A) \[a=0,\ b=0\]
B) \[a=1,\ b=1\]
C) \[a=-1,\ b=1\]
D) \[a=1,\ b=-1\]
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 \[a=1.\]
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