Functions Question 73
Question: The function $ f(x)= \begin{cases} & x,\text{if 0}\le x\le 1 \\ & \text{1,} \text{ if} 1 \lt x\le 2 \\ \end{cases} $ is
[SCRA 1996]
Options:
A) Continuous at all x, $ 0\le x\le 2 $ and differentiable at all x, except $ 1 $ in the interval [0,2]
B) Continuous and differentiable at all x in [0,2]
C) Not continuous at any point in [0,2]
D) Not differentiable at any point [0,2]
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Answer:
Correct Answer: A
Solution:
$ f(x)= \begin{cases} x\text{ ,} & 0\le x\le 1 \\ 1\text{ ,} & 1\ lt x\le 2 \\ \end{cases} . $ $ \underset{x\to {1^{-}}}{\mathop{\lim }} f(x)=\underset{h\to 0}{\mathop{\lim }} f(1-h) $ $ =\underset{h\to 0}{\mathop{\lim }} (1-h)=1 $ $ \underset{x\to {1^{+}}}{\mathop{\lim }} f(x)=\underset{h\to 0}{\mathop{\lim }}f(1+h)=1 $ Hence function is continuous in (0, 2). Now $ \underset{x\to {0^{+}}}{\mathop{\lim }} f(x)=\underset{h\to 0}{\mathop{\lim }}(0+h)=0=f(0) $ $ \underset{x\to {2^{-}}}{\mathop{\lim }} f(x)=\underset{h\to 0}{\mathop{\lim }}(2-h)=1=f(2) $ Hence function is continuous in [0, 2] Clearly, from graph it is not differentiable at $ x=1. $
BETA
