Question
If $$2a + 3b + 6c = 0,$$ then at least one root of the equation $$a{x^2} + bx + c = 0$$ lies in the interval
A.
(1, 3)
B.
(1, 2)
C.
(2, 3)
D.
(0, 1)
Answer :
(0, 1)
Solution :
Let us define a function
$$f\left( x \right) = \frac{{a{x^3}}}{3} + \frac{{b{x^2}}}{2} + cx$$
Being polynomial, it is continuous and differentiable, also,
$$\eqalign{
& f\left( 0 \right) = 0\,{\text{and}}\,f\left( 1 \right) = \frac{a}{3} + \frac{b}{2} + c \cr
& \Rightarrow f\left( 1 \right) = \frac{{2a + 3b + 6c}}{6} = 0\,\left( {{\text{given}}} \right) \cr
& \therefore f\left( 0 \right) = f\left( 1 \right) \cr
& \therefore f\left( x \right)\,{\text{satisfies all conditions of Rolle}}\,{\text{theorem}}\,{\text{therefore}}\,f'\left( x \right) = 0\,{\text{has a root in}}\left( {{\text{0,1}}} \right) \cr
& {\text{i}}{\text{.e}}{\text{.}}\,a{x^2} + bx + c = 0\,{\text{has at lease one root in }}\left( {{\text{0,1}}} \right) \cr} $$