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Let I be set of integers, $N =$ the set of non-negative integers, $N_{p} =$ the set of non-positive integers. Then
the sets $A$ and $B$ satisfying $A \cap B = \emptyset$ are

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$A = I ~ N p, B = N ~ N p$

$A = I ~ N, B = I ~ N p$

$A = N ∆ N p, B = I ~ N p$

$A = N ∆ N p, B = (I ~ N) \cup \{0\}$

answer is B.

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Detailed Solution

$I ~ N = {\dots - 3, - 2, - 1}, I ~ N p = {1, 2, 3 \dots}$ ,

$\begin{array}{l} N ~ N_{p} = {1, 2, \dots}, N Δ N_{p} = (N ~ N_{p}) \cup (N_{p} ~ N) \\ = {0, 1, 2, \dots} \\ (I ~ N) \cup {0} = {\dots, - 3, - 2, - 1, 0} \end{array}$

The disjoint sets are $I ~ N$ and $I ~ N p .$

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