If `0 lt x lt 1000 and [x/2]+[x/3]+[x/5]=31/30x`, (where `[.]` denotes the greatest integer function then number of possible values of x.

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If f(x)=[2x], where [.] denotes the greatest integer function,then

If [log_2 (x/[[x]))]>=0 . where [.] denotes the greatest integer function, then :

If [log_(2)((x)/([x]))]>=0, where [.] denote the greatest integer function,then

If 0 lt x lt 1000 and [(x)/(2)]+[(x)/(3)]+[(x)/(5)]=(31)/(30)x, where [x] is the greatest integer less than or equal to x ,the number of possible values of x is

the value of int_(0)^([x]) dx (where , [.] denotes the greatest integer function)

If 0 lt x lt 1000 and [(x)/(2)] +[(x)/(3)]+[(x)/(5)] =(31)/(30)x , where [x] is the greatest integer less than or equal to x, the number of possible values of x is

Number of solutions of sin x=[x] where [.] denotes the greatest integer function is

If (x)={x[x],0<=x<=2 and (x-1)[x],2<=x<=3,where [x] denotes the greatest integer function, then

lim_(x -> 0)[sin[x-3]/([x-3])] where [.] denotes greatest integer function is