Question

Let $f(x)=\{\begin{array}{ll}\frac{x-4}{|x-4|}+a,& x<4\\ a+b,& x=4\\ \frac{x-4}{|x-4|}+b,& x>4\end{array}$
Then $f(x)$ 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

Answer: (D)

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