Question

Let $y=\frac{1}{u^2+u-2}$where $u=\frac{1}{x-1}$ then $y$ is discontinuous at $x$ =__.

• A. $1,2$
• B. $1,-2$
• C. $1,\frac{1}{2},2$
• D. $1,\frac{1}{2}$

Answer 1

The correct option is C $1,\frac{1}{2},2$

$y$ will be discontinuous at all those points where it is undefined.
Since, $y=\frac{1}{u^2+u-2}$
It is undefined at the points where denominator is 0 ie $u=-2$ and $u=1$
Also, $u$ itself is not defined at $x=1$ since, $\frac{1}{x-1}$ will get undefined,
Thus, $x=1$ is a point of discontinuity.
Now $u=\frac{1}{x-1}=-2$
Hence,$x=\frac{1}{2}$
$u=\frac{1}{x-1}=1$ Hence, $x=2$
So in all there are 3 points of discontinuity.
Those are $x=1,x=2$ and $x=\frac{1}{2}$