If the function f(x),defined below,is continu | vedclass

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(D) Since $f(x)$ is continuous in $[0, \pi]$,it must be continuous at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$.At $x = \frac{\pi}{4}$,$\lim_{x 	o

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$a = \frac{\pi}{6}, b = \frac{-\pi}{12}$

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(D) Since $f(x)$ is continuous in $[0, \pi]$,it must be continuous at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$.At $x = \frac{\pi}{4}$,$\lim_{x 	o \frac{\pi}{4}^-} f(x) = \lim_{x 	o \frac{\pi}{4}^+} f(x) = f(\frac{\pi}{4})$.$\frac{\pi}{4} + a\sqrt{2}(\sin \frac{\pi}{4}) = 2(\frac{\pi}{4})(\cot \frac{\pi}{4}) + b$$\frac{\pi}{4} + a = \frac{\pi}{2} + b \Rightarrow a - b = \frac{\pi}{4} \dots (I)$.At $x = \frac{\pi}{2}$,$\lim_{x 	o \frac{\pi}{2}^-} f(x) = \lim_{x 	o \frac{\pi}{2}^+} f(x) = f(\frac{\pi}{2})$.$2(\frac{\pi}{2})(\cot \frac{\pi}{2}) + b = a(\cos \pi) - b(\sin \frac{\pi}{2})$.Since $\cot \frac{\pi}{2} = 0$,we get $b = -a - b \Rightarrow a = -2b \dots (II)$.Substituting $(II)$ into $(I)$: $-2b - b = \frac{\pi}{4} \Rightarrow -3b = \frac{\pi}{4} \Rightarrow b = -\frac{\pi}{12}$.Then $a = -2(-\frac{\pi}{12}) = \frac{\pi}{6}$.

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