let the parabolas y=x^(2)+ax+band y=x(c-x)touch each other at the point (1,0).Then

a)a= -3 b)b= 1 c)c=2 d)b+c=3

from the first equation, a+b+1=0, from the second equation, c-1=0,c=1,

the quadratic equation in 'x': x^2+ax+b=x(c-x), has only one solution, thus discriminant is zero, (a-c)^2 =8b,c=1,

(a-1)^2 =8b, now combining this with a+b+1=0, and solving we get a=-3,b=2,c=1,b+c=3

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y=x^2+ax+b .......1

y=cx-x^2..........2

if both parabola touch each other then they will have a common tangent , we can find dy/dx/solpe  of tangent  from any of the curve 

dy/dx for curve 1=2x +a at 1,0

                        =a+2....eq3

dy/dx for curve 2=c-2x at 1,0

                        =c-2.....eq4

 since 1,0 is point of contact so it will satisfy both the curves

    a+b+2=0....eq5

    c-1=0......eq6

  after solving we get  a=-3,b=1