# Solve this

Question:

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) $a=1, b=-1$.

Solution:

(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 $a=1$