If $1+\left(1-2^2\cdot 1\right)+\left(1-4^2\cdot 3\right)+\left(1-6^2\cdot 5\right)+\dots \dots .+\left(1-{20}^2\cdot 19\right)=\alpha -220\beta$ , then an ordered pair $(\alpha ,\beta )$ is equal to:

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

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

$a_1,a_2,a_3......$ are in A.P. (Let common difference is d₁)

$b_1,b_2,b_3....$  are in A.P. (Let common difference is d₂)

and a₁2, a₁₀ = 3, a₁b₁ = 1 = a₁₀b₁₀

$?a_1{b}_1=1$               

$\therefore b_1=\frac{1}{2}$

$a_{10}{b}_{10}=1$               

$\therefore b_{10}=\frac{1}{3}$              2026 Info edge India Ltd. All rights reserved.