Question

If $y=\frac{1}{t^2+t-2}$ where $t=\frac{1}{x-1},$ then which of the following is/are true about continuity of $y=f\left(x\right)$ in $R$?

• A. $f\left(x\right)$ is discontinuous at $x=\frac{1}{2}$
• B. $f\left(x\right)$ is continuous at $x=1$
• C. $f\left(x\right)$ is discontinuous at $x=2$
• D. $f\left(x\right)$ is continuous at $x=\frac{1}{2}$

Answer 1

Answer: (A), (C)

$f\left(x\right)$ is discontinuous at $x=2$

Given: $y=\frac{1}{t^2+t-2},t=\frac{1}{x-1}$
Now,
$y=f\left(x\right)$ will be discontinuous where $t$ is discontinuous i.e. at $x=1$
Also, when
$(t+2)(t-1)=0\Rightarrow t=-2,1\Rightarrow \frac{1}{x-1}=-2,1\Rightarrow x=\frac{1}{2},2$
Hence, $y=f\left(x\right)$ is discontinuous at $x=\frac{1}{2},1,2$