Functions Question 209
Question: A function $ f(x)= \begin{cases} 1+x, & x\le 2 \\ 5-x, & x>2 \\ \end{cases} ,$ is
[AMU 2001]
Options:
A) Not continuous at $ x=2 $
B) Differentiable at $ x=2 $
C) Continuous but not differentiable at $ x=2 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \underset{h\to {0^{-}}}{\mathop{\lim }}1+(2-h)=3 $ , $ \underset{h\to {0^{+}}}{\mathop{\lim }}5-(2+h)=3 $ , $ f(2)=3 $ Hence, f is continuous at $ x=2 $
Now $ R{f}’(x)=\underset{h\to 0}{\mathop{\lim }}\frac{5-(2+h)-3}{h}=-1 $
$ L{f}’(x)=\underset{h\to 0}{\mathop{\lim }}\frac{1+(2-h)-3}{-h}=1 $ $ \because R{f}’(x)\ne L{f}’(x) $ ; f is not differentiable at $ x=2 $ .
BETA
