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(C) For the function $f(x)$ to be continuous everywhere,it must be continuous at the transition points $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$.At

Vedclass · Accepted Answer

$A = -1, B = 1$

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(C) For the function $f(x)$ to be continuous everywhere,it must be continuous at the transition points $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$.At $x = -\frac{\pi}{2}$:$\lim_{x 	o -\frac{\pi}{2}^-} f(x) = \lim_{x 	o -\frac{\pi}{2}^-} (-2\sin x) = -2\sin(-\frac{\pi}{2}) = -2(-1) = 2$.$\lim_{x 	o -\frac{\pi}{2}^+} f(x) = \lim_{x 	o -\frac{\pi}{2}^+} (A\sin x + B) = A\sin(-\frac{\pi}{2}) + B = -A + B$.Since the function is continuous,$-A + B = 2$ ..... $(i)$.At $x = \frac{\pi}{2}$:$\lim_{x 	o \frac{\pi}{2}^-} f(x) = \lim_{x 	o \frac{\pi}{2}^-} (A\sin x + B) = A\sin(\frac{\pi}{2}) + B = A + B$.$\lim_{x 	o \frac{\pi}{2}^+} f(x) = \lim_{x 	o \frac{\pi}{2}^+} (\cos x) = \cos(\frac{\pi}{2}) = 0$.Since the function is continuous,$A + B = 0$ ..... $(ii)$.Adding equations $(i)$ and $(ii)$:$(-A + B) + (A + B) = 2 + 0 \implies 2B = 2 \implies B = 1$.Substituting $B = 1$ into $(ii)$:$A + 1 = 0 \implies A = -1$.Thus,the values are $A = -1$ and $B = 1$.

The values of $A$ and $B$ such that the function $f(x) = \begin{cases} -2\sin x, & x \le -\frac{\pi}{2} \\ A\sin x + B, & -\frac{\pi}{2} < x < \frac{\pi}{2} \\ \cos x, & x \ge \frac{\pi}{2} \end{cases}$ is continuous everywhere are

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