Question

The greatest and least value of $\sin x\cos x$are

• A. $1,-1$
• B. $\frac{1}{2},-\frac{1}{2}$
• C. $\frac{1}{4},-\frac{1}{4}$
• D. $2,-2$

Answer 1

The correct option is B

$\frac{1}{2},-\frac{1}{2}$

Explanation of the correct answer.

Compute the required values:

Given: $\sin x\cos x$

Multiplying and dividing by $2$,

$\Rightarrow \frac{1}{2}\left(2\sin x\cos x\right)$

$\Rightarrow \frac{\left(\sin 2x\right)}{2}$

We know that,

$-1\le \left(\sin 2x\right)\le 1$

Dividing by $2$,

$-\frac{1}{2}\le \left(\frac{\sin 2x}{2}\right)\le \frac{1}{2}$

Hence, the greatest and least values are $\frac{1}{2}$ and $-\frac{1}{2}$ respectively.

Option $B$ is the correct answer.