Limits Continuity And Differentiability Question 5
Question: Consider the function $ f(x)= \begin{cases} x^{2}, & x>2 \\ 3x-2, & x\le 2 \\ \end{cases} . $ . Which one of the following statements is correct in respect of the above function?
Options:
A) f(x) is derivable but not continuous at x = 2.
B) f(x) is continuous but not derivable at x = 2.
C) f(x) is neither continuous nor derivable at x = 2.
D) f(x) is continuous as well as derivable at x = 2.
Show Answer
Answer:
Correct Answer: B
Solution:
First we check continuity at $ x=2 $
L.H.L. = $ \underset{h\to 0}{\mathop{\lim }}f(2-h)=\underset{h\to 0}{\mathop{\lim }}3(2-h)-2 $
$ =\underset{h\to 0}{\mathop{\lim }}4-3h=4 $
R.H.L. $ =\underset{h\to 0}{\mathop{\lim }}f(2+h)=\underset{h\to 0}{\mathop{\lim }}{{(2+h)}^{2}}=4 $
Also, $ f(2)={{(2)}^{2}}=4 $
Since, L.H.L = R.H.L. = f(2)
$ \therefore $ f(x) is continuous at 2.
Now, we check for differentiability
L.H.D at x = 2 R.H.D at x = 2
$ f’(x)=3x-2 $
$ f’(x)=x^{2} $ .
$ f’(x)=3 $
$ f’(x)=2x $
$ f’(x)| _{x=2} .=3 $
$ f’(x)| _{x=2} .=4 $
Since L.H.D $ \ne $ R.H.D
$ \therefore f(x) $ is not derivable at $ x=2 $
BETA
