A) \[1\]
B) \[2\]
C) \[\sqrt{2}\]
D) \[\sqrt{3}\]
Correct Answer: A
Solution :
Let \[y={{\sin }^{3}}x+{{\cos }^{3}}x\] \[\frac{dy}{dx\,}=3{{\sin }^{2}}\,x\,\cos x-3\,{{\cos }^{2}}x\,\sin x\] \[=3\sin x\,\cos x(\sin x-\cos x)\] Put \[\frac{dy}{dx}=0\] \[\Rightarrow \] \[3\,\sin x\,\cos x(sin\,x-cos\,x)=0\] \[\Rightarrow \] \[x=0,\,\frac{\pi }{2}\] or \[\frac{\pi }{4}\] Now, y has its maximum value at \[x=0\] or \[\frac{\pi }{2},\] and \[{{y}_{\max }}=1\]
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