Maths

 $z^5+{\left(\overline{z}\right)}^5$

$={\left(2+3i\right)}^5+{\left(2-3i\right)}^5$

$=2\left(32-720+810\right)=244$

R = { (a, b): b = pq, where p, q $\ge 3$ are prime}

$\Rightarrow 60*11=660$

p, q $\in$  {3, 5, 7, 11, 13, 17, 19, 23, 29, 3, 37, 41, 43, 47, 53, 59} total 16

p, q $\in$  {3, 5, 7, 11, 13, 17, 19}

{3, 5, 7, 11, 13, 17, 19}

7 + 3 + 1 = 11

We have,  $1-$  (probability of all shots result in failure)  $>\frac{1}{4}$

$\begin{array}{rr}& \Rightarrow 1-{\left(\frac{9}{10}\right)}^{\mathrm{n}}>\frac{1}{4}\\ & \Rightarrow \frac{3}{4}>{\left(\frac{9}{10}\right)}^{\mathrm{n}}\Rightarrow \mathrm{n}\ge 3\end{array}$

 Since,  ${\mathrm{l}\mathrm{i}\mathrm{m}}_{x\to 0}?\frac{f\left(x\right)}{x}$ exist $\Rightarrow f\left(0\right)=0$

Now,  $f^{\mathrm{\text{'}}}\left(x\right)={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f(x+h)-f\left(x\right)}{h}={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f\left(h\right)+x{h}^2+x^{2}h}{h}($ take $y=h)$

$={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f\left(h\right)}{h}+{\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to .0}?\left(xh\right)+x^2$

$\Rightarrow {\mathrm{f}}^{\mathrm{\text{'}}}\left(\mathrm{x}\right)=1+0+{\mathrm{x}}^2$

$\Rightarrow f^{\mathrm{\text{'}}}\left(3\right)=10$

$D=\left|\begin{array}{ccc}1& -2& 3\\ 2& 1& 1\\ 1& -7& a\end{array}\right|=0\Rightarrow a=8$

also,  $D_1=\left|\begin{array}{ccc}9& -2& 3\\ b& 1& 1\\ 24& -7& 8\end{array}\right|=0\Rightarrow b=3$

hence,  $a-b=8-3=5$

$\begin{array}{rr}& D_1=\left|\begin{array}{ccc}-7& 4& -1\\ 8& 1& 5\\ 15& b& 6\end{array}\right|=0\Rightarrow b=-3\\ & D=\left|\begin{array}{ccc}1& 4& -1\\ 3& 1& 5\\ a& b& 6\end{array}\right|=0\Rightarrow 21a-8b-66=0\\ & P:2x-3y+6z=15\end{array}$

so required distance $=\frac{21}{7}=3$

Given ${\left(2x^2+3x+4\right)}^{10}={\sum }_{r=0}^{20}?a_r{x}^r$

replace $x$ by $\frac{2}{x}$ in above identity :-

$\begin{array}{rr}& \frac{2^{10}{\left(2x^2+3x+4\right)}^{10}}{x^{20}}=\sum _{r=0}^{20}??\frac{a_r{2}^r}{x^r}\\ & \Rightarrow 2^{10}\sum _{r=0}^{20}??a_r{x}^r=\sum _{r=0}^{20}??a_r{2}^r{x}^{(20-r)}\left(\text{from}\right(i\left)\right)\end{array}$

now, comparing coefficient of $x^7$ from both sides

(take $r=7$ in L.H.S. and $r=13$ in R.H.S.)

$2^{10}{a}_7=a_{13}{2}^{13}\Rightarrow \frac{a_7}{a_{13}}=2^3=8$

$x\frac{dy}{dx}-y=x^2(x\mathrm{c}\mathrm{o}\mathrm{s}?x+\mathrm{s}\mathrm{i}\mathrm{n}?x),x>0$

$\frac{dy}{dx}-\frac{y}{x}=x(x\mathrm{c}\mathrm{o}\mathrm{s}?x+\mathrm{s}\mathrm{i}\mathrm{n}?x)\Rightarrow \frac{dy}{dx}+Py=Q$

So, I.F. $=e^{\int -\frac{1}{x}dx}\frac{1}{\left|x\right|}=\frac{1}{x}(x>0)$

Thus, $\frac{y}{x}=\int \frac{1}{x}\left(x\right(x\mathrm{c}\mathrm{o}\mathrm{s}?x+\mathrm{s}\mathrm{i}\mathrm{n}?x\left)\right)dx$

$\Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x}\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{x}+\mathrm{C}$

$?\mathrm{y}\left(\pi \right)=\pi \Rightarrow \mathrm{C}=1$

So, $y=x^2\mathrm{s}\mathrm{i}\mathrm{n}?x+x\Rightarrow (y{)}_{\pi /2}=\frac{{\pi }^2}{4}+\frac{\pi }{2}$

Also, $\frac{dy}{dx}=x^2\mathrm{c}\mathrm{o}\mathrm{s}?x+2x\mathrm{s}\mathrm{i}\mathrm{n}?x+1$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-x^2\mathrm{s}\mathrm{i}\mathrm{n}?x+4x\mathrm{c}\mathrm{o}\mathrm{s}?x+2\mathrm{s}\mathrm{i}\mathrm{n}?x$

 $\mathrm{m}\mathrm{a}\mathrm{x}\left\{n\right(A),n(B\left)\right\}\le n(A\cup B)\le n\left(U\right)$

$\Rightarrow 76\le 76+63-x\le 100$

$\Rightarrow -63\le -x\le -39$

$\Rightarrow 63\ge x\ge 39$

Gain in surface energy, $\Delta U=T\Delta A$  

from volume centenary, $\frac{4}{3}\pi R^3=64*\frac{4}{3}\pi r^3$  

$\Rightarrow r=\frac{R}{4}$                                            

Initial surface area, Ai = 4pR²

final surface area, $A_f=64*4\pi r^2$  

$\therefore \Delta U=12\pi R^2.T=0.075*12*3.14*10^{-4}=2.8*10^{-4}J$