Chemistry NCERT Exemplar Solutions Class 12th Chapter Five

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

  $e_1^2=1-\frac{b^2}{a^2}$ -(i)

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

B > a

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

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

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

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.$  

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

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

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

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

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

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

2ⁿ = 4096 -> n = 12

¹²C₆ = 924 (the greatest coefficient)

$\Rightarrow \frac{6}{3}=\frac{2}{-t}\Rightarrow t=-1$

 ->R (-1,0)

    $P{R}^2+R{Q}^2=20+5=25$