मात्रात्मक योग्यता प्रश्न 637
प्रश्न: यदि $ \sin \theta +\cos \theta =x, $ तब $ {{\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 $ $ {{\sin }^{2}}+{{\cos }^{2}}\theta +2\sin \theta \cos =x^{2} $ $ \sin \theta \times \cos \theta =\frac{x^{2}-1}{2} $ अब, $ {{\cos }^{6}}\theta +{{\sin }^{6}}\theta ={{({{\sin }^{2}}\theta )}^{3}}+{{({{\cos }^{2}}\theta )}^{3}} $ $ =({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )({{\cos }^{4}}\theta +{{\sin }^{4}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta ) $ $ ={{({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )}^{2}}-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta $ $ ={{(cos^{2}\theta +{{\sin }^{2}}\theta )}^{2}}-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta $ $ =1-3{{( \frac{x^{2}-1}{2} )}^{2}}=\frac{1}{4}(1+6x^{2}-3x^{4}) $
BETA
