সৃজনশীলঃ ‘যেখানেf(x)=2p2+2(p+q)x+3p−2P; যেখানে p,q∈R…’ ২০২৪ – বহুপদী ও বহুপদী সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $f\left(x\right)=2p{x}^2+2(p+a)x+3q-2p$

$\therefore f\left(x\right)=0$

বা, $2p{x}^2+2(p+\omega )x+3q-2p=0$

ধরি, সমীকরণের মূলদ্বয় $\alpha$ এবং $2\alpha$

$\therefore \alpha +2\alpha =-\frac{2(Q+p)}{2p}$

বা, $3\alpha =-\frac{(q+p)}{p}$

বা. $\alpha =-\frac{(\alpha +P)}{3P}$

আবার, $2\alpha =\frac{3q-2p}{2p}$

বা, ${\alpha }^2=\frac{3q-2P}{4P}$

বা, ${\{-\frac{(q+p)}{3p}\}}^2=\frac{3q-2p}{4p}$

বা, $4P(q+p{)}^2=9P^2(3q-2P)$

বা, $4(q^2+2pq+p^2)=9p(3q-2p)$

বা, $4q^2+8pq+4p^2=27pq-18p^2$

বা, $4q^2-19pq+22p^2=0$

বা, $4q^2-11pq-8pq+22p^2=0$

বা, $q(4q-11p)-2p(4q-11p)=0$

বা, $(q-2p)(4q-11p)=0$

$\therefore q=2p$ overor, $4q=11p$ [প্রমানিত]

$p=1$এবং$q=2$হলে,

$f\left(x\right)=2p{x}^2+2(p+q)x+3q-2p$

$=2\cdot 1\cdot x^2+2(1+2)x+3\cdot 2-2\cdot 1$

$=2x^2+6x+6-2$

$\therefore f\left(x\right)=2x^2+6x+4$

$\alpha \beta =\frac{4}{2}=2$

উক্ত সমীকরণটির মূলদ্বয় $\alpha ,\beta$ হলে, $\alpha +\beta =\frac{-6}{2}=-3$

$\alpha \beta =\frac{4}{2}=2$

$\begin{array}{cc}& \frac{1}{1-\alpha }+\frac{1}{1-\beta }\\ =& \frac{1-\beta +1-\alpha }{(1-\alpha )(1-\beta )}\\ =& \frac{2-(\alpha +\beta )}{(1+\alpha )(1-\beta )}\\ =& \frac{2-(\alpha +\beta )}{1-(\alpha +\beta )+\alpha \beta }\\ =& \frac{2-(-3)}{1-(-3)+2}\\ =& \frac{5}{6}\end{array}$

আবার, $\frac{1}{1-\alpha }\cdot \frac{1}{1-\beta }=\frac{1}{1-(\alpha +\beta )+\alpha \beta }$

$=\frac{1}{1-(-3)+2}=\frac{1}{6}-$

সুতরাং, ও$\frac{1}{1-\alpha }ও\frac{1}{1-\beta }$ মূলবিশিষ্ট সমীকরণ, 

$x^2-(\frac{1}{1-\alpha }+\frac{1}{1-\beta })x+\left(\frac{1}{1-\alpha }\right)\left(\frac{1}{1-\beta }\right)=0$

$x^2-\frac{5}{6}x+\frac{1}{6}=0$

বা, $6x^2-5x+1=0$ (Ans.)