Maths

  $l={\displaystyle \underset{-2}{\overset{2}{\int }}\frac{\left|x^3+x\right|}{e^{x\left|x\right|}+1}dx}$ ……. (i)

$l={\displaystyle \underset{-2}{\overset{2}{\int }}\frac{\left|x^3+x\right|}{e^{-x\left|x\right|}+1}dx}$ …. (ii)

$=\left(\frac{16}{4}+\frac{4}{2}\right)$ - 0

= 4 + 2 = 6

  $l={\displaystyle \int \frac{e^x\left(x^2+1\right)}{{\left(x+1\right)}^2}dx=f\left(x\right)e^X+c}$

$l=\text{?}{\displaystyle \int \frac{e^x\left(x^2-1+1+1\right)}{{\left(x+1\right)}^2}dx}$

 $={\displaystyle \int e^x}\left[\frac{x-1}{x+1}+\frac{2}{{\left(x+1\right)}^2}\right]dx$

 for x = 1

$f\text{'}\text{'}\text{'}\left(1\right)=\frac{12}{24}=\frac{12}{16}=\frac{3}{4}$                

$co{s}^{-1}\left(\frac{y}{2}\right)=lo{g}_e{\left(\frac{x}{5}\right)}^5,\left|y\right|<2$  

 Differentiating on both side

$-\frac{1}{\sqrt[]{1-{\left(\frac{y}{2}\right)}^2}}*\frac{y\text{'}}{2}=\frac{5}{\frac{x}{5}}*\frac{1}{5}$  

$\frac{-xy\text{'}}{2}=5\sqrt[]{1-{\left(\frac{y}{2}\right)}^2}$

Square on both side

$\frac{x^{2}y{\text{'}}^2}{4}=25\left(\frac{4-y^2}{4}\right)$

Diff on both side

$xy\text{'}+y\text{'}\text{'}{x}^2+25y=0$  

  $CD=\sqrt[]{{\left(10+x^2\right)}^2-{\left(10-x^2\right)}^2}=2\sqrt[]{10}\left|x\right|$

$=\frac{1}{2}*CD*AB=\frac{1}{2}*2\sqrt[]{10}\left|x\right|\left(20-2x^2\right)$

$\Rightarrow 10-x^2=2x$              

 3x² = 10

x = k

3k² = 10

  $f\left(x\right)=x^4-4x+1=0$              

$f\text{'}\left(x\right)=4x^3-4$

$=4\left(x-1\right)\left(x^2+1+x\right)$              

$\Rightarrow$ Two solution

  $\underset{x\to 7}{lim}\frac{18-\left[1-x\right]}{\left[x-3a\right]}$

exist  & $a\in I.$  

$=\underset{x\to 7}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}$      

exist

RHL = $\underset{x\to 7^{+}}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}=\frac{25}{7-3a}\left[a\ne \frac{7}{3}\right]$  

$LHL=\underset{x\to 7^{-}}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}=\frac{24}{6-3a}\left[a\ne 2\right]$

LHL = RHL

$\Rightarrow \frac{25}{7-3a}=\frac{8}{2-a}$

$\therefore a=-6$  

  $\frac{dy}{dx}=\frac{ax-by+a}{bx+cy+a}$

$=bxdy+cydy+ady=axdx-bydx+adx$                

$=\frac{c{y}^2}{2}+ay-\frac{a{x}^2}{2}-ax+bxy=k$              

$a{x}^2+a{y}^2+2ax-2ay=k$            

$\Rightarrow x^2+y^2+2x-2y=\lambda$              

Short distance of (11,6)

= $\sqrt[]{12^2+5^2}-5$

= 13 – 5

= 8

  $x={\displaystyle \sum _{n=0}^{\infty }{a}^n}=\frac{1}{1-a}y={\displaystyle \sum _{n=0}^{\infty }{b}^n=\frac{1}{1-b}{\displaystyle \sum _{n=0}^{\infty }{c}^n=\frac{1}{1-c}}}$               

Now,

a, b, c -> AP

1 – a, 1 – b, 1 – c -> AP

$\frac{1}{1-a},\frac{1}{1-b},\frac{1}{1-c}\to HP$

x, y, z -> HP

x + 2y + z = 2

$\alpha x+3y-z=\alpha$

$-\alpha x+y+2z=-\alpha$            

$\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill \alpha \hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill -\alpha \hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right|=1\left(6+1\right)-2\left(2\alpha -\alpha \right)+1\left(\alpha +3\alpha \right)=7+2a$            

$\alpha =-\frac{7}{2}$                

  $\overline{z}=i{z}^2$

Let z = x + iy

x – iy = I (x² – y² + 2xiy)

Case-I

x = 0

-y² = -y

y = 0, 1

Case – II

$y=-\frac{1}{2}$

$\Rightarrow x^2-\frac{1}{4}=\frac{1}{2}\Rightarrow x=\pm \frac{\sqrt[]{3}}{2}$

Area of polygon

$=\frac{1}{2}\left|\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \frac{\sqrt[]{3}}{2}\hfill & \hfill \frac{-1}{2}\hfill & \hfill 1\hfill \\ \hfill \frac{-\sqrt[]{3}}{2}\hfill & \hfill \frac{-1}{2}\hfill & \hfill 1\hfill \end{array}\right|=\frac{1}{2}\left|-\sqrt[]{3}-\frac{\sqrt[]{3}}{2}\right|=\frac{3\sqrt[]{3}}{4}$