38:T810,

\Rightarrow {m^2} - 2m - 15 \ge 0

2f:["$","li","2",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":3}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$38"}}]]}] 39:T831,

\Rightarrow

(m - 5)(m + 3) \ge 0

30:["$","li","3",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":4}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$39"}}]]}] 3a:Tbca,

m\, \in \,

(

-

\propto

-

\cup

[5,

\propto

)31:["$","li","4",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":5}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$3a"}}]]}] 3b:Ta43,

A= ( -

\propto

-

\cup

[5,

\propto

)32:["$","li","5",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":6}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$3b"}}]]}] 33:["$","li","6",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":7}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"Given B = [

-

3, 5)\nNow, let's examine the options."}}]]}] 34:["$","li","7",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":8}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"Option A : A ∩ B = {–3}\nThe intersection of sets A and B would be the set of elements common to both sets. In this case, the only common element is -3. So, option A is true."}}]]}] 35:["$","li","8",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":9}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"Option B : B – A = (–3, 5)\nThe subtraction (or difference) of sets A from B is the set of elements that are in B but not in A. B is [–3, 5), and A is (-∞, -3] U [5, ∞). Subtracting A from B would leave an open interval (-3, 5), not including -3 and 5. So, option B is also true."}}]]}] 36:["$","li","9",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":10}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"Option C : A ∪ B = R\nThe union of sets A and B is the set of elements that are in A, or B, or both. Here, A U B would cover all real numbers. So, option C is true."}}]]}] 37:["$","li","10",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":11}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"Option D : A - B = (-∞, -3) ∪ (5, ∞)\nThe subtraction (or difference) of set B from A is the set of elements that are in A but not in B. B is [–3, 5), and A is (-∞, -3] U [5, ∞). Subtracting B from A would leave (-∞, -3) U [5, ∞), not including -3 and 5. But according to the convention for writing intervals, it should be (-∞, -3) U (5, ∞). So, option D is not true."}}]]}]

More on Sets and Relations

More on Sets and Relations