Let `f(x)={(x-4)/(|x-4|a+b)+a ,x<0 (x-4)/(|x-4|)+b ,x >0` Then, `f(x)` is continuous at `x=4` when (a)`a=0,b=0` (b) `a=1,b=1` (c)`a=-1,b=1` (d) `a=1,b=-1`

Answer

`=> b=-1` and ` a=1`

Explanation

$$ \begin{aligned} \lim _{x \rightarrow 4^{-}} f(x) &=\lim _{h \rightarrow 0} f(4-h)=\lim _{h \rightarrow 0} \frac{4-h-4}{|4-h-4|}+a \\ &=\lim _{h \rightarrow 0}-\frac{h}{h}+a=a-1 \\ \lim _{x \rightarrow 4+} f(x) &=\lim _{h \rightarrow 0} f(4+h)=\lim _{h \rightarrow 0} \frac{4+h-4}{|4+h-4|}+b=b+1 \end{aligned} $$
`f(4)=a+b`

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