Let f(x)= (e^x-a )(3 a x+1) . Number of possible values of a satisfying f(x) 0 for x R .

Step-by-step solution

Case-I: If a 0 and 3 a x+1 is sometimes + ve /- ve both f(x) cannot be positive always because of second bracket. Hence in this case no possible values of a . Case-II: If a=0 then f(x)=e^x which is positive x R either both brackets must be positive or both brackets must be negative.Hence both the critical points must coincide. e^x=a x= a and 3 a x+1=0 x= -1 3 a a= -1 3 a a a= -1 3 No. of values of a satisfying above equation is 2 . Hence total number of possible values of a are 3 .