Maths

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

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

$\therefore c^2=a^2{m}^2-b^2=20$        

$x\left(1-x^2\right)\frac{dy}{dx}+\left(3x^{2}y-y-4x^3\right)=0$

$\Rightarrow x\left(1-x^2\right)\frac{dy}{dx}+\left(3x^2-1\right)y=4x^3$

$\Rightarrow y=2x-1\left(x^3-x\right)\therefore y\left(3\right)=-18$

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

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

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

Integrating

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

Passes (1, 0)

1 = c

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

For x < 0 0 < e^x < 1 [e^x] = 0

$0\le x<1a{e}^x+\left[x-1\right]$               

$=a{e}^x+\left[x\right]-1$               

= a e^x – 1             b + [sin px]

$\therefore f\left(x\right)=\left[\begin{array}{l}0x<0\\ a{e}^x-10\le x<1\\ b-11\le x<2\\ -cx\ge 2\end{array}\right]$               

For f to be continuous at x = 0

a – 1 = 0 ⇒ a = 1

$\begin{array}{l}\therefore a+b+c=1+e+1-e=2\\ a+b+c\ne 1\end{array}$