26. A man wants to cut three lengths from a single piece of board of length 91cm.The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5cm longer than the second?

[Hint: If x is the length of the shortest board, then x , (x + 3) and 2x are the lengths of the second and third piece, respectively. Thus, x + (x + 3) + 2x ≤ 91 and2x ≥ (x + 3) + 5].

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

 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

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