Solutions to 2009 Mathematics Screening Test

(Questions 1–10)
Generated by Grok 3
June 23, 2025

Introduction
This document provides detailed solutions to questions 1–10 from the 2009 Mathematics
Screening Test (Serial No. 137041). The test comprises 120 multiple-choice questions
covering advanced mathematical topics. Solutions are presented with explanations to aid
understanding. To generate a PDF, compile this LaTeX code using a LaTeX editor such
as Overleaf.

Solutions

1/2
1. Question: 2 + 51/3 is: (a) an integer, (b) a rational number of the form pq , p, q
integers, gcd(p, q) = 1, q ̸= 0, ±1, (c) an algebraic number, (d) a transcendental
number.
√
Solution: Let x = 2 + 51/3 and a = 51/3 , so x2 = 2 + a. Since 51/3 ≈ 1.709975, we
have x2 ≈ 3.709975, not a perfect square, so x is not an integer (rule out (a)). To
2
check rationality, assume x = pq , gcd(p, q) = 1, q ̸= 0, ±1. Then x2 = pq2 = 2 + a, so
2
3
a = pq2 − 2. If a = 51/3 is rational, say a = mn
, then 5 = mn
=⇒ 5n3 = m3 . Since
5 divides m3 , 5 divides m, so m = 5k. Then 5n3 = 125k 3 =⇒ n3 = 25k 3 , so 5
divides n, contradicting gcd(m, n) = 1. Thus, a is irrational, and 2 + a is irrational,
implying x2 is irrational, so x is not rational (rule out (b)). For algebraic number,
note a3 = 5, and x2 = 2 + a =⇒ a = x2 − 2. Then a3 = (x2 − 2)3 = 5 =⇒
x6 − 6x4 + 12x2 − 13 = 0, a polynomial with rational coefficients, so x is algebraic
(choose (c)). Since x is algebraic, it is not transcendental (rule out (d)).
Answer: (c) an algebraic number.

2. Question: The infimum of the set {n(−1) : n ∈ N} is: (a) -1, (b) 0, (c) − 12 , (d) not
n

defined in the real numbers.

Solution: Compute the set: for n = 1, (−1)1 = −1, so 1−1 = 1; for n = 2, (−1)2 = 1,
so 21 = 2; for n = 3, 3−1 = 13 ; for n = 4, 41 = 4, etc. The set is {1, 2, 13 , 4, 51 , 6, . . .}.
For odd n = 2k − 1, n(−1) = 2k−1
n 1 n
; for even n = 2k, n(−1) = 2k. All elements are
positive, so 0 is a lower bound. As k → ∞, 2k−1 1
→ 0, so the infimum is 0. Negative
values like −1 or − 2 are smaller but unnecessary, and the infimum is defined in R.
1

Answer: (b) 0.

1
 an
3. Question: For any real number a, limn→∞ n2
equals: (a) a, (b) −a, (c) infinity, (d)
0.
n n
Solution: Consider limn→∞ an! . (Note: The
question in the source lists nan , but
n n n
context and options suggest an! , as nan = na → 0 for all a, making options (a, b,
n+1 |a| n
c) inconsistent.) Use the ratio test: a an/(n+1)!
/n!
= n+1 → 0 < 1. Thus, an! → 0.
For a = 0, it’s trivially 0. The factorial growth dominates any exponential.
Answer: (d) 0.


4. Question: The sequence cos nπ 3
is: (a) unbounded, (b) convergent and hence
bounded, (c) divergent, (d) oscillatory.
Solution: Compute terms: n = 1, cos π3 = 12 ; n = 2, cos 2π 3
= − 12 ; n = 3, cos π = −1;
n = 4, cos 4π3
= − 12 ; n = 5, cos 5π 3
= 12 ; n = 6, cos 2π = 1. The sequence repeats
every 6 terms: { 2 , − 2 , −1, − 2 , 2 , 1, . . .}. It’s bounded ([−1, 1]), so not unbounded
1 1 1 1

(rule out (a)). It doesn’t converge to a single limit (e.g., subsequences give 1, -1),
so not convergent (rule out (b)). It’s divergent due to multiple limit points and
oscillatory due to periodic fluctuation.
Answer: (d) oscillatory.

5. Question: The value of b for which 1 + eb + e2b + e3b + · · · = 9 is: (a) 3 log 2 − 2 log 3,
(b) 2 log 3 − 3 log 2, (c) 3 log 2 − log 3, (d) log 3 − 3 log 2.
P P
Solution: The series is ∞ n=0 e
nb
= ∞ n=0 (e ) = 1−eb = 9, for |e | < 1 =⇒ b < 0.
b n 1 b

Solve: 1 − eb = 91 =⇒ eb = 89 =⇒ b = ln 98 = ln 8 − ln 9 = 3 ln 2 − 2 ln 3 =
3 log 2 − 2 log 3.
Answer: (a) 3 log 2 − 2 log 3.

6. Question: The norm of the partition {−2, −1.6, −0.5, 0, 0.8, 1} of [−2, 1] is: (a) 0.8,
(b) 0.5, (c) 0.2, (d) 1.1.
Solution: The norm is the length of the largest subinterval. Intervals are: [−2, −1.6]
(0.4), [−1.6, −0.5] (1.1), [−0.5, 0] (0.5), [0, 0.8] (0.8), [0.8, 1] (0.2). The largest is
1.1.
Answer: (d) 1.1.
 
7. Question: The average value of cos x on 0, π2 is: (a) π/2, (b) π/4, (c) 2/π, (d)
4/π.
Rπ π
Solution: Average value = π 1−0 02 cos x dx = π2 [sin x]02 = π2 (1 − 0) = π2 .
2

Answer: (c) 2/π.

8. Question: Let ⟨bn ⟩ be a sequence with bn = n(1 + (−1)n ), and A = lim sup bn ,
B = lim inf bn . Then: (a) A = ∞, B = 0, (b) A = 0, B = ∞, (c) A = B = 0, (d)
A = B = ∞.
Solution: For even n = 2k, b2k = 2k · 2 = 4k; for odd n = 2k + 1, b2k+1 =
(2k + 1) · 0 = 0. Sequence: 0, 4, 0, 8, 0, . . .. Even terms → ∞, odd terms = 0. Thus,
lim sup bn = ∞, lim inf bn = 0.
Answer: (a) A = ∞, B = 0.

2
 9. Question: limn→∞ n1 [1 + 21/2 + 31/3 + · · · + n1/n ] equals: (a) 0, (b) 1/2, (c) 1, (d) 2.
Pn ln k
Solution: The limit is limn→∞ n1 P k=1 k
1/k
. Since kP1/k
= e k → 1P as k → ∞,
approximate k ≈ 1 + k . Then k=1 k ≈ n + k=1 k , and n nk=1 k 1/k ≈
1/k ln k n 1/k n ln k 1
P P 2
1 + n1 nk=1 lnkk . Since nk=1 lnkk ∼ (ln2n) , the second term → 0. Thus, the limit is
1.
Answer: (c) 1.

10. Question: Let f, g be Riemann-integrable on [a, b]. Which is not necessarily Riemann-
integrable? (a) f g, (b) f ◦ g, (c) max{f, g}, (d) min{f, g}.
Solution: For Riemann-integrable f, g, f g is integrable (discontinuities have mea-
−g| −g|
sure zero). Similarly, max{f, g} = f +g+|f 2
and min{f, g} = f +g−|f
2
are inte-
grable since |f − g| is. However, f ◦ g may not be. Example: Let f (x) = 0 for x
irrational, f (x) = 1q for x = pq , Riemann-integrable. Let g(x) map to rationals on a
set of positive measure; f ◦ g may have discontinuities on a set of positive measure,
making it non-integrable.
Answer: (b) f ◦ g.

Conclusion
This document solves questions 1–10. For questions 11–120, please specify a range or
topic. Compile this LaTeX code in Overleaf to generate a PDF.