If the function f(x) = begin{cases} -2 sin x, | vedclass

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(A) Since $f(x)$ is continuous everywhere,it must be continuous at $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$.At $x = -\frac{\pi}{2}$:$\lim_{x \righ

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$(-1, 1)$

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(A) Since $f(x)$ is continuous everywhere,it must be continuous at $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$.At $x = -\frac{\pi}{2}$:$\lim_{x \rightarrow -\frac{\pi}{2}^-} f(x) = \lim_{x \rightarrow -\frac{\pi}{2}^+} f(x)$$-2 \sin(-\frac{\pi}{2}) = A \sin(-\frac{\pi}{2}) + B$$-2(-1) = A(-1) + B$$2 = -A + B$  ... $(i)$At $x = \frac{\pi}{2}$:$\lim_{x \rightarrow \frac{\pi}{2}^-} f(x) = \lim_{x \rightarrow \frac{\pi}{2}^+} f(x)$$A \sin(\frac{\pi}{2}) + B = \cos(\frac{\pi}{2})$$A(1) + B = 0$$A + B = 0$  ... $(ii)$Adding $(i)$ and $(ii)$:$(-A + B) + (A + B) = 2 + 0$$2B = 2 \Rightarrow B = 1$Substituting $B = 1$ in $(ii)$:$A + 1 = 0 \Rightarrow A = -1$Thus,the values are $A = -1$ and $B = 1$.

If the function $f(x) = \begin{cases} -2 \sin x, & x \leq \frac{-\pi}{2} \\ A \sin x+B, & \frac{-\pi}{2} < x < \frac{\pi}{2} \\ \cos x, & x \geq \frac{\pi}{2} \end{cases}$ is continuous everywhere,then the values of $A$ and $B$ are respectively

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