for the quadratic polynomial f(x) = 4x2-8ax+a, the statements which hold good is/are(multiple correct answers) 
give explaination
a.) there is only one integral 'a' for which f(x) is non-negative for all x belongs to R
b.) for a< 0, the number lies between the zeroes of the polynomial
c.) f(x) =0 has two distinct solutions in (0,1) for 'a' belonging to (1/7,4/7)
d.) the minimum value of f(x) for minimum value of 'a' for which f(x) is non-negative for all x in R is zero

Dear student,
fx  = 4x2-8ax+a D = b2-4ac  = 64a2-16aD = 16a4a-1for two distinct solution D>0then a4a-1>0a-,014,for two exact solution  and non negative fx  for all x in RD=0then a4a-1=0a0,14for no solution and positve fx  for all x in RD<0then a4a-1<0a0,14answer d  and I do not understand option b
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