Limits Continuity And Differentiability Question 21
Question: Let $ f(x)= \begin{matrix} \frac{x-4}{| x-4 |}+a,x4 \\ \end{matrix} . $ Then $ f(x) $ is continuous at x=4
Options:
A) $ a=0,b=0 $
B) $ a=1,b=1 $
C) $ a=-1,b=1 $
D) $ a=1,b=-1 $
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Answer:
Correct Answer: D
Solution:
We have L.H.L.= $ \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 |}+\alpha $
$ =\underset{h\to 0}{\mathop{\lim }}( -\frac{h}{h}+\alpha )=a-1 $ R.H.L. $ =\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 $
$ \therefore f(4)=a+b $ Since $ f(x) $ is continuous at x=4, $ \underset{x\to {4^{-}}}{\mathop{\lim }}f(x)=f(4)=\underset{x\to {4^{+}}}{\mathop{\lim }}f(x) $
or $ a-1=a +b=b+1 $ or $ b=-1 $ and $ a=1 $
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