4d:I[546941,["/_next/static/chunks/b9d2e297d487bf63.js","/_next/static/chunks/37c597db36e01703.js","/_next/static/chunks/0b6593f63ecb0bcb.js","/_next/static/chunks/013d4e90104cf322.js","/_next/static/chunks/d5ba0d44895fc7e6.js","/_next/static/chunks/f87111da6e6b682e.js","/_next/static/chunks/afed86511920eee0.js","/_next/static/chunks/209f67b6047e9901.js","/_next/static/chunks/ac2da80bd2e688db.js","/_next/static/chunks/626e97a0add685c0.js","/_next/static/chunks/f4547e8fbc6e6129.js","/_next/static/chunks/5a0da50be06b37ff.js","/_next/static/chunks/dbfb90c1456067cb.js","/_next/static/chunks/422dea29c3172ef8.js","/_next/static/chunks/81c709a569c04f23.js","/_next/static/chunks/8d3945c9ea1274d1.js"],"default"] 4e:T407,Given: - $$M = \{x: 3 \le x < 8\}$$ means all numbers from 3 up to (but not including) 8. - $$N = \{x: 8 < x \le 12\}$$ means all numbers greater than 8 up to (and including) 12. I. Check if $$8 \in M \cap N$$ - For $$8\in M$$, we need $$3 \le 8 < 8$$, which is false because $$8 < 8$$ is not true. - For $$8\in N$$, we need $$8 < 8 \le 12$$, which is also false because $$8 < 8$$ is not true. So $$8\notin M$$ and $$8\notin N$$, hence $$8\notin M\cap N$$. Statement I is false. II. Check if $$8 \in M \cup N$$ The union $$M \cup N$$ contains elements in $$M$$ or in $$N$$. Since $$8$$ is in neither $$M$$ nor $$N$$, $$8\notin M\cup N$$. Statement II is false. III. Check if $$M \cap N = \varnothing$$ To be in the intersection, a number must satisfy both: $$ 3 \le x < 8 \quad \text{and} \quad 8 < x \le 12 $$ No number can be less than 8 and also greater than 8 at the same time. So there are no common elements. Thus $$M \cap N = \varnothing$$. Statement III is true. Therefore, only III is true, so the correct option is A.4f:T407,Given: - $$M = \{x: 3 \le x < 8\}$$ means all numbers from 3 up to (but not including) 8. - $$N = \{x: 8 < x \le 12\}$$ means all numbers greater than 8 up to (and including) 12. I. Check if $$8 \in M \cap N$$ - For $$8\in M$$, we need $$3 \le 8 < 8$$, which is false because $$8 < 8$$ is not true. - For $$8\in N$$, we need $$8 < 8 \le 12$$, which is also false because $$8 < 8$$ is not true. So $$8\notin M$$ and $$8\notin N$$, hence $$8\notin M\cap N$$. Statement I is false. II. Check if $$8 \in M \cup N$$ The union $$M \cup N$$ contains elements in $$M$$ or in $$N$$. Since $$8$$ is in neither $$M$$ nor $$N$$, $$8\notin M\cup N$$. Statement II is false. III. Check if $$M \cap N = \varnothing$$ To be in the intersection, a number must satisfy both: $$ 3 \le x < 8 \quad \text{and} \quad 8 < x \le 12 $$ No number can be less than 8 and also greater than 8 at the same time. So there are no common elements. Thus $$M \cap N = \varnothing$$. Statement III is true. Therefore, only III is true, so the correct option is A.