A) 0
B) 1
C) x
D) None of these
Correct Answer: D
Solution :
\[f(x)=|x|,\] we have \[f(0)=|0|=0\] \[f(0+0)=\underset{h\to 0}{\mathop{\lim }}\,|0+h|=0\] and \[f(0-0)=\underset{h\to 0}{\mathop{\lim }}\,|0-h|=0\] \[Rf'(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{|h|-0}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{h}{h}\](h being positive)=1 \[Lf'(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0-h)-f(0)}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{|h|-0}{-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{h}{-h}(h\]being positive) = ?1. \[\therefore Rf'(0)\ne Lf'(0)\]. The function f is not differentiable.
You need to login to perform this action.
You will be redirected in
3 sec
