Ncert Solutions Maths class 12th

$\underset{x\to 7}{lim}\frac{18-\left[1-x\right]}{\left[x-3a\right]}$

exist &   $a\in I.$

$=\underset{x\to 7}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}$

exist

RHL =   $\underset{x\to 7^{+}}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}=\frac{25}{7-3a}\left[a\ne \frac{7}{3}\right]$

$LHL=\underset{x\to 7^{-}}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}=\frac{24}{6-3a}\left[a\ne 2\right]$

LHL = RHL

$\Rightarrow \frac{25}{7-3a}=\frac{8}{2-a}$

$\therefore a=-6$

$A={\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)}_{3*3}$

 B is a matrix of same order with entries from {1,2,3,4,5}. and satisfying AB = BA.

$LetB=\left(\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right)$           

$\therefore \left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$

$a_{21}=a_{12},a_{22}=a_{11},a_{23}=a_{13},a_{31}=a_{32}$

$\therefore$ there exist only 5 distinct entries in the matrix B so that possible case = 5⁵ = 3125

$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

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

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

(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

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

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.$  

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

&  $\overrightarrow{r}.\left(2\widehat{i}+\widehat{j}-\widehat{k}\right)=2$ 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.  $\overrightarrow{a}*\left(\overrightarrow{c}*\overrightarrow{b}-\overrightarrow{b}*\overrightarrow{c}\right)$