Question

If $a+b+c=0$, then the equation $3a{x}^2+2bx+c=0$ has

• A. imaginary roots
• B. real and equal roots lying in $(0,1)$
• C. real and different roots of which atleast one lies in $(0,1)$
• D. rational roots

Answer 1

The correct option is C real and different roots of which atleast one lies in $(0,1)$

Given $a+b+c=0$ and

Let $f^{{}^{\prime }}\left(x\right)=3a{x}^2+2bx+c$

$⟹f\left(x\right)=a{x}^3+b{x}^2+cx+d$ (on integration)

Rolle's theorem states that if $f\left(x\right)$ be continuous on $[a,b]$, differentiable on $(a,b)$ and $f\left(a\right)=f\left(b\right)$ then there exists some $c$ between $a$ and $b$ such that $f^{{}^{\prime }}\left(c\right)=0$

from the options let $(a,b)=(0,1)$

$f\left(0\right)=d$ and $f\left(1\right)=a+b+c+d=d$ (since, $a+b+c=0$)

Therefore, $f\left(0\right)=f\left(1\right)$. Hence, there exists some $c$ between $0$ and $1$ such that $f^{{}^{\prime }}\left(c\right)=0$

Therefore, the equation has atleast one root lying on $(0,1)$