Let the parabolas `y=x(c-x)a n dy=x^2+a x+b` touch each other at the point (1,0). Then `a+b+c=0` `a+b=2` `b-c=1` (d) `a+c=-2`

Answer

1

Explanation

`y=x(c-x) y=x^(2)+ax+b`
`(dy)/(dx):|_("(1,0)")` `=c-2m_(1)(dy)/(dx):|_("(1,0)")=2x+a=2+a=m_(2)`
Curve are touching at `(1,0)` so
`m_(1)=m_(2)`
`2+a=c-2`………..i
Also `(1,0)` lies both the curve
so `c=1`……….ii
and `a+b=-1`……….iii
by i, ii, iii `b=2,c=1,a=-3`