\( y = \frac{1}{2} \tan^{-1} \left[ \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}} \right] \) এর সঠিক সংক্ষিপ্ত রূপ কোনটি?

Explanation: \(f(x) = \frac{x}{\log x} \implies f'(x) = \frac{\log x \cdot 1 - x \cdot \frac{1}{x}}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2}\) লঘু-গুরু মানের জন্য \(f'(x) = 0 \implies \frac{\log x - 1}{(\log x)^2} = 0 \implies \log x = 1 \implies x = e\) \(f''(x) = \frac{(\log x)^2 \cdot \frac{1}{x} - (\log x - 1) 2 \log x \cdot \frac{1}{x}}{(\log x)^4}\) \(\implies f''(e) = \frac{1^2 \cdot \frac{1}{e} - (1 - 1) \cdot 2 \log e \cdot \frac{1}{e}}{1^4} = \frac{\frac{1}{e}}{1} = \frac{1}{e} > 0\) \(\implies x = e\) হলে লঘুমান হবে, \(\therefore \text{লঘুমান} = \frac{e}{\log e} = e\)

Another Explanation (5): ```html

প্রশ্ন: \( y = \frac{1}{2} \tan^{-1} \left[ \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}} \right] \) এর সঠিক সংক্ষিপ্ত রূপ কোনটি?

উত্তর: \( x=4y \)

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ব্যাখ্যা:

আমরা জানি, \( 1 + \sin x = \left( \cos \frac{x}{2} + \sin \frac{x}{2} \right)^2 \) এবং \( 1 - \sin x = \left( \cos \frac{x}{2} - \sin \frac{x}{2} \right)^2 \)।

সুতরাং, \( \sqrt{1 + \sin x} = \left| \cos \frac{x}{2} + \sin \frac{x}{2} \right| \) এবং \( \sqrt{1 - \sin x} = \left| \cos \frac{x}{2} - \sin \frac{x}{2} \right| \)।

ধরি, \( 0 < x < \frac{\pi}{2} \), তাহলে \( \cos \frac{x}{2} > \sin \frac{x}{2} > 0 \)।

অতএব, \( \sqrt{1 + \sin x} = \cos \frac{x}{2} + \sin \frac{x}{2} \) এবং \( \sqrt{1 - \sin x} = \cos \frac{x}{2} - \sin \frac{x}{2} \)।

এখন,

\( \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}} = \frac{\cos \frac{x}{2} + \sin \frac{x}{2} - \left( \cos \frac{x}{2} - \sin \frac{x}{2} \right)}{\cos \frac{x}{2} + \sin \frac{x}{2} + \cos \frac{x}{2} - \sin \frac{x}{2}} \)

\( = \frac{2 \sin \frac{x}{2}}{2 \cos \frac{x}{2}} = \tan \frac{x}{2} \)

সুতরাং, \( y = \frac{1}{2} \tan^{-1} \left( \tan \frac{x}{2} \right) \)

যেহেতু \( 0 < x < \frac{\pi}{2} \), \( 0 < \frac{x}{2} < \frac{\pi}{4} \), তাই \( \tan^{-1} \left( \tan \frac{x}{2} \right) = \frac{x}{2} \)।

সুতরাং, \( y = \frac{1}{2} \cdot \frac{x}{2} = \frac{x}{4} \)।

অতএব, \( x = 4y \)। 🎉

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