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If $f(x) = \dfrac{\sqrt{3}\sin x}{2+\cos x}$, then the range of $f(x)$ is

  1. the interval $[-1, \sqrt{3}/2]$
  2. the interval $[- \sqrt{3}/2, 1]$
  3. the interval $[-1, 1]$
  4. none of the above
in Calculus
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is $C$ the answer.
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Yes, you are right.

1 Answer

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Note that  $y=f(x)=\dfrac{\sqrt 3 \sin x}{2+\cos x}$.

Note that  $y^2=\dfrac{3 \sin ^2 x}{(2+\cos ^2 x)^2}=\dfrac{3 (1- \cos ^2 x)}{(2+\cos ^2 x)^2}$

Put $\cos x =u$, we have 

\begin{align}
y^2(2+u)^2 =3(1-u^2) \\ \implies (3+y^2)u^2+4y^2u+4y^2-3=0
\end{align}
For this to have a solution, we need discriminant of the quadratic to be non-negative. Hence, we get
$$(4y)^2-4(3+y^2)(4y^2-3) \ge 0 \\ \implies y^2  \le 1\\ \implies-1 \le y \le 1$$ 

So, the correct answer is $(C)$.

Source from: https://math.stackexchange.com/questions/382136/a-problem-on-range-of-a-trigonometric-function-what-is-the-range-of-frac-sqr/383172#383172

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