(a) 5+px-x2=9-(x+q)2,
5+px-x2=9-x2-2qx-q2,
5+px=9-2qx-q2,
q2+2qx+px-4=0.
If this is to be true for all x then the x terms must cancel out:
2q+p=0, p=-2q, and q2-4=(q-2)(q+2)=0.
Since q>0 then q=2 and p=-4.
(b) f(x)=9-(x+2)2.
When x=-2 f(x)=9 (maximum).
(c) 4x>x2, x2-4x<0, x(x-4)<0.
To satisfy the inequality, x and x-4 must be of opposite signs. If x<0 then x-4 is also negative;
so if x>0, x-4<0, that is, x<4, therefore 0<x<4 (x is in the interval (0,4)).