A) \[\frac{1}{2\sqrt{2}}\]
B) \[\frac{1}{4}\]
C) \[\frac{-1}{2}\]
D) \[\frac{1}{2}\]
Correct Answer: D
Solution :
Given \[f(x)={{\sin }^{4}}x+{{\cos }^{4}}x\] \[={{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x\] \[\Rightarrow \] \[f(x)=1-\frac{1}{2}{{\sin }^{2}}2x\] Also, \[0<si{{n}^{2}}2x\le 1\] \[\therefore \]Minimum value of\[f(x)\]is\[1-\frac{1}{2}=\frac{1}{2}\]You need to login to perform this action.
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