If the line ` y=2x` touches the curve `y=ax^(2)+bx+c ` at the point where x=1 and the curve passes through the point (-1,0), then

Options

A. `a=(1)/(2), b=1, c=(1)/(2) `
B. `a=1, b=(1)/(2), c=(1)/(2) `
C. `a=(1)/(2), c=(1)/(2), b=1 `
D. none of these

Answer

A

Explanation

We have,
` y= ax^(2)+bx+c " "...(i) `
` rArr (dy)/(dx)=2ax+b rArr ((dy)/(dx))_(x=1)= 2a+b `
Since, the line `y=2x ` touches (i) at the point where x=1.
` therefore ("Slope of the tangent at x=1 " )=(" Slope of the line " y=2x) `
` rArr 2a+b=2 " "...(ii)`
Putting x=1 in y=2x, we get y=2.
Clearly, (1,2) and (-1,0) lie on (i). Therefore,
` 2=a+b+c " "...(iii)`
and, ` 0=a-b+c " "...(iv) `
Solving (i), (ii) and (iii), we get `a=(1)/(2), b=1 " and " c=(1)/(2) `