This problem appeared in one of the exams which I appeared in. I am mentioning it here only because I could not solve it. Here goes the problem:

Let **f(x)** =\(\frac{1}{x+2 cos(x)}\), x \(\geq\)0

Find the set A={y| y \(\in\)**R**, y=**f(x)**

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## Comments

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TopNewestAll they are asking for is the range of the given function, within the domain, right?

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Thank you all for your suggestions and solutions. Though I had solved the problem(in the exam in a

clumsymanner) but didn't get the minimum as 0, which now I can see very easily. That was a silly mistake on my part.Log in to reply

The minimum will never be zero.

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Yeah you are right. Actually I should have written it as approaching 0, since it is not attained at a finite value of x.

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A=[6/(5Pi-6sqrt{3)),0)

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just consider the function g(x)= x +2 cosx, now f(x)= 1/g(x), so if g(x1)> g(x2), then f(x1)< f(x2), now g(pi/6)=(pi/6+ sqrt3), g(5pi/6)= (5pi/6 + sqrt3), it is obvious g(pi/6)<g(5pi/6), therefore f(pi/6)>f(5pi/6)

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g(5pi/6)=(5pi/6-sqrt(3)) pls check the minus sign....

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hey....may...i ask a simple doubt....what's this ISI 2013 ? :|

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an exam for indian statistical institute

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*for admission in

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The answer to all the questions in the universe

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f(pi/6)= 6/(pi+ 6sqrt3, now you decide which is larger, 6/(5Pi+6sqrt{3), or 6/(pi+ 6sqrt3

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I think the problem is that sinx = sin(pi/6) has various solutions.You have simply taken the minimum value of x.Hence the wrong answer (verified using graphing calculator).Maximum is at x = 5pi/6

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its obvious as x tends to become larger, f(x) tends loser to 0, but never 0, so the infimum of the function is 0.

next step is to see that f(x), is continuous from 0 to infinity, so f(x), varies continously from its infimum to supremum, next step is to find the supremum, so we find the minima of the function x+ 2cosx., which occurs at x= pi/6, so maxima of f(x)= f(pi/3)= 6/(pi+ 6

sqrt3), so A=( 0,6/(pi+ 6sqrt3) ].Log in to reply

At pi/6 it reaches minimum not maximum. Only at 5pi/6 it reaches maximum plot the graph and see.

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At pi/6 x+2cos(x) reaches maximum therefore 1/(x+2cos(x)) reaches local minimum. but at 5pi/6 x+2cos(x) reaches minimum therefore 1/(x+2cos(x)) reaches maximum...

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