Maths

IF =e^x

$y{e}^x={\displaystyle \int \frac{e^x}{1+e^{2x}}dx}$

${\displaystyle \int \frac{dt}{1+t^2}}$

= tan^-1 (t) + c

$\underset{x\to \infty }{lim}{e}^{x}y=\underset{x\to \infty }{lim}ta{n}^{-1}{e}^x+\frac{\pi }{4}=\frac{3\pi }{4}$

 ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{dx}{3+2sinx+cosx}Puttan\frac{x}{2}=t\Rightarrow \frac{1}{2}se{c}^2\frac{x}{2}dx=dt}$

$={\displaystyle \underset{0}{\overset{1}{\int }}\frac{2dt}{2t^2+4t+4}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{{\left(t+1\right)}^2+1}={\left[ta{n}^{-1}\left(t+1\right)\right]}_0^1}=ta{n}^{-1}2-ta{n}^{-1}1}$

 $\underset{x\to 0}{lim}\frac{\alpha e^x+\beta e^{-x}+\gamma sinx}{xsi{n}^{2}x}$

$=\underset{x\to 0}{lim}\frac{\alpha e^x+\beta e^{-x}+\gamma sinx}{x^3}$

$\Rightarrow \alpha +\beta =0,\alpha -\beta +\gamma =0,\frac{\alpha +\beta }{2}=0,\frac{\alpha }{6}-\frac{\beta }{6}-\frac{\gamma }{6}=\frac{2}{3}$

$\Rightarrow \beta =-\alpha \gamma =-2\alpha \alpha -\beta -\gamma =4\Rightarrow \alpha +\alpha +2\alpha =4\Rightarrow \gamma =1$

$\alpha =1,\beta =-1,\gamma =-2$

 $\frac{1}{\left(20-a\right)\left(40-a\right)}+\frac{1}{\left(40-a\right)\left(60-a\right)}+\frac{1}{\left(60-a\right)\left(80-a\right)}+.....+\frac{1}{\left(180-a\right)\left(200-a\right)}=\frac{1}{256}$

LHS = $\frac{1}{20}\left(\frac{1}{20-a}-\frac{1}{40-a}\right)+\frac{1}{20}\left(\frac{1}{40-a}-\frac{1}{60-a}\right)+...+\frac{1}{20}\left(\frac{1}{180-a}-\frac{1}{200-a}\right)$

$=\frac{1}{20}\left(\frac{1}{20-a}-\frac{1}{200-a}\right)=\frac{1}{20}.\frac{180}{\left(20-a\right)\left(200-a\right)}$

$\Rightarrow a^2-220a+4000-2304=0\Rightarrow a^{2}220a+1696=0$

$a=\frac{220\pm 204}{8}=212^2$

 $A_{3*3},B_{3*3}\&AB=0$

$\ne 0\ne 0\Rightarrow \left|A\right|=0\&\left|B\right|=0$ only .

 $z^5+{\left(\overline{z}\right)}^5$

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

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

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,  $a-b=8-3=5$