If a function y = f(x) is defined as $y=\frac{1}{t^2-t-6}$ and $t=\frac{1}{x-2}, t \in R$. Then, f(x) is discontinuous at
Answer & explanation
Correct answer: option 2
We have,
$y=\frac{1}{t^2-t-6}=\frac{1}{(t-3)(t+2)}$ and, $t=\frac{1}{x-2}$
We observe that t is not defined at x = 2 and y is not defined at t = -2, 3.
Now,
$t=-2 \Rightarrow-2=\frac{1}{x-2} \Rightarrow x=\frac{3}{2}$
and, $t=3 \Rightarrow 3=\frac{1}{x-2} \Rightarrow x=\frac{7}{3}$
Hence, f(x) is discontinuous at $x=2, \frac{3}{2}$ and $\frac{7}{3}$.