8. (2 – x) ≥ 2 (1 – x)

  $x^2+y^2-2\sqrt[]{2}x-6\sqrt[]{2}y+14=0$

 centre $\left(\sqrt[]{2},3\sqrt[]{2}\right)$

radius  ${\left({\left(\sqrt[]{2}\right)}^2+{\left(3\sqrt[]{2}\right)}^2-14\right)}^{1/2}$

$={\left(2+18-14\right)}^{1/2}=\left(\sqrt[]{6}\right)$

${\left(x-2\sqrt[]{2}\right)}^2+{\left(y-2\sqrt[]{2}\right)}^2=r^2$

centre  $\left(2\sqrt[]{2},2\sqrt[]{2}\right)$

OA = $\sqrt[]{{\left(2\sqrt[]{2}-\sqrt[]{2}\right)}^2+{\left(2\sqrt[]{2}-3\sqrt[]{2}\right)}^2}=\sqrt[]{2+2}=2$                  

r² = ${\left(\sqrt[]{6}\right)}^2+{\left(2\right)}^2=6+4=10$  

Let y = mx + c is the common tangent

$soc=\frac{1}{m}=\pm \frac{3}{2}\sqrt[]{1+m^2}\Rightarrow m^2=\frac{1}{3}$

so equation of common tangents will be

$y=\pm \frac{1}{\sqrt[]{3}}x\pm \sqrt[]{3}$

which intersects at Q (3, 0)

Major axis and minor axis of ellipse are 12 and 6. So eccentricity

$e^2=1-\frac{1}{4}=\frac{3}{4}$

and length of latus rectum

$=\frac{2b^2}{a}=3$

Hence

$\frac{\mathcal{l}}{e^2}=\frac{3}{3/4}=4$

$z^2+z+1=0\Rightarrow z=w,w^2$

$\left|{\displaystyle \sum _{n=1}^{15}{\left(z^n+{\left(-1\right)}^n\frac{1}{z^n}\right)}^2}\right|=\left|{\displaystyle \sum _{n=1}^{15}\left(z^{2n}+\frac{1}{z^{2n}}+2{\left(-1\right)}^n\right)}\right|=\left|{\displaystyle \sum _{n=1}^{15}{w}^{2n}+\frac{1}{w^{2n}}+2{\left(-1\right)}^n}\right|$

$=\left|\frac{w^2\left(1-1\right)}{1-w^2}+\frac{\frac{1}{w^2}\left(1-1\right)}{1-\frac{1}{w^2}}-2\right|=2$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l}\left(i\right)-4x\ge 12\Rightarrow x\le -3\\ Hence,thefilleris\left(\le \right)\\ \left(ii\right)If-\frac{3}{4}x\le -3\Rightarrow x\ge 3x\times \frac{4}{3}\Rightarrow x\ge 4\\ Hence,thefilleris\left(\ge \right)\\ \left(iii\right)If\frac{2}{x+2}>0\Rightarrow x>-2\\ Hence,thefilleris\left(>\right)\\ \left(iv\right)Ifx>-5\Rightarrow 4x>-20\\ Hence,thefilleris\left(>\right)\\ \left(v\right)Ifx>yandz<0,then\\ xz<yz\Rightarrow -xz>-yz\\ Hence,thefilleris\left(>\right)\\ \left(vi\right)Ifp>0andq<0,then\\ p-q>p\\ Hence,thefilleris\left(>\right)\\ \left(vii\right)If\left|x+2\right|>5then\\ x+2<-5orx+2>5\\ \Rightarrow x<-5-2orx>5-2\\ \Rightarrow x<-7orx>3\\ So,x\in \left(-\infty ,-7\right)\cup \left(3,\infty \right)\\ Hence,thefilleris\left(<\right)or\left(>\right)\\ \left(viii\right)If-2x+1\ge 9then\\ \Rightarrow -2x\ge 9-1\Rightarrow -2x\ge 8\Rightarrow 2x\le -8\Rightarrow x\le -4\\ Hence,thefilleris\left(\le \right)\end{array}$

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l}\left(i\right)Ifx<yandb<0\\ \Rightarrow \frac{x}{b}>\frac{y}{b}\\ Hence,statement\left(i\right)isFalse.\\ \left(ii\right)Ifx,y>0thenx>0,y>0orx<0,y<0\\ Hence,statement\left(ii\right)isFalse.\\ \left(iii\right)Ifxy>0thenx<0,andy<0\\ Hence,statement\left(iii\right)isTrue.\\ \left(iv\right)Ifxy<0thenx<0,y>0orx>0,y<0\\ Hence,statement\left(iv\right)isFalse.\\ \left(v\right)Ifx<-5andx<-2\Rightarrow x\in \left(-\infty ,-5\right)\\ Hence,statement\left(v\right)isTrue.\\ \left(vi\right)Ifx<-5andx>2thenxhasnovalue.\\ Hence,statement\left(vi\right)isFalse.\\ \left(vii\right)Ifx>-2andx<9\Rightarrow x\in \left(-2,9\right)\\ Hence,statement\left(vii\right)isTrue.\\ \left(viii\right)If\left|x\right|>5thenx<-5orx>5\\ \Rightarrow x\in \left(-\infty ,-5\right)\cup \left(5,\infty \right)\\ Hence,statement\left(viii\right)isFalse.\\ \left(ix\right)If\left|x\right|\le 4then-4\le x\le 4\\ \Rightarrow x\in \left[-4,4\right]\\ Hence,statement\left(ix\right)isTrue.\\ \left(x\right)Thegivengraphrepresentsx\le 3\\ Hence,statement\left(x\right)isFalse.\\ \left(xi\right)Thegivengraphrepresentsx\ge 0\\ Hence,statement\left(xi\right)isTrue.\\ \left(xii\right)Thegivengraphrepresentsy\ge 0\\ Hence,statement\left(xii\right)is\text{\&thins}\end{array}$