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Proofs

Practice exam-style IB Math AA questions for Proofs, aligned with the syllabus and grouped by topic.

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Question 1
SL • Paper 1
Easy
Non Calculator
SL • Paper 1
Easy
Non Calculator

Let n∈Zn\in\mathbb{Z}.

A

(2n+3)2−(2n−1)2≡16n+8(2n+3)^2-(2n-1)^2\equiv 16n+8.

[2]
B

that the difference of the squares of any two odd integers which differ by 44 is divisible by 88.

[3]
Question 2
SL • Paper 1
Easy
Non Calculator
SL • Paper 1
Easy
Non Calculator

A

the values of xx for which the expression 1x−2−1x+2\frac{1}{x-2}-\frac{1}{x+2} is not defined.

[1]
B

1x−2−1x+2≡4x2−4\frac{1}{x-2}-\frac{1}{x+2}\equiv \frac{4}{x^2-4}, for all allowed values of xx.

[3]
C

why substituting x=0x=0 into both sides is not a proof of the identity in part (b).

[1]
Question 3
SL • Paper 2
Easy
Calculator Permitted
SL • Paper 2
Easy
Calculator Permitted

Consider the statement

(2x−1)2−(x+3)(x−3)≡3x2−4x+10(2x-1)^2-(x+3)(x-3)\equiv 3x^2-4x+10

A

the identity by starting with the left-hand side.

[2]
B

Check the identity using x=−2x=-2.

[1]
C

why the calculation in part (b) is not a proof of the identity.

[2]
Question 4
SL • Paper 2
Easy
Calculator Permitted
SL • Paper 2
Easy
Calculator Permitted

A

exactly that

16+115+110=13\frac{1}{6}+\frac{1}{15}+\frac{1}{10}=\frac{1}{3}

[2]
B

the identity

1k−1k+3≡3k(k+3)\frac{1}{k}-\frac{1}{k+3}\equiv \frac{3}{k(k+3)}

where k≠0,−3k\neq 0,-3.

[2]
C

why the decimal calculation 0.166666…+0.066666…+0.1=0.333333…0.166666\ldots+0.066666\ldots+0.1=0.333333\ldots is not the preferred proof for part (a).

[1]

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Question 5
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A

16+112+120=310\frac{1}{6}+\frac{1}{12}+\frac{1}{20}=\frac{3}{10}.

[2]
B

1m(m+1)+1(m+1)(m+2)+1(m+2)(m+3)≡3m(m+3)\frac{1}{m(m+1)}+\frac{1}{(m+1)(m+2)}+\frac{1}{(m+2)(m+3)}\equiv \frac{3}{m(m+3)}, where m≠0,−1,−2,−3m\neq 0,-1,-2,-3.

[4]
Question 6
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Let n∈Zn\in\mathbb{Z}.

A

n(n+1)(n+2)(n+3)+1≡(n2+3n+1)2n(n+1)(n+2)(n+3)+1\equiv (n^2+3n+1)^2.

[4]
B

that the product of four consecutive integers, increased by 11, is a perfect square.

[2]
Question 7
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A

(3+2)2=5+26(\sqrt{3}+\sqrt{2})^2=5+2\sqrt{6} and (3−2)2=5−26(\sqrt{3}-\sqrt{2})^2=5-2\sqrt{6}.

[3]
B

5+26−5−26=22\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=2\sqrt{2}.

[2]
Question 8
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let aa be a non-zero rational number and let bb be an irrational number.

A

by contradiction that abab is irrational.

[4]

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Question 9
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A student makes the following claim.

For every n∈Nn\in\mathbb{N}, the number n2+n+17n^2+n+17 is prime.

A

Disprove the claim by giving a counterexample. You must why your example is a counterexample.

[4]
Question 10
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Consider the statement:

The sum of any two irrational numbers is irrational.

A

Disprove the statement by giving a counterexample. You must why your example is a counterexample.

[4]
Question 11
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The following identity is defined for suitable real values of xx:

1x−2−1x+2≡4x2−4\frac{1}{x-2}-\frac{1}{x+2}\equiv \frac{4}{x^2-4}

A

the values of xx that must be excluded.

[1]
B

the identity is true.

[3]
C

exactly

1x−2−1x+2=12\frac{1}{x-2}-\frac{1}{x+2}=\frac{1}{2}

[2]
Question 12
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

Consider the identity

x2−9x2+6x+9≡x−3x+3\frac{x^2-9}{x^2+6x+9}\equiv \frac{x-3}{x+3}

A

the restriction on xx.

[1]
B

the identity.

[2]
C

exactly

x2−9x2+6x+9=14\frac{x^2-9}{x^2+6x+9}=\frac{1}{4}

[2]

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Question 13
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The function ff is defined by

f(x)=(x+1)3−(x−1)3f(x)=(x+1)^3-(x-1)^3

A

Use your GDC to f(2)f(2) and f(5)f(5).

[1]
B

that f(x)≡6x2+2f(x)\equiv 6x^2+2.

[2]
C

A student says that the two values found in part (a) the result in part (b). why the student is incorrect.

[2]
Question 14
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A

Let rr be a non-zero rational number and let ss be an irrational number. by contradiction that rsrs is irrational.

[5]
Question 15
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Consider the statement:

For every n∈Nn\in\mathbb{N}, the number n2+n+17n^2+n+17 is prime.

A

Use your GDC, or otherwise, to a value of nn that may be used as a counterexample.

[1]
B

your value is a counterexample.

[2]
C

why this disproves the statement.

[1]
Question 16
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Consider the expression x3−1x−1\frac{x^3-1}{x-1}.

A

x3−1x−1≡x2+x+1\frac{x^3-1}{x-1}\equiv x^2+x+1, for x≠1x\neq 1.

[2]
B

the solution of x3−1x−1=3\frac{x^3-1}{x-1}=3, giving a reason why any rejected value is not a solution.

[3]

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Question 17
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A

Use mathematical induction to that ∑r=1nr(r+1)=n(n+1)(n+2)3\sum_{r=1}^{n} r(r+1)=\frac{n(n+1)(n+2)}{3} for all n∈Z+n\in\mathbb{Z}^+.

[6]
Question 18
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A

by contradiction that 6\sqrt{6} is irrational. You may use the fact that if a prime divides a2a^2, then it divides aa.

[3]
B

that 2+3\sqrt{2}+\sqrt{3} is irrational.

[3]
Question 19
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

For all real values of xx,

a(x−1)2+b(x+2)≡3x2−4x+ca(x-1)^2+b(x+2)\equiv 3x^2-4x+c

where aa, bb and cc are constants.

A

Expand the left-hand side and collect like terms.

[2]
B

the values of aa, bb and cc.

[3]
C

Check your result by substituting x=2x=2 into both sides of the identity.

[1]
Question 20
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A

Use mathematical induction to that

∑r=1nr(r+1)=n(n+1)(n+2)3\sum_{r=1}^{n} r(r+1)=\frac{n(n+1)(n+2)}{3}

for all n∈Z+n\in\mathbb{Z}^+.

[6]

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Question 21
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A

Use mathematical induction to that 77 divides

32n+1+2n+23^{2n+1}+2^{n+2}

for all n∈Z+n\in\mathbb{Z}^+.

[6]
Question 22
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A

by contradiction that 7\sqrt{7} is irrational.

[6]
Question 23
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

For real values of xx, consider

A(x)=x2+x−6x2+4x+3A(x)=\frac{x^2+x-6}{x^2+4x+3}
A

Algebraic form and domain

I.

the values of xx for which A(x)A(x) is not defined.

[2]
II.

A(x)≡x−2x+1A(x)\equiv \frac{x-2}{x+1}

for all allowed values of xx.

[3]
B

Further algebraic manipulation

I.

x−2x+1−1≡−3x+1\frac{x-2}{x+1}-1\equiv -\frac{3}{x+1}
[2]
II.

A(x)=12A(x)=\dfrac12.

[2]
C

both sides when x=0x=0 and whether this check proves the identity in part (a)(ii).

[3]
Question 24
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Let n∈Zn\in\mathbb{Z} and define

F(n)=(n+2)3−(n−1)3F(n)=(n+2)^3-(n-1)^3
A
I.

F(n)≡9(n2+n+1)F(n)\equiv 9(n^2+n+1)
[4]
II.

that F(n)F(n) is divisible by 99 for every integer nn.

[2]
B

A student calculates F(0)F(0) and F(1)F(1) to support the result in part (a).

I.

F(0)F(0) and F(1)F(1).

[2]
II.

why these two calculations are not a proof of the result in part (a)(ii).

[2]

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Question 25
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Consider the exact surd expressions

P=7+43,Q=7−43P=\sqrt{7+4\sqrt3},\qquad Q=\sqrt{7-4\sqrt3}
A
AI.

P=2+3P=2+\sqrt3 and Q=2−3Q=2-\sqrt3.

[3]
AII.

the exact value of P−QP-Q.

[2]
B
BI.

PQ=1PQ=1.

[2]
BII.

the exact value of P+QP+Q and

(P−Q)y+(P+Q)=8(P-Q)y+(P+Q)=8
[3]
C

why a decimal approximation to P−QP-Q is not a proof of the exact value found in part (a)(ii).

[2]
Question 26
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A

Use mathematical induction to that 77 divides 32n+1+2n+23^{2n+1}+2^{n+2} for all n∈Z+n\in\mathbb{Z}^+.

[6]
Question 27
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

For real values of xx, consider the expression

E(x)=2x−1+3x+2E(x)=\frac{2}{x-1}+\frac{3}{x+2}
A
I.

the values of xx for which E(x)E(x) is not defined.

[2]
II.

that

2x−1+3x+2≡5x+1x2+x−2\frac{2}{x-1}+\frac{3}{x+2}\equiv \frac{5x+1}{x^2+x-2}

for all allowed values of xx.

[3]
B

exactly

2x−1+3x+2=1\frac{2}{x-1}+\frac{3}{x+2}=1
[3]
C
I.

Use your GDC to check the identity in part (a)(ii) when x=4x=4.

[1]
II.

why the calculation in part (c)(i) is not a proof of the identity.

[1]
Question 28
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

For all real values of xx,

A(x−2)2+B(x+1)+C≡4x2−7x+9A(x-2)^2+B(x+1)+C\equiv 4x^2-7x+9

where AA, BB and CC are constants.

A
I.

Expand the left-hand side and collect like terms.

[2]
II.

the values of AA, BB and CC.

[3]
B

Let F(x)=4(x−2)2+9(x+1)−16F(x)=4(x-2)^2+9(x+1)-16 and G(x)=4x2−7x+9G(x)=4x^2-7x+9.

I.

Use your GDC to F(−3)F(-3) and G(−3)G(-3).

[2]
II.

why the result in part (b)(i) is consistent with, but does not , the identity.

[2]

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Question 29
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The function ff is defined by

f(x)=(2x+1)3−(2x−3)3f(x)=(2x+1)^3-(2x-3)^3
A
I.

Use your GDC to f(0)f(0) and f(2)f(2).

[2]
II.

that f(x)≡48x2−48x+28f(x)\equiv 48x^2-48x+28.

[2]
B

f(x)=200f(x)=200, giving exact answers.

[3]
C

A student claims that the two values found in part (a)(i) the identity in part (a)(ii). why this claim is incorrect.

[2]
Question 30
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A

Use mathematical induction to that, for x>0x>0,

dndxn(ln⁡x)=(−1)n−1(n−1)!xn\frac{d^n}{dx^n}(\ln x)=\frac{(-1)^{n-1}(n-1)!}{x^n}

for all n∈Z+n\in\mathbb{Z}^+.

[6]
Question 31
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

A sequence is defined by u1=1u_1=1 and un+1=5un+4u_{n+1}=5u_n+4 for n∈Z+n\in\mathbb{Z}^+. This question investigates a closed form and divisibility properties of the sequence.

nn

unu_n

un+1u_n+1

un2−1u_n^2-1

1

2

3

4

A
I.

u2u_2, u3u_3 and u4u_4.

[2]
II.

Use your values to conjecture a formula for un+1u_n+1.

[1]
B
I.

Use mathematical induction to that

un=2⋅5n−1−1u_n=2\cdot 5^{n-1}-1

for all n∈Z+n\in\mathbb{Z}^+.

[5]
C
I.

that 88 divides un2−1u_n^2-1 for every integer n≥2n\geq 2.

[2]
II.

A student claims that 3232 divides un2−1u_n^2-1 for every integer n≥2n\geq 2. Disprove the claim by giving a counterexample.

[1]
Question 32
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

For each n∈Z+n\in\mathbb{Z}^+, define

an=n2+1−na_n=\sqrt{n^2+1}-n

This question investigates exact forms and irrationality.

n

ana_n exact

ana_n decimal approx.

1

2−1\sqrt{2}-1

0.414214

2

5−2\sqrt{5}-2

0.236068

3

0.162278

4

17−4\sqrt{17}-4

0.123106

5

26−5\sqrt{26}-5

0.099020

6

37−6\sqrt{37}-6

0.082763

7

50−7\sqrt{50}-7

0.071068

8

65−8\sqrt{65}-8

0.062258

A
I.

an≡1n2+1+na_n\equiv \frac{1}{\sqrt{n^2+1}+n}
[2]
II.

Use the identity in part (a)(i) to an exact expression for a3a_3.

[2]
B

by contradiction that n2+1\sqrt{n^2+1} is irrational for every n∈Z+n\in\mathbb{Z}^+.

[4]
C
I.

that ana_n is irrational for every n∈Z+n\in\mathbb{Z}^+.

[1]
II.

Disprove the statement: n2+7\sqrt{n^2+7} is irrational for every n∈Z+n\in\mathbb{Z}^+.

[1]

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Question 33
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

A student investigates the quadratic expression

Tn=n2+n+11T_n=n^2+n+11

where n∈Nn\in\mathbb{N}. The student claims that TnT_n is prime for every n∈Nn\in\mathbb{N}.

n

TnT_n

Prime?

0

11

Yes

1

13

Yes

2

17

Yes

3

23

Yes

4

31

Yes

5

41

Yes

6

53

Yes

7

67

Yes

8

83

Yes

9

101

Yes

A
I.

T0T_0, T1T_1 and T2T_2.

[2]
II.

why the calculations in part (a)(i) do not the student's claim.

[1]
B

and a counterexample to the student's claim.

[2]
C
I.

, for every k∈Nk\in\mathbb{N},

T11k+10=11(11k2+21k+11)T_{11k+10}=11(11k^2+21k+11)
[3]
II.

why there are infinitely many values of nn for which TnT_n is not prime.

[2]
Question 34
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

Let f(x)=xexf(x)=xe^x. For n∈Z+n\in\mathbb{Z}^+, let f(n)(x)f^{(n)}(x) denote the nnth derivative of ff. This question investigates a formula for f(n)(x)f^{(n)}(x).

Derivative order

Simplified expression

0

xexxe^x

1

(x+1)ex(x+1)e^x

2

(x+2)ex(x+2)e^x

3

(x+3)ex(x+3)e^x

A
I.

f′(x)f'(x) and f′′(x)f''(x).

[2]
II.

Conjecture a formula for f(n)(x)f^{(n)}(x).

[1]
B

Use mathematical induction to your conjecture from part (a)(ii). If you did not obtain a conjecture, that f(n)(x)=(x+n)exf^{(n)}(x)=(x+n)e^x.

[5]
C
I.

the value of nn for which f(n)(0)=10f^{(n)}(0)=10.

[1]
II.

why finding f′(x)f'(x), f′′(x)f''(x) and f′′′(x)f'''(x) would not the formula for every n∈Z+n\in\mathbb{Z}^+.

[1]
Question 35
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

Let xx be a real number for which sec⁡x\sec x and tan⁡x\tan x are defined.

A

Part (a)

I.

sec⁡x−tan⁡x≡1−sin⁡xcos⁡x\sec x-\tan x\equiv \frac{1-\sin x}{\cos x}
[2]
II.

(sec⁡x−tan⁡x)(sec⁡x+tan⁡x)≡1(\sec x-\tan x)(\sec x+\tan x)\equiv 1
[3]
B

Part (b)

I.

write 1sec⁡x+tan⁡x\dfrac{1}{\sec x+\tan x} in the form sec⁡x−tan⁡x\sec x-\tan x.

[1]
II.

exactly

sec⁡x+tan⁡x=3\sec x+\tan x=\sqrt3

for 0≤x<2π0\leq x<2\pi.

[3]
C

why verifying the identity in part (a)(ii) at x=0x=0 is not a proof of the identity.

[2]
Question 36
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

The function ff is defined by

f(x)=(x+1)4−(x−1)48,x∈Rf(x)=\frac{(x+1)^4-(x-1)^4}{8},\qquad x\in\mathbb{R}
A
I.

f(x)≡x3+xf(x)\equiv x^3+x
[4]
II.

that f(n)f(n) is even for every integer nn.

[3]
B
I.

exactly f(x)=6xf(x)=6x.

[3]
II.

why the identity in part (a)(i) justifies using x3+xx^3+x in part (b)(i).

[1]
C

Check one of the non-zero solutions from part (b)(i) by substituting it into x3+x=6xx^3+x=6x.

[2]

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Question 37
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

For real values of xx, let

B(x)=xx−1−2x+1B(x)=\frac{x}{x-1}-\frac{2}{x+1}
A
I.

the values of xx for which B(x)B(x) is not defined.

[1]
II.

B(x)≡x2−x+2x2−1B(x)\equiv \frac{x^2-x+2}{x^2-1}

for all allowed values of xx.

[3]
B
I.

x2−x+2≡(x−12)2+74x^2-x+2\equiv \left(x-\frac12\right)^2+\frac74
[2]
II.

that B(x)>0B(x)>0 for x>1x>1.

[2]
C

A student attempts to B(x)=0B(x)=0 by using only the numerator of the expression in part (a)(ii). why B(x)=0B(x)=0 has no real solutions.

[2]
Question 38
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For n∈Z+n\in\mathbb{Z}^+, define

Sn=∑r=1nr(2r+1)S_n=\sum_{r=1}^{n} r(2r+1)
A
I.

S3S_3.

[2]
II.

that S3=3(3+1)(4⋅3+5)6S_3=\dfrac{3(3+1)(4\cdot3+5)}{6}.

[1]
B

Use mathematical induction to that

Sn=n(n+1)(4n+5)6S_n=\frac{n(n+1)(4n+5)}{6}

for all n∈Z+n\in\mathbb{Z}^+.

[7]
C

∑r=4nr(2r+1),n≥4\sum_{r=4}^{n} r(2r+1),\qquad n\geq4
[2]
Question 39
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Consider the statement P(n)P(n): n3+5nn^3+5n is divisible by 66, where n∈Z+n\in\mathbb{Z}^+.

A
I.

P(1)P(1) and P(2)P(2).

[2]
II.

Use mathematical induction to P(n)P(n) for all n∈Z+n\in\mathbb{Z}^+.

[6]
B

A student claims that n3+5nn^3+5n is divisible by 1212 for every n∈Z+n\in\mathbb{Z}^+. Disprove this claim by giving a counterexample.

[3]
Question 40
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A sequence (un)(u_n) is defined by

u1=4,un+1=2un+3u_1=4,\qquad u_{n+1}=2u_n+3

for n∈Z+n\in\mathbb{Z}^+.

A
I.

u2u_2 and u3u_3.

[2]
II.

that un=7⋅2n−1−3u_n=7\cdot2^{n-1}-3 for n=1,2,3n=1,2,3.

[1]
B

Use mathematical induction to that

un=7⋅2n−1−3u_n=7\cdot2^{n-1}-3

for all n∈Z+n\in\mathbb{Z}^+.

[6]
C
I.

the value of nn for which un=109u_n=109.

[2]
II.

A student claims that unu_n is divisible by 55 for every odd positive integer nn. Disprove this claim by giving a counterexample.

[2]

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Question 41
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

For x≥1x\geq 1, define

R(x)=x+1−x−1R(x)=\sqrt{x+1}-\sqrt{x-1}
A
I.

R(x)≡2x+1+x−1R(x)\equiv \frac{2}{\sqrt{x+1}+\sqrt{x-1}}
[3]
II.

Use your GDC to check the identity when x=10x=10. Give both values to three significant figures.

[2]
B

exactly R(x)=0.4R(x)=0.4.

[4]
Question 42
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

For n∈Z+n\in\mathbb{Z}^+, let

Sn=∑r=1n1r(r+2)S_n=\sum_{r=1}^{n}\frac{1}{r(r+2)}
A
I.

1r(r+2)≡12(1r−1r+2)\frac{1}{r(r+2)}\equiv \frac{1}{2}\left(\frac{1}{r}-\frac{1}{r+2}\right)
[3]
II.

Use your GDC to S20S_{20}, giving your answer to three significant figures.

[1]
B

Sn≡34−12(n+1)−12(n+2)S_n\equiv \frac{3}{4}-\frac{1}{2(n+1)}-\frac{1}{2(n+2)}
[4]
C

that Sn<34S_n<\frac{3}{4} for all n∈Z+n\in\mathbb{Z}^+.

[2]
Question 43
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

For real θ\theta, define

T(θ)=1−cos⁡θsin⁡θ+sin⁡θ1−cos⁡θT(\theta)=\frac{1-\cos \theta}{\sin \theta}+\frac{\sin \theta}{1-\cos \theta}
A
I.

the values of θ\theta in the interval 0≤θ≤2π0\leq \theta\leq 2\pi for which T(θ)T(\theta) is not defined.

[2]
II.

that

T(θ)≡2sin⁡θT(\theta)\equiv \frac{2}{\sin \theta}

for all allowed values of θ\theta.

[3]
B

T(θ)=3T(\theta)=3 for 0≤θ≤2π0\leq \theta\leq 2\pi. Give your answers in radians to three significant figures.

[2]
C

Use your GDC to check the identity in part (a)(ii) at θ=1.2\theta=1.2, and why this check is not a proof.

[2]
Question 44
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A student investigates the numbers

P(n)=n2−n+11,n∈NP(n)=n^2-n+11,\qquad n\in\mathbb{N}

The student claims that P(n)P(n) is prime for every n∈Nn\in\mathbb{N}.

nn

P(n)P(n)

Prime?

1

11

yes

2

13

yes

3

17

yes

4

23

yes

5

31

yes

6

41

yes

7

53

yes

8

67

yes

9

83

yes

10

not shown

not shown

11

not shown

not shown

A
I.

Use your GDC to P(10)P(10) and P(11)P(11).

[2]
II.

which input value of nn gives a counterexample.

[1]
B

n=11n=11 is a counterexample.

[3]
C

For an integer a>1a>1, define Qa(n)=n2−n+aQ_a(n)=n^2-n+a. the statement “Qa(n)Q_a(n) is prime for every n∈Nn\in\mathbb{N}” is false whenever aa is prime.

[3]
D

why the values of P(n)P(n) for 1≤n≤101\leq n\leq 10 cannot the student’s claim.

[1]

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Question 45
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

This question investigates a matrix representation of the Fibonacci sequence. The Fibonacci numbers are defined by F0=0F_0=0, F1=1F_1=1 and Fn+1=Fn+Fn−1F_{n+1}=F_n+F_{n-1} for n≥1n\geq 1. Let

A=(1110)A=\begin{pmatrix}1&1\\1&0\end{pmatrix}

n

Fibonacci terms

AnA^{n}

1

1, 1, 0

(1110)\begin{pmatrix}1&1\\1&0\end{pmatrix}

2

1, 1, 1

(2111)\begin{pmatrix}2&1\\1&1\end{pmatrix}

3

2, 1, 1

(3221)\begin{pmatrix}3&2\\2&1\end{pmatrix}

4

3, 2, 1

(5332)\begin{pmatrix}5&3\\3&2\end{pmatrix}

5

5, 3, 2

(8553)\begin{pmatrix}8&5\\5&3\end{pmatrix}

6

8, 5, 3

(13885)\begin{pmatrix}13&8\\8&5\end{pmatrix}

A
I.

A2A^2 and A3A^3.

[2]
II.

, using your answers to part (a)(i), that

An=(Fn+1FnFnFn−1)A^n=\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}

for n=2n=2 and n=3n=3.

[2]
B

Use mathematical induction to that

An=(Fn+1FnFnFn−1)A^n=\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}

for all n∈Z+n\in\mathbb{Z}^+.

[5]
C
I.

F8F_8.

[1]
II.

By considering determinants, Cassini's identity

Fn+1Fn−1−Fn2=(−1)nF_{n+1}F_{n-1}-F_n^2=(-1)^n

for all n∈Z+n\in\mathbb{Z}^+.

[3]
Question 46
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Let z=cos⁡θ+isin⁡θz=\cos\theta+i\sin\theta, where θ∈R\theta\in\mathbb{R}. This question investigates powers of complex numbers on the unit circle.

A
I.

z2=cos⁡2θ+isin⁡2θz^2=\cos 2\theta+i\sin 2\theta.

[2]
II.

z3z^3 in the form cos⁡a+isin⁡a\cos a+i\sin a.

[1]
B

Use mathematical induction to De Moivre's theorem,

(cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta

for all n∈Z+n\in\mathbb{Z}^+.

[5]
C
I.

Let ω=cos⁡2π5+isin⁡2π5\omega=\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}. that

1+ω+ω2+ω3+ω4=01+\omega+\omega^2+\omega^3+\omega^4=0
[2]
II.

that

cos⁡2π5+cos⁡4π5=−12\cos\frac{2\pi}{5}+\cos\frac{4\pi}{5}=-\frac12
[3]
Question 47
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For n∈Z+n\in\mathbb{Z}^+, define

Sn=∑r=1n1r(r+2)S_n=\sum_{r=1}^{n}\frac{1}{r(r+2)}

This question investigates a telescoping sum and two different proof methods.

n

Exact S_n

Decimal approx.

1

1/3

0.333333

2

11/24

0.458333

4

17/30

0.566667

8

29/45

0.644444

16

106/153

0.692810

32

404/561

0.720143

64

1576/2145

0.734732

A
I.

1r(r+2)≡12(1r−1r+2)\frac{1}{r(r+2)}\equiv \frac12\left(\frac1r-\frac{1}{r+2}\right)

for all allowed values of rr.

[2]
II.

S3S_3 exactly.

[2]
B

Use part (a)(i) to

Sn=34−12(n+1)−12(n+2)S_n=\frac34-\frac{1}{2(n+1)}-\frac{1}{2(n+2)}
[3]
C

the result in part (b) by mathematical induction. If you did not obtain the formula in part (b), use

Sn=34−12(n+1)−12(n+2)S_n=\frac34-\frac{1}{2(n+1)}-\frac{1}{2(n+2)}
[5]
Question 48
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

This question investigates identities involving binomial coefficients. The notation (nr)\binom{n}{r} has its usual meaning, and n∈Z+n\in\mathbb{Z}^+.

nn

r=0r=0

r=1r=1

r=2r=2

r=3r=3

r=4r=4

r=5r=5

r=6r=6

1

1

1

—

—

—

—

—

2

1

2

1

—

—

—

—

3

1

3

3

1

—

—

—

4

1

4

6

4

1

—

—

5

1

5

10

10

5

1

—

6

1

6

15

20

15

6

1

A
I.

∑r=03(3r)2\sum_{r=0}^{3}\binom{3}{r}^2
[2]
II.

(63)\binom{6}{3}.

[1]
B

By considering the coefficient of xnx^n in (1+x)n(1+x)n(1+x)^n(1+x)^n, that

∑r=0n(nr)2=(2nn)\sum_{r=0}^{n}\binom{n}{r}^2=\binom{2n}{n}
[4]
C
I.

Use mathematical induction to that

∑r=0n(nr)=2n\sum_{r=0}^{n}\binom{n}{r}=2^n

for all n∈Z+n\in\mathbb{Z}^+.

[4]
II.

A student claims that (2nn)=2n\binom{2n}{n}=2^n for all n∈Z+n\in\mathbb{Z}^+. Disprove the claim.

[1]

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Question 49
HL • Paper 3
Hard
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HL • Paper 3
Hard
Calculator Permitted

This question investigates Bernoulli's inequality:

(1+x)n≥1+nx(1+x)^n\geq 1+nx

Assume throughout parts (a), (b) and (d) that x≥−1x\geq -1 and n∈Z+n\in\mathbb{Z}^+.

A
I.

Bernoulli's inequality for n=2n=2 by simplifying the difference (1+x)2−(1+2x)(1+x)^2-(1+2x).

[2]
II.

the value of xx for which equality holds in part (a)(i).

[1]
B

Use mathematical induction to Bernoulli's inequality for all n∈Z+n\in\mathbb{Z}^+.

[6]
C

Disprove the statement that Bernoulli's inequality is true for all real xx and all n∈Z+n\in\mathbb{Z}^+.

[1]
D
I.

Use Bernoulli's inequality to

(1+1n)n≥2\left(1+\frac1n\right)^n\geq 2

for all n∈Z+n\in\mathbb{Z}^+.

[1]
II.

why checking the inequality for n=1,2,3n=1,2,3 would not the result in part (d)(i).

[1]
Question 50
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

This question investigates a divisibility result involving powers. Define

Dn=52n−3nD_n=5^{2n}-3^n

for n∈Z+n\in\mathbb{Z}^+.

nn

Dn=52n−3nD_n = 5^{2n} - 3^n

Dn/11D_n / 11

5

9765382

887762

6

244139896

22194536

3

15598

1418

4

390544

35504

A
I.

D1D_1 and D2D_2.

[2]
II.

that 1111 divides both values found in part (a)(i).

[1]
B

Use mathematical induction to that 1111 divides 52n−3n5^{2n}-3^n for all n∈Z+n\in\mathbb{Z}^+.

[5]
C
I.

A student claims that 121121 divides 52n−3n5^{2n}-3^n for all n∈Z+n\in\mathbb{Z}^+. Disprove this claim.

[1]
II.

why part (b) does not imply the student's claim in part (c)(i).

[1]
Question 51
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Let p1,p2,…,pkp_1,p_2,\ldots,p_k denote the first kk prime numbers, and define

Nk=p1p2⋯pk+1N_k=p_1p_2\cdots p_k+1

This question investigates Euclid's argument for the infinitude of primes.

kk

p1⋯pkp_1\cdots p_k

NkN_k

Nature of NkN_k

1

2

3

Prime

2

6

7

Prime

3

30

31

Prime

4

210

211

Prime

5

2310

2311

Prime

6

30030

30031

Composite

A
I.

N3N_3.

[1]
II.

whether N3N_3 is prime.

[1]
B

Use proof by contradiction to that there are infinitely many prime numbers.

[5]
C
I.

A student claims that NkN_k is prime for every k∈Z+k\in\mathbb{Z}^+. Use the case k=6k=6 to disprove the claim.

[2]
Question 52
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let

z=cos⁡θ+isin⁡θz=\cos\theta+{\mathrm{i}}\sin\theta

where θ∈R\theta\in\mathbb{R}.

A
I.

zz‾=1z\overline z=1, where z‾\overline z is the complex conjugate of zz.

[1]
II.

z2z^2 in the form a+iba+{\mathrm{i}}b.

[2]
B

Use mathematical induction to that

zn=cos⁡(nθ)+isin⁡(nθ)z^n=\cos(n\theta)+{\mathrm{i}}\sin(n\theta)

for all n∈Z+n\in\mathbb{Z}^+.

[6]
C
I.

By expanding z3z^3, the real part of z3z^3 is cos⁡3θ−3cos⁡θsin⁡2θ\cos^3\theta-3\cos\theta\sin^2\theta.

[2]
II.

cos⁡(3θ)=4cos⁡3θ−3cos⁡θ\cos(3\theta)=4\cos^3\theta-3\cos\theta
[2]

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Question 53
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

This question concerns irrational numbers.

A
I.

by contradiction that 43\sqrt[3]{4} is irrational. You may use the fact that if a prime pp divides a3a^3, then pp divides aa.

[5]
II.

that 5−435-\sqrt[3]{4} is irrational.

[3]
B

Disprove the statement: the product of any two irrational numbers is irrational.

[2]
Question 54
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

This question considers Euclid's argument about prime numbers. Assume, for contradiction, that p1,p2,…,pkp_1,p_2,\ldots,p_k is a complete finite of all primes, with the primes distinct, and define

N=p1p2⋯pk+1N=p_1p_2\cdots p_k+1
A

Assume, for contradiction, that there are only finitely many primes, p1,p2,…,pkp_1,p_2,\ldots,p_k. Define

N=p1p2⋯pk+1N=p_1p_2\cdots p_k+1
I.

none of p1,p2,…,pkp_1,p_2,\ldots,p_k divides NN.

[3]
II.

why NN has a prime divisor.

[2]
III.

complete the proof by contradiction that there are infinitely many primes.

[2]
B

A student claims that for every finite set of primes, the product of the primes plus 11 is itself prime.

[3]
C

why the counterexample in part (b) does not invalidate the proof in part (a).

[2]
Question 55
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For n∈Z+n\in\mathbb{Z}^+, let

Sn=∑r=1nr2rS_n=\sum_{r=1}^{n} r2^r
A

The formula for SnS_n is assumed to have the form

Sn=(an+b)2n+1+cS_n=(an+b)2^{n+1}+c
I.

Use your GDC to S1S_1, S2S_2 and S3S_3.

[1]
II.

aa, bb and cc.

[2]
B

Use mathematical induction to that

∑r=1nr2r=(n−1)2n+1+2\sum_{r=1}^{n} r2^r=(n-1)2^{n+1}+2

for all n∈Z+n\in\mathbb{Z}^+.

[6]
C

Use the formula to the least value of nn for which Sn>106S_n>10^6.

[2]
Question 56
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For n∈Z+n\in\mathbb{Z}^+, define

An=52n−(−1)nA_n=5^{2n}-(-1)^n
A
I.

Use your GDC to A1A_1, A2A_2 and A3A_3.

[1]
II.

a conjecture about the divisibility of AnA_n.

[2]
B

Use mathematical induction to your conjecture.

[5]
C

the remainder when 520265^{2026} is divided by 1313.

[2]

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Question 57
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let y=x2exy=x^2e^x.

A
I.

dydx\frac{dy}{dx} and d2ydx2\frac{d^2y}{dx^2}.

[2]
II.

Use your answers to a formula for dnydxn\frac{d^ny}{dx^n}.

[2]
B

Use mathematical induction to that

dndxn(x2ex)=(x2+2nx+n(n−1))ex\frac{d^n}{dx^n}(x^2e^x)=(x^2+2nx+n(n-1))e^x

for all n∈Z+n\in\mathbb{Z}^+.

[6]
C

Use the formula and your GDC to the minimum value of d5ydx5\frac{d^5y}{dx^5} for −8≤x≤1-8\leq x\leq 1, giving your answer to three significant figures.

[2]
Question 58
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

You may use the fact that if a prime number divides a3a^3, where a∈Za\in\mathbb{Z}, then it divides aa.

A
I.

Use your GDC to a decimal approximation to 43\sqrt[3]{4}.

[1]
II.

by contradiction that 43\sqrt[3]{4} is irrational.

[5]
B

Let pp and qq be rational numbers with q≠0q\ne 0. by contradiction that p+q43p+q\sqrt[3]{4} is irrational.

[4]
Question 59
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let

Sn=∑r=0nirS_n=\sum_{r=0}^{n} i^r

where i2=−1i^2=-1 and nn is a non-negative integer.

A
I.

Use your GDC, or otherwise, to S0S_0, S1S_1, S2S_2 and S3S_3.

[2]
II.

a formula for SnS_n in terms of in+1i^{n+1}.

[1]
B

Use mathematical induction to that

Sn=1−in+11−iS_n=\frac{1-i^{n+1}}{1-i}

for all integers n≥0n\geq 0.

[5]
C

all integers nn such that 0≤n≤500\leq n\leq 50 and Sn=0S_n=0.

[2]
Question 60
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A sequence is defined by

x1=2,xn+1=2+xnx_1=\sqrt2,\qquad x_{n+1}=\sqrt{2+x_n}

for n∈Z+n\in\mathbb{Z}^+. This question investigates bounds and the limiting behaviour suggested by the sequence.

A
I.

x2x_2 and x3x_3 exactly in nested radical form.

[2]
II.

Use your GDC to give x5x_5 to three significant figures.

[1]
B

Use mathematical induction to that 1<xn<21<x_n<2 for all n∈Z+n\in\mathbb{Z}^+.

[5]
C

the sequence is increasing.

[3]
D
I.

Assuming the sequence converges to a limit LL, LL.

[1]
II.

why the induction proof in part (b) alone does not that the sequence converges.

[1]

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