Maths

kindly consider the following Image 

$P:\left(x+3y-z-6\right)=\lambda \left(-6x+5y-z-7\right)=0$

passes $\left(2,3,\frac{1}{2}\right)$

$\Rightarrow \left(2+9-\frac{1}{2}-6\right)=\lambda \left(-12+15-\frac{1}{2}-7\right)=0$

$\Rightarrow \lambda =1$

$\frac{{\left|13\overrightarrow{a}\right|}^2}{d^2}=\frac{{\left(13\right)}^2\left(93\right)}{{\left(13\right)}^2}=93$

$\left(5x+8y+13z-29\right)+\lambda \left(8x-7y+z-20\right)=0$  

P₁ passing (2, 1, 3)

(10 + 8 + 39 – 29) + $\lambda \left(16-7+3-20\right)=0$  

$28-8\lambda =0\lambda =7/2$  

$\Rightarrow$  2X – Y + Z – 6 = 0      ….(i)

For P₂ passes (0, 1, 2)

$\left(8+26-29\right)+\lambda \left(-7+2-20\right)=0$

$5-25\lambda =0\Rightarrow \lambda =\frac{1}{5}$  

$P_2:x+y+2z-5=0........(ii)$  

Acute angle between the planes

  $cos\theta =\frac{2-1+2}{\sqrt[]{4+1+1}\sqrt[]{1+1+4}}=\frac{1}{2}$  

$\Rightarrow \theta =\frac{\pi }{3}$  

$a^2\left(e^2-1\right)=b^2$

$e=\sqrt[]{\frac{5}{2}}\Rightarrow b^2=\frac{3a^2}{2}$

Length of latus rectum $\frac{2b^2}{a}=6\sqrt[]{2}$  

$3a=6\sqrt[]{2}\Rightarrow a=2\sqrt[]{2}$  

$b=2\sqrt[]{3}$  

y = 2x + c is tangent to hyperbola

$\therefore c^2=a^2{m}^2-b^2=20$  

$f\left(x\right)=x^7+5x^3+3x+1$

$f\text{'}\left(x\right)=7x^6+15x^4+3>0\forall x\in R$

$\therefore f\left(x\right)$ is increasing

 for x $\infty$ - , f (x) $\infty$

x = 0, f (x) = 1

$\therefore f\left(x\right)$ = 0 has only one real root.

$\left[\frac{x}{\sqrt[]{x^2-y^2}}+e^{y/x}\right]x\frac{dy}{dx}=x+\left[\frac{x}{\sqrt[]{x^2-y^2}}+e^{y/x}\right]y$

$\Rightarrow e^{y/x}\left[xdy-ydx\right]=xdx+\frac{x}{\sqrt[]{x^2-y^2}}\left(ydx-xdy\right)$

$\Rightarrow e^{y/x}d\left(y/x\right)=\frac{dx}{x}-\frac{d\left(y/x\right)}{\sqrt[]{1-{\left(y/x\right)}^2}}$

Integrating

$e^{y/x}=\mathcal{l}nx-si{n}^{-1}\left(\frac{y}{x}\right)+c$

Passes (1, 0)

⇒ 1 = c

$\Rightarrow \alpha =\frac{1}{2}exp\left(\sqrt[]{e}-1+\frac{\pi }{6}\right)$

$y^2\le 8xy>\sqrt[]{2}x$

$2x^2=8x\Rightarrow x\left(x-4\right)=0\Rightarrow x=0,x=4x=0,x=y$

Required area = ${\displaystyle \underset{1}{\overset{4}{\int }}\left(2\sqrt[]{2}\sqrt[]{x}-\sqrt[]{2}x\right)}$ dx

$=\frac{28\sqrt[]{2}}{3}-\frac{15\sqrt[]{2}}{2}=\frac{11\sqrt[]{2}}{6}$