উচ্চতর গণিত ২য় পত্র প্রশ্ন

উত্তর ও ব্যাখ্যা

1. ক)

   AI-তৈরি নমুনা উত্তর

   প্রদত্ত রাশি = \sec^{-1} \frac{\sqrt{34}}{5} + \sin^{-1} \frac{1}{\sqrt{17}}

   আমরা জানি,
   \sec^{-1} \frac{\sqrt{34}}{5} = \tan^{-1} \sqrt{\left(\frac{\sqrt{34}}{5}\right)^2 - 1} = \tan^{-1} \sqrt{\frac{34}{25} - 1} = \tan^{-1} \sqrt{\frac{9}{25}} = \tan^{-1} \frac{3}{5}

   এবং
   \sin^{-1} \frac{1}{\sqrt{17}} = \tan^{-1} \left(\frac{\frac{1}{\sqrt{17}}}{\sqrt{1 - \frac{1}{17}}}\right) = \tan^{-1} \left(\frac{1}{\sqrt{16}}\right) = \tan^{-1} \frac{1}{4}

   \therefore \sec^{-1} \frac{\sqrt{34}}{5} + \sin^{-1} \frac{1}{\sqrt{17}}
   = \tan^{-1} \frac{3}{5} + \tan^{-1} \frac{1}{4}
   = \tan^{-1} \left( \frac{\frac{3}{5} + \frac{1}{4}}{1 - \frac{3}{5} \cdot \frac{1}{4}} \right)
   = \tan^{-1} \left( \frac{\frac{12 + 5}{20}}{\frac{20 - 3}{20}} \right)
   = \tan^{-1} \left( \frac{17}{17} \right)
   = \tan^{-1}(1) = \frac{\pi}{4} (Ans.)

2. খ)

   AI-তৈরি নমুনা উত্তর

   দৃশ্যকল্প-১ এর সমকোণী ত্রিভুজ থেকে পাই,
   অতিভুজ = 13, লম্ব = 12
   \therefore \text{ভূমি} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5
   \therefore \cos \theta = \frac{5}{13} \implies \theta = \cos^{-1}\left(\frac{5}{13}\right)

   আমরা জানি, \cos^{-1} x = 2\tan^{-1}\sqrt{\frac{1-x}{1+x}}
   \therefore \frac{1}{2} \theta = \frac{1}{2} \cos^{-1}\left(\frac{5}{13}\right) = \tan^{-1}\sqrt{\frac{1 - \frac{5}{13}}{1 + \frac{5}{13}}} = \tan^{-1}\sqrt{\frac{8}{18}} = \tan^{-1}\sqrt{\frac{4}{9}} = \tan^{-1}\left(\frac{2}{3}\right)

   আবার, \operatorname{cosec}^{-1}\left(\frac{5}{3}\right) = \tan^{-1}\left(\frac{3}{\sqrt{5^2 - 3^2}}\right) = \tan^{-1}\left(\frac{3}{4}\right)

   \therefore \text{L.H.S.} = \frac{1}{2}\theta + \operatorname{cosec}^{-1}\left(\frac{5}{3}\right)
   = \tan^{-1}\left(\frac{2}{3}\right) + \tan^{-1}\left(\frac{3}{4}\right)
   = \tan^{-1}\left(\frac{\frac{2}{3} + \frac{3}{4}}{1 - \frac{2}{3} \cdot \frac{3}{4}}\right)
   = \tan^{-1}\left(\frac{\frac{17}{12}}{\frac{6}{12}}\right) = \tan^{-1}\left(\frac{17}{6}\right)

   এখন, \text{R.H.S.} = \cot^{-1} 2 + \cot^{-1}\left(\frac{29}{28}\right)
   = \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{28}{29}\right)
   = \tan^{-1}\left(\frac{\frac{1}{2} + \frac{28}{29}}{1 - \frac{1}{2} \cdot \frac{28}{29}}\right)
   = \tan^{-1}\left(\frac{\frac{29 + 56}{58}}{\frac{58 - 28}{58}}\right)
   = \tan^{-1}\left(\frac{85}{30}\right) = \tan^{-1}\left(\frac{17}{6}\right)

   \therefore \text{L.H.S.} = \text{R.H.S.} (প্রমাণিত)

3. গ)

   AI-তৈরি নমুনা উত্তর

   দেওয়া আছে, f(x) = \cos x
   \therefore \sqrt{3} f(x) + f\left(\frac{\pi}{2} + x\right) = 1
   \implies \sqrt{3}\cos x + \cos\left(\frac{\pi}{2} + x\right) = 1
   \implies \sqrt{3}\cos x - \sin x = 1

   উভয়পক্ষকে \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3+1} = 2 দ্বারা ভাগ করে পাই,
   \frac{\sqrt{3}}{2}\cos x - \frac{1}{2}\sin x = \frac{1}{2}
   \implies \cos x \cos\frac{\pi}{6} - \sin x \sin\frac{\pi}{6} = \frac{1}{2}
   \implies \cos\left(x + \frac{\pi}{6}\right) = \cos\frac{\pi}{3}

   \therefore x + \frac{\pi}{6} = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z}
   \implies x = 2n\pi \pm \frac{\pi}{3} - \frac{\pi}{6}

   ধনাত্মক চিহ্নের জন্য:
   x = 2n\pi + \frac{\pi}{3} - \frac{\pi}{6} = 2n\pi + \frac{\pi}{6}
   ঋণাত্মক চিহ্নের জন্য:
   x = 2n\pi - \frac{\pi}{3} - \frac{\pi}{6} = 2n\pi - \frac{\pi}{2}

   প্রদত্ত ব্যবধি: -2\pi < x < 2\pi

   n = 0 হলে:
   x = \frac{\pi}{6}
   x = -\frac{\pi}{2}

   n = 1 হলে:
   x = 2\pi + \frac{\pi}{6} > 2\pi (গ্রহণযোগ্য নয়)
   x = 2\pi - \frac{\pi}{2} = \frac{3\pi}{2}

   n = -1 হলে:
   x = -2\pi + \frac{\pi}{6} = -\frac{11\pi}{6}
   x = -2\pi - \frac{\pi}{2} < -2\pi (গ্রহণযোগ্য নয়)

   \therefore নির্ণেয় সমাধান: x = -\frac{11\pi}{6}, -\frac{\pi}{2}, \frac{\pi}{6}, \frac{3\pi}{2}।