A) \[a=0,\text{ }b=0\]
B) \[a=1,\text{ }b=1\]
C) \[a=-\,1,\text{ }b=1\]
D) \[a=1,\text{ }b=-1\]
Correct Answer: D
Solution :
[d] We have L.H.L.=\[\underset{x\to 4}{\mathop{\lim }}\,f(x)\] =\[\underset{h\to 0}{\mathop{\lim }}\,f(4-h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{4-h-4}{\left| 4-h-4 \right|}+\alpha \] \[=\underset{h\to 0}{\mathop{\lim }}\,\left( -\frac{h}{h}+\alpha \right)=a-1\] R.H.L.\[=\underset{x\to 4}{\mathop{\lim }}\,f(x)\] \[=\underset{h\to 0}{\mathop{\lim }}\,f(4+h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{4+h-4}{\left| 4+h-4 \right|}+b=b+1\] \[\therefore f(4)=a+b\] 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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