If the function given by f(x) = begin{cases} | vedclass

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(C) For $f(x)$ to be continuous in $[-\pi, \pi]$,it must be continuous at $x = -\pi/2$ and $x = \pi/2$. At $x = -\pi/2$: $\lim_{x 	o -\pi/2^-} f(x) =

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-$1$

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(C) For $f(x)$ to be continuous in $[-\pi, \pi]$,it must be continuous at $x = -\pi/2$ and $x = \pi/2$. At $x = -\pi/2$: $\lim_{x 	o -\pi/2^-} f(x) = -2 \sin(-\pi/2) = -2(-1) = 2$. $\lim_{x 	o -\pi/2^+} f(x) = a \sin(-\pi/2) + b = -a + b$. Since it is continuous,$-a + b = 2$ (Equation $1$). At $x = \pi/2$: $\lim_{x 	o \pi/2^-} f(x) = a \sin(\pi/2) + b = a + b$. $\lim_{x 	o \pi/2^+} f(x) = \cos(\pi/2) = 0$. Since it is continuous,$a + b = 0$ (Equation $2$). Adding Equation $1$ and Equation $2$: $(-a + b) + (a + b) = 2 + 0 \implies 2b = 2 \implies b = 1$. Substituting $b = 1$ in Equation $2$: $a + 1 = 0 \implies a = -1$. Now,calculate $(3a + 2b)^3$: $(3(-1) + 2(1))^3 = (-3 + 2)^3 = (-1)^3 = -1$.

If the function given by $f(x) = \begin{cases} -2 \sin x & -\pi \leq x < -\pi/2 \\ a \sin x + b & -\pi/2 \leq x \leq \pi/2 \\ \cos x & \pi/2 < x \leq \pi \end{cases}$ is continuous in $[-\pi, \pi]$,then the value of $(3a + 2b)^3$ is

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