Question 1Level 2 · ExamSingle correct
Let A={⌊60n2⌋:n∈Z, 1≤n≤60}, where ⌊t⌋ denotes the greatest integer less than or equal to t, and let B={0,1,2,…,60}. Let n(X) denote the number of elements of a finite set X. Then n(B−A) equals ______. Exactly one option is correct.
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Answer: C
Write f(n)=⌊n2/60⌋. Every value lies in B since 0≤f(n)≤60, so A⊆B and n(B−A)=61−n(A); only distinct values count in the roster form of A.
For 1≤n≤29: (n+1)2−n2=2n+1≤59<60, so f(n+1)−f(n)∈{0,1}. Since f(1)=0 and f(30)=15, the values for 1≤n≤30 are exactly 0,1,…,15 (16 values).
For n≥30: 2n+1≥61>60, so f(n+1)>f(n); the 31 values f(30),…,f(60) are distinct, from 15 to 60.
These two lists share only 15, so n(A)=16+31−1=46 and n(B−A)=61−46=15. Answer (C).
Difficulty: HardQuestion 2Level 3 · AdvancedSingle correct
Which one of the following sets is finite and non-empty? Exactly one option is correct.
A{(x,y)∈Z×Z:x2−y2=2026}
B{(x,y)∈Z×Z:x2−y2=2025}
C{(x,y)∈Q×Q:x2+y2=1}
D{(x,y)∈Z×Z:x3=y2}
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Answer: B
(A) For integers, x2−y2≡0,1 or 3(mod4) (squares are 0,1mod4). Since 2026≡2(mod4), the set is empty.
(B) x2−y2=(x−y)(x+y)=2025. Each solution gives a factorisation 2025=d⋅e with d=x−y, e=x+y; 2025 has finitely many (positive and negative) divisors and each pair (d,e) gives at most one (x,y), so the set is finite. It contains (45,0), so it is non-empty.
(C) For every t∈Q, (1+t21−t2,1+t22t) is a rational point on the circle. The map t↦1+t22t is strictly increasing on [0,1] (its derivative (1+t2)22(1−t2) is positive on [0,1)), so distinct rationals t∈[0,1] give distinct points. The set is infinite.
(D) (t2,t3) lies in it for every t∈Z, so it is infinite.
Only (B) is finite and non-empty.
Difficulty: Very hardQuestion 3Level 3 · AdvancedSingle correct
Let A={x∈R:x2−5x+4≤0} and, for each real number λ, let Bλ={x∈R:λ≤x≤λ2}. The set of all real λ for which A∪Bλ=A is ______. Exactly one option is correct.
A[1,2]
B[−2,−1]∪[1,2]
C(0,1)∪(1,2]
D(0,2]
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Answer: D
A=[1,4], and A∪Bλ=A holds exactly when Bλ⊆A.
Case 0<λ<1: then λ2<λ, so Bλ=∅, and the empty set is a subset of every set. All such λ work.
Case λ≤0 or λ≥1: then Bλ=[λ,λ2]=∅, and Bλ⊆[1,4] iff λ≥1 and λ2≤4, i.e. 1≤λ≤2 (λ=1 gives B1={1}). For λ≤0, λ∈Bλ but λ∈/A.
Hence the required set is (0,1)∪[1,2]=(0,2]. Answer (D).
Trap: forgetting that Bλ=∅ for 0<λ<1 gives (A); dropping λ=1 gives (C).
Difficulty: Very hard