#### Please solve RD Sharma class 12 chapter 8 Continuity exercise Fill in the blanks question 5 maths textbook solution

Answer: $b=-1, a=1,a+b=0$

Hint: $\left | x \right |=x$, When $x< 0$ and $\left | x \right |=x$ , when $x> 0$
Given:

$f(x)=\left\{\begin{array}{l} \frac{x-4}{|x-4|}+a, x<4 \\ a+b, x=4 \quad \text { is continuous at } x=4 \\ \frac{x-4}{|x-4|}+b, x>4 \end{array}\right.$

Solution:

As the function is continuous at $x=4$,

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

$\lim _{h \rightarrow 0} \frac{(4-h)-4}{|(4-h)-4|}+a=\lim _{h \rightarrow 0} \frac{(4+h)-4}{|(4+h)-4|}+b=\lim _{x \rightarrow 4} a+b$

\begin{aligned} &\lim _{h \rightarrow 0} \frac{-h}{|h|}+a=\lim _{x \rightarrow 4^{+}} \frac{h}{|h|}+b=a+b \\ &-1+a=1+b=a+b \end{aligned}

\begin{aligned} &a+b=-1+a \quad, \quad a+b=1+b \\ &b=-1 \quad, \quad a=1 \end{aligned}

Now, $a+b=1-1=0$