If the function f(x) = begin{cases} x + a sqr | vedclass

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(A) For $f(x)$ to be continuous at $x = \frac{\pi}{4}$,the left-hand limit $(LHL)$ must equal the right-hand limit $(RHL)$: $\lim_{x 	o \frac{\pi}{4}

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$\frac{\pi}{4}$

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(A) For $f(x)$ to be continuous at $x = \frac{\pi}{4}$,the left-hand limit $(LHL)$ must equal the right-hand limit $(RHL)$: $\lim_{x 	o \frac{\pi}{4}^-} (x + a \sqrt{2} \sin x) = \lim_{x 	o \frac{\pi}{4}^+} (2x \cot x + b)$ $\frac{\pi}{4} + a \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 2 \cdot \frac{\pi}{4} \cdot 1 + b$ $\frac{\pi}{4} + a = \frac{\pi}{2} + b \implies a - b = \frac{\pi}{4}$. For $f(x)$ to be continuous at $x = \frac{\pi}{2}$,the $LHL$ must equal the $RHL$: $\lim_{x 	o \frac{\pi}{2}^-} (2x \cot x + b) = \lim_{x 	o \frac{\pi}{2}^+} (a \cos 2x - b \sin x)$ $2 \cdot \frac{\pi}{2} \cdot 0 + b = a \cos \pi - b \sin \frac{\pi}{2}$ $b = -a - b \implies a = -2b$. Substituting $a = -2b$ into $a - b = \frac{\pi}{4}$: $-2b - b = \frac{\pi}{4} \implies -3b = \frac{\pi}{4} \implies b = -\frac{\pi}{12}$. Then $a = -2(-\frac{\pi}{12}) = \frac{\pi}{6}$. Thus,$a - b = \frac{\pi}{6} - (-\frac{\pi}{12}) = \frac{2\pi + \pi}{12} = \frac{3\pi}{12} = \frac{\pi}{4}$.

If the function $f(x) = \begin{cases} x + a \sqrt{2} \sin x & \text{if } 0 \leq x \leq \frac{\pi}{4} \\ 2x \cot x + b & \text{if } \frac{\pi}{4} < x \leq \frac{\pi}{2} \\ a \cos 2x - b \sin x & \text{if } \frac{\pi}{2} < x \leq \pi \end{cases}$ is continuous in $[0, \pi]$,then $a - b = $

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