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Check the monotonicity of f(x) = [x÷sin x] ,in (0,p/2).
f(x)= x/sinxdf/dx = (sinx - xcosx)/ (sinx)^2A 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)^2tan 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 & RegardsBharat BajajaskIITians FacultyIIT 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
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