Question

The values of A and B so that $f\left(x\right)=\{\begin{array}{ccc}-2\sin x& \text{if}& x\le -\pi /2\\ A\sin x+B& \text{if}& -\pi /2<x<\pi /2\\ \cos x,& \text{if}& x\ge \pi /2\end{array}$

is continuous everywhere are

• A. $A=0,B=1$
• B. $A=1,B=1$
• C. $A=-1,B=1$
• D. $A=-1,B=0$

Answer 1

The correct option is C $A=-1,B=1$

$f\left(x\right)=\{\begin{array}{ccc}-2\sin x& \text{if}& x\le -\pi /2\\ A\sin x+B& \text{if}& -\pi /2<x<\pi /2\\ \cos x,& \text{if}& x\ge \pi /2\end{array}$
Since $\sin x$ and $\cos x$ are continuous functions, $f\left(x\right)$ is continuous except possibily at $x=-\pi /2$ and $x=\pi /2$. For f to be continuous at $x=-\pi /2$, we must have
$f(-\pi /2)=\underset{x\to \pi /2^{+}}{lim}f\left(x\right)=\underset{x\to \pi /2^{+}}{lim}(A\sin x+B)=-A+B$
$\therefore -A+B=f(-\pi /2)=-2\sin (-\pi /2)=2$
For f to be continuous at $x=\pi /2$, we must have
$0=\cos \pi /2=f\left(\pi /2\right)=\underset{x\to \pi /2^{-}}{lim}f\left(x\right)$
$=\underset{x\to \pi /2}{lim}(A\sin x+B)=A+B$
Hence $B=1$ and $A=-1$