Limits Continuity And Differentiability Question 74
Question: Which of the following function 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) | $
$ =(x^{2}-1)| (x-1)(x-2) | $
$ =(x+1)[(x-1)| x-1 |]| x-2 | $
Which is differentiable at x=1. For $ f(x)=sin(| x-1 |)-| x-1 |, $
$ f’({1^{+}})=\underset{h\to 0}{\mathop{\lim }}\frac{\sin h-h-0}{h}=0 $
$ f’({1^{-}})=\underset{h\to 0}{\mathop{\lim }}\frac{\sin h| -h |-| -h |}{-h}=\underset{h\to 0}{\mathop{\lim }}\frac{\sin h-h}{-h}=0 $
Hence, $ f(x) $ is differentiable at x=1.
For $ f(x)=tan(| x-1 |)+| x-1 | $ , $ f’({1^{+}})=\underset{h\to 0}{\mathop{\lim }}\frac{tanh+h-0}{h}=2 $
$ f’({1^{-}})=\underset{h\to 0}{\mathop{\lim }}\frac{\sin h| -h |-| -h |}{-h} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{\tan h+h}{-h}=-2 $
Hence, f(x) is non-differentiable at x=1.
BETA
