Question

Find the value of \( a \) and \( b \) so that the function \( f \) defined as: \[ f(x) = \begin{cases} \frac{x - 2}{|x - 2|} + a, & x < 2, \\ a + b, & x = 2, \\ \frac{x - 2}{|x - 2|} + b, & x > 2, \end{cases} \] is a continuous function.

Hint:

For piecewise functions, equate LHL, RHL, and the value at the point to ensure continuity.

Answer 1

For \( f(x) \) to be continuous at \( x = 2 \), the left-hand limit (LHL), right-hand limit (RHL), and the value of \( f(2) \) must all be equal. 1. Left-hand limit (LHL): For \( x<2 \): \[ f(x) = \frac{x - 2}{|x - 2|} + a = -1 + a. \] As \( x \to 2^- \): \[ {LHL} = -1 + a. \] 2. Right-hand limit (RHL): For \( x>2 \): \[ f(x) = \frac{x - 2}{|x - 2|} + b = 1 + b. \] As \( x \to 2^+ \): \[ {RHL} = 1 + b. \] 3. Value at \( x = 2 \): \[ f(2) = a + b. \] 4. Continuity condition: \[ {LHL} = {RHL} = f(2). \] Substitute: \[ -1 + a = 1 + b = a + b. \] From \( -1 + a = a + b \): \[ b = -1. \] From \( 1 + b = a + b \): \[ a = 1. \] Final Answer: \( \boxed{a = 1, b = -1} \)