Let a function f:RR be defined as f(x)=cases x-e^x, & x0+[-x], & 0

Question

Let a function $f:\mathbb{R}	o\mathbb{R}$ be defined as $f(x)=\begin{cases}\sin x-e^x, & x\le0\a+[-x], & 0<x<1\2x-b, & x\ge1\end{cases}$, where $[x]$ is the greatest integer $\le x$. If $f$ is continuous on $\mathbb{R}$, then $(a+b)$ is

Accepted Answer

At $x=0$: $\sin0-e^0=-1$; RHL as $x	o0^+$, $[-x]	o-1$ (since $-x$ is a small negative number, $[-x]=-1$), giving $a-1$. Continuity: $a-1=-1\Rightarrow a=0$.

At $x=1$: LHL $=\displaystyle\lim_{h	o0}a+[-1+h]=a-1=-1$. RHL $=2-b$. Continuity: $2-b=-1\Rightarrow b=3$.

$a+b=0+3=3$.

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