Maths

 l + m – n = 0 Þ n = l + m

$3l^2+m^2+cnl=0$  

$3l^2+m^2+cl\left(l+m\right)=0$    

$=\left(3+c\right){\left(\frac{l}{m}\right)}^2+c\left(\frac{l}{m}\right)+1=0$    

$?$    Lines are parallel

D = 0

$c=4$     (as c > 0)

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow \frac{{\left(-4\sqrt[]{\frac{2}{5}}\right)}^2}{a^2}+\frac{32}{b^2}=1$                

$\Rightarrow \frac{32}{5a^2}+\frac{9}{b^2}=1$ ……. (i)

From (i)

$\frac{6}{b^2}+\frac{9}{b^2}=1\Rightarrow b^2=15\&a^2=16$

$a^2+b^2=15+16=31$

  $\begin{array}{l}2x+y=4\\ 2x+6y=14\end{array}\}y=2,x=3$              

B (1, 2)

Let C (k, 4 – 2k)

Now AB² = AC²

->5k² – 24k + 19 = 0

$\alpha =\frac{6+1+\frac{10}{5}}{3}=\frac{18}{5}$    

Now 15 (a + b)

$15\left(\frac{17}{5}\right)=51$                

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

x, y > 0, y(1) = 1

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

${\displaystyle \int \frac{2^y}{2^y-1}dy=-{\displaystyle \int \frac{2^x}{2^x-1}dx}}$           

$=\frac{lo{g}_e\left(2^y-1\right)}{lo{g}_{e}2}=-\frac{lo{g}_e\left(2^x-1\right)}{lo{g}_{e}2}+\frac{lo{g}_{e}c}{lo{g}_{e}2}$  

Taking log of base 2.

$\therefore$  y = 2 – log₂ 

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

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

$\therefore a=-6$