গণিত প্রশ্ন

উত্তর ও ব্যাখ্যা

1. ক)

   $\log_7 \sqrt{7} = \log_7 7^{\frac{1}{2}} = \dfrac{1}{2}\log_7 7 = \mathbf{\dfrac{1}{2}}$

2. খ)

   $f(x) = \dfrac{x^3 - 3x^2 + 1}{x(1-x)}$

   বামপক্ষ: $f\left(\dfrac{1}{x}\right) = \dfrac{\frac{1}{x^3} - \frac{3}{x^2} + 1}{\frac{1}{x}\left(1 - \frac{1}{x}\right)} = \dfrac{\frac{1 - 3x + x^3}{x^3}}{\frac{x-1}{x^2}} = \dfrac{x^3 - 3x + 1}{x(x-1)}$

   ডানপক্ষ: $f(1-x) = \dfrac{(1-x)^3 - 3(1-x)^2 + 1}{(1-x)\{1-(1-x)\}}$

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

   তাহলে $f(1-x) = \dfrac{-(x^3 - 3x + 1)}{x(1-x)} = \dfrac{x^3 - 3x + 1}{x(x-1)}$

   সুতরাং $f\left(\dfrac{1}{x}\right) = f(1-x)$। (দেখানো হলো)

3. গ)

   দেওয়া আছে, $a = \sqrt{5} + \sqrt{3}$

   $\dfrac{1}{a} = \dfrac{1}{\sqrt{5} + \sqrt{3}} = \dfrac{\sqrt{5} - \sqrt{3}}{5 - 3} = \dfrac{\sqrt{5} - \sqrt{3}}{2}$

   তাহলে $\dfrac{2}{a} = \sqrt{5} - \sqrt{3}$

   এখন, $a + \dfrac{2}{a} = (\sqrt{5} + \sqrt{3}) + (\sqrt{5} - \sqrt{3}) = 2\sqrt{5}$

   $a^3 + \dfrac{8}{a^3} = a^3 + \left(\dfrac{2}{a}\right)^3 = \left(a + \dfrac{2}{a}\right)^3 - 3 \cdot a \cdot \dfrac{2}{a}\left(a + \dfrac{2}{a}\right)$

   $= (2\sqrt{5})^3 - 3 \cdot 2 \cdot 2\sqrt{5}$
   $= 40\sqrt{5} - 12\sqrt{5} = 28\sqrt{5}$

   সুতরাং $a^3 + \dfrac{8}{a^3} = 28\sqrt{5}$। (প্রমাণিত)