Functions Question 160
Question: Let $ f(x)= \begin{cases} & \frac{x-4}{|x-4|}+a,\ x<4 \\ & a+b,,x=4 \\ & \frac{x-4}{|x-4|}+b,,x>4 \\ \end{cases} . $ . Then $ f(x) $ is continuous at $ x=4 $ when
Options:
A) $ a=0,\ b=0 $
B) $ a=1,\ b=1 $
C) $ a=-1,\ b=1 $
D) $ a=1,\ b=-1 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \underset{x\to 4-}{\mathop{\lim }},f(x)=\underset{h\to 0}{\mathop{\lim }},f(4-h)=\underset{h\to 0}{\mathop{\lim }},\frac{4-h-4}{|4-h-4|}+a $ $ =\underset{h\to 0}{\mathop{\lim }},-\frac{h}{h}+a=a-1. $ $ =\underset{x\to 4+}{\mathop{\lim }},f(x)=\underset{h\to 0}{\mathop{\lim }},f(4+h)=\underset{h\to 0}{\mathop{\lim }}\frac{4+h-4}{|4+h-4|}+b=b+1 $ and $ f(4)=a+b $ Since $ f(x) $ is continuous at $ x=4 $ Therefore $ \underset{x\to 4-}{\mathop{\lim }},f(x)=f(4)=\underset{x\to 4+}{\mathop{\lim }},f(x) $
$ \Rightarrow a-1=a+b=b+1\Rightarrow b=-1 $ and $ a=1. $
BETA
