Maths

$e_H=\sqrt[]{1+\frac{64}{49}}=\frac{\sqrt[]{113}}{7}$

$\Rightarrow e_H.e_E=\frac{1}{2}$

$\Rightarrow \frac{113}{49}.\frac{\left(64-a^2\right)}{64}=\frac{1}{4}\Rightarrow a^2-64=\frac{32^2}{113}$

$\mathcal{l}=\frac{2a^2}{b}=2\left(64+\frac{32^2}{113}\right).\frac{1}{8}$

$113\mathcal{l}=1552$

Plane through 4ax – y + 5z – 7a = 0 = 2x – 5y – z – 3

is    $x\left(4a+2\lambda \right)+y\left(-1-5\lambda \right)+z\left(5-\lambda \right)=7a+3\lambda$

This plane contains 4, -1, 0

->9a + 1 + 10l = 0          …… (i)

Plane contains the line $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$  

-> $4a+11\lambda +7=0$ ……. (ii)

From (i) & (ii) a = 1,   $\lambda$  =-1

Equation of plane $\pi \equiv x+2y+3z-2=0$  

$\Rightarrow 7P+3-2P+4-12P+9-2=0\Rightarrow P=2$

$\overline{x}=\frac{{\displaystyle \sum _{i=1}^{10}{x}_i}}{10}=15;\frac{{\displaystyle \sum _{i=1}^{10}{x_i}^2}}{10}-{\left(\overline{x}\right)}^2=15$

$\Rightarrow \Sigma x_i=150;\Sigma {x_i}^2=2400$

Actual mean $\overline{x}=\frac{\Sigma x_i+15-25}{10}=\frac{140}{10}=14$

Actual variance =  $\frac{\Sigma {x_i}^2+15^2-25^2}{10}-{\left(14\right)}^2$

$=\frac{2400-400}{10}-196$

$\begin{array}{l}{\sigma }^2=4\\ \sigma =2\end{array}$

use $si{n}^{-1}x=co{s}^{-1}\sqrt[]{1-x^2}$  

$ta{n}^{-1}\sqrt[]{1-x^2}=co{t}^{-1}\frac{1}{\sqrt[]{1-x^2}}$               

$si{n}^{-1}\frac{x}{\sqrt[]{1-x^2}}=co{s}^{-1}\frac{\sqrt[]{1-2x^2}}{\sqrt[]{1-x^2}}$               

Sum of roots ->b = -1 + 2   $\frac{\left(k^2-1\right)}{k^2-2}$

Product of roots -> -5 = -2  $\left(\frac{k^2-1}{k^2-2}\right)$  

b = 4, k² = $\frac{1}{3}$  

  $f\left(x\right)\{\begin{array}{l}{\displaystyle \underset{0}{\overset{3}{\int }}\left(2-t\right)dt+{\displaystyle \underset{3}{\overset{x}{\int }}\left(8-t\right)dt;x>4}}\\ x^2+bx;x\le 4\end{array}$

f(x) is continuous at x = 4

$16+4b={\displaystyle \underset{0}{\overset{3}{\int }}\left(2-t\right)dt+{\displaystyle \underset{3}{\overset{4}{\int }}\left(8-t\right)dt}}$           

16 + 4b = 15

$\Rightarrow b=\frac{-1}{4}$               

$forx\le 4,$ $\begin{array}{l}f\left(x\right)=x^2-\frac{x}{4}\\ f\text{'}\left(x\right)=2x-\frac{1}{4}\end{array}$                 

f(x) is increasing in  $\left(\frac{1}{8},\infty \right)$  

 rate change at $x=\frac{1}{8},$  from -ve to +ve. So minima occurs.

Circle passes through (6, 1)

12 g – 19 c = 43               …. (i)

Centre lies on x – 2xy = 8

->g + 6c = 8                     …. (ii)

From (i) & (ii), c = 1, 9 = 2

Length of x – intercept -  $2\sqrt[]{g^2-C}$

$\frac{{\Delta }_1}{{\Delta }_2}=\frac{\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill x\hfill & \hfill -4x-x\hfill & \hfill 1\hfill \\ \hfill -4\hfill & \hfill 3\hfill & \hfill 1\hfill \end{array}\right|}{\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill -4\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill -5\hfill & \hfill 1\hfill \end{array}\right|}=\frac{4}{7}$

->14 x – 35 y = -95        …. (ii)

Solve (i) & (ii), x =    $\frac{-20}{7},y=\frac{-11}{7}$

$ar\Delta AQR$

$=\frac{1}{2}*1*1=\frac{1}{2}$                

${\pi }_1\equiv 2x+ky-5z=1$  

${\pi }_2\equiv 3kx-ky+z=5$                

Given  ${\pi }_1\perp {\pi }_2\Rightarrow 6k-k^2-5=0$  

k = 1, 5

Now required plane is p₁ +   $\lambda {\pi }_2=0$

y intercept = $\frac{1+5\lambda }{1-\lambda }$  

Solve tan 2a = $\frac{h}{b}$  

$tan\alpha =\frac{2h}{\sqrt[]{7}h+b}$