উচ্চতর গণিত ২য় পত্র প্রশ্ন

উত্তর ও ব্যাখ্যা

1. ক)

   AI-তৈরি নমুনা উত্তর

   ধরি, $\tan^{-1}x = \theta \implies x = \tan\theta$
   আমরা জানি,
   $\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} = \frac{2x}{1 - x^2}$
   $\implies 2\theta = \tan^{-1}\left(\frac{2x}{1 - x^2}\right)$
   $\implies 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1 - x^2}\right)$ (দেখানো হলো)।

2. খ)

   AI-তৈরি নমুনা উত্তর

   দেওয়া আছে,
   $A = \sec^{-1}\sqrt{5} = \tan^{-1}\sqrt{(\sqrt{5})^2 - 1} = \tan^{-1}\sqrt{4} = \tan^{-1}2$
   $B = \frac{1}{2}\sin^{-1}\left(\frac{p}{q}\right) = \frac{1}{2}\sin^{-1}\left(\frac{3}{5}\right)$
   আমরা জানি, $\sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x$ অথবা ধরি $\frac{1}{2}\sin^{-1}\left(\frac{3}{5}\right) = \theta \implies \sin 2\theta = \frac{3}{5}$
   $\implies \frac{2\tan\theta}{1+\tan^2\theta} = \frac{3}{5} \implies 3\tan^2\theta - 10\tan\theta + 3 = 0$
   $\implies (3\tan\theta - 1)(\tan\theta - 3) = 0$
   যেহেতু $\theta = \frac{1}{2}\sin^{-1}(3/5) < \frac{\pi}{4}$, তাই $\tan\theta = \frac{1}{3} < 1$
   $\therefore B = \tan^{-1}\left(\frac{1}{3}\right)$
   আবার, $C = \sin^{-1}r = \sin^{-1}\left(\frac{1}{\sqrt{10}}\right) = \tan^{-1}\left(\frac{\frac{1}{\sqrt{10}}}{\sqrt{1 - \frac{1}{10}}}\right) = \tan^{-1}\left(\frac{1/\sqrt{10}}{3/\sqrt{10}}\right) = \tan^{-1}\left(\frac{1}{3}\right)$

   এখন,
   $A - B + C = \tan^{-1}2 - \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{3}\right)$
   $= \tan^{-1}2 = \cot^{-1}\left(\frac{1}{2}\right)$
   $\therefore A - B + C = \cot^{-1}\left(\frac{1}{2}\right)$ (প্রমাণিত)।

3. গ)

   AI-তৈরি নমুনা উত্তর

   দেওয়া আছে, $\alpha = 1, \beta = 3$
   $\therefore f(x) = \sin x$ এবং $g(x) = \sin 3x$
   প্রদত্ত সমীকরণ:
   $2f(x)g(x) = 1$
   $\implies 2\sin 3x \sin x = 1$
   $\implies \cos(3x - x) - \cos(3x + x) = 1$
   $\implies \cos 2x - \cos 4x = 1$
   $\implies \cos 2x - (2\cos^2 2x - 1) = 1$
   $\implies \cos 2x - 2\cos^2 2x + 1 = 1$
   $\implies 2\cos^2 2x - \cos 2x = 0$
   $\implies \cos 2x (2\cos 2x - 1) = 0$

   হয়, $\cos 2x = 0$
   $\implies 2x = (2n + 1)\frac{\pi}{2} \implies x = (2n + 1)\frac{\pi}{4}$, যেখানে $n \in \mathbb{Z}$
   $- \pi \le x \le \pi$ ব্যবধিতে:
   $n = -2 \implies x = -\frac{3\pi}{4}$
   $n = -1 \implies x = -\frac{\pi}{4}$
   $n = 0 \implies x = \frac{\pi}{4}$
   $n = 1 \implies x = \frac{3\pi}{4}$

   অথবা, $2\cos 2x - 1 = 0 \implies \cos 2x = \frac{1}{2} = \cos\left(\frac{\pi}{3}\right)$
   $\implies 2x = 2n\pi \pm \frac{\pi}{3} \implies x = n\pi \pm \frac{\pi}{6}$, যেখানে $n \in \mathbb{Z}$
   $- \pi \le x \le \pi$ ব্যবধিতে:
   $n = 0 \implies x = \pm \frac{\pi}{6}$
   $n = 1 \implies x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ (যেহেতু $\pi + \frac{\pi}{6} > \pi$)
   $n = -1 \implies x = -\pi + \frac{\pi}{6} = -\frac{5\pi}{6}$ (যেহেতু $-\pi - \frac{\pi}{6} < -\pi$)

   অতএব, নির্ণেয় সমাধানসমূহ:
   $x = \pm \frac{\pi}{6}, \pm \frac{\pi}{4}, \pm \frac{3\pi}{4}, \pm \frac{5\pi}{6}$।