Question

Let $f\left(x\right)=\{\begin{array}{cc}-2\sin x,& \text{if}x\le -\frac{\pi }{2}\\ A\sin x+B,& \text{if}-\frac{\pi }{2}<x<\frac{\pi }{2}\\ \cos x,& \text{if}x\ge \frac{\pi }{2}\end{array}$
Then

• A. $f$ is discontinuous for all $A$ and $B$
• B. $f$ is continuous for all $A=–1$ and $B=1$
• C. $f$ is continuous for all $A=1$ and $B=-1$
• D. $f$ is continuous for all real values of $A,B$

Answer 1

The correct option is B $f$ is continuous for all $A=–1$ and $B=1$

For continuity at $x=-\frac{\pi }{2}$
$L.H.L\underset{x\to (-\pi /2{)}^{-}}{lim}-\sin x=2$
$R.H.L\underset{x\to (-\pi /2{)}^{+}}{lim}A\sin x+B=-A+B$
$\Rightarrow -A+B=2\cdots \left(1\right)$

For continuity at $x=\frac{\pi }{2}$
$L.H.L\Rightarrow \underset{x\to (\pi /2{)}^{-}}{lim}A\sin x+B=A+B$
$R.H.L\underset{x\to (\pi /2{)}^{+}}{lim}\cos x=0$
$\Rightarrow A+B=0\cdots \left(2\right)$

From eqn(1) and (2),
$A=–1,B=1$