#### Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 28 Maths Textbook Solution.

$a=1, b=-1$

Hint:

$f\left ( x \right )$  must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$  must exist.

Given:

$f(x)=\left\{\begin{array}{r} \frac{x-4}{|x-4|}+a, \text { if } x<4 \\\\ a+b, \text { if } x=4 \\\\ \frac{x-4}{|x-4|}+b, \text { if } x>4 \end{array}\right.$

Solution:

$f(x)=\left\{\begin{array}{r} \frac{x-4}{|x-4|}+a, \text { if } x<4 \\\\ a+b, \text { if } x=4 \\\\ \frac{x-4}{|x-4|}+b, \text { if } x>4 \end{array}\right.$

We observe

(LHL at $x=4$ )

\begin{aligned} \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 \end{aligned}

(RHL at $x=4$ )

\begin{aligned} &\lim _{x \rightarrow 4^{+}} f(x)=\lim _{h \rightarrow 0} f(4+h) \\\\ &\quad=\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 \\\\ &f(4)=a+b \end{aligned}

If $f\left ( x \right )$ is continuous at  $x=4$

\begin{aligned} &\lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4^{-}} f(x)=f(4) \\ &a-1=b+1=a+b \\ &a-1=a+b, b+1=a+b \\ &b=-1, a=1 \end{aligned}