Question

The points of discontinuity of the function \(f\) defined by \(f(x) = \begin{cases} x+2 & x≤1 \\ x-2 &1<x<2\\ 0& x≥2\end{cases}\) are:

• A. 0 and 1
• B. 1 and 2
• C. 1
• D. 2

Answer 1

The Correct Option is C

The given function \( f(x) \) is defined piecewise as: \[ f(x) = \begin{cases} x+2 & x \leq 1 \\ x-2 & 1 < x < 2 \\ 0 & x \geq 2 \end{cases} \] We need to find the points of discontinuity. A function is continuous at a point \( c \) if the following conditions are met:

• \( f(c) \) is defined.
• \(\lim_{{x \to c^-}} f(x) = \lim_{{x \to c^+}} f(x)\).
• \(\lim_{{x \to c}} f(x) = f(c)\).

Let's check at \( x = 1 \):

• \( f(1) = 1 + 2 = 3 \) (from \( x \leq 1 \) branch).
• \(\lim_{{x \to 1^-}} f(x) = 1 + 2 = 3\).
• \(\lim_{{x \to 1^+}} f(x) = 1 - 2 = -1\).

Since \(\lim_{{x \to 1^-}} f(x) \neq \lim_{{x \to 1^+}} f(x)\), \( f(x) \) is discontinuous at \( x = 1 \).

Now, check at \( x = 2 \):

• \( f(2) = 0 \) (from \( x \geq 2 \) branch).
• \(\lim_{{x \to 2^-}} f(x) = 2 - 2 = 0\).
• \(\lim_{{x \to 2^+}} f(x) = 0\) (same as \( f(x) = 0 \) for \( x \geq 2 \)).

Since \(\lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^+}} f(x) = f(2)\), \( f(x) \) is continuous at \( x = 2 \).

Therefore, the point of discontinuity is \( x = 1 \).

The correct answer is: 1