If roots of x2−(a−3)x+a=0 are such that at least one of them is greater than 2, then

x2−(a−3)x+a=0⇒ D=(a−3)2−4a=a2−10a+9=(a−1)(a−9)Case I:Both the roots are greater than 2.D≥0,f(2)>0,−B2A>2⇒(a−1)(a−9)≥0;4−(a−3)2+a>0;a−32>2⇒a∈(−∞,1]∪[9,∞);a<10;a>7⇒a∈[9,10)                                                        (1)Case II:One root is greater than 2 and the other root is less than or. equal to 2. Hence,f(2)≤0⇒4−(a−3)2+a≤0⇒ a≥10                                              (2)From (1) and (2),a∈[9,10)∪[10,∞)⇒ a∈[9,∞)