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Practice · Oxbridge maths · TMUA

TMUA practice questions

30 questions with worked solutions. Both papers: applied mathematical knowledge, and logic and proof, including contrapositives, counterexamples and necessary and sufficient conditions.

10 easy · 18 medium · 2 hard · TMUA Paper 2 · TMUA Paper 1

Practise timed: 20 questions · 75 min All Oxbridge maths topics

1 Easy TMUA Paper 2
What is the contrapositive of the statement "if n2n^2 is even, then nn is even"?
  1. If nn is even, then n2\displaystyle n^2 is even.
  2. If n2\displaystyle n^2 is odd, then nn is odd.
  3. If nn is odd, then n2\displaystyle n^2 is odd.
  4. If nn is odd, then n2\displaystyle n^2 is even.
  5. n2\displaystyle n^2 is even and nn is odd.
Show answer
Answer: C. If nn is odd, then n2\displaystyle n^2 is odd.
The contrapositive of "P⇒QP\Rightarrow Q" is "not Q⇒Q\Rightarrow not PP": if nn is not even, then n2n^2 is not even. The first option is the converse, the second the inverse, and the last is the negation.
2 Easy TMUA Paper 2
Which value of xx is a counterexample to the claim "for all real xx, if x2>4x^2>4 then x>2x>2"?
  1. x=3x=3
  2. x=−3x=-3
  3. x=2x=2
  4. x=0x=0
  5. x=−2x=-2
Show answer
Answer: B. x=−3x=-3
A counterexample makes the hypothesis true and the conclusion false. x=−3x=-3: x2=9>4x^2=9>4 but x<2x<2. For x=−2x=-2 the hypothesis fails (4≯44\not>4).
3 Easy TMUA Paper 2
For real xx, the condition x>1x>1 is which of the following for x2>1x^2>1?
  1. Necessary but not sufficient
  2. Necessary and sufficient
  3. Neither necessary nor sufficient
  4. Sufficient but not necessary
Show answer
Answer: D. Sufficient but not necessary
If x>1x>1 then x2>1x^2>1, so it is sufficient. But x=−2x=-2 has x2>1x^2>1 without x>1x>1, so it is not necessary.
4 Medium TMUA Paper 2
What is the negation of "every student passed at least one exam"?
  1. Every student failed at least one exam.
  2. No student passed any exam.
  3. Some student failed at least one exam.
  4. Some student passed no exam.
  5. Every student failed every exam.
Show answer
Answer: D. Some student passed no exam.
Negating "for all ss, there exists an exam ee that ss passed" gives "there exists ss such that for all ee, ss did not pass ee": some student passed no exam.
5 Easy TMUA Paper 1
How many real solutions does 32x−4⋅3x+3=03^{2x}-4\cdot3^{x}+3=0 have?
  1. 0
  2. 1
  3. 2
  4. 3
  5. 4
Show answer
Answer: C. 2
Put y=3x>0y=3^x>0: y2−4y+3=0y^2-4y+3=0, so y=1y=1 or y=3y=3, giving x=0x=0 or x=1x=1.
6 Medium TMUA Paper 1
∫14x+2x dx=\displaystyle\int_1^4\frac{x+2}{\sqrt x}\,dx=
  1. 203\displaystyle \frac{20}3
  2. 8
  3. 263\displaystyle \frac{26}3
  4. 283\displaystyle \frac{28}3
  5. 10
Show answer
Answer: C. 263\displaystyle \frac{26}3
Split: x1/2+2x−1/2x^{1/2}+2x^{-1/2}, integrating to 23x3/2+4x1/2\tfrac23x^{3/2}+4x^{1/2}. At 4: 163+8=403\tfrac{16}3+8=\tfrac{40}3. At 1: 23+4=143\tfrac23+4=\tfrac{14}3. Difference 263\tfrac{26}3.
7 Medium TMUA Paper 1
What is the sum of all real solutions of log⁡2x+log⁡2(x−2)=3\log_2x+\log_2(x-2)=3?
  1. −2
  2. 2
  3. 4
  4. 6
  5. 8
Show answer
Answer: C. 4
x(x−2)=8x(x-2)=8 gives x=4x=4 or x=−2x=-2. But log⁡2x\log_2x needs x>0x>0 (and x−2>0x-2>0), so only x=4x=4 is a solution. The sum of the quadratic's roots, 2, is the trap.
8 Easy TMUA Paper 1
What is the minimum value of x2+4x\dfrac{x^2+4}{x} for x>0x>0?
  1. 1
  2. 2
  3. 3
  4. 4
  5. There is no minimum
Show answer
Answer: D. 4
x+4x≥2x⋅4x=4x+\tfrac4x\ge2\sqrt{x\cdot\tfrac4x}=4 by AM–GM, with equality at x=2x=2.
9 Medium TMUA Paper 2
A "proof" that 1=21=2: Let a=ba=b. (Step 1) a2=aba^2=ab. (Step 2) a2−b2=ab−b2a^2-b^2=ab-b^2. (Step 3) (a+b)(a−b)=b(a−b)(a+b)(a-b)=b(a-b). (Step 4) a+b=ba+b=b. (Step 5) 2b=b2b=b. (Step 6) 2=12=1. Which is the first incorrect step?
  1. Step 2
  2. Step 3
  3. Step 4
  4. Step 5
  5. Step 6
Show answer
Answer: C. Step 4
Step 4 divides both sides by a−ba-b, which is 0 since a=ba=b. Steps 1 to 3 are valid identities. Step 6 would also be invalid if b=0b=0, but the first error is Step 4.
10 Medium TMUA Paper 1
For how many integers nn is 12n−3\dfrac{12}{n-3} an integer?
  1. 6
  2. 8
  3. 11
  4. 12
  5. 24
Show answer
Answer: D. 12
n−3n-3 must divide 12, and it may be negative: ±1,±2,±3,±4,±6,±12\pm1,\pm2,\pm3,\pm4,\pm6,\pm12. That is 12 values of nn.
11 Medium TMUA Paper 1
For 0≤θ≤π0\le\theta\le\pi, the series ∑k=0∞(2cos⁡θ)k\displaystyle\sum_{k=0}^{\infty}(2\cos\theta)^k converges exactly when
  1. 0<θ<π3\displaystyle 0<\theta<\frac\pi3
  2. π6<θ<5π6\displaystyle \frac\pi6<\theta<\frac{5\pi}6
  3. π3<θ<2π3\displaystyle \frac\pi3<\theta<\frac{2\pi}3
  4. π2<θ<π\displaystyle \frac\pi2<\theta<\pi
  5. 0≤θ≤π0\le\theta\le\pi
Show answer
Answer: C. π3<θ<2π3\displaystyle \frac\pi3<\theta<\frac{2\pi}3
A geometric series converges iff ∣2cos⁡θ∣<1|2\cos\theta|<1, i.e. −12<cos⁡θ<12-\tfrac12<\cos\theta<\tfrac12. On [0,π][0,\pi] that is π3<θ<2π3\tfrac\pi3<\theta<\tfrac{2\pi}3.
12 Easy TMUA Paper 1
What is the area of the triangle with vertices (0,0)(0,0), (4,1)(4,1) and (1,3)(1,3)?
  1. 92\displaystyle \frac92
  2. 5
  3. 112\displaystyle \frac{11}2
  4. 6
  5. 132\displaystyle \frac{13}2
Show answer
Answer: C. 112\displaystyle \frac{11}2
With one vertex at the origin, area =12∣x1y2−x2y1∣=12∣4⋅3−1⋅1∣=112=\tfrac12|x_1y_2-x_2y_1|=\tfrac12|4\cdot3-1\cdot1|=\tfrac{11}2.
13 Medium TMUA Paper 1
If f(x)=2x+1f(x)=2x+1 and g(x)=x2g(x)=x^2, how many real xx satisfy f(g(x))=g(f(x))f(g(x))=g(f(x))?
  1. 0
  2. 1
  3. 2
  4. 3
  5. 4
Show answer
Answer: C. 2
2x2+1=(2x+1)2=4x2+4x+12x^2+1=(2x+1)^2=4x^2+4x+1, so 2x2+4x=02x^2+4x=0 and x=0x=0 or x=−2x=-2.
14 Medium TMUA Paper 1
∑k=110log⁡10 ⁣(k+1k)=\displaystyle\sum_{k=1}^{10}\log_{10}\!\left(\frac{k+1}{k}\right)=
  1. 1
  2. log⁡1011\log_{10}11
  3. log⁡10(10!)\log_{10}(10!)
  4. 10
  5. 11
Show answer
Answer: B. log⁡1011\log_{10}11
The sum of logs is the log of the product, and ∏k=110k+1k=111\prod_{k=1}^{10}\tfrac{k+1}k=\tfrac{11}1 telescopes.
15 Medium TMUA Paper 2
Which of these statements are true for every real xx? I: x2+1≥2xx^2+1\ge2x. II: x3≥xx^3\ge x. III: ∣x∣≥x|x|\ge x.
  1. I only
  2. II only
  3. III only
  4. I and III only
  5. I, II and III
Show answer
Answer: D. I and III only
I is (x−1)2≥0(x-1)^2\ge0: true. II fails at x=−2x=-2 (−8<−2-8<-2). III is true because ∣x∣|x| is xx or −x≥0>x-x\ge0>x. So I and III only.
16 Easy TMUA Paper 1
What is the gradient of y=x3−3xy=x^3-3x at the point where it crosses the positive xx-axis?
  1. 0
  2. 3
  3. 33\displaystyle 3\sqrt3
  4. 6
  5. 9
Show answer
Answer: D. 6
x3−3x=x(x2−3)x^3-3x=x(x^2-3) is zero at x=3x=\sqrt3 on the positive axis. y′=3x2−3=9−3=6y'=3x^2-3=9-3=6.
17 Medium TMUA Paper 1
How many integers nn with 1≤n≤501\le n\le50 have no common factor with 50 other than 1?
  1. 10
  2. 16
  3. 20
  4. 25
  5. 40
Show answer
Answer: C. 20
Exclude multiples of 2 or 5: 50−25−10+5=2050-25-10+5=20. This is Euler's φ(50)=50⋅12⋅45\varphi(50)=50\cdot\tfrac12\cdot\tfrac45.
18 Medium TMUA Paper 2
The claim "for every integer n≥0n\ge0, n2+n+41n^2+n+41 is prime" is false. Which value of nn is a counterexample?
  1. n=1n=1
  2. n=10n=10
  3. n=39n=39
  4. n=40n=40
  5. There is none
Show answer
Answer: D. n=40n=40
At n=40n=40: 1600+40+41=1681=4121600+40+41=1681=41^2. (The polynomial gives primes for n=0,…,39n=0,\dots,39; at n=39n=39 it is 16011601, which is prime.) The factorisation is easy to see by writing 402+40+41=40⋅41+41=41240^2+40+41=40\cdot41+41=41^2.
19 Medium TMUA Paper 1
The curve y=xe−xy=xe^{-x} has exactly one stationary point. What is it?
  1. (0,0)(0,0), a minimum
  2. (1,1e)\displaystyle (1,\tfrac1e), a minimum
  3. (1,1e)\displaystyle (1,\tfrac1e), a maximum
  4. (−1,−e)(-1,-e), a maximum
  5. (e,1)(e,1), a maximum
Show answer
Answer: C. (1,1e)\displaystyle (1,\tfrac1e), a maximum
y′=(1−x)e−xy'=(1-x)e^{-x} vanishes at x=1x=1, where y=e−1y=e^{-1}. y′y' changes from positive to negative there, so it is a maximum.
20 Hard TMUA Paper 2
Consider the statement "if nn is prime, then 2n−12^n-1 is prime" and its converse. Which is correct?
  1. Both are true.
  2. Both are false.
  3. The statement is true and its converse is false.
  4. The statement is false and its converse is true.
Show answer
Answer: D. The statement is false and its converse is true.
The statement fails at n=11n=11: 211−1=2047=23⋅892^{11}-1=2047=23\cdot89. The converse holds: if n=abn=ab with a,b>1a,b>1, then 2a−12^a-1 divides 2ab−12^{ab}-1, so 2n−12^n-1 is composite.
21 Medium TMUA Paper 2
Which statement is logically equivalent to "PP only if QQ"?
  1. Q⇒PQ\Rightarrow P
  2. P⇒QP\Rightarrow Q
  3. P⇔QP\Leftrightarrow Q
  4. (not PP) ⇒\Rightarrow (not QQ)
  5. QQ only if PP
Show answer
Answer: B. P⇒QP\Rightarrow Q
"PP only if QQ" says PP cannot hold without QQ, which is P⇒QP\Rightarrow Q. "PP if QQ" is the other direction.
22 Medium TMUA Paper 1
How many real solutions does x4+x2−2=0x^4+x^2-2=0 have?
  1. 0
  2. 1
  3. 2
  4. 3
  5. 4
Show answer
Answer: C. 2
With y=x2y=x^2: y2+y−2=0y^2+y-2=0, so y=1y=1 or y=−2y=-2. Only y=1y=1 is possible for real xx, giving x=±1x=\pm1.
23 Medium TMUA Paper 1
What is the sum of the coefficients of (2x−3)5(2x-3)^5?
  1. −243
  2. −1
  3. 1
  4. 32
  5. 243
Show answer
Answer: B. −1
The sum of the coefficients is the value at x=1x=1: (2−3)5=−1(2-3)^5=-1.
24 Easy TMUA Paper 1
If tan⁡θ=34\tan\theta=\tfrac34 and θ\theta is acute, what is sin⁡θ+cos⁡θ\sin\theta+\cos\theta?
  1. 1
  2. 75\displaystyle \frac75
  3. 54\displaystyle \frac54
  4. 125\displaystyle \frac{12}5
  5. 74\displaystyle \frac74
Show answer
Answer: B. 75\displaystyle \frac75
The 3-4-5 triangle gives sin⁡θ=35\sin\theta=\tfrac35 and cos⁡θ=45\cos\theta=\tfrac45.
25 Medium TMUA Paper 2
Which of these statements is false?
  1. Every prime greater than 2 is odd.
  2. If n2\displaystyle n^2 is divisible by 4 then nn is divisible by 4.
  3. The sum of two odd integers is even.
  4. Every integer is rational.
  5. There are infinitely many primes.
Show answer
Answer: B. If n2\displaystyle n^2 is divisible by 4 then nn is divisible by 4.
Take n=2n=2: n2=4n^2=4 is divisible by 4 but nn is not. The correct statement is that nn must be even.
26 Easy TMUA Paper 1
What is the remainder when x3−2x+5x^3-2x+5 is divided by x−2x-2?
  1. 5
  2. 7
  3. 9
  4. 11
  5. 13
Show answer
Answer: C. 9
By the remainder theorem, substitute x=2x=2: 8−4+5=98-4+5=9.
27 Medium TMUA Paper 1
The quadratic x2+bx+9x^2+bx+9 has a repeated root. What is the sum of the possible values of bb?
  1. −6
  2. 0
  3. 6
  4. 12
  5. 36
Show answer
Answer: B. 0
b2=36b^2=36, so b=±6b=\pm6 and the two values cancel.
28 Hard TMUA Paper 2
A "proof" by induction claims that in any set of nn horses all the horses have the same colour. The base case n=1n=1 is fine, and the inductive step removes one horse, applies the hypothesis, and puts it back. Where does the argument fail?
  1. The base case
  2. The step from n=1n=1 to n=2n=2
  3. The step from n=2n=2 to n=3n=3
  4. The inductive hypothesis is not stated
  5. It does not fail
Show answer
Answer: B. The step from n=1n=1 to n=2n=2
The step needs the two smaller sets to overlap. For n=2n=2 they do not, so nothing forces the two horses to match. Every later step is fine, which is why the flaw is easy to miss.
29 Easy TMUA Paper 1
How many integers xx satisfy 2<3x−4<202<3x-4<20?
  1. 4
  2. 5
  3. 6
  4. 7
  5. 8
Show answer
Answer: B. 5
6<3x<246<3x<24, so 2<x<82<x<8: the integers 3, 4, 5, 6, 7.
30 Medium TMUA Paper 2
The claim "for all real xx and yy, if x2=y2x^2=y^2 then x=yx=y" is false. Which single change makes it true?
  1. Replace == in the conclusion by ≥\ge
  2. Add the condition x>0x>0 and y>0y>0
  3. Replace x2=y2\displaystyle x^2=y^2 by x3=y3\displaystyle x^3=y^3
  4. Either of the second or third changes
  5. None of these changes works
Show answer
Answer: D. Either of the second or third changes
Restricting to positives removes the case x=−yx=-y, and cubing is one-to-one on the reals, so either change repairs it.

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