Q. Let $f(x) = \left\{ \begin{array}{r} \frac{x - 4}{|x - 4|} + a,\ \ x < 4 \\ a + b,\ \ \ x = 4 \\ \frac{x - 4}{|x - 4|} + b,\ \ \ x > 4 \end{array} \right.\$ 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) $a = 1,\ b = - 1$
Explanation
(d)
$f(x) = \left\{ \begin{array}{r}
\frac{x - 4}{|x - 4|} + a,\ \ x < 4 \\
a + b,\ \ \ x = 4 \\
\frac{x - 4}{|x - 4|} + b,\ \ \ x > 4
\end{array} \right.\ = \left\{ \begin{array}{r}
- 1 + a,\ x < 4 \\
a + b \\
1 + b,\ x > 4
\end{array} \right.\$
LHL$= \lim_{x \rightarrow 4^{\mp}}\ f(x) = a - 1$
RHL$= \lim_{x \rightarrow 4^{\mp}}\ f(x) = 1 + b$
Since, LHL=RHL$= f(4)$
$\Rightarrow \ \ a - 1 = a + b = b + 1$
$a = 1$ and $b = - 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.