in Calculus edited by
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If $\textit{f}(x)=\frac{\sqrt{3}\sin x}{2+\cos x}$ then the range of $\textit{f}(x)$ is

  1. the interval $\left[-1,\frac{\sqrt{3}}{2}\right]$
  2. the interval $\left[\frac{-\sqrt{3}}{2},1\right]$
  3. the interval $\left[-1,1\right]$
  4. none of the above
in Calculus edited by
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1 Answer

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Best answer
$\frac{\sqrt{3}\sin x}{2+\cos x}$ This is of the form $\frac{u}{v}$

Differentiating above equation and equating to $0$ we get

$\frac{(2+\cos x)\sqrt{3}\cos x - \sqrt{3}\sin x(-\sin x)}{(2+\cos x)^{2}}= 0$

$\implies 2\sqrt{3}\cos x + \sqrt{3}\cos^{2}x + \sqrt{3}\sin^{2}x = 0$

$\implies 2\sqrt{3}\cos x + \sqrt{3} = 0$

$\implies 2\cos x + 1 = 0$

$\implies x = \frac{2π}{3} , \frac{-2π}{3}$

for $x= \frac{2π}{3}$ ,   $\frac{\sqrt{3}\sin x}{2+\cos x} = 1 $ and

for $x=  \frac{-2π}{3}$ , $\frac{\sqrt{3}\sin x}{2+\cos x} = -1$

Hence, option $C$ is correct.
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if anybody is wondering why he has differentiated and equated to zero here is explaination.
It is because Range[Maxima , Minima]  , he has just found out maxima and minima
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