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]$
- ✓$f$ is not differentiable on $ (-1,1)$
- C$f( - 1) \ne f(1)$
- D$f( - 1) = f(1) \ne 0$
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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