- Home
- Standard 12
- Mathematics
સરેરાશ મૂલ્ય પ્રમેયના અનુસાર $x \in $ [$0, 1$] અંતરાલમાં કયું વિધેય અનુસરતું નથી ?
${f}(x)\, = \left\{ {\begin{array}{*{20}{c}}{\frac{1}{2}\,\, - \,\,x,}&{x\,\, < \,\,\frac{1}{2}}\\{{{\left( {\frac{1}{2}\, - \,\,x} \right)}^2},}&{x\,\, \ge \,\,\frac{1}{2}}\end{array}} \right.$
${f}(x)\, = \,\frac{{\sin x}}{x};\,\,\,x\, \ne \,0,\,1\,;\,\,x\, = \,\,0$
$f(x) = x | x |$
$f(x) = | x |$
Solution
$\,{f}\left( {\frac{1}{2}} \right)\,\, = \,\,0\,$ આપેલ છે
$\begin{array}{l}
{\rm{LHD}}\,\,\,\,\mathop {{\rm{lim}}}\limits_{x \to \frac{1}{2} – 0} \,\frac{{{f}(x)\,\, – \,\,{f}\left( {\frac{1}{2}} \right)}}{{x\,\, – \,\,\frac{1}{2}}}\,\, = \,\, – \,1\\
RHD\,\,\,\mathop {\lim }\limits_{x \to \frac{1}{2} + 0} \,\frac{{{f}(x)\,\, – \,\,{f}\left( {\frac{1}{2}} \right)}}{{x\,\, – \,\,\frac{1}{2}}}\,\,\, = \,\,0
\end{array}$
$\therefore \,\,{f}$ એ $x\,\, = \,\,\frac{1}{2}$ આગળ વિકલનીય નથી
જેથી $\,\,{\rm{x}}\,\, = \,\,\frac{{\rm{1}}}{{\rm{2}}}\,\,{\rm{ }}$ આગળ ${\rm{LMV}}\,$ લાગુ પાડી શકાય નહીં
