Question

If $f\left(x\right)=\left(\frac{a^2-1}{3}\right)x^3+(a-1)x^2+2x+1$ is monotonic increasing for every $xϵR$, then let the range of values of $a$ be $aϵ(-\infty ,k]\cup (m,\infty )$. Find $k+6m$?

Answer 1

$f\left(x\right)=\left(\frac{a^2-1}{3}\right)x^3+(a-1)x^2+2x+1$
For $a=1$, $f\left(x\right)=2x+1$
$\Rightarrow$ $f$ is monotonic increasing .
If $a\ne 1$, $f^{\prime }\left(x\right)=(a^2-1)x^2+2(a-1)x+2$
$f^{\prime }\left(x\right)\ge 0$ ($\because$ '$f$' is monotonic increasing)
$\Rightarrow D\le 0\&a^2-1>0$
$4(a-1{)}^2-8(a-1\left)\right(a+1)\le 0$
$\Rightarrow (a-1)(a-1-2a-2)\le 0$
$\Rightarrow (a-1)(-a-3)\le 0$
$\Rightarrow (a-1)(a+3)\ge 0$
$\Rightarrow aϵ(-\infty ,-3]\cup [1,\infty )$ ....(1)
And $a^2-1>0$
$\Rightarrow (a-1)(a+1)>0$
$\Rightarrow a\in (-\infty ,-1)\cup (1,\infty )$ ....(2)
From (1) and (2), we have
$a\in (-\infty ,-3]\cup (1,\infty )$