# Function Problem

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)

Note by Nishant Sharma
5 years, 2 months ago

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Thank you all for your suggestions and solutions. Though I had solved the problem(in the exam in a clumsy manner) but didn't get the minimum as 0, which now I can see very easily. That was a silly mistake on my part.

- 5 years, 2 months ago

The minimum will never be zero.

- 5 years, 2 months ago

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

- 5 years, 2 months ago

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

- 5 years, 2 months ago

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

- 5 years, 2 months ago

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

- 5 years, 2 months ago

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

- 5 years, 2 months ago

an exam for indian statistical institute

- 5 years, 2 months ago

- 5 years, 2 months ago

that's...ok...but..is it something like iit,nit,engineering college.....? i mean what kind of study..courses...? act. i dont know abt. this college...:)

- 5 years, 2 months ago

thanksss :)

- 5 years, 2 months ago

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

- 5 years, 2 months ago

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)

- 5 years, 2 months ago

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

- 5 years, 2 months ago

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+ 6sqrt3), so A=( 0,6/(pi+ 6sqrt3) ].

- 5 years, 2 months ago

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

- 5 years, 2 months ago

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...

- 5 years, 2 months ago

pi/6 is untrue. if you differentiate f'(x) again and put x=pi/6. f''(x) at x=pi/6 is more than 0. So pi/6 is local minimum.

- 5 years, 2 months ago

Please plot the graph using a graphing utility and see

- 5 years, 2 months ago

check your graphs once more, or atleast try to see the proof that i just postd

- 5 years, 2 months ago

$$x=\frac{\pi}{6}$$ is the local minimum.

- 5 years, 2 months ago

Which would make x=5pi/6 maximum

- 5 years, 2 months ago