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

$\frac{1}{2},2$

The correct answer is Option (3) → $\frac{1}{2},2$ ##

We have, $f(t) = \frac{1}{t^2 + t - 2}$ and $t = \frac{1}{x - 1}$.

On putting $t = \frac{1}{x - 1}$ in $f(t)$, we get:

$f(t) = \frac{1}{\left( \frac{1}{x - 1} \right)^2 + \left( \frac{1}{x - 1} \right) - \frac{2}{1}}$

$= \frac{1}{\frac{1}{(x - 1)^2} + \frac{1}{(x - 1)} - \frac{2}{1}}$

$= \frac{1}{\frac{1 + x - 1 - 2(x - 1)^2}{(x - 1)^2}}$

$= \frac{(x - 1)^2}{1 + x - 1 - 2(x^2 + 1 - 2x)} \quad [∵(a - b)^2 = a^2 + b^2 - 2ab]$

$= \frac{(x - 1)^2}{1 + x - 1 - 2x^2 - 2 + 4x} = \frac{(x - 1)^2}{-2x^2 + 5x - 2}$

$= \frac{(x - 1)^2}{-(2x^2 - 5x + 2)} = \frac{(x - 1)^2}{-(2x - 1)(x - 2)}$

$= \frac{(x - 1)^2}{(2x - 1)(2 - x)}$

So, $f(t)$ is discontinuous at:

$2x - 1 = 0 \Rightarrow x = 1/2$

and $2 - x = 0 \Rightarrow x = 2$