IIT-JEE 1993Question 1
Let $f(x)=\begin{cases}-x^{3}+\dfrac{b^{3}-b^{2}+b-1}{b^{2}+3b+2}, & 0\le x<1\\ 2x-3, & 1\le x\le3\end{cases}$ Find all possible real values of $b$ such that $f(x)$ has the smallest value at $x=1$.
Functions
Answer$b\in(-2,-1)\cup[1,\infty)$
Solution
On $[1,3]$, $f=2x-3\ge-1=f(1)$. On $[0,1)$, $f=-x^3+c>c-1$ with $c=\frac{(b-1)(b^2+1)}{(b+1)(b+2)}$, so $f\ge-1$ iff $c\ge0$, i.e. $\frac{b-1}{(b+1)(b+2)}\ge0$, giving $b\in(-2,-1)\cup[1,\infty)$.