Maths

$l=\underset{n\to \infty }{lim}{\left(u_n\right)}^{\frac{-4}{n^2}}=\underset{n\to \infty }{lim}{\left\{\left(1+\frac{1}{n^2}\right){\left(1+\frac{2^2}{n^2}\right)}^2{\left(1+\frac{3^2}{n^2}\right)}^3......{\left(1+\frac{n^2}{n^2}\right)}^n\right\}}^{\frac{-4}{n^2}},$ Taking log on both the sides

$\underset{n\to \infty }{lim}\frac{-4}{n^2}{\displaystyle \sum _{r=1}^{n}rlog\left(1+\frac{r^2}{n^2}\right)}=\underset{n\to \infty }{lim}\frac{-4}{n}{\displaystyle \sum _{r=1}^n\frac{r}{n}log\left(1+{\left(\frac{r}{n}\right)}^2\right)}$

$\therefore logl=-4{\displaystyle \underset{0}{\overset{1}{\int }}xlog\left(1+x^2\right)dx}$

$logl=-2\left[log4-1\right]=-2log\frac{4}{e}=log\frac{e^2}{16}$       

$l=\frac{e^2}{16}$

x + y + z = 4

3x + 2y + 5z = 3

$9x+4y+\left(28+\left[\lambda \right]\right)z=\left[\lambda \right]$           

$\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill 5\hfill \\ \hfill 9\hfill & \hfill 4\hfill & \hfill 28+\left[\lambda \right]\hfill \end{array}\right|=56+2\left[\lambda \right]-20-\left(84+3\left[\lambda \right]-45\right)+\left(-6\right)$

$=-\left[\lambda \right]-9$              

$If\left[\lambda \right]+9\ne 0$ then unique solution

& if  $\left[\lambda \right]+9=0then{\Delta }_1={\Delta }_2={\Delta }_3=0$ so infinite solution will exist

Hence $\lambda \in R$  

Required probability of obtaining a (sum = 7) = $\frac{13}{96}\left(given\right)$  

    $i.e.2\left(\left(\frac{1}{6}+x\right)\left(\frac{1}{6}-x\right)+\frac{1}{6}*\frac{1}{6}+\frac{1}{6}*\frac{1}{6}\right)=\frac{13}{96}$

$x=\frac{1}{8}$           

$ar\left(\Delta ABC\right)=\frac{1}{2}\left|\begin{array}{ccc}\hfill a\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill b\hfill & \hfill 2b+1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill b\hfill & \hfill 1\hfill \end{array}\right|=\frac{1}{2}\left[a\left(2b+1-b\right)+\left(b^2\right)\right]=\frac{1}{2}\left[ab+a+b^2\right]$  

Given ar (  $\Delta ABC)$ = 1 so |ab + a + b²| = 2

  $\Rightarrow a\left(b+1\right)+b^2=\pm 2$             

a = $\frac{-b^2+2}{b+1},\frac{-b^2-2}{b+1}$  

So sum = $\frac{-2b^2}{b+1}$  

Equation of line PQ :

$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=k$

So, let Q (2k + 1, 3k – 2, -6k + 3) Q lies on given plane so

2k + 1 – 3k + 2 – 6k + 3 = 5

$-7k=-1ork=\frac{1}{7}$

So, $Q\left(\frac{9}{7},\frac{-11}{7},\frac{15}{7}\right)$

Now required distance = PQ = $\sqrt[]{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=1$

Equation of tangent at P (2, -4)

y (-4) = 4 (x + 2)

x + y + 2 = 0

So, A (-2, 0)

Equation of normal at P:

y + 4 = 1 (x – 2)

x – y = 6

So, B (-2, -8)

For square mid-point of AB = mid-point of PQ

$\Rightarrow \frac{a+2}{2}=-2\Rightarrow a=-6$        

$\frac{b-4}{2}=\frac{-8}{2}\Rightarrow b=-4$   

So, 2a + b = -16

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

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

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

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

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