If q²-4pr=0, p>0, then find the domain of the function f(x) = log(px³+(p+q)x²+(q+r)x+r)=0.

Also explain your answer.

csc

May be i have become sloppy in years ..but your approach should go like this.. px^3+(p+q)x^2+(q+r)x+r)>0 (Why ? i think you should know this).. x(px^2+(p+q)x^1+(q+r))+r>0 what is minumum value of px^2+(p+q)x^1+q+r.....put x=-b/2a and get this value which comes to be (-p^2+2pq) and also use fact q^2=4pr; => x (-p^2+2pq)+r> => x>(r/p^2-2pq)

f(x)=log{(px^2+qx+r)(x+1)}=log(px^2+qx+r)+log(x+1) we Know that px^2+qx+r>0 & x+1>0 since p>0 & q^2-4pr=0 therefore px^2+qx+r always greater than zero except -q/2p and x>-1 so domain of the function x>-1/{q/2p} Thanks & Regards Sunil Raikwar askIITians faculty