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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
4 years, 3 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. Nishant Sharma · 4 years, 3 months ago

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@Nishant Sharma The minimum will never be zero. Aditya Parson · 4 years, 3 months ago

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@Aditya Parson Yeah you are right. Actually I should have written it as approaching 0, since it is not attained at a finite value of x. Nishant Sharma · 4 years, 3 months ago

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All they are asking for is the range of the given function, within the domain, right? Rohan Rao · 4 years, 3 months ago

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f(pi/6)= 6/(pi+ 6sqrt3, now you decide which is larger, 6/(5Pi+6sqrt{3), or 6/(pi+ 6sqrt3 Nidhin Kurian · 4 years, 3 months ago

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@Nidhin Kurian 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 Sridhar Thiagarajan · 4 years, 3 months ago

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hey....may...i ask a simple doubt....what's this ISI 2013 ? :| Riya Gupta · 4 years, 3 months ago

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@Riya Gupta an exam for indian statistical institute Superman Son · 4 years, 3 months ago

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@Superman Son *for admission in Superman Son · 4 years, 3 months ago

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@Superman Son 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...:) Riya Gupta · 4 years, 3 months ago

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@Abhishek De thanksss :) Riya Gupta · 4 years, 3 months ago

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A=[6/(5Pi-6sqrt{3)),0) Aniruddha S Prasad · 4 years, 3 months ago

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@Aniruddha S Prasad 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) Nidhin Kurian · 4 years, 3 months ago

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@Nidhin Kurian g(5pi/6)=(5pi/6-sqrt(3)) pls check the minus sign.... Aniruddha S Prasad · 4 years, 3 months ago

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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+ 6sqrt3), so A=( 0,6/(pi+ 6sqrt3) ]. Nidhin Kurian · 4 years, 3 months ago

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@Nidhin Kurian At pi/6 it reaches minimum not maximum. Only at 5pi/6 it reaches maximum plot the graph and see. Aniruddha S Prasad · 4 years, 3 months ago

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@Aniruddha S Prasad 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... Aniruddha S Prasad · 4 years, 3 months ago

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@Aniruddha S Prasad 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. Aditya Parson · 4 years, 3 months ago

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@Aditya Parson Please plot the graph using a graphing utility and see Aniruddha S Prasad · 4 years, 3 months ago

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@Aniruddha S Prasad check your graphs once more, or atleast try to see the proof that i just postd Nidhin Kurian · 4 years, 3 months ago

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@Nidhin Kurian \(x=\frac{\pi}{6}\) is the local minimum. Aditya Parson · 4 years, 3 months ago

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@Aditya Parson Which would make x=5pi/6 maximum Aniruddha S Prasad · 4 years, 3 months ago

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