Question

lf $a+b+c=0$, then the equation $3a{x}^2+2bx+c=0$ has at least one root in

• A. $(1,2)$
• B. $(0,1)$
• C. $(-1,1)$
• D. $(2,3)$

Answer 1

Answer: (B), (C)

The correct options are
B $(0,1)$
C $(-1,1)$

Let $f\left(x\right)=a{x}^3+b{x}^2+cx$.............(1)
Substitute $x=0$ in equation (1), we get
$f\left(0\right)=0$
Now, substitute $x=1$ in equation (1), we get
$f\left(1\right)=a+b+c$
Since, it is given that $a+b+c=0$
i.e. $f\left(1\right)=0$
Since, $f\left(x\right)$ is a cubic expression. So it is continuous and differentiable for $\forall x\in R.$
Now, according to the Rolle's Theorem, there exists at least one $x$ between $0$ and $1$ for which $f^{{}^{\prime }}\left(x\right)=0$,
$f^{{}^{\prime }}\left(x\right)=3a{x}^2+2bx+c$
Hence, $3a{x}^2+2bx+c=0$ has at least one root in the interval $(0,1)$.
Since, $3a{x}^2+2bx+c=0$ posses at least one root in the interval $(0,1)$. So, automatically one root will be in the interval $(-1,1)$.
Hence, options (B) and (C) are correct.