Let f:RR be defined as [f(x)=cases[e^x], & x

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be defined as
$$f(x)=\begin{cases}[e^x], & x<0\ ae^x+[x-1], & 0\le x<1\ b+[\sin(\pi x)], & 1\le x<2\ [e^{-x}]-c, & x\ge2\end{cases}$$
where $a,b,c\in\mathbb{R}$ and $[t]$ denotes the greatest integer $\le t$. Which of the following statements is true?

Accepted Answer

Simplifying each piece: $f(x)=\begin{cases}0, & x<0\ ae^x-1, & 0\le x<1\ b, & x=1\ b-1, & 1<x<2\ -c, & x\ge2\end{cases}$

Continuity at $x=0$: requires $a-1=0\implies a=1$.

Continuity at $x=1$: requires $ae^{1}-1=b=b-1$, which is impossible (no value of $b$ satisfies $b=b-1$) --- so $f$ can never be continuous at $x=1$ regardless of $a,b,c$. Hence option A and the "at most one discontinuity from all points" claims must be checked around $x=1$ and $x=2$ only.

Continuity at $x=2$: requires $b-1=-c\implies b+c=1$.

If $a=1$ and $b+c=1$, then $f$ is discontinuous at exactly one point (namely $x=1$, since the jump there is unavoidable while $x=0$ and $x=2$ are both continuous). In that case, $a+b+c=1+1=2
e1$.

So if $f$ is discontinuous at exactly one point, then $a+b+c
e1$.

Solutions · 1