The values of a and b,so that the function f( | vedclass

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(D) Since the function is continuous at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$,we have:At $x = \frac{\pi}{4}$:$\lim_{x 	o \frac{\pi}{4}^-} (x +

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$\frac{\pi}{6}, -\frac{\pi}{12}$

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(D) Since the function is continuous at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$,we have:At $x = \frac{\pi}{4}$:$\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}$ . . . $(i)$At $x = \frac{\pi}{2}$:$\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 = 0$ . . . $(ii)$Solving $(i)$ and $(ii)$:From $(ii)$,$a = -2b$. Substituting in $(i)$:$-2b - b = \frac{\pi}{4} \implies -3b = \frac{\pi}{4} \implies b = -\frac{\pi}{12}$Then $a = -2(-\frac{\pi}{12}) = \frac{\pi}{6}$.Thus,$a = \frac{\pi}{6}$ and $b = -\frac{\pi}{12}$.

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