Question

If $a,b,c,d$ are real numbers such that $\frac{a+2c}{b+3d}+\frac{4}{3}=0$, then the equation $a{x}^3+bx{x}^2+cx+d=0$ has

• A. at least one root in $(-1,0)$
• B. at least one root in $(0,1)$
• C. at least two roots in $(-1,1)$
• D. no root in $(-1,1)$

Answer 1

Answer: (B)

$\frac{a+2c}{b+3d}+\frac{4}{3}=0\Rightarrow 3a+4b+6c+12d=0$
$\underset{0}{\overset{1}{\int }}\left(a{x}^3+b{x}^2+cx+d\right)dx=\frac{1}{12}(3a+4b+6c+12d)=0.$
Hence, $a{x}^3+b{x}^2+cx+d=0$ has at least one root in $(0,1)$.