Browse Questions

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

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

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.