Question

The function $f\left(x\right)=a{x}^3-3x^2-12x+1$ is monotonically decreasing in the open interval $(-1,2)$. Then one value of $a$ is

• A. 2
• B. 12
• C. 1
• D. none of these

Answer 1

Answer: (A)

2

$f^{\prime }\left(x\right)=3a{x}^2-6x-12$
Now
$f^{\prime }\left(x\right)<0$ for $xϵ(-1,2)$
Or
$(x+1)(x-2)<0$
Or
$x^2-x-2<0$
Or
$6x^2-6x-12<0$
Comparing with $f^{\prime }\left(x\right)$ , we get
$3a=6$
Or
$a=2$.