Question

If $2a+3b+6c=0$, then atleast one root of the equation $a{x}^2+bx+c=0$ lies in the interval

• A. $(0,1)$
• B. $(1,2)$
• C. $(2,3)$
• D. None of these

Answer 1

Answer: (A)

Consider the function $f(x)=\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx+d$
We have, $f(0)=d$
and $f(1)=\frac{a}{3}+\frac{b}{2}+c+d=\frac{2a+3b+6c}{6}+d$
$=\frac{0}{6}+d[\because 2a+3b+6c=0]$
Therefore, $0$ and $1$ are roots of the polynomial $f(x)$.
Hence, according to Rolle's theorem, there exists atleast one root of the polynomial $f^{\prime }(x)=a{x}^2+bx+c$ lying between $0$ and $1.$