Question

Let $f\left(x\right)=\{\begin{array}{cc}-x^3+\frac{b^3-b^2+b-1}{b^2+3b+2}& ,0\le x<1\\ 2x-3& ,1\le x\le 3\end{array}$.

The all possible real values of b such that f(x) has the smallest value at $x=1$ are

• A. $b\in (-2,-1)\cup (1,\infty )$
• B. $b\in [-2,-1)\cup (1,\infty )$
• C. $b\in [-2,-1)\cup (0,\infty )$
• D. $b\in [-2,-1]\cup [0,\infty )$

Answer 1

Answer: (A)

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

Given $f\left(x\right)=\{\begin{array}{c}{-x}^3+\frac{(b^3-b^2+b-1)}{(b^2+3b+2)},0\le x\le 1\\ 2x-3,1\le x\le 3\end{array}$
is, $f\left(x\right)$ is decreasing on $[0,1]$and increasing on $[1,3]$
Here, $f\left(1\right)=-1$ is the smallest value at $x=1$
$\therefore$ Its smallest value as.
$\underset{x\to 1-}{lim}f\left(x\right)=\underset{x\to 1-}{lim}\left({-x}^3\right)+\frac{(b^3-b^2+b-1)}{b^2+3b+2}$
$\frac{b^3+b^2+b-1}{b^2+3b+2}\ge 0$
$\Rightarrow \frac{(b^2+1)(b-1)}{(b+1)(b+2)}\ge 0$
$\therefore b\in (-2,-1)\cup [1,\infty )$