Let f:RR be defined by a piecewise sine-type rule (matched to x near x=1 and x=-1).

Question

Let $f:\mathbb{R}	o\mathbb{R}$ be defined by a piecewise sine-type rule (matched to $\sin\pi x$ near $x=1$ and $x=-1$). If $f(x)$ is continuous on $\mathbb{R}$, then $a+b$ equals

Accepted Answer

Continuity at $x=1$ gives $|a+1+b|=\sin\pi=0\Rightarrow a+b=-1$ ... (i)

Continuity at $x=-1$ gives $2\sin\left(\frac{-\pi}{2}\right)(-1)=a-1+b\Rightarrow2=|a+b-1|$ ... (ii), which is satisfied by (i).

Hence $a+b=-1$.

Solutions · 1