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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Discuss the continuity of the function f, where f is defined by: $\text{f(x)}= \begin{cases}\ 2\text{x},\ \ \text{if}\ \text{x}<0 \\0,\ \ \ \ \text{if}\ 0\leq\text{x}\leq1\\4\text{x},\ \ \ \text{if}\ \text{x}>1\end{cases}$
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Answer

The given function is $\text{f(x)}= \begin{cases}\ 2\text{x},\ \ \text{if}\ \text{x}<0 \\0,\ \ \ \ \text{if}\ 0\leq\text{x}\leq1\\4\text{x},\ \ \ \text{if}\ \text{x}>1\end{cases}$ The function f is defined at all points of the real line. Then, we have 5 cases i.e. k < 0, k = 0, 0 < k < 1, k = 1 or k < 1. Now, Case I: k < 0 Then, f(k) = 2k $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}(\text{2x}) = \text{2k} = \text{f(k)}$ Thus, $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = \text{f(k)}$ Hence, f is continuous at all points x, s.t. x < 0. Case II: k = 0 f(0) = 0 $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}(\text{2x}) = 2\times 0 = 0$ $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{+}}(0) = 0$ $\Rightarrow\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{-}}\text{f(x)} =^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{+}}\text{f(x)}=\text{f(k)}$ Hence, f is not continuous at x = 0. Case III: 0 < k < 1 Then, f(k) = 0 $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}(0) = 0 = \text{f(k)}$ Thus, $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = \text{f(k)}$ Hence, f is continuous in (0, 1). Case IV: k = 1 Then f(k) = f(1) = 0 $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{-}}(0) = 0$ $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{+}}(\text{4x}) = 4\times 1 = 4$ $\Rightarrow\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{-}}\text{f(x)} \neq ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{+}}\text{f(x)}$ Hence, f is not continuous at x = 1.Case V: k < 1
Then, f(k) = 4k
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}(\text{4x}) = 4\text{k} = \text{f(k)}$ Thus $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = \text{f(k)}$ Hence, f is contionuous at all points x, s.t. x > 1. Therefore, x = 1 is the only point of discontinuity of f.

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