recategorized by
558 views

3 Answers

1 votes
1 votes

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.

0 votes
0 votes

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.

edited by
0 votes
0 votes
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 ]

Related questions

2 votes
2 votes
2 answers
1
Arjun asked Sep 23, 2019
507 views
If $f(x)$ is a real valued function such that $2f(x)+3f(-x)=15-4x$, for every $x \in \mathbb{R}$, then $f(2)$ is$-15$$22$$11$$0$
1 votes
1 votes
1 answer
2
Arjun asked Sep 23, 2019
459 views
Suppose that the function $h(x)$ is defined as $h(x)=g(f(x))$ where $g(x)$ is monotone increasing, $f(x)$ is concave, and $g’’(x)$ and $f’’(x)$ exist for all $x$....
0 votes
0 votes
1 answer
3
Arjun asked Sep 23, 2019
410 views
Let $f(x) = \dfrac{2x}{x-1}, \: x \neq 1$. State which of the following statements is true.For all real $y$, there exists $x$ such that $f(x)=y$For all real $y \neq 1$, ...
1 votes
1 votes
0 answers
4
Arjun asked Sep 23, 2019
461 views
Let $f(x)$ be a continuous function from $[0,1]$ to $[0,1]$ satisfying the following properties.$f(0)=0$,$f(1)=1$, and$f(x_1)<f(x_2)$ for $x_1 < x_2$ with $0 < x_1, \: x_...