The greatest and least value of sinx cosx are

[MNR 1975]

1,-1

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2,-2

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Solution

The correct option is B

Let f(x) = sin x cos x = $\frac{1}{2}$ sin 2x

We know -1 $\leq$ sin2x $\leq$ 1 $\Rightarrow$ - $\frac{1}{2}$ $\leq$ $\frac{1}{2}$ sin2x $\leq$ $\frac{1}{2}$

Thus the greatest and least value of f(x) are $\frac{1}{2}$ and - $\frac{1}{2}$ respectively.

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