The set of values of a for which each one of the roots of x2−4ax+2a2−3a+5=0 is greater than 2, is

Let f(x)=x2−4ax+2a2−3a+5.Clearly, y = f (x) represents a parabola opening upward. So, its both roots will be greater than 2, if(i) Discriminant ≥ 0 (ii) x-coordinate of vertex > 2 (iii) 2 lies outside the roots i.e. f (2) > 0Now,(i) Discriminant ≥ 0 ⇒ 16a2−42a2−3a+5≥0⇒ 2a2+3a−5≥0⇒ (2a+5)(a−1)≥0⇒a≤−52 or, a≥1                         …(i)(ii)  x-coordinate of vertex > 2⇒ 2a>2⇒a>1                       …(ii)(iii) f (2) > 0⇒ 2a2−11a+9>0⇒ (a−1)(2a−9)>0⇒a<1 or, a>92         …(iii)From (i), (ii) and (iii), we get a>92.