Question

For all real values of $a_0,a_1,a_2,a_3$ satisfying $a_0+\frac{a_1}{2}+\frac{a_2}{3}+\frac{a_3}{4}=0$, the equation $a_0+a_{1}x+a_2{x}^2+a_3{x}^3=0$ has a real root in the interval

• A. $[0,1]$
• B. $[-1,0]$
• C. $[1,2]$
• D. $[-2,-1]$

Answer 1

The correct option is A $[0,1]$

Let $f\left(x\right)=\frac{a_3{x}^4}{4}+\frac{a_2{x}^3}{3}+\frac{a_1{x}^2}{2}+a_{0}x$
$f\left(0\right)=0$
$f\left(1\right)=a_0+\frac{a_1}{2}+\frac{a_2}{3}+\frac{a_3}{4}=0$

So, $f\left(0\right)=f\left(1\right)=0$
So, by Rolle's theorem,
$f^{\prime }\left(x\right)=0$ has real root in $[0,1]$.