The points of discontinuity of the function f | vedclass

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(B) The function $f(x)$ is defined as:$f(x) = \frac{1}{x-1}$ for $0 \leq x \leq 2$$f(x) = \frac{x+5}{x+3}$ for $2 < x \leq 4$Step $1$: Check for disco

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$x=1, x=2$

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(B) The function $f(x)$ is defined as:$f(x) = \frac{1}{x-1}$ for $0 \leq x \leq 2$$f(x) = \frac{x+5}{x+3}$ for $2 Step $1$: Check for discontinuity within the intervals.For $0 \leq x \leq 2$,the function $f(x) = \frac{1}{x-1}$ is undefined at $x=1$. Since $1 \in [0, 2]$,$x=1$ is a point of discontinuity.For $2 Step $2$: Check for continuity at the boundary point $x=2$.$\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^-} \frac{1}{x-1} = \frac{1}{2-1} = 1$$\lim_{x 	o 2^+} f(x) = \lim_{x 	o 2^+} \frac{x+5}{x+3} = \frac{2+5}{2+3} = \frac{7}{5}$Since $\lim_{x 	o 2^-} f(x) 
eq \lim_{x 	o 2^+} f(x)$,the function is discontinuous at $x=2$.Conclusion: The points of discontinuity are $x=1$ and $x=2$.

The points of discontinuity of the function $f(x) = \begin{cases} \frac{1}{x-1} & 0 \leq x \leq 2 \\ \frac{x+5}{x+3} & 2 < x \leq 4 \end{cases}$ in its domain are:

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