A) \[\frac{-\pi }{2},\,0\]
B) \[\frac{-\pi }{2},\,0\,\frac{\pi }{2}\]
C) \[\frac{\pi }{2},\,0\]
D) \[0,\frac{\pi }{2},\,\frac{\pi }{2}\]
Correct Answer: B
Solution :
\[\because \] \[{{\cos }^{7}}x\le {{\cos }^{2}}x\] and \[{{\sin }^{4}}x\le \,{{\sin }^{2}}x\] On adding Eqs. (i) and (ii), we get \[\Rightarrow \] \[{{\cos }^{7}}x+{{\sin }^{4}}x\le 1\] But given, \[{{\cos }^{7}}x+{{\sin }^{4}}x=1\] Equality holds only if \[{{\cos }^{7}}x={{\cos }^{2}}x\] and \[{{\sin }^{4}}x={{\sin }^{2}}x\] Both are satisfies by \[x=\pm \,\frac{\pi }{2},\,0\]
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