Maths

$\frac{1}{\alpha \left(\alpha +1\right)\left(\alpha +2\right)..........\left(\alpha +20\right)}=\frac{A_0}{\alpha }+\frac{A_1}{\alpha +1}+\frac{A_2}{\alpha +2}+......+\frac{A_{20}}{\alpha +20}$

Solving by partial fraction, we get

$A_{13}=\frac{-1}{14!7!},A_{14}=\frac{-1}{14!6!}and{A}_{15}=\frac{-1}{14!5!}$

$\therefore \frac{A_{14}+A_{15}}{A_{13}}=\frac{3}{10}\Rightarrow {\left(\frac{A_{14}+A_{15}}{A_{13}}\right)}^2={\left(\frac{3}{10}\right)}^2\Rightarrow 100{\left(\frac{A_{14}+A_{15}}{A_{13}}\right)}^2=9$

$\frac{log\left(\left(2x+5\right)\left(x+1\right)\right)}{log\left(x+1\right)}+\frac{log\left(x+1\right)}{log\left(2x+5\right)}-4=0\Rightarrow \frac{log\left(2x+5\right)}{log\left(x+1\right)}+\frac{log\left(x+1\right)}{log\left(2x+5\right)}-3=0$

$\Rightarrow x\in \left(-1,0\right)\cup \left(0,\infty \right)$

$\frac{log\left(2x+5\right)}{log\left(x+1\right)}=1\Rightarrow x=-4$ (not possible) and $\frac{log\left(2x+5\right)}{log\left(x+1\right)}=2\Rightarrow x=2$

Hence only one solution is possible.

${\displaystyle \int dy}={\displaystyle \int \frac{cos\left(\frac{1}{2}co{s}^{-1}\left(e^{-x}\right)\right)}{e^x\sqrt[]{1-{\left(e^{-x}\right)}^2}}}dxputcos2\theta =e^{-x}\Rightarrow -2sin2\theta d\theta =-e^{-x}dx$

$\therefore {\displaystyle \int dy}={\displaystyle \int \frac{2cos\theta sin2\theta d\theta }{\sqrt[]{1-co{s}^{2}2\theta }}\Rightarrow y=2sin\theta +c\Rightarrow y=-1,\theta =0,c=0}$

$\therefore y=2\sqrt[]{\frac{1-e^{-x}}{2}}-1at\left(\alpha ,0\right),e^{\alpha }=2$

$?\left|{\overrightarrow{v}}_1\right|=\left|{\overrightarrow{v}}_2\right|\Rightarrow p^2-p-2=0\Rightarrow p-1,2$ we take only p = 2 (p > 0)

$\therefore cos\theta =\frac{\overrightarrow{v_2}.\overrightarrow{v_2}}{\left|\overrightarrow{v_1}\right|\left|\overrightarrow{v_2}\right|}=\frac{4\sqrt[]{3}+3}{13}\Rightarrow tan\theta =\frac{6\sqrt[]{3}-2}{4\sqrt[]{3}+3}=\frac{\alpha \sqrt[]{3}-2}{4\sqrt[]{3}+3}\therefore \alpha =6$

$A=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \end{array}\right]\Rightarrow \left|A\right|=4\Rightarrow \left|3adj\left(2A^{-1}\right)\right|=\left|3.2^{2}adj\left(A^{-1}\right)\right|$

$\Rightarrow 12^3\left|adj\left(A^{-1}\right)\right|=12^3{\left|A^{-1}\right|}^2=\frac{12^3}{{\left|A\right|}^2}=\frac{12*12*12}{16}=108$

Let equation of normal PQ is

$y=-tx+3t+\frac{3}{2}{t}^3$ and it is passing through

$P\left(3,\frac{3}{2}\right)\therefore t=1$

$\therefore Q\left(\frac{3}{2},3\right)\Rightarrow a=\frac{3}{2}andb=3$     

2 (a + b) = 9

${\left(\sqrt[]{50}\right)}^2={\left(\sqrt[]{45}\right)}^2+{\left(\sqrt[]{5}\right)}^2$

$\therefore \angle B=90°$ circum centre = $O\left(\frac{1}{2},\frac{11}{2}\right)$

Mid point of BC = $D\left(2,\frac{17}{2}\right)$

Equation of OD is y = 2x + $\frac{9}{2}.$ This line passes through

$\left(0,\frac{\alpha }{2}\right)\therefore \alpha =9$

$\underset{x\to 0}{lim}\frac{\alpha x\left(1+x+\frac{x^2}{2}+....\right)-\beta \left(x-\frac{x^2}{2}+\frac{x^3}{3}+....\right)+\gamma x^2\left(1-x\right)}{x^3}=10$

For limit to exist a - b = 0 . (i), $\alpha +\frac{\beta }{2}+\gamma =0.........(ii)$

and $\frac{\alpha }{2}-\frac{\beta }{2}-\gamma =10$ . (iii)

Solving (i), (ii) and (iii) we get α = 6, β = 6 and $\lambda =-9\Rightarrow \alpha +\beta +\lambda =3$

$a_{n+1}=\frac{a_{n+2}-a_n}{2}letp={\displaystyle \sum _{n=1}^{\infty }\frac{a_n}{8^n}}$

$\Rightarrow 64\frac{a_{n+2}}{8^{n+2}}=\frac{16a_{n+1}}{8^{n+1}}+\frac{a_n}{8^n}$

$\Rightarrow 64{\displaystyle \sum _{n=1}^{\infty }\frac{a_{n+2}}{8^{n+2}}=}{\displaystyle \sum _{n=1}^{\infty }\frac{16a_{n+1}}{8^{n+1}}+}{\displaystyle \sum _{n=1}^{\infty }\frac{a_n}{8^n}}\Rightarrow 64\left(p-\frac{a_1}{8}-\frac{a_2}{8^2}\right)=16\left(p-\frac{a_1}{8}\right)+p$

$\Rightarrow 64\left(p-\frac{1}{8}-\frac{1}{8^2}\right)=16\left(p-\frac{1}{8}\right)+p\Rightarrow 64p-8-1=16p-2+p$