Functions Question 32
Question: If $ f(x)=|x-3|, $ then f is
[SCRA 1996; RPET 1997]
Options:
A) Discontinuous at $ x=2 $
B) Not differentiable $ x=2 $
C) Differentiable at $ x=3 $
D) Continuous but not differentiable at $ x=3 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \underset{x\to {3^{-}}}{\mathop{\lim }}f(x)=\underset{h\to 0}{\mathop{\lim }},f(3-h)=\underset{h\to 0}{\mathop{\lim }},|3-h-3|=0 $ $ \underset{x\to {3^{+}}}{\mathop{\lim }}f(x)=\underset{h\to 0}{\mathop{\lim }},f(3+h)=\underset{h\to 0}{\mathop{\lim }},|3+h-3|=0 $ $ \because \underset{x\to {3^{-}}}{\mathop{\lim }}f(x)=\underset{x\to {3^{+}}}{\mathop{\lim }},f(x)=f(3) $ Hence f is continuous at $ x=3 $ Now $ L,{f}’(3)=\underset{h\to 0}{\mathop{\lim }},\frac{f(3-h)-f(3)}{-h} $ $ =\underset{h\to 0}{\mathop{\lim }},\frac{|3-h-3|-0}{-h}=\underset{h\to 0}{\mathop{\lim }}\frac{h}{-h}=-1 $ $ R,{f}’(3)=\underset{h\to 0}{\mathop{\lim }},\frac{f(3+h)-f(3)}{h} $ $ =\underset{h\to 0}{\mathop{\lim }},\frac{|3+h-3|-0}{h}=1 $ $ \because L,{f}’(3),\ne ,R,{f}’(3) $ . Hence f is not differentiable at $ x=3 $ . Trick : Can be seen by graph it is continuous but tangent is not defined at $ x=3 $ .
BETA
