Maths

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

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

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

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

$\therefore$ 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. $e^{\left(e^x-1\right)}-1$

  $A^2=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

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

Similarly we get A¹⁹ =   $=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^{19}\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\&A^{20}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^{20}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

=  $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$\Rightarrow 1+\alpha +\beta =1gives\alpha +\beta =0........\left(i\right)$

$2^{20}+\left(2^{19}-2\right)\alpha =4from\left(i\right)$

$\Rightarrow \alpha =\frac{4-2^{20}}{2^{19}-2}=\frac{4\left(1-2^{18}\right)}{-2\left(1-2^{18}\right)}=-2$

So, β = 2

Hence β - α = 4

Let the equation of normal is Y – y = - $\frac{1}{m}\left(X-x\right)$  

where m is slope of tangent to the given curve then

  $Y-y=-\frac{dx}{dy}\left(X-x\right)$         

It passes through (a, b) so b – y = $\frac{-dx}{dy}\left(a-x\right)$

=> (a – x) dx = (y – b) dy

On integration     $ax-\frac{x^2}{2}=\frac{y^2}{2}-by+c.........\left(i\right)$  

(ii) passes through (3, -3) &  then

3a – 3b – c = 9       .(ii)

& 4a - $2\sqrt[]{2}b$ - c = 12           .(iii)

also given  $a-2\sqrt[]{2}b=3............(iv)$

Solve (ii), (iii) & (iv) b = 0, a = 3

Hence a² + b² + ab = 9

Given $l_{m,n}={\displaystyle {\int }_0^1{x}^{m-1}{\left(1-x\right)}^{n-1}dx............\left(i\right)}$  

put 1 - x = $t\{\begin{array}{l}x=0,t=1\\ x=1,t=0\end{array}$

dx = -dt

From (i) $l_{m,n}={\displaystyle {\int }_1^0{\left(1-t\right)}^{m-1}.t^{n-1}\left(-dt\right)}$

$l_{m,n}={\displaystyle {\int }_0^1{t}^{n-1}{\left(1-t\right)}^{m-1}dt={\displaystyle {\int }_0^1{x}^{n-1}{\left(1-x\right)}^{m-1}dx..........\left(ii\right)}}$    

$Putx=\frac{1}{1+y}in\left(i\right)thendx=-\frac{1}{{\left(1+y\right)}^2}dy$                          

(i)  $l_{m,n}={\displaystyle {\int }_{\infty }^0\frac{1}{{\left(1+y\right)}^{m-1}}}{\left(1-\frac{1}{1+y}\right)}^{n-1}\left(\frac{-dy}{{\left(1+y\right)}^2}\right)={\displaystyle {\int }_0^{\infty }\frac{y^{n-1}}{{\left(1+y\right)}^{m+n}}dy}..........\left(iii\right)$

Similarly by (ii) $l_{m,n}={\displaystyle {\int }_0^{\infty }\frac{y^{m-1}}{{\left(y+1\right)}^{m+n}}dy..........\left(iv\right)}$

Adding (iii) & (iv) $2l_{m,n}={\displaystyle {\int }_0^{\infty }\frac{y^{n-1}+y^{m+1}}{{\left(y+1\right)}^{m+n}}}$

Putting  $\frac{1}{z}\{\begin{array}{l}y=1,z=1\\ y=\infty ,z=0\end{array}$ $\Rightarrow dy=\frac{-1}{z^2}dz$  

Hence $l_{m,n}={\displaystyle {\int }_0^1\frac{x^{m-1}+x^{n-1}}{{\left(1+x\right)}^{m+n}}}$ dx = α lm, n

=> α = 1

Given   ${\displaystyle \sum _{i=1}^{18}\left(x_i-\alpha \right)=36i.e}{\displaystyle \sum _{i=1}^{18}{x}_i-18\alpha =36..........\left(i\right)}$

&       ${\displaystyle \sum _{i=1}^{18}{\left(x_i-\beta \right)}^2=90i.e{\displaystyle \sum _{i=1}^{18}{x_i}^2}}-2\beta {\displaystyle \sum _{}^{}{x}_i+18{\beta }^2=90.............\left(ii\right)}$         

(i) & (ii)  ${\displaystyle \sum _{i=1}^{18}{x_i}^2}=90-18\beta +36\beta \left(\alpha +2\right).............\left(iii\right)$

Now variance ${\sigma }^2=\frac{\sum {x_i}^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2$ = 1 given

=> (α - β) (α - β + 4) = 0

Since   $\alpha \ne \beta so\left|\alpha -\beta \right|=4$

Given $f\left(x\right)=2x^5+5x^4+10x^3+10x^2+10x+10$ ${}^{}{}^{}{}^{}{}^{}$

$f\left(-1\right)=3>0\&f\left(-2\right)=-34<0$

So at least one root will lie in (2, 1)

now $f\text{'}\left(x\right)=10x^4+20x^3+30x^2+20x+10$ ${}^{}{}^{}{}^{}$

$=10\left[x^4+2x^3+3x^2+2x+1\right]$

$=10x^2{\left(x+\frac{1}{x}+1\right)}^2>0\forall x\in R$

So, f (x) be purely increasing function so exactly one root of f (x) that will lie in (-2, 1). Hence |a| = 2

$Given\left|z+5\right|\le 4..........\left(i\right)$

->Represent a circle ${\left(x+5\right)}^2+y^2\le 16$

$=z\left(1+i\right)+\overline{z}\left(1-i\right)\ge -10..........\left(ii\right)$

->Represent a line X – y $\ge -5$

So max |z + 1|² = AQ²

$={\left(-4-2\sqrt[]{2}\right)}^2+8$

$\begin{array}{l}=32+16\sqrt[]{2}\\ =\alpha +\beta \sqrt[]{2}\end{array}$  

Hence α + β) = 48

Given $P_n={\alpha }^n+{\beta }^n,P_{n-1}=11\&P_{n+1}=29$

    $P_n={\alpha }^{n-2}.{\alpha }^2+{\beta }^{n-2}.{\beta }^2.........\left(i\right)$                          

Now quadratic equation having roots α & β will be x² – (α + β)x + αβ = 0

i.e.         x² – x – 1 = 0,     put x = α and put x = β

So          α² = α + 1           & β² = β + 1

(i)     $P_n={\alpha }^{n-2}\left(\alpha +1\right)+{\beta }^{n-2}\left(\beta +1\right)$

$P_n=P_{n-1}+P_{n-2}$          

= > ${P_n}^2=234$