Maths

$\frac{3x^2-9x+17}{x^2+3x+10}=\frac{5x^2-7x+19}{3x^2+5x+12}$

$1+\frac{2x^2-12x+7}{x^2+3x+10}=1+\frac{2x^2-12x+7}{3x^2+5x+12}$

$\left(2x^2-12x+7\right)\left(\frac{1}{x^2+3x+10}-\frac{1}{3x^2+5x+12}\right)=0$

$\underset{x\to 10}{lim}{\left(\frac{{\left(x+2cosx\right)}^3+2{\left(x+2cosx\right)}^2+3sin\left(x+2cosx\right)}{{\left(x+2\right)}^3\pm 2{\left(x+2\right)}^2+3sin\left(x+2\right)}\right)}^x$

$e^{\frac{100}{16+3sin2}\left[\frac{12-3\left(4\right)+8*1-8+3cos2-3cos2}{1}\right]}$ , using L.H.L Rule, e0 = 1

$2x^2-12x+7=0or3x^2+5x+12=x^2+3x+10$

$x^2+x+1=0$

Sum of roots = 6, no solution.

${\displaystyle \sum _{}^{}{x}_0^1=\frac{3{\left(1-\left(\frac{1}{2}\right)\right)}^{20}}{1-\frac{1}{2}}=6\left(1-\frac{1}{2^{20}}\right)}$

$={\displaystyle \sum _{i=1}^{20}{\left(x_i\right)}^2+{\left(i\right)}^2-2x_{i}i}$

Now, ${\displaystyle \sum _{i=1}^{20}{\left(x_i\right)}^2=\frac{9{\left(1-\left(\frac{1}{4}\right)\right)}^{20}}{\left(1-\frac{1}{4}\right)}=12\left(1-\frac{1}{2^{40}}\right)}$

$\overline{x}=\frac{2858}{20}+\left(\frac{-12}{2^{40}}+\frac{22}{2^{20}}\right)*\frac{1}{20}$

$\left[\overline{x}\right]=142$

$e_E=\sqrt[]{1-\frac{b^2}{a^2}},e_H=\sqrt[]{2}$

$If\Rightarrow e_E=\frac{1}{e_H}\Rightarrow \frac{a^2-b^2}{a^2}=\frac{1}{2}$

$k^2=a^2*\frac{5}{2}+b^2=\frac{3}{2}$

$6b^2=\frac{3}{2}\Rightarrow b^2=\frac{1}{4}and{a}^2=\frac{1}{2}$

$\therefore 4\left(a^2+b^2\right)=3$

$f\left(x\right)=0\Rightarrow {\left(x-p\right)}^2-q=0$  

Roots are  $p+\sqrt[]{q},p-\sqrt[]{q}$

Now, $\left|f\left(a_i\right)\right|=500$  

$Let{a}_1,a_2,a_3...ara,a+d,a+2d,a+3d$               

=>  $\frac{9}{4}{d}^2-q=500$           …….(i)

and ${\left|f\left(a_1\right)\right|}^2={\left|f\left(a_2\right)\right|}^2$

From equation (i)

$\frac{9}{4}.\frac{4q}{5}-q=500$     

$\frac{4q}{5}=500$           

and  $2\sqrt[]{q}=2*\frac{50}{2}=50$

$A+B=\left[\begin{array}{cc}\hfill \beta +1\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill \alpha \hfill \end{array}\right]$

${\left(A+B\right)}^2=\left[\begin{array}{cc}\hfill {\left(\beta +1\right)}^2\hfill & \hfill 0\hfill \\ \hfill 3\left(\beta +1\right)+3\alpha \hfill & \hfill {\alpha }^2\hfill \end{array}\right]$

$\therefore \left[\begin{array}{cc}\hfill 1\hfill & \hfill -\alpha +1\hfill \\ \hfill 2\alpha +4\hfill & \hfill {\alpha }^2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill {\left(\beta +1\right)}^2\hfill & \hfill 0\hfill \\ \hfill 3\left(\alpha +\beta +1\right)\hfill & \hfill {\alpha }^2\hfill \end{array}\right]$

= 1 = 1

$B^2=\left[\begin{array}{cc}\hfill \beta \hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{cc}\hfill \beta \hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$

$\beta =0,\alpha =-1={\alpha }_2$

$\left|{\alpha }_1-{\alpha }_2\right|=\left|1-\left(-1\right)\right|=2$

Put $1+x^2=t^2\Rightarrow 2xdx=2tdt$  

$\therefore {\displaystyle \underset{1}{\overset{2}{\int }}\frac{15\left(t^2-1\right)tdt}{\sqrt[]{t^2+t^3}}dt}$ Put (1 + t) = u²

$30{\displaystyle \underset{\sqrt[]{2}}{\overset{\sqrt[]{3}}{\int }}\left(u^4-2u^2\right)du}$ dt = 2u du

$=-6\sqrt[]{3}+16\sqrt[]{2}=\alpha \sqrt[]{2}+\beta \sqrt[]{3}$

$\begin{array}{l}\alpha =16,\beta =-6\\ \therefore \alpha +\beta =10\end{array}$      

$f\text{'}\left(a\right)=f\text{'}\left(b\right)=f\text{'}\left(c\right)=2$

f'' (x) is zero for atleast $x_1\in \left(a,b\right)\&x_2\in \left(b,c\right)$ ${}_{}{}_{}$

$RS\equiv \left(\alpha ,-,1\beta \right)$

$DRofPQ\equiv \left(\frac{56}{17}+2,\frac{43}{17}+1,\frac{111}{17}-1\right)$

$\equiv \left(\frac{90}{17},\frac{60}{17},\frac{94}{17}\right)$

$90\alpha +94\beta =60$

$\begin{array}{l}\beta =-15,\alpha =-15\\ {\alpha }^2+{\beta }^2=450\end{array}$

Required number = Total – no character from {1, 2, 3, 4, 5}

$=\left(10^6-5^6\right)+\left(10^7-5^7\right)+\left(10^8-5^8\right)$

$=5^6\left(2^6*111-31\right)=5^6*\frac{7073}{\alpha }$

= 7073