Q. If $f(x) = \left\{ \begin{array}{r} x,\ \ \ \ \ \ \ \ \ \ for\ \ \ \ \ \ \ \ 0 < x < 1 \\ 2 - x,\ \ \ for\ \ \ \ \ \ \ \ \ 1 \leq x < 2 \\ x - (1/2)x^{2},\ \ for\ \ x = 2\ \ \ \end{array} \right.\$.Then, $f^{'}(1)$ is equal to
A) -1
B) 1
C) 0
D) None of these
Explanation
(d)
At $x = 1$,
RHD$= \lim_{h{\rightarrow 0}^{+}}\frac{f(1 + h) - f(1)}{h}$
$= \lim_{h \rightarrow 0}\frac{2 - (1 + h) - (2 - 1)}{h} = - 1$
LHD$= \lim_{h{\rightarrow 0}^{-}}\frac{f(1 - h) - f(1)}{- h}$
$= \lim_{h \rightarrow 0}\frac{(1 - h) - (2 - 1)}{- h} = 1$
$\therefore$ LHD$\neq$RHD
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.