On the Venn diagram, shade the region which represents (A∪B′)∩(A′∪B)(A \cup B') \cap (A' \cup B). E={(x,y):x+y≤10}E = \{(x,y): x + y \le 10\} A={(0,1),(1,2),(1,4),(1,5)}A = \{(0,1), (1,2), (1,4), (1,5)\} B={(1,0),(1,2),(1,3),(2,1),(3,7)}B = \{(1,0), (1,2), (1,3), (2,1), (3,7)\} Underline the correct statements from the list below. A ∩ B = (1, 2) {1, 5} ⊂ A ∅ ⊂ B (2, 6) ∈ (A ∨ B)' A ⊂ B

Work through it with AI →

It won't just hand you the answer — it asks where you're stuck.How the checker works

Worked solution

10(a)

1. Identify the set expression as the regions where membership of $A$ and $B$ is the same.

(A∪B′)∩(A′∪B)(A \cup B') \cap (A' \cup B)

For each Venn region:

  • A∩BA \cap B: in AA and in BB, so it is in both A∪B′A \cup B' and A′∪BA' \cup B.
  • A∩B′A \cap B': in AA but not in BB, so it is not in A′∪BA' \cup B.
  • A′∩BA' \cap B: not in AA but in BB, so it is not in A∪B′A \cup B'.
  • A′∩B′A' \cap B': not in AA and not in BB, so it is in both A∪B′A \cup B' and A′∪BA' \cup B.

Hence the required region is

(A∩B)∪(A′∩B′).(A \cap B) \cup (A' \cap B').
Answer

Shade the regions A∩BA \cap B and A′∩B′A' \cap B' only. Leave the two non-overlapping parts of AA and BB unshaded.

10(b)

1. Find the common elements of $A$ and $B$ and check the first statement.

The only ordered pair that appears in both sets is (1,2)(1,2).

So
A∩B={(1,2)}.A \cap B = \{(1,2)\}.

The statement A∩B=(1,2)A \cap B = (1,2) is not correct, since A∩BA \cap B should be a set, not an ordered pair.

2. Check the subset statements involving $A$ and $B$.

For {1,5}⊂A\{1,5\} \subset A, the elements 11 and 55 are not elements of AA; the elements of AA are ordered pairs. So {1,5}⊂A\{1,5\} \subset A is false.

Since BB is non-empty, the empty set is a proper subset of BB.

So ∅⊂B\varnothing \subset B is true.

For A⊂BA \subset B, not every element of AA is in BB. For example, (0,1)∈A(0,1) \in A but (0,1)∉B(0,1) \notin B. So A⊂BA \subset B is false.

3. Check whether $(2,6)$ is in the complement of $A \cup B$ within $E$.

First check that (2,6)(2,6) is in the universal set EE:
2+6=8≤10.2+6=8 \leq 10.

So (2,6)∈E(2,6) \in E.

Also, (2,6)(2,6) is not listed in AA or BB, so (2,6)∉A∪B(2,6) \notin A \cup B.

Therefore
(2,6)∈(A∪B)′.(2,6) \in (A \cup B)'.

Answer

Underline: ∅⊂B\varnothing \subset B and (2,6)∈(A∪B)′(2,6) \in (A \cup B)'.

AI analysis — may contain mistakes.

Worked this one out on paper? Snap your steps and Gigle's AI math checker will circle the step where it went wrong.

More questions on this topic

  1. 1On the Venn diagram, shade the region that represents A′∪BA' \cup B. Use set notation to describe the shaded region.
  2. 2It is given that ξ={x:x is polygon},\xi = \{x: x \text{ is polygon}\}, p={x:x is a parallelogram},p = \{x: x \text{ is a parallelogram}\}, q={x:x is a quadrilateral},q = \{x: x \text{ is a quadrilateral}\},…
  3. 3ξ={integers x:2<x≤12}\xi = \{\text{integers } x: 2 < x \le 12\} A={prime numbers}A = \{\text{prime numbers}\} B={factors of 12}B = \{\text{factors of 12}\} C={greater than 100}C = \{\text{greater than } \sqrt{100}\}…
  4. 4On the Venn Diagram shown in the answer space, shade the set A∩B′A \cap B'. ε=x:x is an integer and 1≤x≤100ε = {x: x \text{ is an integer and } 1 \le x \le 100}…
  5. 5三年级(2)班有56名学生,这个月进行了两次数学测试:第一次得100分的学生的学号是6,9,15,16,27,33,56;第二次得100分的学生的学号是:7,9,16,27,36,40,48,51,53。 1.第一次得100分的有( )人。…