In the given figure, vertices of △ABC lie on y=f(x)=ax2+bx+c. The △ABC is right-angled isosceles triangle whose hypotenuse AC=42 units.y = f (x) is given byMinimum value of y = f(x) isNumber of integral values of k for which one root of f(x) = 0 is more than k and other less than k

∵AC    =42∴     AB=BC=422=4 units     OB=42−(22)2=22∴     A(−22,0),C(22,0),B(0,−22)Since y = ax2 + bx + c passes through A, B, and C, we gety=x222−22Minimum value of y=x2/(22)−22 is −22 at x=0.f(x)=0⇒ x2/(22)−22=0 or x=±22Therefore, number of integral values of k for which k lies in (−22, 22) is 5.