Question

The values of $a$ for which $f\left(x\right)=\frac{a^2{x}^3}{3}+\frac{3a{x}^2}{2}+2x+1$ is strictly decreasing at $x=1$

• A. $a\in (-2,-1)$
• B. $a\in (-1,0)$
• C. $a\in (1,2)$
• D. $a\in (-2,1)$

Answer 1

The correct option is A $a\in (-2,-1)$

Given function is $f\left(x\right)=\frac{a^2{x}^3}{3}+\frac{3a{x}^2}{2}+2x+1$

$f^{\prime }\left(x\right)=a^2{x}^2+3ax+2$

If $f\left(x\right)$ is a strictly decreasing function at $x=1,$ then $f^{\prime }\left(1\right)>0$

$\Rightarrow f^{\prime }\left(1\right)=a^2+3a+2>0$

$\Rightarrow (a+2)(a+1)>0$

$\Rightarrow a>-2$ and $a<-1$

$a$ belongs to interval $(-2,-1)$