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Let $f(x)=\left\{\begin{array}{cc}-2 \sin x, & \text { if } x \leq-\frac{\pi}{2} \\ A \sin x+B, & \text { if }-\frac{\pi}{2} < x < \frac{\pi}{2} \text { . Then, } \\ \cos x, & \text { if } x \geq \frac{\pi}{2}\end{array}\right.$
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Verified Answer
The correct answer is:
$f$ is continuous for all $A=-1$ and $B=1$
At $\quad x=-\frac{\pi}{2}$
$$
\begin{array}{l}
\text { LHL }=-2 \\
\text { RHL }=-A+B \\
\text { For continuity, } \mathrm{LHL}=\mathrm{RHL}=f(-\pi / 2)
\end{array}
$$
$\Rightarrow \quad-A+B=2$
At $x=\pi / 2$
$$
\begin{aligned}
\mathrm{LHL} &=A+B \\
\mathrm{RHL} &=0 \\
\text { For continuity, } \mathrm{LHL} &=\mathrm{RHL}=f(\pi / 2)
\end{aligned}
$$
$\Rightarrow \quad A+B=0$
On solving Eqs. (i) and (ii), we get
$$
A=-1 \text { and } B=1
$$
$$
\begin{array}{l}
\text { LHL }=-2 \\
\text { RHL }=-A+B \\
\text { For continuity, } \mathrm{LHL}=\mathrm{RHL}=f(-\pi / 2)
\end{array}
$$
$\Rightarrow \quad-A+B=2$
At $x=\pi / 2$
$$
\begin{aligned}
\mathrm{LHL} &=A+B \\
\mathrm{RHL} &=0 \\
\text { For continuity, } \mathrm{LHL} &=\mathrm{RHL}=f(\pi / 2)
\end{aligned}
$$
$\Rightarrow \quad A+B=0$
On solving Eqs. (i) and (ii), we get
$$
A=-1 \text { and } B=1
$$
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