Question:
If $4 a+2 b+c=0$, then the equation $3 a x^{2}+2 b x+c=0$ has at least one real root Iying in the interval
(a) (0, 1)
(b) (1, 2)
(c) (0, 2)
(d) none of these
Solution:
(c) (0, 2)
Let
$f(x)=a x^{3}+b x^{2}+c x+d$ ......(1)
$f(0)=d$
$f(2)=8 a+4 b+2 c+d$
$=2(4 a+2 b+c)+d$
$=d$ $(\because(4 a+2 b+c)=0)$
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,
$f^{\prime}(\alpha)=0 \quad$ for $0<\alpha<2$
Now, $f^{\prime}(x)=3 a x^{2}+2 b x+c$
$\Rightarrow f^{\prime}(\alpha)=3 a \alpha^{2}+2 b \alpha+c=0$
Equation (1) has atleast one root in the interval $(0,2)$.
Thus, $f^{\prime}(\mathrm{x})$ must have root in the interval $(0,2)$.