Comprehension

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

p का न्यूनतम मान क्या है?

This question was previously asked in
NDA-I (Mathematics) Official Paper (Held On: 13 Apr, 2025)
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  1. 0
  2. 1/2
  3. \(1/\sqrt{2}\)
  4. 1

Answer (Detailed Solution Below)

Option 1 : 0
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गणना:

दिया गया है,

हमें व्यंजक दिया गया है:

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

ज्या अंतर के लिए सर्वसमिका का उपयोग करने पर:

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

p के लिए इस सर्वसमिका को व्यंजक में प्रतिस्थापित करने पर:

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

\( | \sin \alpha - \cos \alpha | \) का न्यूनतम मान तब होता है जब \(\sin \alpha = \cos \alpha \) है, जो तब होता है जब \(\alpha = 45^\circ \) है। इस बिंदु पर:

\( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \)

इसलिए, \(\alpha = 45^\circ \) पर, \(\sin \alpha \) और \(\cos \alpha \) के बीच का अंतर 0 है, इसलिए:

\( p = 0 \)

∴ p का न्यूनतम मान 0 है।

इसलिए, सही उत्तर विकल्प (a) 0 है।

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