Question

If $2a+3b+6c=0$ then equation $a{x}^2+bx+c=0$ has

• A. At least one solution in [1, 2]
• B. Two solutions in [0, 1]
• C. At least one solution in [0, 1]
• D. At least one solution in [-1, 0]

Answer 1

The correct option is C At least one solution in [0, 1]

Apply Rollel's theorem for $\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx+d$ in [0, 1]

$f\left(0\right)=6d$
$f\left(1\right)=6d$

$\therefore f\left(x\right)=0$ has atleast one solution in [0, 1]