Let $f(x)=\left\{\begin{array}{cl}\frac{x-3}{|x-3|}+a ; & x \lt 3 \\ a-b ; & x=3 \text { and } \\ \frac{x-3}{|x-3|}+b ; & x \gt 3\end{array}\right.$ $f(x)$ be continuous at $x=3$. What is the value of b?

• A

  -1
• Correct answer
  B

  1
• C

  2
• D

  3

Hence, the correct answer is option: 'B'.