  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. MathematicsRS AgarwalStandard XII

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