Limits Continuity And Differentiability Question 32
Question: Which of the following functions is not differentiable at $ x=1 $ ?
Options:
A) $ f(x)=(x^{2}-1)| (x-1)(x-2) | $
B) $ f(x)=\sin (| x-1 |)-| x-1 | $
C) $ f(x)=\tan (| x-1 |)+| x-1 | $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(x)=(x^{2}-1)| (x-1)(x-2) | $
$ f’({1^{+}})=\underset{h\to 0}{\mathop{\lim }}\frac{({{(1+h)}^{2}}-1)| h\cdot (1+h-2) |-0}{h}=0,f’({1^{-}}) $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{({{(1-h)}^{2}}-1)| -h\cdot (1-h-2) |-0}{-h}=0 $
Hence, it is differentiable at x = 0
For, $ f(x)=sin(| x-1 |)-| x-1 | $
$ f’({0^{+}})=\underset{h\to 0}{\mathop{lim}}\frac{\sin h-h-0}{h}=0,f’({0^{-}}) $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{\sin | -h |-| -h |}{-h}=0\underset{h\to 0}{\mathop{\lim }}\frac{\sin h-h}{-h}=0 $
Hence, $ f(x) $ is differentiable at $ x=0 $
For $ f(x)=\tan (| x-1 |)+| x-1 | $
$ f’({0^{+}})=\underset{h\to 0}{\mathop{\lim }}\frac{\tan h+h-0}{h}=2 $ ,
$ f’({0^{-}})=\underset{h\to 0}{\mathop{\lim }}\frac{\tan | -h |+| -h |}{-h}=\underset{h\to 0}{\mathop{\lim }}\frac{\tan h+h}{-h}=-2 $
BETA
