Question

$f\left(x\right)=\{\begin{array}{c}4x-x^3+\ln (a^2-3a+3),0\le x<3\\ x-18,x\ge 3\end{array}$
Complete set of values of $a$ such that $f\left(x\right)$ as a local minima at $x=3$ is

• A. $[-1,2]$
• B. $(-\infty ,1)\cup (2,\infty )$
• C. $[1,2]$
• D. $(-\infty ,-1)\cup (2,\infty )$

Answer 1

Answer: (B)

$(-\infty ,1)\cup (2,\infty )$

Clearly, $f\left(x\right)$ in decreasing just before $x=3$ and increasing after $x=3$. For $x=3$ to be the point of local minima,
$f\left(3\right)\le f\left(3^{-}\right)$.
$\Rightarrow -15\le 12-27+\ln (a^2-3a+3)$
$\Rightarrow a^2-3a+3\ge 1$
$\Rightarrow a\in (-\infty ,1)\cup (2,\infty )$.