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

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Toolbox:
  • For a function to be continuous at any point c $L.H.L = R.H.L = f(c)$
  • $ 1-cosx=2sin^2\large\frac{x}{2}$
  • $ cos^2x=1-sin^2x$
$ LHL \rightarrow \lim\limits_{x \to \large\frac{\pi}{2}} \large\frac{1-sin^3x}{3cos^2x}=\large\frac{1}{2} $
$ \lim\limits_{x \to \large\frac{\pi}{2}} \large\frac{(1-sinx)(1+sin^2x+sinx)}{3(1-sin^2x)}=\large\frac{1}{2}$
 
R.H.L $ \lim\limits_{x \to \large\frac{\pi}{2}} \large\frac{b(1-sinx)}{(\pi-2x)^2}=\large\frac{b}{8}$
Let $ y =\large \frac{\pi}{2}-x$
as $ x \rightarrow \large\frac{\pi}{2} \: y \to 0$
$ \Rightarrow \lim\limits_{y \to 0} \large\frac{b(1-cosy)}{4y^2}$
=$ \lim\limits_{y \to 0} \large\frac{b.2sin^2\large\frac{y}{2}}{16\large\frac{y^2}{4}} =\large \frac{6}{8}$
$ f\bigg(\large \frac{\pi}{2} \bigg)=a$
 
f(x) is continuous
$ \Rightarrow \large\frac{1}{2} = a = \large\frac{b}{8}$
$ \Rightarrow a=\large\frac{1}{2}\: b= 4$

 

answered Mar 9, 2013 by thanvigandhi_1
edited Mar 26, 2013 by thanvigandhi_1
 

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