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IB Mathematics AA HL - Questionbank

Proofs

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

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Question 1

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easy

[Maximum mark: 4]

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

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Question 2

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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.

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Question 3

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easy

[Maximum mark: 4]

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

Prove that the result is a perfect cube.

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Question 4

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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]

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Question 5

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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}^+.

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Question 6

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easy

[Maximum mark: 5]

  1. Prove that 5x2=5x(x2)10x2(x2)\dfrac{5}{x^2} = \dfrac{5}{x(x-2)}-\dfrac{10}{x^2(x-2)}. [3]

  2. Determine the set of numbers xx for which the proof in part (a) is valid. [2]

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Question 7

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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}

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Question 8

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easy

[Maximum mark: 4]

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

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Question 9

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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}

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Question 10

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easy

[Maximum mark: 6]

Prove by contradiction that the equation 3x37x2+5=03x^3-7x^2+5=0 has no integer roots.

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Question 11

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medium

[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}^+.

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Question 12

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[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]

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Question 13

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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]

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Question 14

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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.

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Question 15

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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]

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Question 16

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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]

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Question 17

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hard

[Maximum mark: 28]

This question asks you to explore the sequence defined by

un=36(αnβn)u_n=\dfrac{\sqrt{3}}{6}(\alpha^n-\beta^n)

where α\alpha and β\beta are the roots of the quadratic equation x24x+1=0,α>βx^2-4x+1=0, \, \alpha > \beta and nZ+n \in \mathbb{Z}^+.

  1. Find the value of α\alpha and the value of β\beta. Give your answers in the form a±ba \pm \sqrt{b}, where a,bZ+a,b \in \mathbb{Z^+}.[3]

  2. Hence find the values of u1u_1 and u2u_2. [4]

  3. Show that α2=4α1\alpha^2 = 4\alpha -1 and β2=4β1\beta^2 = 4\beta - 1. [1]

  4. Hence show that un+2=4un+1unu_{n+2} = 4u_{n+1}-u_n.[4]

  5. Suppose that unu_n and un+1u_{n+1} are integers. Show that un+2u_{n+2} is also an integer.[2]

  6. Hence show that unu_n is an integer for all nNn \in \mathbb{N}.[2]

Now consider the sequence defined by

vn=36(αn+βn).v_n = \dfrac{\sqrt{3}}{6}\left(\alpha^n + \beta^n\right).
  1. Find the exact values of v1v_1 and v2v_2.[4]

  2. Express vn+2v_{n+2} in terms of vn+1v_{n+1} and vnv_n.[4]

  3. Hence show that vnv_n is a multiple of 33\dfrac{\sqrt{3}}{3} for all nNn \in \mathbb{N}.[4]

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Question 18

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hard

[Maximum mark: 24]

This question asks you to investigate some properties of hexagonal numbers.

Hexagonal numbers can be represented by dots as shown below where hnh_n denotes the nnth hexagonal number, nNn\in \mathbb{N}.

b2b46cf7a1e76fe3e94b0c5adbb806ad8bd4f38f.svg

Note that 66 points are required to create the regular hexagon h2h_2 with side of length 11, while 1515 points are required to create the next hexagon h3h_3 with side of length 22, and so on.

  1. Write down the value of h5h_5.[1]

  2. By examining the pattern, show that hn+1=hn+4n+1h_{n+1} = h_{n}+4n+1, nNn\in \mathbb{N}. [3]

  3. By expressing hnh_n as a series, show that hn=2n2nh_n = 2n^2-n, nNn\in \mathbb{N}.[3]

  4. Hence, determine whether 20162016 is a hexagonal number.[3]

  5. Find the least hexagonal number which is greater than 8000080\hspace{0.10em}000.[5]

  6. Consider the statement:

    4545 is the only hexagonal number which is divisible by 99.

    Show that this statement is false.[2]

Matt claims that given h1=1h_1 = 1 and hn+1=hn+4n+1h_{n+1} = h_n + 4n + 1, nNn \in \mathbb{N}, then

hn=2n2n,nN.\begin{aligned} h_n &= 2n^2 - n, \quad n\in\mathbb{N}. \end{aligned}
  1. Show, by mathematical induction, that Matt's claim is true for all nNn\in \mathbb{N}.[7]

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Question 19

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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]

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Question 20

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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]

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Question 21

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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]

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Question 22

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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}^{+}.

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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.