A) \[\frac{1}{2}\]
B) \[\frac{3}{4}\]
C) 1
D) \[\frac{4}{3}\]
Correct Answer: A
Solution :
Let \[f(A)=\cos A\cos B=\cos A\cos \left( \frac{\pi }{2}-A \right)=\cos A\sin A\] \ \[{f}'(A)={{\cos }^{2}}A-{{\sin }^{2}}A=\cos 2A\] Now, \[{f}'(A)=0\Rightarrow \cos 2A=0\]Þ \[2A=\frac{\pi }{2}\Rightarrow A=\frac{\pi }{4}\] Now \[{f}''(A)=-2\sin 2A=-2\sin \frac{\pi }{2}=-2\,\,\,\,\,\,(-ve)\] Hence \[f(A)\] is maximum at \[\frac{\pi }{4}\] \ Maximum value = \[\cos \frac{\pi }{4}\sin \frac{\pi }{4}=\frac{1}{2}\].
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