A) \[1,\,\,-1\]
B) \[\frac{1}{2},\,\,-\frac{1}{2}\]
C) \[\frac{1}{4},\,\,-\frac{1}{4}\]
D) \[2,\,\,-\,2\]
Correct Answer: B
Solution :
Let \[f(x)=sin\,x\,cos\,x=\frac{1}{2}sin2x\] We know \[-1\le \sin \,2x\le 1\,\,\Rightarrow \,\,-\frac{1}{2}\le \frac{1}{2}\sin 2x\,\,\le \,\,\frac{1}{2}\] Thus the greatest and least value of f(x) are \[\frac{1}{2}\,\,and\,\,-\frac{1}{2}\] respectively
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