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.

Calculus , Function 

Difficulty Level


Suggested Book


by I.A. Maron

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 \inftythe 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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