Question

The maximum value of $\sin x.\cos x$ is

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

Answer 1

The correct option is B $\frac{1}{2}$

Given: $f\left(x\right)=\sin x.\cos x$
$f\left(x\right)=\frac{1}{2}(2\sin x.\cos x)$
$f\left(x\right)=\frac{1}{2}\sin 2x$
$f(x{)}_{\text{max}}=\frac{1}{2}\times 1$
$[\because$ Max.value of $\sin 2x=1$ ]
$f(x{)}_{\text{max}}=\frac{1}{2}$
The maximum value of $\sin x.\cos x$ is $\frac{1}{2}$
Hence , option B is correct