Maths

$\frac{sinA}{sinB}=\frac{sin\left(A-C\right)}{sin\left(C-B\right)}$

sin A sin C cos B – sin A cos C sin B = sin B sin A cos C – sin B cos A sin C

2 sin A sin B cos C = sin A sin C cos B + sin B sin C cos A

By sine rule $\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}=k$

$2.ak.bk.\frac{\left(a^2+b^2-c^2\right)}{2ab}=ak.ck.\frac{\left(c^2+a^2-b^2\right)}{2ac}+bk.ck.\frac{\left(b^2+c^2-a^2\right)}{2bc}$

$\Rightarrow b^2,c^2,a^{2}areinA.P.$

Locus of point of intersection of perpendicular tangent will be its director circle & director circle of parabola be its directrix.

Given parabola y² = 16 (x – 3) so equation of directrix be x – 3 = - 4 i.e., x + 1 = 0

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

$y\left(x\right)=co{t}^{-1}\left(\frac{\sqrt[]{1+sinx}+\sqrt[]{1-sinx}}{\sqrt[]{1+sinx}-\sqrt[]{1-sinx}}\right),x\in \left(\frac{\pi }{2},\pi \right)$

$=co{t}^{-1}\left(\frac{sin\frac{x}{2}+cos\frac{x}{2}+sin\frac{x}{2}-cos\frac{x}{2}}{sin\frac{x}{2}+cos\frac{x}{2}-sin\frac{x}{2}+cos\frac{x}{2}}\right)=co{t}^{-1}tan\frac{x}{2}=co{t}^{-1}cot\left(\frac{\pi }{2}-\frac{\pi }{2}\right)=\frac{\pi }{2}-\frac{x}{2}$

$\frac{dy}{dx}=\frac{-1}{2}$

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

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

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

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

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

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

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

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

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

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

=>α = 1, β = -16

|α - β| = 17

${\left(3x^2+4x+3\right)}^2-\left(k+1\right)\left(3x^2+4x+3\right)\left(3x^2+4x+2\right)+k{\left(3x^2+4x+2\right)}^2=0$

 Let $3x^2+4x+3=a\&3x^2+4x+2=b\Rightarrow a-1$  

So, (i) becomes a² – (k + 1)ab + kb² = 0

(a – kb) (a – b) = 0 Þ a = kb or a = b ® not possible

->3x² + 4x + 3 = k (3x² + 4x + 2)

For real roots $D\ge 0$  

$16{\left(k-1\right)}^2-12\left(k-1\right)\left(2k-3\right)\ge 0$     

So, $k\in \left(1,\frac{5}{2}\right]$  

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

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

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

Variance = $\frac{\sum {\left(x_i-\overline{x}\right)}^2}{n}=\frac{\sum \left({x_i}^2+{\overline{x}}^2-2\overline{x}xi\right)}{n}$  

$=\frac{\sum {x_i}^2+{\overline{x}}^2\sum 1-2\overline{x}\sum x_i}{n}$

$=\frac{\frac{n\left(n+1\right)\left(2n+1\right)}{6}+{\left(\frac{n\left(n+1\right)}{2n}\right)}^2.n-\frac{2\left(n+1\right)}{2n}.\frac{n\left(n+1\right)}{2}}{n}$ $=\frac{n^2-1}{12}$

Now, $\frac{n^2-1}{12}=14son=13$