Maths

Now equation of line OA be

$\frac{x-1}{4}=\frac{y-3}{-5}=\frac{z-5}{2}=\lambda$           

direction cosines of plane are 4, -5, 2

Equation of any point on OA be

$O\left(4\lambda +1,-5\lambda +3,2\lambda +5\right)$           

Since O lies on given plane so

$4\left(4\lambda +1\right)-5\left(-5\lambda +3\right)+2\left(2\lambda +5\right)=8$           

So, O (9/5,2,27/5). Hence by mid-point formula

B $\left(\frac{13}{5},1,\frac{29}{5}\right)\Rightarrow 5\left(\alpha +\beta +\gamma \right)=47$  

  $\underset{x\to a}{lim}\frac{xf\left(a\right)-af\left(x\right)}{x-a}\left(\frac{0}{0}\right)$

By L'hospital Rule

$=\underset{x\to a}{lim}\frac{f\left(a\right)-af\text{'}\left(x\right)}{1}=f\left(a\right)-af\text{'}\left(a\right)$         

$=4-2a$           

Now equation of line OA be

3, 4, 5, 5

In remaining six places you have to arrange

3, 4, 5,5

So no. of ways $=\frac{6!}{2!2!2!}$  

Total no. of seven digits nos. = $\frac{7!}{2!3!2!}*1$  

Hence Req. prob. $\frac{\frac{6!}{2!2!2!}}{\frac{7!}{2!3!2!}}=\frac{6!3!}{2!7!}=\frac{3}{7}$

$g\left(2\right)=\underset{x\to 2}{lim}g\left(x\right)=\underset{x\to 2}{lim}\frac{x^2-x-2}{2x^2-x-6}=\underset{x\to 2}{lim}\frac{\left(x-2\right)\left(x+1\right)}{2x\left(x-2\right)+3\left(x-2\right)}$           

$Nowfog=f\left(g\left(x\right)\right)=si{n}^{-1}g\left(x\right)=si{n}^{-1}\left(\frac{x^2-x-2}{2x^2-x-6}\right)$

$\frac{3x^2-2x-8}{2x^2-x-6}\ge 0\&\frac{-x^2+4}{2x^2-x-6}\le 0$          

On solving we get   $x\in \left(-\infty ,-2\right)\cup \left[\frac{-4}{3},\infty \right)$

As x = 2 also lies in domain since g(2) $=\underset{x\to 2}{lim}g\left(x\right)$  

  $h=\frac{cos\theta +3}{2}$          

$k=\frac{sin\theta +2}{2}$            

-> $cos\theta =2h-3\&sin\theta =2k-2$

-> ${\left(h-3/2\right)}^2+{\left(k-1\right)}^2=\frac{1}{4}$

$\therefore$ circle of radius r = $\frac{1}{2}$  

${\overrightarrow{a}}_1=x\widehat{i}-\widehat{j}+\widehat{k}\&{\overrightarrow{a}}_2=\widehat{i}+y\widehat{j}+z\widehat{k}$           

given  ${\overrightarrow{a}}_1\&{\overrightarrow{a}}_2$ are collinear then ${\overrightarrow{a}}_1=\lambda {\overrightarrow{a}}_2$  

$\Rightarrow \left(x\widehat{i}-\widehat{j}+\widehat{k}\right)=\lambda \left(\widehat{i}+y\widehat{j}+z\widehat{k}\right)$         

Since $\widehat{i},\widehat{j}\&\widehat{k}$ are not collinear so

  $Sox\widehat{i}+y\widehat{j}+z\widehat{k}=\lambda \widehat{i}-\frac{1}{\lambda }\widehat{j}+\frac{1}{\lambda }\widehat{k}$         

Hence possible unit vector parallel to it be  $\frac{1}{\sqrt[]{3}}\left(\widehat{i}-\widehat{j}+\widehat{k}\right)$ for $\lambda$ =

Given f(x) = ${\displaystyle {\int }_e^x{e}^{t}f\left(t\right)dt+e^x..........\left(i\right)}$  

using Leibniz rule then

f'(x) = e^xf(x) + e^x

  _{$\Rightarrow \frac{dy}{dx}=e^{x}y+e^{x}wherey=f\left(x\right)then\frac{dy}{dx}=f\text{'}\left(x\right)$}                

P = -e^x, Q = e^x

Solution be y. (I.F.) =   ${\displaystyle \int Q\left(I.F.\right)dx+c}$

I. f. =   $e^{{\displaystyle \int -e^{x}dx}=e^{-e^x}}$

$\Rightarrow y.\left(e^{-e^x}\right)={\displaystyle \int e^x.e^{-e^x}dx+c}$          

  $y.e^{-e^x}=-{\displaystyle \int dt+c=-t+c=-e^{-e^x}+c..........\left(ii\right)}$          

Put x = 0 , in (i) f (0) = 1

$From\left(ii\right),\frac{1}{e}=-\frac{1}{e}+cgivenc=\frac{2}{e}$        

Hence f(x) = 2. $e^{\left(e^x-1\right)}-1$  

$\frac{dy}{dx}=\frac{1}{1+sin2x}$

$dy=\frac{se{c}^{2}xdx}{{\left(1+tanx\right)}^2}$

$\Rightarrow y=-\frac{1}{1+tanx}+c$

when   $x=\frac{\pi }{4},y=\frac{1}{2}$ gives c = 1

so $x+\frac{\pi }{4}=\frac{5\pi }{6}or\frac{13\pi }{6}\Rightarrow x=\frac{7\pi }{12}or\frac{23\pi }{12}$

sum of all solutions =

$\pi +\frac{7\pi }{12}+\frac{23\pi }{12}=\frac{42\pi }{12}$

Hence k = 42

Each element of ordered pair (i, j) is either present in A or in B.

              So, A + B = Sum of all elements of all ordered pairs {i, j} for $1\le i\le 10$ and $1\le j\le 10$  

              = 20 (1 + 2 + 3 + … + 10) = 1100