If x2−2x+a=0, where [ ·] represents greatest integer  function,has no solution then,

x2−2x+a=0⇒ x2−2x+a∈[0,1)For no solution case (i):  x2−2x+a<0,∀x∈Rwhich is not possible ,since x2 coefficient is 1 is positive and expression is negative. case (ii): x2−2x+a≥1,∀x∈R⇒x2−2x+a−1≥0,∀x∈R⇒ Disc. ≤0⇒4−4(a−1)≤0⇒2≤a