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Question Number 133994 by benjo_mathlover last updated on 26/Feb/21

$$\mathrm{Given}\mathrm{a}\mathrm{function}\mathrm{f}\mathrm{satisfies}$$ $$\mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}x+a\sqrt{2}\sin x;0⩽x<\frac{\pi }{4}\\ 2x\cot x+b;\frac{\pi }{4}⩽x⩽\frac{\pi }{2}\\ a\cos 2x-b\sin x;\frac{\pi }{2}<x⩽\pi \end{array}$$ $$\mathrm{continuous}\mathrm{in}[0,\pi ],\mathrm{then}\mathrm{find}$$ $$\mathrm{the}\mathrm{value}\mathrm{of}a\mathrm{and}b.$$

Answered by EDWIN88 last updated on 26/Feb/21

$$\left(1\right)\mathrm{f}\left(\mathrm{x}\right)\mathrm{must}\mathrm{be}\mathrm{continuous}\mathrm{at}\mathrm{x}=\frac{\pi }{4}$$ $$\underset{x\to \pi /4^{-}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{x\to \pi /4^{+}}{\lim f}\left(\mathrm{x}\right)$$ $$\Rightarrow \frac{\pi }{4}+a\sqrt{2}.\frac{\sqrt{2}}{2}=2.\frac{\pi }{4}.1+b$$ $$\Rightarrow a-b=\frac{\pi }{4}...\left(\mathrm{i}\right)$$ $$\left(2\right)\mathrm{f}\left(\mathrm{x}\right)\mathrm{also}\mathrm{must}\mathrm{be}\mathrm{continuous}\mathrm{at}\mathrm{x}=\frac{\pi }{2}$$ $$\underset{x\to \pi /2^{-}}{\lim f}\left(\mathrm{x}\right)=\underset{x\to \pi /2^{+}}{\lim f}\left(\mathrm{x}\right)$$ $$\Rightarrow 0+b=a(-1)-b;a+2b=0...\left(ii\right)$$ $$\left(i\right)-\left(ii\right):-3b=\frac{\pi }{4};\underset{―}{b=-\frac{\pi }{12}\mathrm{and}a=\frac{\pi }{6}}$$ $$$$

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