Comprehension

निम्नलिखित दो (02) प्रश्नों के लिए इस पर विचार कीजिए: मान लीजिए 

p का अधिकतम मान क्या है?

This question was previously asked in
NDA-I (Mathematics) Official Paper (Held On: 13 Apr, 2025)
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  1. 1
  2. \(\sqrt{2}\)
  3. \(\sqrt{3}\)
  4. 2

Answer (Detailed Solution Below)

Option 2 : \(\sqrt{2}\)
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गणना:

दिया गया है,

p के लिए व्यंजक है:

\( p = |\sin \alpha - \sin(\alpha - 90^\circ)| \)

\(\sin(\alpha - 90^\circ) \) के लिए सर्वसमिका का उपयोग करने पर,

\( \sin(\alpha - 90^\circ) = \cos \alpha \)

p के व्यंजक में इसे प्रतिस्थापित करने पर:

\( p = |\sin \alpha - \cos \alpha| \)

व्यंजक \(|\sin \alpha - \cos \alpha|\) अपना अधिकतम मान तब प्राप्त करता है जब \(\sin \alpha \) और \(\cos \alpha \) के बीच का अंतर सबसे बड़ा होता है।

हम \(\sin \alpha - \cos \alpha \) को इस प्रकार फिर से लिख सकते हैं:

\( \sin \alpha - \cos \alpha = \sqrt{2} \left( \sin \left( \alpha - 45^\circ \right) \right) \)

चूँकि \(\sin(\alpha - 45^\circ) \) का अधिकतम मान 1 है, \( \sqrt{2} \left( \sin(\alpha - 45^\circ) \right) \) का अधिकतम मान: \( \sqrt{2} \) है। 

इसलिए, सही उत्तर विकल्प 2 है। 

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