Maths

  $f\left(x\right)=\{\begin{array}{l}-2x^2+3,-2<x<-1\\ x^2,-1\le x<0\\ \frac{\sqrt[]{3}}{2}-1,0\le x<1\\ x^2-3+\frac{1}{\sqrt[]{2}},1\le x<2\end{array}$

$f\left(-1^{-}\right)=f\left(-1^{+}\right)=f\left(-1\right)=1$                

$f\left(1^{+}\right)=f\left(1\right)=\frac{1}{\sqrt[]{2}}-2$                

Points of discontinuity

x = 0, 1

$\overrightarrow{v}=l\overrightarrow{a}+m\overrightarrow{b}$  

$=\left(2l+m,-l+2m,2l=m\right)$               

  $\overrightarrow{v}.\left(3,2,-1\right)=0\Rightarrow l=-4m.......\left(i\right)$              

  $\left|\overrightarrow{v}.\widehat{a}\right|=19\Rightarrow 9l-2m=57........\left(ii\right)$              

(i) & (ii) Þ l = 6, m =   $\frac{-3}{2}$

$\Rightarrow 2\overrightarrow{v}=\left(21,-18,27\right)$              

${\left|2\overrightarrow{v}\right|}^2=1494$            

$\underset{x\to 1}{lim}\frac{x^{n}f\left(1\right)-f\left(x\right)}{x-1}$  

$=\underset{x\to 1}{lim}\frac{9x^n-\left(x^6+2x^4+x^3+2x+3\right)}{x-1}$                

= 9n – 19 = 44 -> n = 7

ar (ABC) = 4ar (DEF)

$=4*\frac{1}{2}*\left|2\left(2-5\right)+1\left(5-3\right)+7\left(3-2\right)\right|=2\left|-6+2+7\right|=6$

  $\overline{x}=\sum *p\left(x\right)$  

$\Rightarrow \frac{23}{10}=\frac{-2}{5}-a+1+\frac{4}{5}+6b$            

  $\Rightarrow 6b-a=\frac{9}{10}.........\left(i\right)$             

Also,   $\frac{1}{5}+a+\frac{1}{3}+\frac{1}{5}+b=1$

$\Rightarrow a+b=\frac{4}{15}.......\left(ii\right)$       

(i) & (ii) -> $a=\frac{1}{10},b=\frac{1}{6}$

$\Rightarrow 100{\sigma }^2=781$          

  $\sim \left(\sim p\vee q\right)=p\wedge \sim q$

$I-A^3-3A+3A^2=I-A^3$

=> 3A² – 3A = 0

=> 3A (A – I) = 0

=>A² = A

$\left[\begin{array}{cc}\hfill a^2\hfill & \hfill ab+bd\hfill \\ \hfill 0\hfill & \hfill d^2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill d\hfill \end{array}\right]$    

Total number of ways = 8

For every a, there  must be a² – 2. So, there will be infinitely many pairs (a, b)