Question

If $a,b,c$ are real numbers such that $a+b+c=0$, then the quadratic equation $3a{x}^2+2bx+c=0$ has

• A. At least one root in $[0,1]$
• B. At least one root in $[1,2]$
• C. At least one root in $[-1,0]$
• D. $Noneofthese$

Answer 1

The correct option is A At least one root in $[0,1]$

Given,

$a,b,c$ are real numbers such that $a+b+c=0$

We will moved a head using the Rolle's theorem which states that if $f\left(x\right)$ is continuous on $[a,b]$ and differentiable on $(a,b)$ such that $f\left(a\right)=f\left(b\right)$. Then there is some $c$ with $c\in (a,b)$ such that $f^1\left(c\right)=0$

Now,

Let $f^1\left(x\right)={3x}^{2}a+2bx+c$

$\Rightarrow f\left(x\right)={ax}^3+{bx}^2+cx+k$ ; $k$ is an integration constant

$\therefore$ $f(x=0)=0+k$

$f(x=0)=a+b+c+k=0+k$ ($\because$ $a+b+c=0$ is given)

$\therefore$ $f(x=0)=f(x=1)$

Using rolle's theorem, we have $f\left(0\right)=f\left(1\right)$ then three exists at least root ${}^{\prime }{c}^{\prime }\in (0,1))$ such that $f^1\left(x\right)={3ax}^2+2bx+c=0$

$\therefore$ We can say that ${3ax}^2+2bx+c$ has at least one root between $(0,1)$

Answer : Option A.