গণিত প্রশ্ন

উত্তর ও ব্যাখ্যা

1. ক)

   $4P^2-1+2R-R^2 = 4P^2-(1-2R+R^2) = (2P)^2-(1-R)^2 = \mathbf{(2P+1-R)(2P-1+R)}$

2. খ)

   (i) নং থেকে $x^4-38x^2+1=0$; $x^2$ দিয়ে ভাগ করে পাই, $x^2+\frac{1}{x^2}=38$।

   এখন $\frac{x^8-1}{x^4}=x^4-\frac{1}{x^4}=\left(x^2+\frac{1}{x^2}\right)\left(x^2-\frac{1}{x^2}\right)$।

   $\left(x^2-\frac{1}{x^2}\right)^2=\left(x^2+\frac{1}{x^2}\right)^2-4=38^2-4=1440$, সুতরাং $x^2-\frac{1}{x^2}=\pm\sqrt{1440}=\pm12\sqrt{10}$।

   $\therefore \frac{x^8-1}{x^4}=38\times(\pm12\sqrt{10})=\mathbf{\pm456\sqrt{10}}$

   (সমীকরণের দুইটি ধনাত্মক বীজ $x$ ও $\frac{1}{x}$; $x>1$ হলে মান $+456\sqrt{10}$, $0<x<1$ হলে $-456\sqrt{10}$।)

3. গ)

   (ii) নং থেকে $a^2=17+12\sqrt{2}=9+8+2\cdot3\cdot2\sqrt{2}=(3+2\sqrt{2})^2$।

   $\therefore a=3+2\sqrt{2}$ [$a>0$], তাই $\frac{1}{a}=\frac{1}{3+2\sqrt{2}}=\frac{3-2\sqrt{2}}{9-8}=3-2\sqrt{2}$।

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

   $a^2+\frac{1}{a^2}=\left(a+\frac{1}{a}\right)^2-2=36-2=34$

   $a^3+\frac{1}{a^3}=\left(a+\frac{1}{a}\right)^3-3\left(a+\frac{1}{a}\right)=216-18=198$

   এখন, $\left(a^2+\frac{1}{a^2}\right)\left(a^3+\frac{1}{a^3}\right)=a^5+\frac{1}{a^5}+a+\frac{1}{a}$

   $\therefore a^5+\frac{1}{a^5}=34\times198-6=6732-6=\mathbf{6726}$ (প্রমাণিত)