Limits Continuity And Differentiability Question 104
Question: Which one of the following is correct in respect of the function $ f(x)=| x |+x^{2} $
Options:
A) $ f(x) $ is not continuous at x = 0
B) $ f(x) $ is differentiable at x = 0
C) $ f(x) $ is continuous but not differentiable at x = 0
D) None of the above
Show Answer
Answer:
Correct Answer: C
Solution:
$ \because f(x)=| x |+x^{2} $
$ \Rightarrow f(x)= \begin{matrix} x^{2}+x, & x\ge 0 \\ x^{2}-x, & x<0 \\ \end{matrix} . $ LHL $ =\underset{x\to {0^{-}}}{\mathop{\lim }}f(x) $
$ =\underset{h\to {0^{-}}}{\mathop{\lim }}f(0-h)=\underset{h\to 0}{\mathop{\lim }}{{(0-h)}^{2}}-(0-h) $
$ =\underset{h\to 0}{\mathop{\lim }}h^{2}+h=0 $ and RHL $ =\underset{x\to 0}{\mathop{\lim }}f(x)=\underset{h\to 0}{\mathop{\lim }}f(0+h) $
$ =\underset{h\to 0}{\mathop{\lim }}{{(0+h)}^{2}}+(0+h) $
$ =\underset{h\to 0}{\mathop{\lim }}h^{2}+h=0 $
$ \Rightarrow LHL=RHL=f(0) $
$ \Rightarrow f(x) $ is continuous at $ x=0 $ Now, $ LHD=\underset{h\to 0}{\mathop{\lim }}\frac{f(0-h)-f(0)}{-h} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{h^{2}+h}{-h}=-\underset{h\to 0}{\mathop{\lim }}h+1=-1 $ and, $ RHD=\underset{h\to 0}{\mathop{\lim }}\frac{f(0+h)-f(0)}{h} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{h^{2}+h}{h}=\underset{h\to 0}{\mathop{\lim }}h+1=1 $ Thus, $ LHD\ne RHD $
$ \Rightarrow f(x) $ is not differentiable at $ x=0 $
BETA
