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
(a) $a=0, b=0$
(b) $a=1, b=1$
(c) $a=-1, b=1$
(d) $a=1, b=-1$.
(d) $a=1, b=-1$
Given: $f(x)=\left\{\begin{array}{c}\frac{x-4}{|x-4|}+a, \text { if } \mathrm{x}<4 \\ a+b, \text { if } \mathrm{x}=4 \\ \frac{x-4}{|x-4|}+b, \text { if } \mathrm{x}>4\end{array}\right.$
We have
$(\mathrm{LHL}$ at $x=4)=\lim _{x \rightarrow 4^{-}} f(x)=\lim _{h \rightarrow 0} f(4-h)$
$=\lim _{h \rightarrow 0}\left(\frac{4-h-4}{|4-h-4|}+a\right)=\lim _{h \rightarrow 0}\left(\frac{-h}{|-h|}+a\right)=a-1$
$(\mathrm{RHL}$ at $x=4)=\lim _{x \rightarrow 4^{+}} f(x)=\lim _{h \rightarrow 0} f(4+h)$
$=\lim _{h \rightarrow 0}\left(\frac{4+h-4}{|4+h-4|}+b\right)=\lim _{h \rightarrow 0}\left(\frac{h}{|h|}+b\right)=b+1$
Also,
$f(4)=a+b$
If $f(x)$ is continuous at $x=4$, then
$\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{+}} f(x)=f(4)$
$\Rightarrow a-1=b+1=a+b$
$\Rightarrow a-1=a+b$ and $b+1=a+b$
$\Rightarrow b=-1$ and $a=1$