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Question

Find all points of discontinuity of $$f$$, where $$f$$ is defined by 
f(x) = $$\begin{cases} x +1, \quad \text {if} \quad  x \ge 1 \\ x^2 + 1, \quad \text{if}  \quad x <1 \end{cases}$$


Solution

The given function is $$f(x) = \begin{cases} x +1, \text{if}  \,  x \ge 1 \\ x^2 + 1, \text{if} \,     x <1 \end{cases}$$

The given function is defined at all the points of the real line.
Let $$c$$ be a point on the real line. 

Case I: $$c < 1$$, then $$f(c) = c^2 + 1$$ and $$\underset{x \rightarrow c}{\lim} f(x) = \underset{x \rightarrow c}{\lim} (x^2 + 1) = c^2 + 1$$

$$\therefore$$ $$\underset{x \rightarrow c}{\lim} f(x) = f(c) $$

Therefore, $$f$$ is continuous at all points $$x$$, such that $$x < 1$$

Case II :  $$c = 1$$, then $$f(c) = f(1) = 1 + 1 = 2$$

The left hand limit of $$f$$ at $$x = 1 $$is, 

$$\underset{x \rightarrow 1}{lim}$$ $$f(x) =$$ $$\underset{x \rightarrow 1}{lim}$$ ($$x^2+ 1$$ ) = $$1^2+1 = 2$$

The right hand limit of $$f$$ at $$x = 1$$ is, 

$$\underset{x \rightarrow 1}{\lim} f(x) = \underset{x \rightarrow 1}{\lim} (x + 1) = 1 +1 = 2$$

$$\therefore$$ $$\underset{x \rightarrow 1}{\lim} f(x) = f(1)$$

Therefore, $$f$$ is continuous at $$x = 1$$

Case III :  $$c > 1$$, then $$f(c) = c + 1$$

$$\underset{x \rightarrow c}{\lim} f(x) = \underset{x \rightarrow c}{\lim} (x + 1) = c + 1$$

$$\therefore$$ $$\underset{x \rightarrow c}{\lim} f(x) = f(c)$$

Therefore, $$f$$ is continuous at all points $$x$$, such that $$x > 1$$
Hence, the given function $$f$$ has no point of discontinuity.

Mathematics
RS Agarwal
Standard XII

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