Straight Lines

S  $\equiv sint,C\equiv cost$ $$

Let orthocenter be (h, k)

Since it if an equilateral triangle hence orthocenter coincides with centroid.

$\therefore a+s+c=3h,b+s-c=3k$

$\frac{a}{3}=1,\frac{b}{3}=\frac{1}{3}\Rightarrow a=3,b=1$

$\therefore a^2-b^2-8$

L₁ : 3x – 4y + 12 = 0

L₂ : 8x + 6y + 11 = 0

$\left(\alpha ,\beta \right)$ lies on that angle which contain origin

$\therefore$ Equation of angle bisector of that angle which contain origin is

$\frac{3x-4y+12}{5}=\frac{8x+6y+11}{10}$

$\Rightarrow 2x+14y-13=0$

$\left(\alpha ,\beta \right)$ lies on it

$\Rightarrow 2\alpha +14\beta -13=0$ …… (i)

$\Rightarrow 3\alpha -4\beta +7=0$ ……. (ii)

Solving (i) & (ii)

$\alpha =-\frac{23}{25}\&\beta =\frac{53}{50}$

$\therefore \alpha +\beta =\frac{7}{50}$

$\Rightarrow 100\left(\alpha +\beta \right)=14$

$B\left(-\frac{3}{\sqrt[]{a}},\sqrt[]{a}\right)andc\left(-\frac{3}{\sqrt[]{a}},-\sqrt[]{a}\right)$

$\therefore Areaof\Delta ACD=\frac{1}{2}\left|\begin{array}{ccc}\hfill \frac{3}{\sqrt[]{a}}\hfill & \hfill \sqrt[]{a}\hfill & \hfill 1\hfill \\ \hfill \frac{-3}{\sqrt[]{a}}\hfill & \hfill -\sqrt[]{a}\hfill & \hfill 1\hfill \\ \hfill 3cos\theta \hfill & \hfill asin\theta \hfill & \hfill 1\hfill \end{array}\right|$  = 12 

$\Rightarrow \Delta =3\sqrt[]{a}\left|cos\theta +sin\theta \right|=12$

$\therefore \Delta max=3\sqrt[]{a}.\sqrt[]{2}=12\Rightarrow a=8$

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

$\frac{1}{26}+{\displaystyle \sum _{n=1}^{50}\left(S_n+\frac{2}{\left(n+1\right)}-\left(n+1\right)\right)=\frac{1}{26}+{\displaystyle \sum _{n=1}^{50}{\left(n+1\right)}^2}-3\left(n+1\right)+2+2\left(\frac{1}{\left(n+1\right)}-\frac{1}{\left(n+2\right)}\right)}$

$=\frac{1}{26}+45525-3*1325+2*50+2\left(\frac{1}{2}-\frac{1}{52}\right)$

$=\frac{1}{26}+41550+100+1-\frac{1}{26}=41651$

  $y=3?\left|x?\frac{1}{2}\right|=\left|x+1\right|$

Graph

Area =   $\left(\frac{1}{2}*\frac{3}{2}*\frac{3}{4}\right)*2+\frac{3}{2}*\frac{3}{2}=\frac{3}{2}*\frac{3}{2}*\frac{3}{2}=\frac{27}{8}$

$\left(sin10°.sin50°.sin70°\right).\left(sin10°.sin20°.sin40°\right)$

$=\left(\frac{1}{4}sin30°\right).\left[\frac{1}{2}sin10°\left(cos20°-cos60°\right)\right]$

$=\frac{1}{32}\left[sin30°-sin10°-sin10°\right]$

$\frac{1}{64}-\frac{1}{16}sin10°$

Clearly $\alpha =\frac{1}{64}$

Hence 16 + a^-1 = 80

36 = 2 * 2 * 3 * 3

Number should be odd multiple of 2 and does not having factor 3 and 9

Odd multiple of 2 are

102, 106, 110, 114….998 (225 no.)

No. of multiplies of 3 are

102, 114, 126 ….990 (75 no.)

Which are also included multiple of 9

Hence,

Required = 225 – 75 = 150

  $x\frac{dy}{dx}+2y=x{e}^x$ $\frac{}{}{}^{}$

$\frac{dy}{dx}+\frac{2y}{x}=e^x$

$\frac{dz}{dx}=e^x.2\left(x-1\right)+e^x{\left(x-1\right)}^2=0$

$e^x\left(x-1\right)\left(2+x-1\right)=0$

$x=-1,1$

x = 1 local maxima. Then maximum value is

$z\left(-1\right)=\frac{4}{e}-e$

${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$

$\Rightarrow \frac{n}{a}{\left(\frac{x}{a}\right)}^{n-1}+\frac{n}{b}{\left(\frac{y}{b}\right)}^{n-1}\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=-\frac{b}{a}{\left(\frac{bx}{ay}\right)}^{n-1}$

$\frac{dy}{d{x}_{\left(a,b\right)}}=-\frac{b}{a}$

So line always touches the given curve.

Let A, A' be (, 2) AB and A'B subtends $\frac{\pi }{4}$ angle at (0, 0) slope of OA = $\frac{2}{\alpha }$

slope of OB = $\frac{3}{2}$

$tan\frac{\pi }{4}=\left|\frac{\frac{2}{\alpha }-\frac{3}{2}}{1+\frac{2}{\alpha }\cdot \frac{3}{2}}\right|$

$\Rightarrow 1+\frac{3}{\alpha }=\pm \left(\frac{4-3\alpha }{2\alpha }\right)$

$\Rightarrow \frac{\alpha +3}{\alpha }=\pm \left(\frac{4-3\alpha }{2\alpha }\right)\Rightarrow \alpha =10,-\frac{2}{5}$

now distance between A'A, (10, 2) & $\left(\frac{-2}{5},2\right)is\frac{52}{5}$