The function defined by $f(x)=\left\{\begin{array}{cc} \frac{x-4}{|x-4|}+a & \text { if } x 4…
The function defined by
$f(x)=\left\{\begin{array}{cc}
\frac{x-4}{|x-4|}+a & \text { if } x<4 \\
a+b & \text { if } x=4 \\
\frac{x-4}{|x-4|}+b & \text { if } x>4
\end{array}\right.$
is continuous at $x=4$, are
$a=0, b=1$
$\mathrm{a}=1, \mathrm{~b}=0$
$\mathrm{a}=1, \mathrm{~b}=-1$
$\mathrm{a}=-1, \mathrm{~b}=0$
Solution
L.H.L at $x=4$
$\lim _{h \rightarrow 0} \frac{4-h-4}{|4-h-4|}+a=\lim _{h \rightarrow 0} \frac{-h}{|-h|}+a=\lim _{h \rightarrow 0} \frac{-h}{h}+a=-1+a$
R.H.L at $x=4$
$\lim _{h \rightarrow 0} \frac{4+h-4}{|4+h-4|}+b=\lim _{h \rightarrow 0} \frac{h}{|h|}+b=\lim _{h \rightarrow 0} \frac{h}{h}+b=1+b$
For continuity at $\mathrm{x}=4$
$-1+\mathrm{a}=\mathrm{a}+\mathrm{b}=1+\mathrm{b} \Rightarrow \mathrm{a}=1$ and $\mathrm{b}=-1$