Question

Let $f\left(x\right)=\{\begin{array}{cc}\frac{b^3+b-2b^2-2}{b^2+5b+6}-x^2;& 0\le x<1\\ 3x-4;& 1\le x\le 3\end{array}$
where $b\in R$. If $f\left(x\right)$ has minimum value at $x=1,$ then the least integral value of $b$ is

Answer 1

$f^{\prime }\left(x\right)=\{\begin{array}{cc}-2x;& 0\le x<1\\ 3;& 1\le x\le 3\end{array}$

$f$ is not differentiable at $x=1$.
As $f^{\prime }\left(x\right)<0$ for $x\in (0,1)$ and $f^{\prime }\left(x\right)>0$ for $x\in (1,3)$,
$\Rightarrow f\left(x\right)$ is strictly decreasing on $(0,1)$ and strictly increasing on $(1,3)$.
$\Rightarrow f\left(x\right)\ge f\left(1\right)$ for $x\in [1,3]$

For $f\left(x\right)$ to have the smallest value at $x=1$ for $x\in [0,3]$, we must have $\underset{x\to 1^{-}}{lim}f\left(x\right)\ge f\left(1\right)$
$\Rightarrow \underset{x\to 1^{-}}{lim}-x^2+\frac{b^3+b-2b^2-2}{b^2+5b+6}\ge -1$

$\Rightarrow -1+\frac{b^3+b-2b^2-2}{b^2+5b+6}\ge -1$

$\Rightarrow \frac{(b-2)(b^2+1)}{(b+2)(b+3)}\ge 0$
$\Rightarrow \frac{(b-2)}{(b+2)(b+3)}\ge 0$ as $b^2+1>0$
$\Rightarrow b\in (-3,-2)\cup [2,\infty )$
Therefore, least integral value of $b$ is $2.$