Discussion Forum

1)

 The points of discontinuity of the function

$f(x)= \frac{1}{x-1}, if 0\leq x\leq2$

   $= \frac{x+5}{x+3}, if 2< x\leq4$

 in its domain are

Answer:

Option A

Explanation:

 We have

$f(x)=\begin{cases}\frac{1}{x-1} & if 0\leq x\leq2,\\\frac{x+5}{x+3}, & if 2<x \leq4\end{cases}$

 Clearly , f(x) is not defined at x=1

 Hence , f(x) is discontinuous at x=1

 At x=2

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

$\lim_{x \rightarrow {2^{+}}}f(x)=\lim_{x \rightarrow {2^{+}}}\frac{x+5}{x+3}=\frac{2+5}{2+3} =\frac{7}{5}$ 

$\therefore$   $\lim_{x \rightarrow {2^{-}}}f(x)\neq\lim_{x \rightarrow {2^{+}}}f(x)$

 $\therefore$     f(x) is discontinuous at x=2