A) \[\frac{{{\pi }^{4}}}{2}\]
B) \[\frac{{{\pi }^{2}}}{2}\]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{2}\]
Correct Answer: D
Solution :
Let\[f(x)={{\sin }^{4}}x+{{\cos }^{4}}x\] \[={{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x\] \[=1-\frac{1}{4}\cdot 2{{(\sin 2x)}^{2}}\] \[=1-\frac{1}{4}(1-\cos 4x)\] \[=\frac{3}{4}+\frac{\cos 4x}{4}\] \[\therefore \] Period of\[f(x)=\frac{2\pi }{4}=\frac{\pi }{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
