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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

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.

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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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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]

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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

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