Applications Of Derivatives Question 127
Question: The function $ f(x)={{\sin }^{4}}x+{{\cos }^{4}}x $ increases, if
[IIT 1999; Pb. CET 2001]
Options:
A) $ 0<x<\frac{\pi }{8} $
B) $ \frac{\pi }{4}<x<\frac{3\pi }{8} $
C) $ \frac{3\pi }{8}<x<\frac{5\pi }{8} $
D) $ \frac{5\pi }{8}<x<\frac{3\pi }{4} $
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Answer:
Correct Answer: B
Solution:
$ f(x)={{\sin }^{4}}x+{{\cos }^{4}}x $
$ ={{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x $
$ =1-\frac{4{{\sin }^{2}}x{{\cos }^{2}}x}{2}=1-\frac{{{\sin }^{2}}2x}{2} $
$ =1-\frac{1}{4}(2{{\sin }^{2}}2x) $
$ =1-( \frac{1-\cos 4x}{4} )=\frac{3}{4}+\frac{1}{4}\cos 4x $
Hence function $ f(x) $ is increasing when $ f’(x)>0 $
$ f’(x)=-\sin 4x>0\Rightarrow \sin 4x<0 $
Hence $ \pi <4x<\frac{3\pi }{2} $ or $ \frac{\pi }{4}<x<\frac{3\pi }{8} $ .
BETA
