সৃজনশীলঃ ‘f(x)=x3−2×2+3x−4 এবং g(x)=ax2+bx+c…’ নটরডেম কলেজ ২০২৪ – বহুপদী ও বহুপদী সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $g\left(x\right)=a{x}^2+bx+c$

$\therefore g\left(\frac{1}{x}\right)=a{\left(\frac{1}{x}\right)}^2+b\left(\frac{1}{x}\right)+c=a\cdot \frac{1}{x^2}+b\cdot \frac{1}{x}+c$

এবং $g\left(\frac{1}{x}\right)=0$

বা, $a\cdot \frac{1}{x^2}+b\cdot \frac{1}{x}+c=0$

বা, $a+bx+c{x}^2=0$

$\therefore c{x}^2+bx+c=0$

ধরি, ${\mathrm{cx}}^2+\mathrm{bx}+a=0$ এর একটি মূল $\alpha$

$\therefore c{\alpha }^2+b\alpha +a=0$ ..... (i)

আবার, $g\left(x\right)=0$

বা, $a{x}^2+bx+c=0$ এর একটি মূল $2\alpha$

$\therefore 4a{\alpha }^2+2b\alpha +c=0$ .... (ii)

(i) ও (ii) হতে বজ্ৰগুণনের সূত্রানুসারে,

$\frac{{\alpha }^2}{bc-2ab}=\frac{\alpha }{4a^2-c^2}=\frac{1}{2bc-4ab}$

$\frac{{\alpha }^2}{bc-2ab}=\frac{\alpha }{4a^2-c^2}$

বা, $\frac{\alpha }{b(c-2a)}=\frac{1}{4a^2-c^2}$

$\therefore \alpha =\frac{b(c-2a)}{4a^2-c^2}$

আবার, $\frac{\alpha }{4a^2-c^2}=\frac{1}{2bc-4ab}$

$\therefore \alpha =\frac{4a^2-c^2}{2b(c-2a)}$

সুতরাং $\frac{b(c-2a)}{4a^2-c^2}=\frac{4a^2-c^2}{2b(c-2a)}$

বা, ${(4a^2-c^2)}^2=2b^2(c-2a{)}^2$

বা, $(2a+c{)}^2(2a-c{)}^2-2b^2(2a-c{)}^2=0$

বা, $(2a-c{)}^2\left\{\right(2a+c{)}^2-2b^2\}=0$

হয়, $(2a-c{)}^2=0$ 

বা, $2a-c=0$

অথবা, $(2a+c{)}^2-2b^2=0$

$\therefore (2a+c{)}^2=2b^2$

অতএব, $2a-c=0$অথবা, $(2a+c{)}^2=2b^2$. (প্রমাণিত)

দেওয়া আছে, $f\left(x\right)=x^3-2x^2+3x-4$

এবং $f\left(x\right)=0$

বা, $x^3-2x^2+3x-4=0$ এর মূল, $\alpha ,\beta ,\gamma$

$\therefore \alpha +\beta +\gamma =-\frac{-2}{1}=2$

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{3}{1}=3$

$\alpha \beta \gamma =-\frac{-4}{1}=4$

প্রদত্ত রাশি$=\frac{1}{\Sigma {\alpha }^3}=\frac{1}{{\alpha }^3}+\frac{1}{{\beta }^3}+\frac{1}{{\gamma }^3}$

$={\left(\frac{1}{\alpha }\right)}^3+{\left(\frac{1}{\beta }\right)}^3+{\left(\frac{1}{\gamma }\right)}^3-\frac{3}{\alpha \beta \gamma }+\frac{3}{\alpha \beta \gamma }$$

$=(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}-\frac{1}{\alpha \beta }-\frac{1}{\beta \gamma }-\frac{1}{\gamma \alpha })+\frac{3}{\alpha \beta \gamma }$

$[\because a^3+b^3+c^3-3abc=(a+b+c\left)(a^2+b^2+c^2-ab-bc-ca)\right]$

$=\left(\frac{\beta \gamma +\gamma \alpha +\alpha \beta }{\alpha \beta \gamma }\right)\{{(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })}^2-2(\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha })-(\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma a})\}+\frac{3}{\alpha \beta \gamma \gamma }$

$[\because a^2+b^2+c^2=(a+b+c{)}^2-2(ab+bc+ca)]$

$=\left(\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right)\{{\left(\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right)}^2-3(\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha })\}+\frac{3}{\alpha \beta \gamma }$

$=\left(\frac{3}{4}\right)\{{\left(\frac{3}{4}\right)}^2-3\cdot \left(\frac{\alpha +\beta +\gamma }{\alpha \beta \gamma }\right)\}+\frac{3}{4}$

$=\left(\frac{3}{4}\right)\{\frac{9}{16}-3\cdot \left(\frac{2}{4}\right)\}+\frac{3}{4}$

$=\frac{3}{4}(\frac{9}{16}-\frac{3}{2})+\frac{3}{4}=\frac{3}{4}\left(\frac{9-24}{16}\right)+\frac{3}{4}$

$=\frac{3}{4}\times (-\frac{15}{16})+\frac{3}{4}=\frac{-45}{64}+\frac{3}{4}=\frac{-45+48}{64}=\frac{3}{64}$

নির্ণেয় মান : $\frac{3}{64}$