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Find the values of a and b such that the given function is continuous at $ x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$. $$f(x)= \left\{\begin{matrix} x + a\sqrt{2} \sin x \ ;& 0\le x < \frac{\pi}{4} \\ 2x\cot x + b \ ; & \frac{\pi}{4} \le x \le \frac{\pi}{2} \\ a\cos 2x - b\sin x \ ; & \frac{\pi}{2} < x \le \pi \end{matrix}\right. $$

This is what I've tried yet :

Finding LHL : $\lim_{x\to {(\pi/4)}^{-}} f(x)$ = $x + a $

FINDING RHL : $\lim_{x \to {(\pi/4)}^{+}} f(x)$ = ?

There will be two functions possible for $x\to (\pi/4)^+$ : $2x\cot x + b$ and $a\cos 2x - b \sin x$. How should I compute RHL here? I just need a hint. Thanks!

Kushashwa Ravi Shrimali
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    When you are coming from the $(\pi/4)^{+}$ side, you must use the second equation when evaluating the limit because you are getting arbitrarily close to $\pi/4$. When you are approaching $\pi/4$, you will not be in the region $[\pi/2, \pi]$. You do not use $a\cos(2x)-b\sin(x)$ for this reason. – JessicaK Apr 29 '15 at 04:03
  • Oh, got it! Thanks a lot, JessicaK. – Kushashwa Ravi Shrimali Apr 29 '15 at 04:05

1 Answers1

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Hint: when $x\to (\pi/4)+$, it is only a little bit bigger than $\pi/4$, i.e. much smaller than $\pi/2$.

vadim123
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