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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:
$\text{f(x)}= \begin{cases}-2,\ \text{if}\ \text{x}\leq-1 \\\text{2x},\text{if}\ -1<\text{x}\leq1\\2,\text{if}\ \text{x}>1\end{cases}$
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Answer

$\text{f(x)}= \begin{cases}-2,\ \text{if}\ \text{x}\leq-1 \\\text{2x},\text{if}\ -1<\text{x}\leq1\\2,\text{if}\ \text{x}>1\end{cases}$The function f is defined at all points of the real line.
When x < -1, we have f(x) = -2, which it is constant and so is continuous.
At x = -1
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{-}}(-{2}) = -2$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{+}}(\text{2x}) = 2(-1)= -2$
Also f(-1) = -2
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{+}}\text{f(x)} = \text{f}(-1)$
$\therefore$ f is continuous at x = -1
in the interval -1 < x < 1, we have f(x) = 2 x, which being a linear polynomial, is continuous.
At x = -1
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{-}}(\text{2x}) = 2(1)= 2$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{+}}({2}) = 2$
Also f(1) = 2(1) = 2
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{+}}\text{f(x)} = \text{f}(1)$
$\therefore$ f is continuous at x = 1
When x > 1, we have f(x) = 2, which is constant and so it is continuous.

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