{c}-\left(i\right)$              

$\overrightarrow{b}*\overrightarrow{c}=\overrightarrow{a}-\left(ii\right)\left|\overrightarrow{a}\right|=2$              

Taking dot product with  $\overrightarrow{c}\&\overrightarrow{a}$ respectively in (i) & (ii) we have 

  $\left[\overrightarrow{c}\overrightarrow{a}\overrightarrow{b}\right]={\left|\overrightarrow{c}\right|}^2$            

Again (i) & (ii) $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0$ Þ    

Opt 1.   Projection of $\overrightarrow{a}on\overrightarrow{b}*\overrightarrow{c}=\frac{\overrightarrow{a}.\left(\overrightarrow{b}*\overrightarrow{c}\right)}{\left|\overrightarrow{b}*\overrightarrow{c}\right|}=\frac{4}{2}=2$  

 Opt 2.   $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]+\left[\overrightarrow{c}\overrightarrow{a}\overrightarrow{b}\right]=2\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=8$ (2)

(1) $\overrightarrow{b}*\left(\overrightarrow{a}*\overrightarrow{b}\right)=\overrightarrow{b}*\overrightarrow{c}$ Þ

Opt 3.   ${\left|3\overrightarrow{a}+\overrightarrow{b}-2\overrightarrow{c}\right|}^2=9{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+4{\left|\overrightarrow{c}\right|}^2+2\left(3\overrightarrow{a}.\overrightarrow{b}-6\overrightarrow{a}.\overrightarrow{c}-2\overrightarrow{b}.\overrightarrow{c}\right)$  

Opt 4.  $\overrightarrow{a}*\left(\overrightarrow{c}*\overrightarrow{b}-\overrightarrow{b}*\overrightarrow{c}\right)$  

Let the total work by LCM of 20, 24 and 30 i.e. 120 units

Work done per day by A and B = a + b = $\frac{120}{20}=6$ $\frac{}{}$

Work done per day by B and C = b + c = $\frac{120}{24}=5$ $\frac{}{}$

Work done per day by A and C = a + c = $\frac{120}{30}=4$ $\frac{}{}$

2 (a + b + c) = 15

So, a + b + c = 7.5

Required number of days = $\frac{120}{7.5}=16\text{?}days$

Let the numbers be x & y

(mean proportion)2 = xy.

So, xy = 784

x = $\frac{784}{y}$ $\frac{}{}$ .… (1)

x, y and 224 are in continued proportion

y2 = 224 x .… (2)

y2 = 224 * $\frac{784}{y}$ $\frac{}{}$

y3 = 14 * 16 * 28 * 28

y = 56 and x = 14

Alternate Method

Use options and check.

Let the speeds be 2Aand A kmph respectively

$\frac{216}{3A}=6$

A = 12 and the speed of B = 3 kmph

$si{n}^{7}x=co{s}^{7}x=1,x\in \left[0,4\pi \right]$            

will satisfy for sin x = 1, cos x = 0

$\Rightarrow x=\frac{\pi }{2}\&\frac{5\pi }{2}.$              

or, cos x = 1, sin x = 0

x = 0, 2π, 4π              total 5 solutions

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

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

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

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

$\therefore \alpha =1$