Question

Let $f\left(x\right)$ be a non-negative function. If $f^{\prime }\left(x\right)\cos x\le f\left(x\right)\sin x,\forall x\ge 0,$ then value of $f\left(\frac{5\pi }{3}\right)$ is

• A. $1$
• B. $\frac{1}{2}$
• C. $0$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is C $0$

$f^{\prime }\left(x\right)\cos x\le f\left(x\right)\sin x$
$\Rightarrow f^{\prime }\left(x\right)\cos x-f\left(x\right)\sin x\le 0$
$\Rightarrow \frac{d}{dx}\left(f\right(x\left)\cos x\right)\le 0$

$\therefore g\left(x\right)=f\left(x\right)\cos x$ is decreasing $\forall x\ge 0$
$\Rightarrow g\left(x\right)\le g\left(0\right)$
$\Rightarrow f\left(x\right)\cos x\le f\left(0\right)\cos 0$
$\Rightarrow f\left(x\right)\cos x\le f\left(0\right)$

We know that
$\frac{5\pi }{3}\ge \frac{\pi }{2}$
$\Rightarrow g\left(\frac{5\pi }{3}\right)\le g\left(\frac{\pi }{2}\right)(\because g\left(x\right)\text{is decreasing})$
$\Rightarrow g\left(\frac{5\pi }{3}\right)\le 0[\because g\left(\frac{\pi }{2}\right)=0]$
$\Rightarrow f\left(\frac{5\pi }{3}\right)\times \cos \left(\frac{5\pi }{3}\right)\le 0$
$\Rightarrow f\left(\frac{5\pi }{3}\right)\times \cos \left(\frac{\pi }{3}\right)\le 0$
$\Rightarrow f\left(\frac{5\pi }{3}\right)\le 0$

but $f\left(x\right)\ge 0$ (Given in the question)

So, $f\left(\frac{5\pi }{3}\right)=0$