# Understand the problem

For x ≥ 0 define
f(x) = $\frac{1}{x+2cos x}$
.
Determine the set {y ∈ R : y = f(x), x ≥ 0}.

##### Source of the problem

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2013. Subjective Problem no. 2.

##### Topic
Calculus , Function

Medium

# Start with hints

Do you really need a hint? Try it first!

This problem simply ask for the range of the function defined by f(x)= $\frac {1}{x+2cosx}$

compute the derivative of the function = $\frac {2sinx-1}{(x+2cosx)^2}$

First extrema occurs at $x$= $\frac{\pi}{6}$

The first derivative is negetive in the interval [ 0, $\frac{\pi}{6}$]

hence the function is decreasing in this interval

f(0)= $\frac{1}{2}$ ; f( $\frac{\pi}{6}$)= $\frac{1}{\frac{\pi}{6}+ {\sqrt{3}}}$

For x> $\frac{\pi}{6}$

the derivative becomes positive , and remain so upto x= $\frac{5\pi}{6}$ after which it becomes negative

thus we have minima at x= $\frac{\pi}{6}$ and maxima at x= $\frac{5\pi}{6}$

f( $\frac{5\pi}{6}$)= $\frac{1}{\frac{5\pi}{6}+\sqrt{3}}$

note that as $x\rightarrow \infty$the denominator of the function increases

Hence we can conclude that $f(x)\rightarrow0$

clearly x= $\frac{5\pi}{6}$ gives the global maxima

so , the range is (0, $\frac{1}{\frac{5\pi}{6}+\sqrt{3}}$]

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