Question
फलन $f(x)\, = \left\{ {\begin{array}{*{20}{c}}{1 + x,}&{x \le 2}\\{5 - x,}&{x > 2}\end{array}} \right.\,$
$\mathop {\lim }\limits_{h \to {0^ + }} 5 - (2 + h) = 3$, $f(2) = 3$
अत: $f$, $x = 2$ पर सतत् है
अब $Rf'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{5 - (2 + h) - 3}}{h} = - 1$
$Lf'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{1 + (2 - h) - 3}}{{ - h}} = 1$
$\because R{f}'(x)\ne L{f}'(x)$
अत: $x = 2$ पर $f$ अवकलनीय नहीं है।
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$\int_0^{\pi / 2} \mathrm{f}(\sin 2 \mathrm{x}) \cdot \sin \mathrm{xdx}+\alpha \int_0^{\pi / 4} \mathrm{f}(\cos 2 \mathrm{x}) \cdot \cos \mathrm{xdx}=0$
है, तो $\alpha$ का मान है
