Rolle's theorem is not applicable to the function f(x) = |x| defi… - Vidyadip

MCQ

Rolle's theorem is not applicable to the function $f(x) = |x|$ defined on $ [-1, 1] $ because

A
$f $ is not continuous on $ [ -1, 1]$
B
$f$ is not differentiable on $ (-1,1)$
C
$f( - 1) \ne f(1)$
D
$f( - 1) = f(1) \ne 0$

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Answer

$f(x) = \left\{ \begin{array}{l} - x,\,{\rm{when -1}} \le x < 0\\{\rm{ }}x,\;{\rm{when}}\;{\rm{0}} \le x \le {\rm{1}}\end{array} \right.$
Clearly $f( - 1) = | - 1| = 1 = f(1)$
But $Rf'(0) = \mathop {\lim }\limits_{h \to 0}\frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{|h|}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{h}{h} = 1$
$Lf'(0) = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 - h) - f(0)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{| - h|}}{{ - h}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{h}{{ - h}} = - 1$
$\therefore Rf'(0) \ne Lf'(0)$
Hence it is notdifferentiable on $( - 1,1)$.

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