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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
Question
Is the function defined by $f(x)=\left\{\begin{array}{ll} {x+5,} & {\text { if } x \leq 1} \\ {x-5,} & {\text { if } x>1} \end{array}\right.$ a continuous function?
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Answer

The given function is $f(x) = \left\{\begin{array}{l} {x+5, \text { if } x \leq 1} \\ {x-5, \text { if } x>1} \end{array}\right.$
The function $f$ is defined at all points of the real line.
Let $k$ be the point on a real line.
Then, we have $3$ cases i.e., $k < 1,$ or $k = 1$ or $k > 1$
Now,
Case $I: k < 1$
Then $, f(k) = k + 5$
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (x + 5) = k + 5 = f(k)$
Thus, $\mathop {\lim }\limits_{x \to k} f(x) = f(k)$
Hence $,f$ is continuous at all real number less than $1$.
Case $II: k = 1$
Then $, f(k) = f(1) = 1 + 5 = 6$
$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} (x + 5) = 1 + 5 = 6$
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(x-5)=1-5=-4$
$ \Rightarrow \mathop {\lim }\limits_{x \to {{{k}}^ - }} {\text{f}}({\text{x}}) \ne \mathop {\lim }\limits_{{\text{x}} \to {{\text{k}}^ + }} {\text{f}}({\text{x}})$
Hence $,f$ is not continuous at $x = 1.$
Case $III: k > 1$
Then $,f(k) = k - 5$
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to {{k}}} (x - 5) = k - 5$
Thus, $\mathop {\lim }\limits_{x \to {{k}}} f(x) = f(k)$
Hence $,f$ is continuous at all real number greater than $1$.
Therefore $, x = 1$ is the only point of discontinuity of $f$.

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