গণিত প্রশ্ন

উত্তর ও ব্যাখ্যা

1. ক)

   $x^3 + 9y^3 + (x+y)^3 = x^3 + 9y^3 + x^3 + 3x^2y + 3xy^2 + y^3 = 2x^3 + 3x^2y + 3xy^2 + 10y^3$

   $x = -2y$ বসালে মান $0$ হয়, তাই $(x + 2y)$ একটি উৎপাদক।

   $2x^3 + 3x^2y + 3xy^2 + 10y^3 = 2x^2(x+2y) - xy(x+2y) + 5y^2(x+2y) = \mathbf{(x+2y)(2x^2 - xy + 5y^2)}$

2. খ)

   $R = \{(x,y) : x, y \in A,\ x - 2y = -1\}$, $A = \{-2,-1,0,1,2\}$। এখানে $x = 2y - 1$।

   • $y = -2 \Rightarrow x = -5 \notin A$ ✗
   • $y = -1 \Rightarrow x = -3 \notin A$ ✗
   • $y = 0 \Rightarrow x = -1$ ✓
   • $y = 1 \Rightarrow x = 1$ ✓
   • $y = 2 \Rightarrow x = 3 \notin A$ ✗

   $\therefore R = \{(-1, 0), (1, 1)\}$

   ডোমেন $= \{-1, 1\}$, রেঞ্জ $= \{0, 1\}$

3. গ)

   $B = 0$ হলে $\sqrt{p^2 - \sqrt{13}p + 1} = 0 \Rightarrow p^2 - \sqrt{13}p + 1 = 0$

   $p \neq 0$ বলে $p$ দ্বারা ভাগ করে পাই, $p + \frac{1}{p} = \sqrt{13}$

   $p^2 + \frac{1}{p^2} = 13 - 2 = 11$
   $p^4 + \frac{1}{p^4} = 11^2 - 2 = 119$
   $\left(p - \frac{1}{p}\right)^2 = 11 - 2 = 9 \Rightarrow p - \frac{1}{p} = \pm 3$

   $p^5 - \frac{1}{p^5} = \left(p - \frac{1}{p}\right)\left(p^4 + p^2 + 1 + \frac{1}{p^2} + \frac{1}{p^4}\right)$
   $= \left(p - \frac{1}{p}\right)(119 + 11 + 1) = 131\left(p - \frac{1}{p}\right)$

   $\therefore p^5 - \frac{1}{p^5} = 131 \times (\pm 3) = \mathbf{\pm 393}$

   ($p = \frac{\sqrt{13}+3}{2}$ হলে $+393$, $p = \frac{\sqrt{13}-3}{2}$ হলে $-393$)