We have f(x)=x2−bx+c and b+c=35 According to the condition b+c=35 Let assume that 2 and 11 be the prime root of above equation, then (x−2)(x−11)=0 ⇒x2−2x−11x+22=0 ⇒x2−13x+22=0 Which satisfies the condition b+c=35. On comparing both equations, we get b = 13 and c =22 Global minimum f'(x)=0 ⇒2x−13=0⇒x=