Question

Find the values of a and b so that the function f(x) defined by \[f\left( x \right) = \begin{cases}x + a\sqrt{2}\sin x , & \text{ if }0 \leq x < \pi/4 \\ 2x \cot x + b , & \text{ if } \pi/4 \leq x < \pi/2 \\ a \cos 2x - b \sin x, & \text{ if }  \pi/2 \leq x \leq \pi\end{cases}\]becomes continuous on [0, π].

Answer 1

Given: f is continuous on  \[\left[ 0, \pi \right]\] .

∴ f is continuous at x =   \[\frac{\pi}{4}\] and  \[\frac{\pi}{2}\]

At x =  \[\frac{\pi}{4}\], we have

\[\lim_{x \to \frac{\pi}{4}^-} f\left( x \right) = \lim_{h \to 0} f\left( \frac{\pi}{4} - h \right) = \lim_{h \to 0} \left[ \left( \frac{\pi}{4} - h \right) + a\sqrt{2}\sin \left( \frac{\pi}{4} - h \right) \right] = \left[ \frac{\pi}{4} + a\sqrt{2} \sin \left( \frac{\pi}{4} \right) \right] = \left[ \frac{\pi}{4} + a \right]\]

\[\lim_{x \to \frac{\pi}{4}^+} f\left( x \right) = \lim_{h \to 0} f\left( \frac{\pi}{4} + h \right) = \lim_{h \to 0} \left[ 2\left( \frac{\pi}{4} + h \right) \cot \left( \frac{\pi}{4} + h \right) + b \right] = \left[ \frac{\pi}{2} \cot \left( \frac{\pi}{4} \right) + b \right] = \left[ \frac{\pi}{2} + b \right]\]

At x =  \[\frac{\pi}{2}\] , we have

\[\lim_{x \to \frac{\pi}{2}^-} f\left( x \right) = \lim_{h \to 0} f\left( \frac{\pi}{2} - h \right) = \lim_{h \to 0} \left[ 2\left( \frac{\pi}{2} - h \right) \cot \left( \frac{\pi}{2} - h \right) + b \right] = b\]

\[\lim_{x \to \frac{\pi}{2}^+} f\left( x \right) = \lim_{h \to 0} f\left( \frac{\pi}{2} + h \right) = \lim_{h \to 0} \left[ a \cos 2\left( \frac{\pi}{2} + h \right) - b \sin \left( \frac{\pi}{2} + h \right) \right] = - a - b\]

Since f is continuous at x = \[\frac{\pi}{4}\]and x = \[\frac{\pi}{2}\]  we get 

\[\lim_{x \to \frac{\pi}{2}^-} f\left( x \right) = \lim_{x \to \frac{\pi}{2}^+} f\left( x \right) \text{ and } \lim_{x \to \frac{\pi}{4}^-} f\left( x \right) = \lim_{x \to \frac{\pi}{4}^+} f\left( x \right)\]

\[\Rightarrow - b - a = b \text{ and } \frac{\pi}{4} + a = \frac{\pi}{2} + b\]
\[ \Rightarrow b = \frac{- a}{2} . . . \left( 1 \right) \text{ and }  \frac{- \pi}{4} = b - a . . . \left( 2 \right)\]
\[ \Rightarrow \frac{- \pi}{4} = \frac{- 3a}{2} \left[ \text{ Substituting the value of b in eq .}  \left( 2 \right) \right]\]
\[ \Rightarrow a = \frac{\pi}{6}\]
\[ \Rightarrow b = \frac{- \pi}{12} \left[ \text{ From eq } . \left( 1 \right) \right]\]