- Home
- Standard 12
- Mathematics
વિધેય $f(x) = |x|$ એ અંતરાલ $[-1, 1]$ માં રોલ ના પ્રમેયનું પાલન કરતું નથી કારણ કે . . . .
$f$ એ $[ -1, 1]$ પર સતત નથી
$f$ એ $[ -1, 1]$ પર વિકલનીય નથી
$f( - 1) \ne f(1)$
$f( - 1) = f(1) \ne 0$
Solution
(b) $f(x) = \left\{ \begin{array}{l} – x,\,{\rm{when\,\, -1}} \le x < 0\\{\rm{ }}x,\;{\rm{when\,\,}}\;{\rm{0}} \le x \le {\rm{1}}\end{array} \right.$
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 $( – 1,\,\,1)$.
