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`

Options

A. 1
B. -1
C. 0
D. none of these

Answer

C

Explanation

We have,
` C_(1): y=x^(2)+ax+b " "...(i)`
` C_(2): y=cx-x^(2) " "...(ii)`
` therefore ((dy)/(dx))_(C_(1))=2x+a " and " ((dy)/(dx))_(C_(2))=c-2x `
At point (1,0), we have
`((dy)/(dx))_(C_(1))=2+a " and " ((dy)/(dx))_(C_(2))=c-2. `
If the two parabolas touch each other at (1,0), then
`2+a=c-2 rArr a-c+4=0 " "...(iii)`
Since (1,0) lies on the two parabolas. Therefore,
`1+a+b=0 " and " c-1=0 `
` rArr c=1 " and " a+b=-1 " "...(iv)`
Solving (iii) and (iv), we get
` a=-3, b=2 " and " c=1 rArr a+b+c=0 `