Question

The maximum value of $sin\, x \cdot cos\, x$ is

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

Answer 1

Answer: (B)

Given $f\left(x\right) = sin\, x \cdot cos\, x = \frac{sin\,2x}{2}$
$f'\left( x\right)= \frac{1}{2} \left(cos\,2 x\right)\cdot2 = cos\, 2x$
For maxima or minima, $f'\left(x\right) = 0$
$\Rightarrow cos\,2 x = 0$
$\Rightarrow 2x = \frac{\pi}{2}$, $\frac{3\pi }{2}$
$\Rightarrow x = \frac{\pi }{4}$, $\frac{3\pi }{4}$
Now, $f''\left(x\right) = - 2\, sin \,2x$
At $x = \frac{\pi }{4}$
$f''\left(x\right) = - 2\, sin \,2 (\frac{\pi }{4}) = -2 < 0$
$\therefore x = \frac{\pi }{4}$ local max point
$\therefore $ max. value is
$f\left(\frac{\pi }{4}\right) = sin\frac{\pi }{4}\cdot cos\frac{\pi }{4}$
$= \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{1}{2}$