I.S.I Entrance-2013 problem 2

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Understand the problem

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

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

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Medium 

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Problems In CALCULUS OF ONE VARIABLE

by I.A. Maron

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Start with hints

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Do you really need a hint? Try it first!

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

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 First extrema occurs at $latex x$= $latex \frac{\pi}{6}$The first derivative is negetive in the interval [ 0, $latex \frac{\pi}{6}$]hence the function is decreasing in this intervalf(0)=$latex \frac{1}{2}$ ; f($latex \frac{\pi}{6}$)=$latex \frac{1}{\frac{\pi}{6}+ {\sqrt{3}}}$  

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 For x>$latex \frac{\pi}{6}$the derivative becomes positive , and remain so upto x=$latex \frac{5\pi}{6}$ after which it becomes negative thus we have minima at x= $latex \frac{\pi}{6}$ and maxima at x= $latex \frac{5\pi}{6}$f($latex \frac{5\pi}{6}$)= $latex \frac{1}{\frac{5\pi}{6}+\sqrt{3}}$note that as $latex x\rightarrow \infty$the denominator of the function increases 

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Hence we can conclude that    $latex f(x)\rightarrow0$clearly x=$latex \frac{5\pi}{6}$ gives the global maxima so , the range is (0,$latex \frac{1}{\frac{5\pi}{6}+\sqrt{3}}$]    

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Connected Program at Cheenta

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Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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Similar Problem

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