If `2a+3b+6c=0,` then prove that at least one root of the equation `a x^2+b x+c=0` lies in the interval (0,1).

Options

A. `(0,1)`
B. `(1,2)`
C. `(2,3)`
D. `(1,3)`

Answer

A

Explanation

Given ,`2a+3b+6c=0`
Let `f'(x)=ax^2+bx+c`
`f(x)=(ax^3)/(3)+(bx^2)/(2)+cx+d`
`rArr f(x)=(2ax^2+3bx^2+6cx+6d)/(6)`
Now , `f(1) =(2a+3b+6c+6d)/(6)=0`
and `f(0)=0`
`therefore f(0)=f(1)`
Therefore , 0 and 1 are the roots of the polynomial `f(x)`, So , by Rolle's