If f(x)= begin{cases}-2 sin x & , quad x leqs | vedclass

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(D) For $f(x)$ to be continuous at $x = -\frac{\pi}{2}$,the left-hand limit must equal the right-hand limit: $\lim_{x 	o -\frac{\pi}{2}^-} (-2 \sin x

Vedclass · Accepted Answer

$-1$

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(D) For $f(x)$ to be continuous at $x = -\frac{\pi}{2}$,the left-hand limit must equal the right-hand limit: $\lim_{x 	o -\frac{\pi}{2}^-} (-2 \sin x) = \lim_{x 	o -\frac{\pi}{2}^+} (a \sin x + b)$ $-2 \sin(-\frac{\pi}{2}) = a \sin(-\frac{\pi}{2}) + b$ $-2(-1) = a(-1) + b \implies 2 = -a + b \quad (1)$ For $f(x)$ to be continuous at $x = \frac{\pi}{2}$,the left-hand limit must equal the right-hand limit: $\lim_{x 	o \frac{\pi}{2}^-} (a \sin x + b) = \lim_{x 	o \frac{\pi}{2}^+} (\cos x)$ $a \sin(\frac{\pi}{2}) + b = \cos(\frac{\pi}{2})$ $a(1) + b = 0 \implies a + b = 0 \quad (2)$ Adding equations $(1)$ and $(2)$: $(-a + b) + (a + b) = 2 + 0 2b = 2 \implies b = 1$ Substituting $b = 1$ into equation $(2)$: $a + 1 = 0 \implies a = -1$ Now,calculate $2a + b$: $2(-1) + 1 = -2 + 1 = -1$ Thus,the value is $-1$.

If $f(x)= \begin{cases}-2 \sin x & , \quad x \leqslant-\frac{\pi}{2} \\ a \sin x+b & , \quad \frac{-\pi}{2} < x < \frac{\pi}{2} \\ \cos x & , \quad x \geqslant \frac{\pi}{2}\end{cases}$ is continuous at $x=-\frac{\pi}{2}$ and $x=\frac{\pi}{2}$,then the value of $2a+b$ is

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