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

1

The correct answer is Option (3) - 1

(i) at x = 1

• Left-hand limit ($x \to 1^-$): We use the definition $f(x) = x + 2$.

  $\lim_{x \to 1^-} (x + 2) = 1 + 2 = 3$

   

• Right-hand limit ($x \to 1^+$): We use the definition $f(x) = x - 2$.

  $\lim_{x \to 1^+} (x - 2) = 1 - 2 = -1$

   

  Since the left-hand limit ($3$) is not equal to the right-hand limit ($-1$), there is a jump discontinuity at $x=1$.

(ii) at x = 2

• Left-hand limit ($x \to 2^-$): We use the definition $f(x) = x - 2$.

  $\lim_{x \to 2^-} (x - 2) = 2 - 2 = 0$

   

• Right-hand limit ($x \to 2^+$): We use the definition $f(x) = 0$.

Function value ($f(2)$): Based on the definition $x \geq 2$, $f(2) = 0$. 

Since the left-hand limit, the right-hand limit, and the function value are all equal ($0$), the function is continuous at $x=2$.

• $x=1$: Discontinuous (limits do not match).

• $x=2$: Continuous.

Therefore, the only point of discontinuity is $1$, making Option 3 the correct answer.