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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 these
in Calculus
recategorized | 49 views

Answer: $\mathbf C$

$f^2(x) =\dfrac{3-3\cos^2x}{\cos^2x+4\cos x+4}$

Now, Let t = $\cos x$

then, $f(t) = \dfrac{3-3t^2}{t^2+4t+4}$

$f_{max}.f'(t) = \frac{-6(2t+1)}{t+2} = 0$, when$t = \frac{-1}{2}$

$\therefore f(-1) =0, f(\frac{-1}{2}) = 1, \;and\; f(1) = 0$

So, $f^2\text{max} = 1 \implies f^2(x) \leq 1 \implies -1\leq f(x) \leq 1$

So, the range of $f$ is $[-1,1]$

$\therefore \mathbf C$ is the right option.

by Boss (18.9k points)
edited by

To find the range of $f(x)$, actually we need to find the maximum and minimum value.

Now

\begin{align}f(x)&=\frac{\sqrt{3}\sin x}{2+\cos x}\\ \Rightarrow f'(x)&=\sqrt{3}\cos x \cdot \frac{1}{2+\cos x}+\sqrt{3}\sin x \cdot \frac{-\sin x}{(2+\cos x)^2}\\ &=\sqrt{3}\cdot \frac{2\cos x+\cos^2 x+\sin^2 x}{(2+\cos x)^2} \end{align}

At the point of which $f(x)$ is maximum or minimum, the slope is $0$. \begin{align}\therefore f'(x)&=0\\ \Rightarrow \sqrt{3}\cdot \frac{2\cos x+\cos^2 x+\sin^2 x}{(2+\cos x)^2} &=0\\ \Rightarrow 2\cos x +1&=0;~[\because \sin^2 x+\cos^2 x=1] \\ \Rightarrow \cos x &= -\frac{1}{2}\\ \Rightarrow x &= \frac{2\pi}{3}, \frac{4\pi}{3}\end{align}

Now putting those values of $x$ to $f(x)$, we obtain

$f(\frac{2\pi}{3})=\frac{\sqrt{3}(\frac{\sqrt{3}}{2})}{2-\frac{1}{2}}=1$ and $f(\frac{4\pi}{3})=\frac{\sqrt{3}(-\frac{\sqrt{3}}{2})}{2-\frac{1}{2}}=-1$. It means $f(x)$ has minimum value $-1$ and maximum $1$.

$$\therefore f(x) \in [-1,1]$$

So the correct answer is C.

by Active (3.5k points)
Max value of fx =( Max val of √3 *sinx)/(Min val of 2+cosx)

= (√3 *1)/(2-1) = √3

Min value of fx =( Min val of √3 *sinx)/(Max val of 2+cosx)

=  (√3 *-1)/(2+1) = -1/ √3

So, Range = [ -1/ √3 , √3 ]
by (253 points)