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- May 26, 2019 at 3:19 pm #27962SARAParticipant
Find the number of points of discontinuity for f(x)=[6 sin x ] ,where x lies between 0 and pi .I have got 13 points ,answer is 11 .Why is the function continuous at f(x)=0 ?

May 27, 2019 at 1:18 am #28014AritraModeratori think [.] denotes the greatest integer function write , other wise the problem has no meaning . now see that a floor function is dis continuous in all those integer value

([6sin(x)]) can take integer value only when (sin (x)) take the value multiple of (\frac{n}{6}) see that n= 0,1,2,3,4,5,6

now (\frac {1}{6}) can be obtained by sin x only once at the point x= (\frac{\pi}{2})

and all the value are attended twice once at 0 to\ (\frac{pi}{2}) and another (\frac{\pi}{2}) to (\pi) so total there are 12 point of discontinuity

i don’t think it is continuous at x=0 because at x=0 it takes a integer value and floor function is not continuous at any integer value

even the graph say that it has 12 point of discontinuity

https://www.desmos.com/calculator/wnacsak92c

please clarify if i am wrong at any point

May 27, 2019 at 5:49 pm #28196SARAParticipantTo me clearly it seems that the number of points should be odd ,so either 11 or 13 .It is indeed 11 ,if the point (pi/2,6) is not counted.But i am confused as to why it isn’t counted ? If we do count it then we will have two more floors at 5 ,and so the answer will be 13 ,which is wrong .I have referred to the desmos graph .

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