Check the monotonicity of f(x) = [x÷sin x] ,in (0,p/2).

f(x)= x/sinx
df/dx = (sinx - xcosx)/ (sinx)^2

A function is monotonic if it is either increasing or decreasing in the given range.
Hence, check the sign of df/dx.

df/dx = (tanx - x)cosx/(sinx)^2

tan x - x> 0 for (0,pi/2) . You may check it by making the graph.
cosx>0 for (0,pi2)
Hence, df/dx>0 for(0,pi/2)
Hence, this function is monotonic in the given range.
Thanks & Regards
Bharat Bajaj
askIITians Faculty
IIT Delhi

f(x)= x/sinx df/dx = (sinx - xcosx)/ (sinx)^2=0 tanx=x this is only possible when x=0 it means there is no critical points in the region (0,p/2). so we conclude that the graph of the function is monotonic in nature