ID#6151 HSC Higher Math 2nd CQ (Barisal 2025)

ক-এর উত্তর:
প্রদত্ত সমীকরণ: $6x^{3} + 0 \cdot x^{2} - x + 13 = 0$
মূলত্রয় $\alpha, \beta, \gamma$ হলে,
$\sum \alpha = \alpha + \beta + \gamma = 0$
$\sum \alpha \beta = \alpha \beta + \beta \gamma + \gamma \alpha = -\frac{1}{6}$
$\alpha \beta \gamma = -\frac{13}{6}$

আমরা জানি, $\sum (\alpha - \beta)^{2} = (\alpha - \beta)^{2} + (\beta - \gamma)^{2} + (\gamma - \alpha)^{2}$
$= 2(\alpha^{2} + \beta^{2} + \gamma^{2}) - 2(\alpha \beta + \beta \gamma + \gamma \alpha)$
$= 2[(\alpha + \beta + \gamma)^{2} - 2(\alpha \beta + \beta \gamma + \gamma \alpha)] - 2(\alpha \beta + \beta \gamma + \gamma \alpha)$
$= 2(0)^{2} - 6(\alpha \beta + \beta \gamma + \gamma \alpha)$
$= -6 \times (-\frac{1}{6}) = 1$
$\therefore \sum (\alpha - \beta)^{2} = 1$

খ-এর উত্তর:
দেওয়া আছে, $ax^{2} + bx + c = 0$ এর মূলদ্বয় $m$ ও $n$।
$\therefore m + n = -\frac{b}{a}$ এবং $mn = \frac{c}{a}$

প্রদত্ত ২য় সমীকরণ: $c^{2}x^{2} - (b^{2} - 2ac)x + a^{2} = 0$
বা, $x^{2} - \frac{b^{2} - 2ac}{c^{2}}x + \frac{a^{2}}{c^{2}} = 0$
এখানে মূলদ্বয়ের যোগফল $S = \frac{b^{2} - 2ac}{c^{2}} = \frac{(b/a)^{2} - 2(c/a)}{(c/a)^{2}} = \frac{(m+n)^{2} - 2mn}{(mn)^{2}} = \frac{m^{2} + n^{2}}{m^{2}n^{2}} = \frac{1}{n^{2}} + \frac{1}{m^{2}}$
মূলদ্বয়ের গুণফল $P = \frac{a^{2}}{c^{2}} = \frac{1}{(c/a)^{2}} = \frac{1}{(mn)^{2}} = \frac{1}{m^{2}} \cdot \frac{1}{n^{2}}$

$\therefore$ $g(x) = 0$ সমীকরণের মূলদ্বয় হবে $\frac{1}{m^{2}}$ ও $\frac{1}{n^{2}}$।

গ-এর উত্তর:
যেহেতু $m, n$ সমীকরণ $ax^{2} + bx + c = 0$ এর মূল,
$\therefore am^{2} + bm + c = 0 \implies am^{2} + c = -bm$
এবং $an^{2} + bn + c = 0 \implies an^{2} + c = -bn$

বামপক্ষ = $(am^{2} + c)^{-2} + (an^{2} + c)^{-2}$
$= \frac{1}{(am^{2} + c)^{2}} + \frac{1}{(an^{2} + c)^{2}}$
$= \frac{1}{(-bm)^{2}} + \frac{1}{(-bn)^{2}}$
$= \frac{1}{b^{2}m^{2}} + \frac{1}{b^{2}n^{2}} = \frac{m^{2} + n^{2}}{b^{2}(mn)^{2}}$
$= \frac{(m + n)^{2} - 2mn}{b^{2}(mn)^{2}}$
$= \frac{(-b/a)^{2} - 2(c/a)}{b^{2}(c/a)^{2}} = \frac{(b^{2} - 2ac)/a^{2}}{b^{2}(c^{2}/a^{2})}$
$= \frac{b^{2} - 2ac}{b^{2}c^{2}}$ = ডানপক্ষ (প্রমাণিত)