Class 12th

Kindly consider the following Image.

Common grandchild of M6 and M8 is D

Please find the answer below

Let square is made with piece of length x metre & hexagon with piece of length y metre

x + y = 20 .(i)

$a=\frac{x}{4}\&b=\frac{y}{6}$       

Now let A = area of square + area of hexagon

$A=\frac{x^2}{16}+6*\frac{\sqrt[]{3}}{4}*\frac{y^2}{36}=\frac{x^2}{16}+\frac{\sqrt[]{3}{y}^2}{24}=\frac{x^2}{16}+\frac{\sqrt[]{3}}{24}{\left(20-x\right)}^{2}from\left(i\right)$ ,

for minimum area

$x=\frac{80\sqrt[]{3}}{6+4\sqrt[]{3}}=\frac{80\sqrt[]{3}}{2\sqrt[]{3}\left(\sqrt[]{3}+2\right)}=\frac{40}{2+\sqrt[]{3}}$

$x=40\left(2-\sqrt[]{3}\right)$

=> side of hexagon = $\frac{y}{6}=\frac{20\sqrt[]{3}\left(2-\sqrt[]{3}\right)}{6}=\frac{20\sqrt[]{3}}{6\left(2+\sqrt[]{3}\right)}=\frac{10\sqrt[]{3}}{3\left(2+\sqrt[]{3}\right)}=\frac{10}{2\sqrt[]{3}+3}$

$\left|A\right|=\left|\begin{array}{ccc}\hfill \left[x+1\right]\hfill & \hfill \left[x+2\right]\hfill & \hfill \left[x+3\right]\hfill \\ \hfill \left[x\right]\hfill & \hfill \left[x+3\right]\hfill & \hfill \left[x+3\right]\hfill \\ \hfill \left[x\right]\hfill & \hfill \left[x+2\right]\hfill & \hfill \left[x+4\right]\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill \left[x\right]+1\hfill & \hfill \left[x\right]+2\hfill & \hfill \left[x\right]+3\hfill \\ \hfill \left[x\right]\hfill & \hfill \left[x\right]+3\hfill & \hfill \left[x\right]+3\hfill \\ \hfill \left[x\right]\hfill & \hfill \left[x\right]+2\hfill & \hfill \left[x\right]+4\hfill \end{array}\right|$

$R_1\to R_1-R_3\&R_2\to R_2-R_3$

$\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill \left[x\right]\hfill & \hfill \left[x\right]+2\hfill & \hfill \left[x\right]+4\hfill \end{array}\right|=192$

$\Rightarrow \left[x\right]=62$

$\Rightarrow x\in [62,63)$              

The SI unit of the electrical power is Watt which is symbolized as W. One watt is equal to one joule per second (J/s). In the context of current and voltage, one watt is the power consumed when one ampere of the current flows through the potential difference of one volt.

$y^{1/4}+\frac{1}{y^{1/4}}=2x$

${\left(y^{1/4}\right)}^2-2x{y}^{1/4}+1=0$    

=> ${\left(y^{1/4}\right)}^2-2x{y}^{1/4}+1=0$

$\Rightarrow \frac{dy}{dx}=\frac{4y}{\sqrt[]{x^2-1}}..........\left(i\right)$

$\Rightarrow \left(x^2-1\right)\frac{d^{2}y}{d{x}^2}+\frac{xdy}{dx}-16y=0from\left(i\right)$      

=>α = 1, β = -16

|α - β| = 17

Let $l={\displaystyle \int \frac{dx}{{\left(x^2+x+1\right)}^2}.............\left(i\right)}$

Now ${\displaystyle \int \frac{dx}{x^2+x+1}={\displaystyle \int \left(\frac{1}{x^2+x+1}\right)dx}}$

by integration by parts

=> ${\displaystyle \int \frac{dx}{{\left(x^2+x+1\right)}^2}=\frac{4}{3\sqrt[]{2}}ta{n}^{-1}\frac{2x+1}{\sqrt[]{3}}+\frac{1}{3}\frac{\left(2x+1\right)}{\left(x^2+x+1\right)}+c}$

$\Rightarrow a=\frac{4}{3\sqrt[]{3}},b=\frac{1}{3}$

Now, $9\left(\sqrt[]{3}a+b\right)=9\left(\frac{4}{3}+\frac{1}{3}\right)=15$

Equation of the plane will be $\left\{\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)-1\right\}+\lambda \left\{\overrightarrow{r}.\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)+4\right\}=0$   

$\overrightarrow{r}.\left\{\left(1+2\lambda \right)i+\left(1+3\lambda \right)\widehat{j}+\left(1-\lambda \right)\widehat{k}\right\}+\left(4\lambda -1\right)=0$               

-> $\left(1+2\lambda \right)x+\left(1+3\lambda \right)y+\left(1-\lambda \right)z+\left(4\lambda -1\right)=0.........\left(i\right)$

(i) is parallel to x-axis so its d.r.s will be (1, 0, 0)

-> $1+2\lambda =0so\lambda =\frac{-1}{2}$

Hence required equation will be

$\overrightarrow{r}\left\{-\frac{1}{2}\widehat{j}.+\frac{3}{2}\widehat{k}\right\}+\left(-3\right)=0$

$\overrightarrow{r}.\left(\widehat{j}-3\widehat{k}\right)+6=0$