मात्रात्मक योग्यता प्रश्न 2343
प्रश्न: यदि $ \sin \theta +\cos \theta =x, $ तो $ {{\cos }^{6}}\theta +{{\sin }^{6}}\theta $ का मान है
विकल्प:
A) $ \frac{1}{4} $
B) $ \frac{1}{4}(1+6x^{2}) $
C) $ \frac{1}{4}(1+6x^{2}-3x^{4}) $
D) $ \frac{1}{2}(5-3x^{2}) $
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उत्तर:
सही उत्तर: C
हल:
- $ \sin \theta +\cos \theta =x $ दोनों पक्षों को वर्ग करने पर, हमें प्राप्त होता है $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta =x^{2} $ $ 1+2\sin \theta \cos \theta =x^{2} $
$ \therefore $ $ \sin \theta \cos \theta =\frac{x^{2}-1}{2} $ $ {{\cos }^{6}}\theta +{{\sin }^{6}}\theta ={{({{\cos }^{2}}\theta )}^{3}}+{{({{\sin }^{2}}\theta )}^{3}} $ $ =({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )(cos^{4}\theta -{{\cos }^{2}}\theta {{\sin }^{2}}\theta +{{\sin }^{4}}\theta ) $ $ =[{{(cos^{2}\theta )}^{2}}+{{({{\sin }^{2}}\theta )}^{2}}-{{\cos }^{2}}\theta {{\sin }^{2}}\theta ] $ $ =[{{(cos^{2}\theta +{{\sin }^{2}}\theta )}^{2}}-2{{\cos }^{2}}\theta {{\sin }^{2}}\theta -{{\cos }^{2}}\theta {{\sin }^{2}}\theta ] $ $ =1-3{{\cos }^{2}}\theta {{\sin }^{2}}\theta =1-3{{( \frac{x^{2}-1}{2} )}^{2}} $ $ =1-\frac{3(x^{4}-2x^{2}+1)}{4}=\frac{4-3x^{4}+6x^{2}-3}{4} $ $ =\frac{1-3x^{4}+6x^{2}}{4}=\frac{1}{4}(1+6x^{2}-3x^{4}) $
BETA
