Question

If 4a + 2b + c = 0, then the equation 3ax² + 2bx + c = 0 has at least one real root lying in the interval
(a) (0, 1)
(b) (1, 2)
(c) (0, 2)
(d) none of these

Answer 1

(c) (0, 2)

$$\begin{gathered} \text{Let} \\ f\left(x\right)=a{x}^3+b{x}^2+cx+d.....\left(1\right) \\ f\left(0\right)=d \\ f\left(2\right)=8a+4b+2c+d \\ =2\left(4a+2b+c\right)+d \\ =d\left(\because \left(4a+2b+c\right)=0\right) \end{gathered}$$

f is continuous in the closed interval [0, 2] and f is derivable in the open interval (0, 2).

Also, f(0) = f(2)

By Rolle's Theorem,
$$\begin{gathered} f\text{'}\left(\alpha \right)=0\mathrm{for}0<\alpha <2 \\ \mathrm{Now},f\text{'}\left(x\right)=3a{x}^2+2bx+c \\ \Rightarrow f\text{'}\left(\alpha \right)=3a{\alpha }^2+2b\alpha +c=0 \\ \mathrm{Equation}\left(1\right)\mathrm{has}\mathrm{atleast}\mathrm{one}\mathrm{root}\mathrm{in}\mathrm{the}\mathrm{interval}\left(0,2\right). \\ \mathrm{Thus},f\text{'}\left(\mathrm{x}\right)\mathrm{must}\mathrm{have}\mathrm{root}\mathrm{in}\mathrm{the}\mathrm{interval}\left(0,2\right). \end{gathered}$$