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. at least one root
B. at most one root
C. no root
D. none of these

Answer

A

Explanation

Let `f(x)=(ax^(3))/(3)+(b)/(2)x^(2)+cx` be a polynomial. Them, `f(0)=0` and , `f(1)=(a)/(3)+(b)/(2)+c=(2a+3b+6c)/(6)=0 [ :. 2 ab+ 3b +6c=0("Given")]`
This shows that 0 and 1 are roots of f(x)=0
Hence, by the algebraic interpretation of Rolle's Therem `f'(x)=0.i.e., ax^(2)+bx+c=0` has a real root between 0 and 1.