If the function f(x) is continuous in 0 leq x | vedclass

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(A) $f(x)$ is continuous in $0 \leq x \leq \pi$,therefore it is continuous at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$.At $x = \frac{\pi}{4}$:$\lim

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(A) $f(x)$ is continuous in $0 \leq x \leq \pi$,therefore it is continuous at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$.At $x = \frac{\pi}{4}$:$\lim_{x \rightarrow \frac{\pi}{4}^-} (x + a \sqrt{2} \sin x) = \lim_{x \rightarrow \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 \Rightarrow a - b = \frac{\pi}{4} \quad (i)$At $x = \frac{\pi}{2}$:$\lim_{x \rightarrow \frac{\pi}{2}^-} (2x \cot x + b) = \lim_{x \rightarrow \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 \Rightarrow a = -2b \quad (ii)$Substituting $(ii)$ into $(i)$:$-2b - b = \frac{\pi}{4} \Rightarrow -3b = \frac{\pi}{4} \Rightarrow b = -\frac{\pi}{12}$Then $a = -2(-\frac{\pi}{12}) = \frac{\pi}{6}$.Finally,$2a + 3b = 2(\frac{\pi}{6}) + 3(-\frac{\pi}{12}) = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}$.

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