Free worked solutions · 2026 Edition

Chapter 18: Paper 3 worked solutions

Worked solutions to the investigations in Chapter 18 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Investigations

The four worked investigations of the chapter, with the full mark scheme.

Investigation 1

[25 marks]

In this question a positive integer N is called polite if it can be written as the sum of two or more consecutive positive integers. Each such sum is a polite representation of N. For example, 9=4+5 and 9=2+3+4, so 9 is polite.

(a) (i) Write 15 as a sum of consecutive positive integers in three different ways. [2] (ii) By considering the smallest possible sum of m consecutive positive integers, show that 8 is not polite.

[2]

(b) Determine which of the integers 1,2,3,\dots,20 are not polite.

[2]

(c) Make a conjecture about the positive integers that are not polite.

[1]

(d) Show that the sum of the m consecutive integers a,\ a+1,\ \dots,\ a+m-1 is \dfrac{m(2a+m-1)}{2}.

[2]

(e) Hence prove that if N is polite, then N has an odd factor greater than 1.

[3]

(f) Suppose that 2N=mL, where m and L are positive integers, one of them odd and the other even, with 2\le m<L. (i) Show that N is the sum of m consecutive positive integers, and find the first of these integers in terms of m and L. [3] (ii) Hence prove that every positive integer with an odd factor greater than 1 is polite.

[2]

(g) Find every polite representation of 45.

[3]

(h) Conjecture a formula for the number of polite representations of N in terms of the odd factors of N, and justify your conjecture.

[3]

(i) Find the smallest positive integer that has exactly seven polite representations.

[2]

Show solutionHide solution

(a)(i) 15=7+8=4+5+6=1+2+3+4+5

(a)(ii) m consecutive positive integers add to at least 1+2+\dots+m=\frac{m(m+1)}{2}, which is 10>8 once m\ge4

m=2: k+(k+1)=2k+1 is odd. m=3: k+(k+1)+(k+2)=3(k+1) is a multiple of 3. Neither is 8, so 8 is not polite

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(b) Not polite: 1,2,4,8,16

(c) N is not polite if and only if N=2^k for some k\in\mathbb{N}

(d) The terms form an arithmetic sequence with u_1=a, d=1 and m terms: S=\frac m2\bigl(2a+(m-1)\bigr)

=\dfrac{m(2a+m-1)}{2}

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(e) N polite \Rightarrow 2N=m(2a+m-1) with m\ge2, a\ge1

The factors m and 2a+m-1 add to 2a+2m-1, which is odd, so one factor is odd and the other even

The odd factor is at least 3: if m is odd then m\ge3; otherwise 2a+m-1\ge m+1\ge3. An odd factor of 2N is a factor of N, so N has an odd factor greater than 1

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(f)(i) Let a=\dfrac{L-m+1}{2}

L-m is odd, so L-m+1 is even and a is an integer; L>m, so a\ge1

The sum of a,\dots,a+m-1 is \dfrac{m(2a+m-1)}{2}=\dfrac{m(L-m+1+m-1)}{2}=\dfrac{mL}{2}=N. The first integer is \dfrac{L-m+1}{2}

(f)(ii) Write N=2^kq with q odd, q\ge3, k\ge0, so 2N=2^{k+1}q. Let m be the smaller and L the larger of q and 2^{k+1}

They differ (one odd, one even) and both are at least 2, so (f)(i) applies and N is polite

(g) 2N=90. Pairs (m,L) with 2\le m<L and opposite parity: (2,45),(3,30),(5,18),(6,15),(9,10)

First terms \frac{L-m+1}{2}=22,14,7,5,1: 45=22+23=14+15+16=7+8+9+10+11=5+6+\dots+10=1+2+\dots+9

(h) Conjecture: the number of polite representations of N is (the number of odd factors of N) -1

Each representation gives a pair 2N=mL of opposite parity with 2\le m<L, and (f)(i) turns each such pair back into a representation

A pair is fixed by its odd member d, which can be any odd factor of N (the other member 2N/d is even). The smaller member is 1 only when d=1; every other d gives a valid pair

(i) Seven representations need exactly 8 odd factors. Odd numbers with 8 factors have the form p^7, p^3q or pqr; the smallest are 3\cdot5\cdot7=105 and 3^3\cdot5=135

N=105

Final answer

7+8, 4+5+6, 1+2+3+4+5; two terms give odd totals, three terms give multiples of 3 (3k+3), m\ge4 terms give at least 10; not polite: 1,2,4,8,16; conjecture: N is not polite iff N=2^k, k\in\mathbb{N}; AP sum \frac m2(2a+m-1); 2N=m(2a+m-1) with factors of opposite parity, the odd one \ge3; a=\frac{L-m+1}{2}; N=2^kq: take \{m,L\}=\{q,2^{k+1}\}; 45: 22+23, 14+15+16, 7+\dots+11, 5+\dots+10, 1+\dots+9; number =d_{\text{odd}}(N)-1; 105.

Investigation 2

[25 marks]

This question investigates the number of tangents to a curve that pass through a given point (a,b). First consider the curve y=e^x.

(a) Show that the tangent to y=e^x at the point where x=t has equation y=e^t(x-t+1).

[2]

(b) (i) Show that this tangent passes through the point (a,b) if and only if b=e^t(a+1-t). [1] (ii) Find the number of tangents to y=e^x that pass through the point (1,2), and find the x-coordinate of each point of contact.

[3]

For a fixed value of a, let h(t)=e^t(a+1-t), where t\in\mathbb{R}.

(c) (i) Show that the maximum value of h is e^a, and state the value of t at which it occurs. [3] (ii) Describe the behaviour of h(t) as t\to-\infty and as t\to\infty. You may assume that \lim_{t\to-\infty}te^t=0. [2] (iii) Sketch the graph of y=h(t).

[2]

(d) Hence find the number of tangents to y=e^x that pass through (a,b) when (i) b>e^a; (ii) b=e^a; (iii) 0<b<e^a; (iv) b\le0.

[4]

(e) Explain why no two tangents to y=e^x are perpendicular.

[1]

Now consider the curve y=\ln x, where x>0.

(f) The curve y=\ln x is the reflection of y=e^x in the line y=x. Use this fact to conjecture the number of tangents to y=\ln x that pass through a point (a,b) with a>0.

[2]

(g) Prove your conjecture from part (f).

[4]

(h) Write down the number of tangents to y=\ln x that pass through a point (a,b) with a\le0.

[1]

Show solutionHide solution

(a) \frac{dy}{dx}=e^x, so the gradient at x=t is e^t. Tangent: y-e^t=e^t(x-t)

y=e^t(x-t+1)

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(b)(i) (a,b) lies on the tangent \iff its coordinates satisfy the equation \iff b=e^t(a-t+1)=e^t(a+1-t)

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(b)(ii) Through (1,2): 2=e^t(2-t)

t=0 (check: e^0\times2=2) and t=1.59 (GDC: 1.59362)

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Two tangents, touching at x=0 and x=1.59

(c)(i) h'(t)=e^t(a+1-t)-e^t=e^t(a-t)

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h'(t)=0\iff t=a; h'>0 for t<a and h'<0 for t>a, so h has a maximum at t=a, and h(a)=e^a

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(c)(ii) As t\to-\infty: h(t)=(a+1)e^t-te^t\to0, and h(t)>0 for t<a+1, so h\to0 from above

As t\to\infty: e^t\to\infty and a+1-t\to-\infty, so h(t)\to-\infty

(c)(iii) Sketch: horizontal asymptote y=0 on the left, rising to the maximum (a,e^a), crossing the t-axis at t=a+1, then decreasing without bound

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(d) The number of tangents equals the number of solutions of h(t)=b: the meets of the line y=b with the graph of h –

(i) b>e^a: 0 (ii) b=e^a: 1 (iii) 0<b<e^a: 2 (iv) b\le0: 1

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(e) The gradients at t_1 and t_2 are e^{t_1} and e^{t_2}; their product is positive, never -1

(f) Reflecting in y=x sends y=e^x to y=\ln x, tangents to tangents, and (a,b) to (b,a). So for y=\ln x through (a,b), a>0, use the e^x count at (b,a)

Conjecture: b>\ln a: two tangents; b=\ln a: one; b<\ln a: none

(g) Tangent at (t,\ln t), t>0: y=\ln t+\frac{x-t}{t}

Through (a,b): b=\frac at+\ln t-1=k(t)

k'(t)=-\frac a{t^2}+\frac1t=\frac{t-a}{t^2}: negative for t<a, positive for t>a; minimum k(a)=\ln a

k(t)\to\infty as t\to0^+ (k=\frac{a+t\ln t}{t}-1 and t\ln t\to0) and as t\to\infty; so two solutions if b>\ln a, one if b=\ln a, none if b<\ln a

(h) a\le0: exactly one tangent, for every b

Final answer

y-e^t=e^t(x-t); substitute (a,b); t=0 by inspection and t=1.59 (GDC), two tangents; h'(t)=e^t(a-t), maximum h(a)=e^a; h\to0^+ as t\to-\infty, h\to-\infty as t\to\infty; 0, 1, 2, 1; gradients e^{t_1}e^{t_2}>0\neq-1; for \ln x: 2 if b>\ln a, 1 if b=\ln a, 0 if b<\ln a; k(t)=\frac at+\ln t-1, k'(t)=\frac{t-a}{t^2}, minimum \ln a, k\to\infty at both ends; a\le0: k increasing from -\infty to \infty, exactly one.

Investigation 3

[29 marks]

An office assistant has n letters, each with its own addressed envelope, where n\in\mathbb{Z}^+. The letters are put into the envelopes at random, one letter in each envelope, so that all n! arrangements are equally likely. Let D_n be the number of arrangements in which no letter is in its own envelope, and let p_n=\dfrac{D_n}{n!}.

(a) (i) Write down the values of D_1 and D_2. [1] (ii) By listing the arrangements of three letters, show that D_3=2. [1] (iii) By considering the envelope that receives letter 1, find D_4.

[2]

(b) Let n\ge3. In an arrangement counted by D_n, letter 1 is in envelope k, where k\neq1. (i) Write down the number of possible values of k. [1] (ii) Show that, for a fixed k, the number of these arrangements in which letter k is in envelope 1 is D_{n-2}. [1] (iii) Explain why, for a fixed k, the number of these arrangements in which letter k is not in envelope 1 is D_{n-1}. [2] (iv) Hence show that D_n=(n-1)(D_{n-1}+D_{n-2}).

[1]

(c) Find D_5 and D_6.

[2]

(d) Use your GDC to find p_n for 2\le n\le8, giving each answer to six decimal places.

[2]

(e) Compare your answers to part (d) with e^{-1}. Hence write down a conjecture about \lim_{n\to\infty}p_n.

[1]

(f) Prove by mathematical induction that D_n-nD_{n-1}=(-1)^n for all integers n\ge2.

[6]

(g) (i) Hence show that p_n-p_{n-1}=\dfrac{(-1)^n}{n!} for n\ge2. [2] (ii) Deduce that p_n=\displaystyle\sum_{r=0}^{n}\frac{(-1)^r}{r!} for n\in\mathbb{Z}^+.

[2]

(h) Use the Maclaurin series for e^x to prove your conjecture from part (e).

[2]

(i) Six letters are put into their six envelopes at random. Find the probability that exactly one letter is in its own envelope.

[3]

Show solutionHide solution

(a)(i) D_1=0, D_2=1

(a)(ii) 123,132,213,231,312,321: only 231 and 312 have no letter in its own envelope, so D_3=2

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(a)(iii) Letter 1 goes into envelope 2, 3 or 4. For envelope 2 the arrangements are 2143, 3142, 4123; the same count, 3, holds for envelopes 3 and 4 by symmetry

D_4=3\times3=9

(b)(i) n-1

(b)(ii) Letters 1 and k swap; the other n-2 letters fill their own n-2 envelopes with none correct: D_{n-2} ways

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(b)(iii) Letters 2,\dots,n go into the envelopes other than k. Each letter j\ne k must avoid envelope j, and letter k must avoid envelope 1

So each of these n-1 letters has exactly one forbidden envelope, all different: this is the no-letter-correct problem for n-1 letters, D_{n-1} ways

(b)(iv) For each of the n-1 values of k there are D_{n-2}+D_{n-1} arrangements, so D_n=(n-1)(D_{n-1}+D_{n-2})

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(c) D_5=4(9+2)=44; D_6=5(44+9)=265

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(d) p_n=\frac{D_n}{n!} (table below)

(e) e^{-1}=0.367879\dots; the values move alternately above and below it and close in. Conjecture: \lim_{n\to\infty}p_n=e^{-1}

(f) Let P(n) be D_n-nD_{n-1}=(-1)^n. n=2: D_2-2D_1=1-0=1=(-1)^2, so P(2) is true

Assume P(k) is true for some k\ge2: D_k-kD_{k-1}=(-1)^k

D_{k+1}-(k+1)D_k=k(D_k+D_{k-1})-(k+1)D_k

=-D_k+kD_{k-1}=-(D_k-kD_{k-1})

=-(-1)^k=(-1)^{k+1}, so P(k+1) is true

P(2) is true and P(k) true \Rightarrow P(k+1) true, so P(n) is true for all integers n\ge2 by induction

(g)(i) Divide by n!: \frac{D_n}{n!}-\frac{nD_{n-1}}{n!}=\frac{(-1)^n}{n!}

\frac{nD_{n-1}}{n!}=\frac{D_{n-1}}{(n-1)!}=p_{n-1}, so p_n-p_{n-1}=\frac{(-1)^n}{n!}

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(g)(ii) Add (g)(i) for 2,3,\dots,n: p_n-p_1=\sum_{r=2}^{n}\frac{(-1)^r}{r!}

p_1=0=1-1=\frac{(-1)^0}{0!}+\frac{(-1)^1}{1!}, so p_n=\sum_{r=0}^{n}\frac{(-1)^r}{r!} (also true for n=1)

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(h) e^x=\sum_{r\ge0}\frac{x^r}{r!} for all x; at x=-1, e^{-1}=\sum\frac{(-1)^r}{r!}

p_n is the partial sum up to r=n, so p_n\to e^{-1} as n\to\infty

(i) Choose the one letter that is correct: 6 ways; the other 5 letters have none correct: D_5=44

P=\dfrac{6\times44}{6!}=\dfrac{264}{720}

=\dfrac{11}{30} (0.367)

Final answer

D_1=0, D_2=1; 231, 312; 3\times3=9; n-1; remaining n-2 letters deranged; treat envelope 1 as letter k's forbidden envelope, a derangement of n-1; D_5=44, D_6=265; 0.5,\ 0.333333,\ 0.375,\ 0.366667,\ 0.368056,\ 0.367857,\ 0.367882; limit e^{-1}; base 1-0=1; step D_{k+1}-(k+1)D_k=-(D_k-kD_{k-1}); divide by n!; telescope from p_1=0=1-1; e^{-1}=\sum\frac{(-1)^r}{r!}; \binom61D_5/6!=\frac{11}{30}.

Investigation 4

[25 marks]

Throughout this question \omega=e^{\frac{2\pi i}{3}}. The points A, B, C are represented by the complex numbers a, b, c, and every triangle ABC is labelled so that A\to B\to C runs anticlockwise.

(a) (i) Show that 1+\omega+\omega^2=0. [2] (ii) Show that -\omega^2=e^{\frac{\pi i}{3}}.

[1]

(b) Explain why triangle ABC is equilateral if and only if c-a=e^{\frac{\pi i}{3}}(b-a).

[2]

(c) Hence show that triangle ABC is equilateral if and only if a+\omega b+\omega^2c=0.

[3]

(d) Verify the result of part (c) for the triangle with vertices 0, 2 and 1+\sqrt3\,i.

[2]

On each side of a triangle ABC an equilateral triangle is drawn outside ABC. For a side running from P to Q in the anticlockwise direction, let g(p,q) be the complex number that represents the centre of the equilateral triangle drawn on PQ. You may use the fact that the centre of an equilateral triangle with side s lies at a distance \frac{s}{2\sqrt3} from each side.

(e) (i) Show that g(p,q)=\dfrac{(3+\sqrt3\,i)p+(3-\sqrt3\,i)q}{6}. [3] (ii) Let a=0, b=6 and c=2+4i, and let g_1=g(b,c), g_2=g(c,a) and g_3=g(a,b). Use your GDC to find |g_1-g_2|, |g_2-g_3| and |g_3-g_1|.

[3]

(f) Make a conjecture about the triangle with vertices g_1, g_2 and g_3 for any triangle ABC.

[1]

(g) (i) Show that \dfrac{3+\sqrt3\,i}{6}=\dfrac{1-\omega^2}{3} and \dfrac{3-\sqrt3\,i}{6}=\dfrac{1-\omega}{3}. [2] (ii) Hence prove your conjecture.

[4]

(h) Show that the triangle with vertices g_1, g_2, g_3 has the same centroid as triangle ABC.

[2]

Show solutionHide solution

(a)(i) \omega^3=e^{2\pi i}=1, so 0=\omega^3-1=(\omega-1)(\omega^2+\omega+1)

\omega\ne1, so 1+\omega+\omega^2=0

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(a)(ii) -\omega^2=e^{\pi i}e^{4\pi i/3}=e^{7\pi i/3}=e^{\pi i/3} (subtract 2\pi)

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(b) Equilateral \iff AC=AB and the angle from AB to AC, turning anticlockwise, is \frac\pi3

Multiplying b-a by e^{\pi i/3} keeps its modulus and turns it by \frac\pi3 anticlockwise, so c-a=e^{\pi i/3}(b-a) says exactly this

(c) By (a)(ii): c-a=-\omega^2(b-a)

c+\omega^2b-(1+\omega^2)a=0, and 1+\omega^2=-\omega, so c+\omega^2b+\omega a=0

Multiply by \omega^2 (\omega^3=1, \omega^4=\omega): \omega^2c+\omega b+a=0, that is a+\omega b+\omega^2c=0. Each step reverses (multiply by \omega), so this is an equivalence

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(d) A=0, B=2, C=1+\sqrt3\,i runs anticlockwise (C is above AB). 2\omega=-1+\sqrt3\,i and \omega^2(1+\sqrt3\,i)=\bigl(-\tfrac12-\tfrac{\sqrt3}{2}i\bigr)(1+\sqrt3\,i)=1-\sqrt3\,i

a+\omega b+\omega^2c=0+(-1+\sqrt3\,i)+(1-\sqrt3\,i)=0, and the sides are all 2

(e)(i) The centre lies on the perpendicular bisector of PQ, at distance \frac{|q-p|}{2\sqrt3} from the midpoint \frac{p+q}{2}

Going anticlockwise round ABC, the inside is on the left of P\to Q, so the outside is on the right: unit direction \frac{-i(q-p)}{|q-p|}

g=\dfrac{p+q}{2}-\dfrac{i(q-p)}{2\sqrt3} =p\Bigl(\dfrac12+\dfrac{\sqrt3}{6}i\Bigr)+q\Bigl(\dfrac12-\dfrac{\sqrt3}{6}i\Bigr) =\dfrac{(3+\sqrt3\,i)p+(3-\sqrt3\,i)q}{6}

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(e)(ii) g_1=5.1547+3.1547i, g_2=-0.1547+2.5774i, g_3=3-1.7321i

|g_1-g_2|=|g_2-g_3|=|g_3-g_1|=5.34 (3 s.f.)

(f) For every triangle ABC, the triangle G_1G_2G_3 is equilateral

(g)(i) \dfrac{1-\omega^2}{3}=\dfrac{1+\frac12+\frac{\sqrt3}{2}i}{3} =\dfrac12+\dfrac{\sqrt3}{6}i=\dfrac{3+\sqrt3\,i}{6}

\dfrac{1-\omega}{3}=\dfrac{1+\frac12-\frac{\sqrt3}{2}i}{3}=\dfrac{3-\sqrt3\,i}{6}

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(g)(ii) 3g_1=(1-\omega^2)b+(1-\omega)c, 3g_2=(1-\omega^2)c+(1-\omega)a, 3g_3=(1-\omega^2)a+(1-\omega)b

In 3(g_1+\omega g_2+\omega^2g_3) the coefficient of a is \omega(1-\omega)+\omega^2(1-\omega^2)=\omega-\omega^4=0

The coefficients of b and c: (1-\omega^2)+\omega^2(1-\omega)=1-\omega^3=0 and (1-\omega)+\omega(1-\omega^2)=1-\omega^3=0

So g_1+\omega g_2+\omega^2g_3=0, which by (c) means G_1G_2G_3 is equilateral (labelled anticlockwise)

(h) g_1+g_2+g_3=\dfrac{\bigl((1-\omega^2)+(1-\omega)\bigr)(a+b+c)}{3} =\dfrac{(2-\omega-\omega^2)(a+b+c)}{3}

2-\omega-\omega^2=3, so \dfrac{g_1+g_2+g_3}{3}=\dfrac{a+b+c}{3}: the same centroid

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Final answer

\omega^3=1, \omega\neq1, (\omega-1)(1+\omega+\omega^2)=0; -\omega^2=e^{\pi i}e^{\frac{4\pi i}{3}}=e^{\frac{7\pi i}{3}}=e^{\frac{\pi i}{3}}; rotation of \vec{AB} by \frac\pi3 about A; c+\omega^2b+\omega a=0, multiply by \omega^2; the two non-zero terms are negatives of each other; g=\frac{p+q}{2}-\frac{i(q-p)}{2\sqrt3}; all three \sqrt{\frac{44}{3}+8\sqrt3}=5.34; equilateral; \sqrt3 i=\omega-\omega^2, 1=-\omega-\omega^2; 3(g_1+\omega g_2+\omega^2g_3)=0 coefficient by coefficient; g_1+g_2+g_3=a+b+c.

Practice investigations

The two practice investigations at the end of the chapter.

Practice A

[16 marks]

Euler's method is applied to the differential equation \dfrac{dy}{dx}=y, with y=1 when x=0, to estimate the value of y when x=1. The method uses n steps, each of length h=\frac1n.

(a) Show that the estimate is y_n=\left(1+\dfrac1n\right)^n.

[3]

(b) Let E_n=e-y_n. Use your GDC to find nE_n for n=10, n=100 and n=1000.

[3]

(c) Conjecture the exact value of \lim_{n\to\infty}nE_n.

[1]

(d) Use the Maclaurin series for \ln(1+x) to show that n\ln\left(1+\dfrac1n\right)=1-\dfrac{1}{2n}+\dfrac{1}{3n^2}-\cdots.

[2]

(e) Hence, using e^u\approx1+u for small u, justify your conjecture.

[3]

(f) A modified method uses y_{k+1}=y_k\left(1+h+\dfrac{h^2}{2}\right). By finding the Maclaurin series for \ln\left(1+h+\dfrac{h^2}{2}\right) up to and including the term in h^3, estimate the error in the modified method's value for y(1) when n is large, giving your answer in terms of n.

[4]

Show solutionHide solution

(a) y_{k+1}=y_k+hy_k=(1+h)y_k

y_0=1, so y_k=(1+h)^k

h=\frac1n and k=n: y_n=\left(1+\frac1n\right)^n

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(b) nE_n=n\left(e-\left(1+\frac1n\right)^n\right): 1.2454, 1.3468, 1.3579 for n=10,100,1000

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(c) \lim nE_n=\frac e2 (=1.3591)

(d) \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots with x=\frac1n

n\ln\left(1+\frac1n\right)=n\left(\frac1n-\frac1{2n^2}+\frac1{3n^3}-\cdots\right)=1-\frac1{2n}+\frac1{3n^2}-\cdots

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(e) y_n=e^{n\ln(1+1/n)}=e\cdot e^{-\frac1{2n}+\frac1{3n^2}-\cdots}

with e^u\approx1+u: y_n\approx e\left(1-\frac1{2n}\right)

E_n\approx\frac{e}{2n}, so nE_n\to\frac e2

(f) u=h+\frac{h^2}{2}: \ln(1+u)=u-\frac{u^2}{2}+\frac{u^3}{3}-\cdots

=h+\frac{h^2}{2}-\frac{h^2+h^3}{2}+\frac{h^3}{3}+\cdots=h-\frac{h^3}{6}+\cdots

n\ln(\cdots)=1-\frac1{6n^2}+\cdots, so y(1)\approx e\left(1-\frac1{6n^2}\right)

error \approx\frac{e}{6n^2}

Final answer

(b) 1.2454, 1.3468, 1.3579
(c) \frac e2
(f) error \approx\frac{e}{6n^2}

Practice B

[18 marks]

Let I_n=\displaystyle\int_0^1\frac{x^n}{1+x}\,dx, where n\in\mathbb{N}.

(a) Find the exact value of I_0.

[2]

(b) Show that I_n+I_{n-1}=\dfrac1n for n\in\mathbb{Z}^+.

[3]

(c) Hence find I_1, I_2 and I_3, giving each answer as p+q\ln2 with p and q rational.

[3]

(d) Show that 0<I_n\le\dfrac{1}{n+1}.

[3]

(e) Prove by mathematical induction that (-1)^nI_n=\ln2-\displaystyle\sum_{r=1}^{n}\frac{(-1)^{r+1}}{r} for n\in\mathbb{Z}^+.

[5]

(f) Hence find the value of \displaystyle\sum_{r=1}^{\infty}\frac{(-1)^{r+1}}{r}.

[2]

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(a) I_0=\int_0^1\frac{dx}{1+x}=\left[\ln(1+x)\right]_0^1

=\ln2

(b) I_n+I_{n-1}=\int_0^1\frac{x^n+x^{n-1}}{1+x}dx=\int_0^1\frac{x^{n-1}(x+1)}{1+x}dx

=\int_0^1x^{n-1}dx=\left[\frac{x^n}{n}\right]_0^1

=\frac1n

A1AG

(c) I_1=1-I_0=1-\ln2; I_2=\frac12-I_1=\ln2-\frac12; I_3=\frac13-I_2=\frac56-\ln2

A1A1A1

(d) for 0<x\le1: 0<\frac{x^n}{1+x}\le x^n since 1+x\ge1

integrate: I_n>0

I_n\le\int_0^1x^ndx=\frac1{n+1}

(e) n=1: -I_1=\ln2-1 and the right side is \ln2-1, so true

assume true for n=k

I_{k+1}=\frac1{k+1}-I_k by (b)

(-1)^{k+1}I_{k+1}=\frac{(-1)^{k+1}}{k+1}+(-1)^kI_k=\frac{(-1)^{k+1}}{k+1}+\ln2-\sum_{r=1}^{k}\frac{(-1)^{r+1}}{r}=\ln2-\sum_{r=1}^{k+1}\frac{(-1)^{r+1}}{r}

true for n=1, and true for k implies true for k+1, so true for all n\in\mathbb{Z}^+

(f) 0<I_n\le\frac1{n+1}\to0, so I_n\to0

\sum_{r=1}^{n}\frac{(-1)^{r+1}}{r}=\ln2-(-1)^nI_n\to\ln2, so the sum is \ln2

Final answer

(a) \ln2
(c) 1-\ln2, \ln2-\frac12, \frac56-\ln2
(f) \ln2

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 19.

Job	TI-84 Plus CE keystrokes
Definite integral	`MATH, 9:fnInt(expression, X, lower, upper)`
Solve an equation	`MATH, Numeric Solver (MATH, 0:Solver on older OS versions): enter the equation, then SOLVE. Or graph both sides and use 2nd TRACE (CALC), 5:intersect`

From the book

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL

Chapter 18 teaches Paper 3 from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book