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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
Question
Discuss the continuity of the function f, where f is defined by: $[f(x) = \left\{ {\begin{array}{*{20}{l}} { - 2,\ {\rm{ if }} \ x \le - 1}\\ { 2x, \ \ {\rm{ if }} - 1 < x \le 1}\\ {{\rm{\ \ \ 2, \ if }} \ x > 1} \end{array}} \right.]$
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Answer

The given function is $f(x)=\left\{\begin{array}{c} {-2, \text { if } x \leq-1} \\ {2 x, \text { if }-1 \leq x \leq 1} \\ {2, \text { if } x>1} \end{array}\right.$
The function f is defined at all points of the real line.
Then, we have 5 cases i.e., k < -1, k = -1, -1 < k < 1, k = 1 or k > 1.
Now,
Case I: k < 0
Then, f(k) = -2
$\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} {\text{f}}({\text{x}}) = \mathop {\lim }\limits_{{\mathbf{x}} \to {\text{k}}} ( - 2)$= -2= f(k)
Thus $\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} {\text{f}}({\text{x}}) $ = f(k)
Hence, f is continuous at all points x, s.t. x < -1.
Case II: k = -1
f(k) = f(=1) = -2
$\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ - }} $(-2) = -2
$\mathop {\lim }\limits_{x \to - {1^ + }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} $(2x) = 2 × (-1) = -2
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\mathbf{k}}^ - }} {\text{f}}({\text{x}}) = \mathop {\lim }\limits_{{\text{x}} \to {{\text{k}}^ + }} {\text{f}}({\text{x}}) = {\text{f}}({\text{k}})$
Hence, f is continuous at x = -1.
Case III: -1 < k < 1
Then, f(k) = 2k
$\mathop {\lim }\limits_{x \to {\mathbf{k}}} f(x) = \mathop {\lim }\limits_{x \to {\mathbf{k}}} (2x) = 2k = f(k)$
Thus $\mathop {\lim }\limits_{x \to {\mathbf{k}}} f(x) = f(k)$
Hence, f is continuous in (-1, 1).
Case IV: k = 1
Then f(k) = f(1) = 2 × 1 = 2
$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} (2x)$ = 2 × 1 = 2
$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} (2) = 2$
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\mathbf{k}}^ - }} f(x) = \mathop {\lim }\limits_{x \to {k^ + }} f(x) = f(x)$
Hence, f is continuous at x = 1.
Case V: k > 1
Then, f(k) = 2
$\mathop {\lim }\limits_{x \to {\mathbf{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (2) = 2 = f(k)$
Thus $\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} {\text{f}}({\text{x}}) = {\text{f}}({\text{k}})$
Hence, f is continuous at all points x, s.t. x > 1.
Therefore, f is continuous at all points of the real line.

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