Let f(x) = begin{cases} -2 sin x, & text{if } | vedclass

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(B) For $f(x)$ to be continuous,it must be continuous at $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$.At $x = -\frac{\pi}{2}$:$LHL = \lim_{x 	o -\fra

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$f$ is continuous for $A = -1$ and $B = 1$

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(B) For $f(x)$ to be continuous,it must be continuous at $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$.At $x = -\frac{\pi}{2}$:$LHL = \lim_{x 	o -\frac{\pi}{2}^-} (-2 \sin x) = -2 \sin(-\frac{\pi}{2}) = -2(-1) = 2$.$RHL = \lim_{x 	o -\frac{\pi}{2}^+} (A \sin x + B) = A \sin(-\frac{\pi}{2}) + B = -A + B$.For continuity,$LHL = RHL \implies -A + B = 2$ (Equation $i$).At $x = \frac{\pi}{2}$:$LHL = \lim_{x 	o \frac{\pi}{2}^-} (A \sin x + B) = A \sin(\frac{\pi}{2}) + B = A + B$.$RHL = \lim_{x 	o \frac{\pi}{2}^+} (\cos x) = \cos(\frac{\pi}{2}) = 0$.For continuity,$LHL = RHL \implies A + B = 0$ (Equation $ii$).Adding Equation $i$ and Equation $ii$: $(-A + B) + (A + B) = 2 + 0 \implies 2B = 2 \implies B = 1$.Substituting $B = 1$ into Equation $ii$: $A + 1 = 0 \implies A = -1$.Thus,$f$ is continuous for $A = -1$ and $B = 1$.

Let $f(x) = \begin{cases} -2 \sin x, & \text{if } x \leq -\frac{\pi}{2} \\ A \sin x + B, & \text{if } -\frac{\pi}{2} < x < \frac{\pi}{2} \\ \cos x, & \text{if } x \geq \frac{\pi}{2} \end{cases}$. For what values of $A$ and $B$ is $f$ continuous?

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