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Let parabolas y=x(c-x) and y=x^(2)+ax+b ...

Let parabolas `y=x(c-x)` and `y=x^(2)+ax+b` touch each other at the point `(1,0)` then `b-c=`

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The correct Answer is:
1

`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`
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