উচ্চতর গণিত ২য় পত্র প্রশ্ন

উত্তর ও ব্যাখ্যা

1. ক)

   AI-তৈরি নমুনা উত্তর

   ধরি, $\cos^{-1} x = \theta$ $\implies x = \cos \theta$
   আমরা জানি, $1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}$
   $\implies 1 + x = 2 \cos^2 \frac{\theta}{2}$
   $\implies \cos^2 \frac{\theta}{2} = \frac{1 + x}{2}$
   $\implies \cos \frac{\theta}{2} = \sqrt{\frac{1 + x}{2}}$
   $\implies \frac{\theta}{2} = \cos^{-1} \sqrt{\frac{1 + x}{2}}$
   $\implies \theta = 2 \cos^{-1} \sqrt{\frac{1 + x}{2}}$
   $\therefore \cos^{-1} x = 2 \cos^{-1} \sqrt{\frac{1 + x}{2}}$ (প্রমাণিত)।

2. খ)

   AI-তৈরি নমুনা উত্তর

   দেওয়া আছে,
   $A = \sin^{-1} \frac{2}{3}$
   $B = \cos^{-1} \frac{3}{4}$
   $C = \tan^{-1} \frac{1}{\sqrt{5}}$

   এখন, $\frac{1}{2} B = \frac{1}{2} \cos^{-1} \frac{3}{4}$
   ধরি, $\frac{1}{2} \cos^{-1} \frac{3}{4} = \theta \implies \cos 2\theta = \frac{3}{4}$
   আমরা জানি, $\tan^2 \theta = \frac{1 - \cos 2\theta}{1 + \cos 2\theta} = \frac{1 - \frac{3}{4}}{1 + \frac{3}{4}} = \frac{1/4}{7/4} = \frac{1}{7}$
   $\implies \tan \theta = \frac{1}{\sqrt{7}}$
   $\implies \theta = \tan^{-1} \frac{1}{\sqrt{7}}$
   $\therefore \frac{1}{2} B = \tan^{-1} \frac{1}{\sqrt{7}}$

   আবার, $A = \sin^{-1} \frac{2}{3} = \tan^{-1} \left( \frac{2}{\sqrt{3^2 - 2^2}} \right) = \tan^{-1} \frac{2}{\sqrt{5}}$

   বামপক্ষ $= A - \frac{1}{2} B + C = \left( A + C \right) - \frac{1}{2} B$
   $= \tan^{-1} \frac{2}{\sqrt{5}} + \tan^{-1} \frac{1}{\sqrt{5}} - \tan^{-1} \frac{1}{\sqrt{7}}$
   $= \tan^{-1} \left( \frac{\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}}}{1 - \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}}} \right) - \tan^{-1} \frac{1}{\sqrt{7}}$
   $= \tan^{-1} \left( \frac{\frac{3}{\sqrt{5}}}{1 - \frac{2}{5}} \right) - \tan^{-1} \frac{1}{\sqrt{7}}$
   $= \tan^{-1} \left( \frac{3/\sqrt{5}}{3/5} \right) - \tan^{-1} \frac{1}{\sqrt{7}}$
   $= \tan^{-1} (\sqrt{5}) - \tan^{-1} \frac{1}{\sqrt{7}}$
   $= \tan^{-1} \left( \frac{\sqrt{5} - \frac{1}{\sqrt{7}}}{1 + \sqrt{5} \cdot \frac{1}{\sqrt{7}}} \right)$
   $= \tan^{-1} \left( \frac{\frac{\sqrt{35} - 1}{\sqrt{7}}}{\frac{\sqrt{7} + \sqrt{5}}{\sqrt{7}}} \right)$
   $= \tan^{-1} \left( \frac{\sqrt{35} - 1}{\sqrt{7} + \sqrt{5}} \right) =$ ডানপক্ষ (প্রমাণিত)।

3. গ)

   AI-তৈরি নমুনা উত্তর

   দেওয়া আছে, $f(x) = \sin x$
   প্রদত্ত সমীকরণ: $f(x) + f(2x) + f(3x) = 0$
   $\implies \sin x + \sin 2x + \sin 3x = 0$
   $\implies (\sin 3x + \sin x) + \sin 2x = 0$
   $\implies 2 \sin \left(\frac{3x + x}{2}\right) \cos \left(\frac{3x - x}{2}\right) + \sin 2x = 0$
   $\implies 2 \sin 2x \cos x + \sin 2x = 0$
   $\implies \sin 2x (2 \cos x + 1) = 0$

   হয়, $\sin 2x = 0$
   $\implies 2x = n\pi$ (যেখানে $n \in \mathbb{Z}$)
   $\implies x = \frac{n\pi}{2}$

   যেহেতু ব্যবধি $0 \leq x \leq \pi$:
   $n = 0 \implies x = 0$
   $n = 1 \implies x = \frac{\pi}{2}$
   $n = 2 \implies x = \pi$

   অথবা, $2 \cos x + 1 = 0$
   $\implies \cos x = -\frac{1}{2} = \cos \left(\pi - \frac{\pi}{3}\right) = \cos \frac{2\pi}{3}$
   $\implies x = 2n\pi \pm \frac{2\pi}{3}$ (যেখানে $n \in \mathbb{Z}$)

   উক্ত ব্যবধিতে $0 \leq x \leq \pi$:
   $n = 0 \implies x = \frac{2\pi}{3}$

   অতএব, নির্ণেয় সমাধান: $x = 0, \frac{\pi}{2}, \frac{2\pi}{3}, \pi$।