Theory of Equations 5 Question 6
6. If $a+b+c=0$, then the quadratic equation $3 a x^{2}+2 b x+c=0$ has
$(1983,1 M)$
(a) at least one root in $(0,1)$
(b) one root in $(2,3)$ and the other in $(-2,-1)$
(c) imaginary roots
(d) None of the above
Show Answer
Answer:
Correct Answer: 6. (a)
Solution:
- Let
$ f(x)=a x^{3}+b x^{2}+c x+d $
$\therefore$
$ f(0)=d \text { and } f(1)=a+b+c+d=d $
$[\because a+b+c=0]$
$ \therefore \quad f(0)=f(1) $
$f$ is continuous in the closed interval $[0,1]$ and $f$ is derivable in the open interval $(0,1)$.
Also,
$ f(0)=f(1) . $
$\therefore$ By Rolle’s theorem, $f^{\prime}(\alpha)=0$ for $0<\alpha<1$
Now,
$ \begin{aligned} & f^{\prime}(x)=3 a x^{2}+2 b x+c \\ & f^{\prime}(\alpha)=3 a \alpha^{2}+2 b \alpha+c=0 \end{aligned} $
$ \Rightarrow $
$\therefore$ Eq. (i) has exist atleast one root in the interval $(0,1)$. Thus, $f^{\prime}(x)$ must have root in the interval $(0,1)$ or $3 a x^{2}+2 b x+c=0$ has root $\in(0,1)$.
