Question

Let\[f\left( x \right) = \left\{ \begin{array}\frac{1 - \sin^3 x}{3 \cos^2 x} , & \text{ if }  x < \frac{\pi}{2} \\ a , & \text{ if }  x = \frac{\pi}{2} \\ \frac{b(1 - \sin x)}{(\pi - 2x )^2}, & \text{ if }  x > \frac{\pi}{2}\end{array} . \right.\] ]If f(x) is continuous at x = \[\frac{\pi}{2}\] , find a and b.

 

Answer 1

Given: 

\[f\left( x \right) = \left\{ \begin{array}\frac{1 - \sin^3 x}{3 \cos^2 x} , & \text{ if }  x < \frac{\pi}{2} \\ a , & \text{ if }  x = \frac{\pi}{2} \\ \frac{b(1 - \sin x)}{(\pi - 2x )^2}, & \text{ if }  x > \frac{\pi}{2}\end{array} . \right.\] ]

We have
(LHL at x = \[\frac{\pi}{2}\]

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

\[= \lim_{h \to 0} \left( \frac{1 - \sin^3 \left( \frac{\pi}{2} - h \right)}{3 \cos^2 \left( \frac{\pi}{2} - h \right)} \right)\]
\[ = \lim_{h \to 0} \left( \frac{1 - \cos^3 h}{3 \sin^2 h} \right)\]
\[ = \frac{1}{3} \lim_{h \to 0} \left( \frac{\left( 1 - \cosh \right)\left( 1 + \cos^2 h + \cosh \right)}{\left( 1 - \cosh \right)\left( 1 + \cosh \right)} \right)\]
\[ = \frac{1}{3} \lim_{h \to 0} \left( \frac{\left( 1 + \cos^2 h + \text{ cos } h \right)}{\left( 1 + \text{ cos }h \right)} \right)\]
\[ = \frac{1}{3}\left( \frac{1 + 1 + 1}{1 + 1} \right) = \frac{1}{2}\]

(RHL at x = \[\frac{\pi}{2}\]

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

\[= \lim_{h \to 0} \left( \frac{b\left[ 1 - \sin\left( \frac{\pi}{2} + h \right) \right]}{\left[ \pi - 2\left( \frac{\pi}{2} + h \right) \right]^2} \right)\]
\[ = \lim_{h \to 0} \left( \frac{b\left( 1 - \text{ cos } h \right)}{\left[ - 2h \right]^2} \right)\]
\[ = \lim_{h \to 0} \left( \frac{2b \sin^2 \frac{h}{2}}{4 h^2} \right)\]
\[ = \lim_{h \to 0} \left( \frac{2b \sin^2 \frac{h}{2}}{16\frac{h^2}{4}} \right)\]
\[ = \frac{b}{8} \lim_{h \to 0} \left( \frac{\sin\frac{h}{2}}{\frac{h}{2}} \right)^2 \]
\[ = \frac{b}{8} \times 1\]
\[ = \frac{b}{8}\]

Also,

\[f\left( \frac{\pi}{2} \right) = a\]

If f(x) is continuous at x =

\[\frac{\pi}{2}\]then

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

\[\Rightarrow \frac{1}{2} = \frac{b}{8} = a\]

\[\Rightarrow a = \frac{1}{2} \text{ and } b = 4\]