If `27a+9b+3c+d=0` then the equation `4ax^(3)+3bx^(2)+2cx+d` has at leat one real root lying between

Options

A. 0 and 1
B. 1 and 3
C. 0 and 3
D. none of these

Answer

C

Explanation

Consider the polynomial f(x) given by
`f(x) =ax^(4)+bx^(3)+cx^(2)+dx`
`rArr g(x)=4ax^(3)+3bx^(2)+2d+d`
We have,
`f(0)=0`
and `f(3)=81a+27b+9x+3d=3(27a+3c+9b+d)=0`
Therefore, 0 and 3 are roots of f(x)=0
Consequently, by Rolle's theorem `f'(x)=0` i.e., `4ax^(3)+3bx^(2)+2cx+d=0` has at least one real root between 0 and 3