Functions Question 654
Question: If $ f(x)= \begin{cases} & 1+x^{2},,when,0\le x\le 1 \\ & 1-x,,whenx>1 \\ \end{cases} . $ , then
Options:
A) $ \underset{x\to {1^{+}}}{\mathop{\lim }},f(x)\ne 0 $
B) $ \underset{x\to {1^{-}}}{\mathop{\lim }},f(x)\ne 2 $
C) $ f(x) $ is discontinuous at $ x=1 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \underset{x\to 1+}{\mathop{\lim }},f(x)=0 $ and $ =\underset{h\to 0}{\mathop{\lim }},h\frac{{e^{1/h}}-{e^{-1/h}}}{{e^{1/h}}+{e^{-1/h}}} $ Hence $ f(x) $ is discontinuous at $ x=1 $ .
BETA
