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:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) - 1
(i) at x = 1
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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
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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 $x=2$.
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$x=1$: Discontinuous (limits do not match).
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$x=2$: Continuous.
Therefore, the only point of discontinuity is $1$, making Option 3 the correct answer.