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Find all points of discontinuity of the function $f(t)=\large\frac{1}{t^2+t-2},$where $\;t=\large\frac{1}{x-1}.$

$\begin{array}{1 1} 1,\frac{1}{2},2 \\ 1,\frac{-1}{2},2 \\ -1,\large\frac{1}{2},2 \\ -1,\large\frac{-1}{2},2 \end{array} $

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Toolbox:
  • A function is said to be discontinuous if the LHL = RHL.
  • A function is discontinuous if both LHL and RHL does not exist.
Step 1:
Given function is $f(t)=\large\frac{1}{t^2+t-2},$where $\;t=\large\frac{1}{x-1}.$
$(t^2+t-2)$ can be factorised as $(t+2)(t-1)$
Therefore $f(t)=\large\frac{1}{(t+2)(t-1)}$
But $t=\large\frac{1}{x-1}$
Substituting for $\large\frac{1}{x-1}$ we get,
$f(x)=\large\frac{1}{\big(\Large\frac{1}{x-1}\normalsize+2\big)\big(\Large\frac{1}{x-1}\normalsize-1\big)}$
$\quad\;\;=\bigg(\large\frac{1+2x-2}{x-1}\bigg)\bigg(\large\frac{1-x+1}{x-1}\bigg)$
$\quad\;\;=\bigg(\large\frac{2x-1}{x-1}\bigg)\bigg(\large\frac{2-x}{x-1}\bigg)$
Clearly when $x=1$ the function goes undefined.Hence one point of discontinuity is $x=1$
Step 2:
When $x=\large\frac{1}{2}$
$f(x)=\bigg(\large\frac{1}{\Large\frac{1}{1/2-1}\normalsize +2}\bigg)$$\times \bigg(\Large\frac{1}{\Large\frac{1}{1/2-1}-\normalsize 1}\bigg)$
$\quad\;\;\;=\infty$
$\quad\;\;\;=$undefined.
Hence $x=\large\frac{1}{2}$ is another point of discontinuity.
Step 3:
When $x=2$
$f(x)=\bigg(\large\frac{1}{\Large\frac{1}{2-1}\normalsize +2}\bigg)$$\times \bigg(\Large\frac{1}{\Large\frac{1}{2-1}-\normalsize 1}\bigg)$
$\quad\;\;\;=\infty$
$\quad\;\;\;=$undefined.
Hence $x=2$ is another point of discontinuity.
Hence $x=1,2,\large\frac{1}{2}$ are the points of discontinuity.
answered Jun 26, 2013 by sreemathi.v
 

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