STD 12 - 5. continuity and differentiation — JEE कक्षा 11 साइन्स — Question

$ = \left\{ {\begin{array}{*{20}{c}}{ - 2x + 1,}&{x < 0}&{}\\{x - x + 1,}&{0 \le x < 1}& = \\{x + x - 1,}&{x \ge 1}&{}\end{array}} \right.\left\{ {\begin{array}{*{20}{c}}{ - 2x + 1,}&{x < 0}\\1&{0 \le x < 1}\\{2x - 1,}&{x \ge 1}\end{array}} \right.$

स्पष्टत: $\mathop {\lim }\limits_{x \to {0^ - }} f(x) = 1,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f(x) = 1,\,\,\mathop {\lim }\limits_{x \to {1^ - }} f(x) = 1$

तथा $\mathop {\lim }\limits_{x \to {1^ + }} f(x) = 1$. अत: $f(x)$, $x = 0,\,\,1$ पर सतत् है

अब $f'(x) = \left\{ {\begin{array}{*{20}{l}}{ - 2,\,\,\,\,x < 0}\\{\,\,0,\,\,\,\,\,0 \le x < 1}\\{\,\,2,\,\,\,\,\,x \ge 1}\end{array}} \right.$

यहाँ $x = 0$, $f'({0^ + }) = 0$ जबकि $f'({0^ - }) = - 2$

तथा $x = 1$, $f'({1^ + }) = 2$ जबकि $f'({1^ - }) = 0$

अत: $f(x)$,$ x = 0$ तथा $1$ पर अवकलनीय नहीं है।

$\therefore $ विकल्प $ (a) $ सही है।