Question

The maximum value of sin x . cos x is ______.

पर्याय

• ${\displaystyle \frac{1}{4}}$

• ${\displaystyle \frac{1}{2}}$

• ${\displaystyle \sqrt{2}}$

• ${\displaystyle 2\sqrt{2}}$

Answer 1

The maximum value of sin x . cos x is ${\displaystyle \frac{1}{2}}$.

Explanation:

We have f(x) = sin x cos x

⇒ f(x) = ${\displaystyle \frac{1}{2}\cdot 2\sin x\cos x}$

= ${\displaystyle \frac{1}{2}\sin 2x}$

f'(x) = ${\displaystyle \frac{1}{2}\cdot 2\cos 2x}$

⇒ f'(x) = cos 2x

Now for local maxima and local minima f'(x) = 0

∴ cos 2x = 0

2x = ${\displaystyle (\text{n}+1)\frac{\pi }{2}}$, n ∈ I

⇒ x = ${\displaystyle (2\text{n}+1)\frac{\pi }{4}}$

∴ x = ${\displaystyle \frac{\pi }{4},\frac{3\pi }{4}}$ .....

f"(x) = – 2 sin 2x

${\displaystyle \text{f''}{\left(x\right)}_{x=\frac{\pi }{4}}}$ = ${\displaystyle -2\sin 2\cdot \frac{\pi }{4}}$

= ${\displaystyle -2\sin \frac{\pi }{2}}$

= – 2 < 0 maxima

${\displaystyle \text{f''}{\left(x\right)}_{x=\frac{3\pi }{4}}=-2\sin 2\cdot \frac{3\pi }{4}}$

= ${\displaystyle -2\sin \frac{3\pi }{4}}$

= 2 > 0 minima

So f(x) is maximum at x = ${\displaystyle \frac{\pi }{4}}$

∴ Maximum value of f(x) = ${\displaystyle \sin \frac{\pi }{4}\cdot \cos \frac{\pi }{4}}$

= ${\displaystyle \frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}}$

= ${\displaystyle \frac{1}{\sqrt{2}}}$.