What is the value of a?

Consider the following for the next two (02) items that follow: Let f(x)=\begin{cases}\frac{x-3}{|x-3|}+a,&x \lt 3\\ a-b,&x=3\\ \frac{x-3}{|x-3|}+b,&x \gt 3\end{cases}. The function f(x) is continuous at x = 3.

  1. A. -1
  2. B. 1
  3. C. 2
  4. D. 3

Correct Answer: D. 3

Explanation

LHL at x=3 is \lim_{x \rightarrow 3-} (\frac{x-3}{-(x-3)} + a) = a-1. RHL at x=3 is \lim_{x \rightarrow 3+} (\frac{x-3}{x-3} + b) = b+1. The function value is f(3) = a-b. For continuity, a-1 = b+1 = a-b. From a-1 = a-b, we get b = 1. Substituting b=1 into b+1 = a-1 gives 1+1 = a-1 \implies a = 3.

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