Functions Question 433
Question: If $ f(x)= \begin{cases} & -x^{2}, \text{when }x\le 0 \\ & 5x-4, when0 \lt x\le 1 \\ & 4x^{2}-3x, \text{when }1 \lt x \lt 2 \\ & 3x+4,\text{when }x\ge 2 \\ \end{cases} $ , then
Options:
A) $ f:R\to R $ is continuous at $ x=0 $
B) $ f(x) $ is continuous $ x=2 $
C) $ f(x) $ is discontinuous at $ x=1 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \underset{x\to 0-}{\mathop{\lim }} f(x)=0 $ $ f(0)=0,\underset{x\to 0+}{\mathop{\lim }} f(x)=-4 $ $ f(x) $ discontinuous at $ x=0. $ and $ \underset{x\to 1-}{\mathop{\lim }} f(x)=1 $ and $ \underset{x\to 1+}{\mathop{\lim }} f(x)=1,f(1)=1 $ Hence $ f(x) $ is continuous at $ x=1 $ . Also $ \underset{x\to 2-}{\mathop{\lim }} f(x)=4{{(2)}^{2}}-3,.,2=10 $ $ f(2)=10 $ and $ \underset{x\to 2+}{\mathop{\lim }},f(x)=3(2)+4=10 $ Hence $ f(x) $ is continuous at $ x=2. $
BETA
