New

Subjects

IB Math AA HL - Questionbank

Proofs

Proof by Mathematical Induction, Contradiction, Counterexample, Simple Deduction…

Paper

Paper 1
Paper 2

Difficulty

Easy
Medium
Hard

View

Question 1

no calculator

easy

[Maximum mark: 4]

Consider two consecutive positive integers, kk and k+1k+1.

Show that the difference of their squares is equal to the sum of the two integers.

easy

Formula Booklet

Mark Scheme

Video

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 2

no calculator

easy

[Maximum mark: 4]

Prove that the sum of three consecutive positive integers is divisible by 33.

easy

Formula Booklet

Mark Scheme

Video

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 3

no calculator

easy

[Maximum mark: 4]

The product of three consecutive integers is increased by the middle integer.

Prove that the result is a perfect cube.

easy

Formula Booklet

Mark Scheme

Video

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 4

no calculator

easy

[Maximum mark: 6]

  1. Show that (2n1)3+(2n+1)3=16n3+12n(2n-1)^3 + (2n+1)^3 = 16n^3+12n for nZn \in \mathbb{Z}. [3]

  2. Hence, or otherwise, prove that the sum of the cubes of any two consecutive odd integers is divisible by four. [3]

easy

Formula Booklet

Mark Scheme

Video (a)

Video (b)

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 5

no calculator

easy

[Maximum mark: 6]

Using mathematical induction, prove that 12+22++n2=n(n+1)(2n+1)61^2 + 2^2 + \cdots + n^2 = \dfrac{n(n+1)(2n+1)}{6} for all nZ+n \in \mathbb{Z}^+.

easy

Formula Booklet

Mark Scheme

Video

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 6

no calculator

easy

[Maximum mark: 7]

Use the principle of mathematical induction to prove that

121+222+323++n2n=2+(n1)2n+1for all nZ+.\begin{aligned} \hspace{4em} 1\cdot2^1 + 2\cdot2^2 + 3\cdot2^3 + \cdots + n\cdot2^n &= 2 + (n-1)2^{n+1} \hspace{2em} \text{for all } n \in \mathbb{Z}^+. \\ \end{aligned}

easy

Formula Booklet

Mark Scheme

Video

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 7

no calculator

easy

[Maximum mark: 6]

Let rR,r1r \in \mathbb{R}, r\neq 1. Use the method of mathematical induction to prove that

1+r+r2++rn=1rn+11rfor all nZ+.\begin{aligned} \hspace{8.3em} 1+r+r^2+\cdots+r^n=\frac{1-r^{n+1}}{1-r} \hspace{2em} \text{for all } n\in \mathbb{Z}^+. \\ \end{aligned}

easy

Formula Booklet

Mark Scheme

Video

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 8

no calculator

easy

[Maximum mark: 4]

Using the method of proof by contradiction, prove that 3\sqrt{3} is irrational.

easy

Formula Booklet

Mark Scheme

Video

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 9

calculator

easy

[Maximum mark: 4]

Prove by contradiction that log47\log_4 7 is an irrational number.

easy

Formula Booklet

Mark Scheme

Video

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 10

no calculator

easy

[Maximum mark: 7]

The Fibonacci sequence is defined as follows:

a0=0,a1=1,a2=1,an=an1+an2forn2.(FS)\begin{aligned} a_0 &= 0,\hspace{0.25em} a_1 = 1,\hspace{0.25em} a_2 = 1, \\[6pt] a_n &= a_{n-1}+a_{n-2} \hspace{0.5em}\text{for}\hspace{0.5em} n \geq 2. \qref{(FS)}\end{aligned}

Prove by mathematical induction that a12+a22++an2=anan+1a_1^2+a_2^2+\cdots+a_n^2=a_na_{n+1}, where nZ+n\in\mathbb{Z}^+.

easy

Formula Booklet

Mark Scheme

Video

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 11

no calculator

medium

[Maximum mark: 8]

Let y=x2exy = x^2 e^x, for xRx \in \mathbb{R}.

  1. Find dydx\dfrac{\mathrm{d}y}{\mathrm{d}x}. [1]

  2. Prove by mathematical induction that

    dndxn(x2ex)=(n(n1)+2nx+x2)exfor all nZ+n2.\hspace{4em} \dfrac{\mathrm{d}^n}{\mathrm{d}x^n}\big(x^2e^x\big) = \big(n(n-1) + 2nx + x^2\big)e^x \hspace{1.5em} \text{for all $n \in \mathbb{Z}^+$, $n\geq2$.}

    [7]

medium

Formula Booklet

Mark Scheme

Video (a)

Video (b)

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 12

no calculator

medium

[Maximum mark: 9]

Let f(x)=(x+1)e2xf(x) = (x+1)e^{-2x}, xRx \in \mathbb{R}.

  1. Find f(x)f'(x). [2]

  2. Prove by induction that dnfdxn=[n(2)n1+(2)n(x+1)]e2x\dfrac{\mathrm{d}^nf}{\mathrm{d}x^n} = \big[n(-2)^{n-1} + (-2)^n(x+1)\big]e^{-2x} for all nZ+n \in \mathbb{Z}^+.[7]

medium

Formula Booklet

Mark Scheme

Video (a)

Video (b)

Video (c)

Video (d)

Video (e)

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 13

no calculator

medium

[Maximum mark: 6]

Using the principle of mathematical induction, prove that n(n2+5)n(n^2+5) is divisible by 66 for all integers n1n \geq 1.

medium

Formula Booklet

Mark Scheme

Video

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 14

calculator

hard

[Maximum mark: 8]

  1. Solve the inequality x22x+3x^2 \geq 2x + 3. [2]

  2. Use mathematical induction to prove that 2n>n222^n > n^2 - 2 for all nZ+n \in \mathbb{Z}^+, n3n \geq 3.[6]

hard

Formula Booklet

Mark Scheme

Video (a)

Video (b)

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 15

calculator

hard

[Maximum mark: 8]

  1. Solve the inequality 4x24x+24x^2 \geq 4x + 2. [2]

  2. Use mathematical induction to prove that 3n>2n23^n > 2n^2 for all nZ+n \in \mathbb{Z}^+, n2n \geq 2.[6]

hard

Formula Booklet

Mark Scheme

Video (a)

Video (b)

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 16

no calculator

hard

[Maximum mark: 14]

  1. Show that 12n+1<n+1n\dfrac{1}{2\sqrt{n+1}} < \sqrt{n+1} - \sqrt{n}, where nZ,n0n \in \mathbb{Z},\hspace{0.1em} n\geq 0. [3]

  2. Hence show that 12<222\dfrac{1}{\sqrt{2}} < 2\sqrt{2} - 2. [2]

  3. Prove by mathematical induction that

    r=2n1r<2n2for all nZ+n2.\hspace{4em} \sum_{r = 2}^n \dfrac{1}{\sqrt{r}} < 2\sqrt{n} - 2 \hspace{2em} \text{for all $n \in \mathbb{Z}^+$, $n \geq 2$.}

    [9]

hard

Formula Booklet

Mark Scheme

Video (a)

Video (b)

Video (c)

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 17

no calculator

hard

[Maximum mark: 23]

Let f(x)=(x1)ex3f(x) = (x-1)e^{\frac{x}{3}}, for xRx \in \mathbb{R}.

  1. Find f(x)f'(x). [2]

  2. Prove by induction that dnfdxn=(3n+x13n)ex3\dfrac{\mathrm{d}^nf}{\mathrm{d}x^n} = \bigg(\dfrac{3n + x - 1}{3^n}\bigg)e^{\frac{x}{3}} for all nZ+n \in \mathbb{Z}^+. [7]

  3. Find the coordinates of any local maximum and minimum points on the graph of y=f(x)y = f(x). Justify whether such point is a maximum or a minimum. [5]

  4. Find the coordinates of any points of inflexion on the graph of y=f(x)y = f(x). Justify whether such point is a point of inflexion. [5]

  5. Hence sketch the graph of y=f(x)y = f(x), indicating clearly the points found in parts (c) and (d) and any intercepts with the axes. [4]

hard

Formula Booklet

Mark Scheme

Video (a)

Video (b)

Video (c)

Video (d)

Video (e)

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 18

no calculator

hard

[Maximum mark: 21]

  1. Use de Moivre's theorem to find the value of [cos(π6)+isin(π6)]12\left[\cos\left(\dfrac{\pi}{6}\right) + {\mathrm{\hspace{0.05em}i}\mkern 1mu}\sin \left(\dfrac{\pi}{6}\right)\right]^{12}. [2]

  2. Use mathematical induction to prove that

    (cosαisinα)n=cos(nα)isin(nα)for all nZ+.\hspace{3.5em} (\cos \alpha - {\mathrm{\hspace{0.05em}i}\mkern 1mu}\sin \alpha)^n = \cos (n\alpha) - {\mathrm{\hspace{0.05em}i}\mkern 1mu}\sin (n\alpha) \hspace{1em} \text{for all } n \in \mathbb{Z}^+.

    [6]

Let w=cosα+isinαw = \cos \alpha + {\mathrm{\hspace{0.05em}i}\mkern 1mu}\sin \alpha.

  1. Find an expression in terms of α\alpha for wn(w)nw^n - (w^\ast)^n, nZ+n \in \mathbb{Z}^+, where ww^\ast is the complex conjugate of ww. [2]

    1. Show that ww=1ww^\ast = 1.

    2. Write down and simplify the binomial expansion of (ww)3(w - w^\ast)^3 in terms of ww and ww^\ast.

    3. Hence show that sin(3α)=3sinα4sin3α\sin (3\alpha) = 3\sin \alpha - 4 \sin^3 \alpha. [5]

  2. Hence solve 4sin3α+(2cosα3)sinα=04\sin^3\alpha + (2 \cos \alpha - 3) \sin \alpha = 0 for 0απ0 \leq \alpha \leq \pi. [6]

hard

Formula Booklet

Mark Scheme

Video (a)

Video (b)

Video (c)

Video (d)

Video (e)

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 19

no calculator

hard

[Maximum mark: 21]

Let f(x)=11xf(x) = \dfrac{1}{\sqrt{1-x}},x<1x < 1.

  1. Show that f(x)=34(1x)5/2f''(x) = \dfrac{3}{4} (1-x)^{-5/2}. [3]

  2. Use mathematical induction to prove that[9]

    f(n)(x)=(14)n(2n)!n!(1x)1/2nnZ,n2.f^{(n)}(x) = \left(\dfrac{1}{4}\right)^n \dfrac{(2n)!}{n!} (1-x)^{-1/2-n} \quad n\in \mathbb{Z},\enskip n\geq 2.

Let g(x)=cos(mx)g(x)=\cos (mx), mQm\in \mathbb{Q}.

Consider the function hh defined by h(x)=f(x)×g(x)h(x)=f(x) \times g(x) for x<1x<1.

The x2x^2 term in the Maclaurin series for h(x)h(x) has a coefficient of 34-\dfrac{3}{4}.

  1. Find the possible values of mm.[9]

hard

Formula Booklet

Mark Scheme

Video (a)

Video (b)

Video (c)

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 20

no calculator

hard

[Maximum mark: 17]

The following diagram shows the graph of y=arctan(2x3)+3π4y=\arctan(2x-3)+\dfrac{3\pi}{4} for xRx\in \mathbb{R},
with asymptotes at y=π4y=\dfrac{\pi}{4} and y=5π4y=\dfrac{5\pi}{4}.

081334c23d135e25efc729fff5cf10e0f4fe7363.svg

  1. Describe a sequence of transformations that transforms the graph of
    y=arctanxy=\arctan x to the graph of y=arctan(2x3)+3π4y=\arctan(2x-3)+\dfrac{3\pi}{4} for xRx\in \mathbb{R}.[3]

  2. Show that arctanparctanqarctan(pq1+pq)\arctan p - \arctan q \equiv \arctan \left(\dfrac{p-q}{1+pq}\right).[3]

  3. Verify that arctan(x+2)arctan(x+1)=arctan(1(x+1)2+(x+1)+1)\arctan(x+2)-\arctan(x+1) = \arctan\left( \dfrac{1}{(x+1)^2+(x+1)+1}\right).[3]

  4. Using mathematical induction and the results from part (b) and (c), prove that[8]

    r=1narctan(1r2+r+1)=arctan(n+1)π4for nZ+.\sum_{r=1}^n \arctan\left(\dfrac{1}{r^2+r+1}\right) = \arctan(n+1)-\dfrac{\pi}{4} \hspace{0.8em} \text{for } n\in \mathbb{Z}^{+}.

hard

Formula Booklet

Mark Scheme

Video (a)

Video (b)

Video (c)

Video (d)

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Question 21

no calculator

easy

[Maximum mark: 6]

The first three terms of an arithmetic sequence are u1,4u19u_1, 4u_1-9, and 3u1+183u_1+18.

  1. Show that u1=9u_1=9. [2]

  2. Prove that the sum of the first nn terms of this arithmetic sequence is a square number. [4]

easy

Formula Booklet

Mark Scheme

Video (a)

Video (b)

Revisit

Check with RV Newton

Formula Booklet

Mark Scheme

Solutions

Revisit

Ask Newton

Thank you Revision Village Members

#1 IB Math Resource

Revision Village was ranked the #1 IB Math Resources by IB Students & Teachers in 2021 & 2022.

34% Grade Increase

Revision Village students scored 34% greater than the IB Global Average in their exams (2021).

70% of IB Students

More and more IB students are using Revision Village to prepare for their IB Math Exams.

Frequently Asked Questions

The IB Math Analysis and Approaches (AA) HL Questionbank is a comprehensive set of IB Mathematics exam style questions, categorised into syllabus topic and concept, and sorted by difficulty of question. The bank of exam style questions are accompanied by high quality step-by-step markschemes and video tutorials, taught by experienced IB Mathematics teachers. The IB Mathematics AA HL Question bank is the perfect exam revision resource for IB students looking to practice IB Math exam style questions in a particular topic or concept in their AA Higher Level course.

The AA HL Questionbank is designed to help IB students practice AA HL exam style questions in a specific topic or concept. Therefore, a good place to start is by identifying a concept that you would like to practice and improve in and go to that area of the AA HL Question bank. For example, if you want to practice AA HL exam style questions that involve Complex Numbers, you can go to AA SL Topic 1 (Number & Algebra) and go to the Complex Numbers area of the question bank. On this page there is a carefully designed set of IB Math AA HL exam style questions, progressing in order of difficulty from easiest to hardest. If you’re just getting started with your revision, you could start at the top of the page with Question 1, or if you already have some confidence, you could start at the medium difficulty questions and progress down.

The AA HL Questionbank is perfect for revising a particular topic or concept, in-depth. For example, if you wanted to improve your knowledge of Counting Principles (Combinations & Permutations), there is a set of full length IB Math AA HL exam style questions focused specifically on this concept. Alternatively, Revision Village also has an extensive library of AA HL Practice Exams, where students can simulate the length and difficulty of an IB exam with the Mock Exam sets, as well as AA HL Key Concepts, where students can learn and revise the underlying theory, if missed or misunderstood in class.

With an extensive and growing library of full length IB Math Analysis and Approaches (AA) HL exam style questions in the AA HL Question bank, finishing all of the questions would be a fantastic effort, and you will be in a great position for your final exams. If you were able to complete all the questions in the AA HL Question bank, then a popular option would be to go to the AA HL Practice Exams section on Revision Village and test yourself with the Mock Exam Papers, to simulate the length and difficulty of an actual IB Mathematics AA HL exam.