Question

Let $f\left(x\right)=\{\begin{array}{cc}\frac{x-4}{|x-4|}+a;& x<4\\ a+b& ;x=4\\ \frac{x-4}{|x-4|}+b& ;x>4\end{array}$ .Then $f\left(x\right)$ is continuous at $x=4$ when

• A. $a=0,b=0$
• B. $a=1,b=1$
• C. $a=-1,b=1$
• D. $a=1,b=-1$

Answer 1

The correct option is D $a=1,b=-1$

We have
$L.H.L.=\underset{x\to 4^{-}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}f(4-h)$
$=\underset{h\to 0}{lim}\frac{4-h-4}{4-h-4}+a$
$=\underset{h\to 0}{lim}(-\frac{h}{h}+a)=a-1$

$R.H.L.=\underset{x\to 4^{=}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}f(4+h)$
$=\underset{h\to 0}{lim}\frac{4+h-4}{|4+h-4|}+b=b+1$
$\therefore f\left(4\right)=a+b$
Since f(x) is continuous at $x=4$,
$\underset{x\to 4^{-}}{lim}f\left(x\right)=f\left(4\right)=\underset{x\to 4^{+}}{lim}f\left(x\right)$
or $a-1=a+b=b+1$ or $b=-1$ and $a=1$