Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

Solve this

Question:

If $f(x)= \begin{cases}\frac{x-4}{|x-4|}+a, & \text { if } \quad x<4 \\ a+b \quad, & \text { if } \quad x=4 \text { is continuous at } x=4, \text { find } a, b . \\ \frac{x-4}{|x-4|}+b, & \text { if } \quad x>4\end{cases}$

Solution:

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 observe

$(\mathrm{LHL}$ 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 $x=4)=\lim _{x \rightarrow 4^{+}} f(x)=\lim _{h \rightarrow 0} f(4+h)$

$=\lim _{h \rightarrow 0}\left(\frac{4+h-4}{|4+h-4|}+b\right)=\lim _{h \rightarrow 0}\left(\frac{h}{|h|}+b\right)=b+1$

And $f(4)=a+b$

If $f(x)$ is continuous at $x=4$, then

$\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{+}} f(x)=f(4)$

$\Rightarrow a-1=b+1=a+b$

$\Rightarrow a-1=a+b, b+1=a+b$

$\Rightarrow b=-1, a=1$