Maths

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

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

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

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

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 Þ a = 1

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

${\displaystyle \underset{0}{\overset{1}{\int }}\left[-8x^2+6x-1\right]dx}$  

Let f (x) = -8x² + 6x – 1

f (0) = -1, f (1) = -3

max f (x) =  $-\frac{D}{4a}=-\frac{36-32}{-32}=\frac{1}{8}$  

= $-\frac{1}{4}-1\left(\frac{3}{4}-\frac{\lambda }{2}\right)-2\left(\frac{\sqrt[]{17}-3}{8}-\frac{3}{4}\right)-3\left(1-\frac{\sqrt[]{17}-3}{8}\right)$  

= $\frac{\sqrt[]{17}-13}{8}$  

Let $\frac{A_2}{A_1}=\frac{A_3}{A_2}=...=r$  

$\therefore A_1{A}_3{A}_5{A}_7=\frac{1}{1296}$

$\Rightarrow A_1{r}^3=\frac{1}{6}.......\left(i\right)$  

Again, A₂ + A₄ = $\frac{7}{36}$  

$A_{1}r=\frac{7}{36}-\frac{1}{6}=\frac{1}{36}........\left(ii\right)$

$\therefore A_6+A_8+A_{10}=A_1{r}^5\left(1+r^2+r^4\right)=43$  

  $\therefore A_6+A_8+A_{10}=A_1{r}^5\left(1+r^2+r^4\right)=43$