Maths

Given P (X = 3) = 5P (X = 4) and n = 7

${\Rightarrow }^7{C}_3{p}^3{q}^4=5^7{C}_4{p}^4{q}^3$

-> q = 5p and also p + q = 1

$\Rightarrow p=\frac{1}{6}andq=\frac{5}{6}$  

Mean = $\frac{7}{6}$  and variance =   $\frac{35}{36}$

Mean + Variance

$=\frac{7}{6}+\frac{35}{36}+\frac{77}{36}$

Total case = ¹⁸C₅

Favourable cases

(Select x₁) (Select x₃)       (Select x₅)

$\overrightarrow{a}=\widehat{i}+\widehat{j}-\widehat{k}$

$\overrightarrow{c}=2\widehat{i}-3\widehat{j}+2\widehat{k}$                

Now,

$\overrightarrow{b}*\overrightarrow{c}=\overrightarrow{a}$      

$\Rightarrow \left(\widehat{i}+\widehat{j}-\widehat{k}\right)\left(2\widehat{i}-3\widehat{j}+2\widehat{k}\right)=0$            

 = 2 – 3 – 2 = 0

->-3 = 0 (Not possible)

->No possible value of

$\overrightarrow{b}$  is possible.

 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

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