Continuity and Differentiability — Class 12 Maths Solution

We have,

$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{ - (x) - [ - (x + 1)],}&{if}&{x < - 1}\\{ - (x) - (x + 1),}&{if}&{ - 1 \le x < 0}\\{(x) - (x + 1),}&{if}&{x \ge 0}\end{array}} \right.$

$\Rightarrow$ $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{1,}&{if}&{x < - 1}\\{ - 2x - 1,}&{if}&{ - 1 \le x < 0}\\{ - 1,}&{if}&{x \ge 0}\end{array}} \right.$

At $x = - 1:$
$\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = 1$

$\mathop {\lim }\limits_{x \to - {1^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to - 1 + h\atop\scriptstyle h \to 0} ( - 2( - 1 + h) - 1) = 1$
$f( - 1) = - 2( - 1) - 1 = 1$

Thus, $\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} f(x) = f( - 1)$

$\Rightarrow$ f is continuous at $x = - 1$

At $x = 0:$

$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} ( - 2x - 1) = \mathop {\lim }\limits_{\scriptstyle x \to 0 - h\atop\scriptstyle h \to 0} ( - 2( - h) - 1) = - 1$

Also, $f(0) = - 1$

Thus, $\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$

$\Rightarrow$ f is continuous at $x = 0.$
Also, f being a constant is continuous when

$x < - 1$ or when x $>$ 0.
therefore, f is continuous for all $x \in R$

Hence we can say that there is no point of discontinuity.

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