The minimum value of function f(x)=2 x+x in interval [0, 2 ] : - Vidyadip

Question

The minimum value of function $f(x)=2 \cos x+x$ in interval $\left[0, \frac{\pi}{2}\right]$ :

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Answer

(C)$f(x)=2 \cos x+x \Rightarrow f^{\prime}(x)=-2 \sin x+1$
now, $f^{\prime}(x)=0 \Rightarrow-2 \sin x+1=0$
$\Rightarrow \quad \sin x=\frac{1}{2}=\sin \frac{\pi}{6} \Rightarrow x=\frac{\pi}{6}$
$\Rightarrow \quad x=\frac{\pi}{6} \in\left[0, \frac{\pi}{2}\right]$
$\therefore \quad f(0)=2+0=2, f\left(\frac{\pi}{6}\right)=2 \cos \frac{\pi}{6}+\frac{\pi}{6}$
$=2\left(\frac{\sqrt{3}}{2}\right)+\frac{\pi}{6}=\sqrt{3}+\frac{\pi}{6}$
and $\quad f\left(\frac{\pi}{2}\right)=0+\frac{\pi}{2}=\frac{\pi}{2}$
The minimum value of $f(x)$ is $\frac{\pi}{2}$ at $x=\frac{\pi}{2}$.

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