উচ্চতর গণিত ২য় পত্র প্রশ্ন

উত্তর ও ব্যাখ্যা

1. ক)

   AI-তৈরি নমুনা উত্তর

   ধরি, $x = \tan \theta \implies \theta = \tan^{-1} x$

   বামপক্ষ $= \sin^{-1} \frac{2x}{1 + x^2}$
   $= \sin^{-1} \left(\frac{2\tan \theta}{1 + \tan^2 \theta}\right)$
   $= \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1} x$

   ডানপক্ষ $= \cos^{-1} \frac{1 - x^2}{1 + x^2}$
   $= \cos^{-1} \left(\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}\right)$
   $= \cos^{-1}(\cos 2\theta) = 2\theta = 2\tan^{-1} x$

   $\therefore \sin^{-1} \frac{2x}{1 + x^2} = \cos^{-1} \frac{1 - x^2}{1 + x^2}$ (প্রমাণিত)।

2. খ)

   AI-তৈরি নমুনা উত্তর

   দেওয়া আছে, $f(x) = \cos x - \cos 7x$
   $\therefore f(\alpha) = \cos \alpha - \cos 7\alpha$

   শর্তানুসারে,
   $f(\alpha) = \sin 4\alpha$
   $\implies \cos \alpha - \cos 7\alpha = \sin 4\alpha$
   $\implies 2 \sin\left(\frac{\alpha + 7\alpha}{2}\right) \sin\left(\frac{7\alpha - \alpha}{2}\right) = \sin 4\alpha$
   $\implies 2 \sin 4\alpha \sin 3\alpha = \sin 4\alpha$
   $\implies 2 \sin 4\alpha \sin 3\alpha - \sin 4\alpha = 0$
   $\implies \sin 4\alpha (2 \sin 3\alpha - 1) = 0$

   হয়, $\sin 4\alpha = 0$
   $\implies 4\alpha = n\pi, \quad [n \in \mathbb{Z}]$
   $\implies \alpha = \frac{n\pi}{4}, \quad [n \in \mathbb{Z}]$

   অথবা,
   $2 \sin 3\alpha - 1 = 0$
   $\implies \sin 3\alpha = \frac{1}{2} = \sin \frac{\pi}{6}$
   $\implies 3\alpha = n\pi + (-1)^n \frac{\pi}{6}, \quad [n \in \mathbb{Z}]$
   $\implies \alpha = \frac{n\pi}{3} + (-1)^n \frac{\pi}{18}, \quad [n \in \mathbb{Z}]$

   $\therefore$ নির্ণেয় সমাধান: $\alpha = \frac{n\pi}{4}$ অথবা $\alpha = \frac{n\pi}{3} + (-1)^n \frac{\pi}{18}$, যেখানে $n \in \mathbb{Z}$।

3. গ)

   AI-তৈরি নমুনা উত্তর

   দেওয়া আছে, $g(x) = \sin x$
   $\therefore g\left(\frac{\pi}{2} - x\right) = \sin\left(\frac{\pi}{2} - x\right) = \cos x$

   প্রদত্ত সমীকরণ:
   $g\left(\pi g\left(\frac{\pi}{2} - x\right)\right) = g\left(\frac{\pi}{2} - \pi g(x)\right)$
   $\implies g(\pi \cos x) = g\left(\frac{\pi}{2} - \pi \sin x\right)$
   $\implies \sin(\pi \cos x) = \sin\left(\frac{\pi}{2} - \pi \sin x\right)$
   $\implies \sin(\pi \cos x) = \cos(\pi \sin x)$
   $\implies \cos(\pi \sin x) = \cos\left(\frac{\pi}{2} - \pi \cos x\right)$
   $\implies \pi \sin x = \pm \left(\frac{\pi}{2} - \pi \cos x\right)$

   $(+)$ চিহ্ন নিয়ে, $\pi \sin x = \frac{\pi}{2} - \pi \cos x \implies \cos x + \sin x = \frac{1}{2}$
   $(-)$ চিহ্ন নিয়ে, $\pi \sin x = -\frac{\pi}{2} + \pi \cos x \implies \cos x - \sin x = \frac{1}{2}$

   একত্রে,
   $\cos x \pm \sin x = \frac{1}{2}$

   উভয় পক্ষে $\frac{1}{\sqrt{2}}$ গুণ করে,
   $\cos x \cdot \frac{1}{\sqrt{2}} \pm \sin x \cdot \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$
   $\implies \cos x \cos\frac{\pi}{4} \pm \sin x \sin\frac{\pi}{4} = \frac{1}{2\sqrt{2}}$
   $\implies \cos\left(x \mp \frac{\pi}{4}\right) = \frac{1}{2\sqrt{2}}$
   $\implies x \mp \frac{\pi}{4} = \cos^{-1}\left(\frac{1}{2\sqrt{2}}\right)$
   $\implies x = \pm \frac{\pi}{4} + \cos^{-1}\left(\frac{1}{2\sqrt{2}}\right)$ (দেখানো হলো)।