यदि $f(x) = \left\{ \begin{array}{l}x\frac{{{e^{(1/x)}} - {e^{( - 1/x)}}}}{{{e^{(1/x)}} + {e^{( - 1/x)}}}},\;\,x \ne 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,\,\;x = 0\end{array} \right.$ तो निम्न कथन सत्य है
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Answer
$f(0 + 0) = \mathop {\lim }\limits_{h \to 0} f(x) = \mathop {\lim }\limits_{h \to 0} f(0 + h)$
$ = \mathop {\lim }\limits_{h \to 0} \,\,(0 + h)\,\frac{{{e^{1/0 + h}} - {e^{ - 1/0 + h}}}}{{{e^{1/0 + h}} + {e^{ - 1/0 + h}}}}$
$ = \mathop {\lim }\limits_{h \to 0} \,\,h\,\,\frac{{{e^{1/h}} - {e^{ - 1/h}}}}{{{e^{1/h}} + {e^{ - 1/h}}}} =0$
तथा $f(0 - 0) = \mathop {\lim }\limits_{h \to 0} f(0 - h) $
$= \mathop {\lim }\limits_{h \to 0} \,\, - h\,\,\frac{{{e^{ - 1/h}} - {e^{1/h}}}}{{{e^{ - 1/h}} + {e^{1/h}}}} = 0$
तथा $f(0) = 0$;
$\therefore \,\,\,f(0 + 0) = f(0 - 0) = f(0)$
$f, x = 0$ पर सतत् है। शेष बिन्दुओं पर फलन $f(x)$ स्पष्टत: सतत् होगा।
इस प्रकार $f(x)$ प्रत्येक बिन्दु पर संतत है।
पुन: $L\,f'(0) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f(0 - h) - f(0)}}{{ - h}}$
$ = \mathop {\lim }\limits_{h \to 0} \,\frac{{h\,.\,\frac{{{e^{ - 1/h}} - {e^{1/h}}}}{{{e^{ - 1/h}} + {e^{1/h}}}} - 0}}{{ - h}} = - 1$
$R\,f'(0) = \mathop {\lim }\limits_{h \to 0} \,\frac{{f\,(0 + h) - f\,(0)}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{h\,\,\frac{{{e^{1/h}} - {e^{ - 1/h}}}}{{{e^{1/h}} + {e^{ - 1/h}}}}}}{h} = 1$
$\because \,\,L\,\,{f}'(0)\ne R\,{f}'(0)$
$\therefore $ फलन $f, x = 0$ पर अवकलनीय नहीं है।
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