# Find the values of $a$ and $b$ such that the function $f$ defined by $f(x)=\left \{\begin{array}{1 1}\large\frac{x-4}{|x-4|}\normalsize+a, & if\;x<4\\a+b, & if\;x=4\\\large\frac{x-4}{|x-4|}\normalsize +b, & if\;x>4\end{array}\right.$ is a continuous function at $x=4.$

$\begin{array}{1 1} a=-1 ; b=1 \\ a=-1 ; b=-1 \\ a=1 ; b=1 \\ a=1 ; b=-1\end{array}$

Toolbox:
• A function is said to be continuous if the LHL = RHL.
• (i.e) $\lim\limits_{\large x\to a^+}f(x)=\lim\limits_{\large x\to a^-}f(x)$
Step 1:
Given : $f(x)=\left \{\begin{array}{1 1}\large\frac{x-4}{|x-4|}\normalsize+a, & if\;x<4\\a+b, & if\;x=4\\\large\frac{x-4}{|x-4|}\normalsize +b, & if\;x>4\end{array}\right.$
$f(x)=\large\frac{x-4}{-( x-4)}$$+a,if x < 4 \quad\;\;\;=-1+a f(x)=-1+a is a constant function,is continuous at each point x < 4 Step 2: f(x)=\large\frac{x-4}{ x-4}$$+b$,if $x > 4$
$\quad\;\;\;=1+b$
$f(x)=1+b$ is a constant function,is continuous at each point $x > 4$
Step 3:
Let us consider the point $x=4$
$f(x)=a+b$
$f(4)=0$
$\Rightarrow a+b=0$------(1)
Since the function is continuous.
LHL = RHL.
(i.e) $-1+a=1+b$
$a-b=2$-----(2)
Step 4:
Solving equ(1) & equ(2)
$a+b=0$
$a-b=2$
______________
$2a=2$
$a=1$
$a+b=0$
$1+b=0$
$b=-1$
Hence the values of $a=1$ and $b=-1$