A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[2\pi \]
D) none of these
Correct Answer: B
Solution :
Key Idea: The period\[{{\sin }^{4}}x\]or\[{{\cos }^{4}}x\]is\[\frac{\pi }{2}\]. \[f(x)={{\sin }^{4}}x+{{\cos }^{4}}x\] \[={{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x\] \[=1-\frac{1}{2}{{(\sin 2x)}^{2}}=1-\frac{1}{2}\left[ \frac{1-\cos 4x}{2} \right]\] \[=\frac{3}{4}+\frac{1}{4}\cos 4x\] Since, \[\cos x\] is periodic with period\[2\pi \]. \[\therefore \]The period of\[f(x)=\frac{2\pi }{4}=\frac{\pi }{2}\]. Alternative Solution: Since the period of\[{{\sin }^{4}}x\]and\[{{\cos }^{4}}x\]is\[\frac{\pi }{2}\]. \[\therefore \]Period of\[f(x)={{\sin }^{4}}x+{{\cos }^{4}}x\]is the \[LCM\] of the period of\[{{\sin }^{4}}x\]and\[{{\cos }^{4}}x\]. \[\therefore \]Required period is\[\frac{\pi }{2}\].
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