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IB Maths HL Paper 3 Chapter 16: Practice papers: Analysis and approaches HL worked solutions

Worked solutions to the practice questions in Chapter 16 of Photon Guidebook: Mathematics HL Paper 3 for the IB Diploma (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Question 1

[27 marks]

Paper A1, Question 1: A cubic over a quadratic [27 marks]

This question asks you to investigate the family of curves y = f_k(x), where f_k(x) = \dfrac{x^3}{x^2 + k} and k is a non-zero real constant.

(a) On separate diagrams, sketch the curves for k = 1 and for k = -1, showing all asymptotes.

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(b) Show that f_k(x) = x - \dfrac{kx}{x^2 + k}, and hence write down the equation of the oblique asymptote.

[3]

(c) Show that f_k'(x) = \dfrac{x^2(x^2 + 3k)}{(x^2 + k)^2}.

[3]

(d) Use your GDC to find the number of stationary points for k = -4, -1, 1 and 4, and suggest how the number depends on k.

[3]

(e) For k < 0, prove that the graph has turning points at x = \pm\sqrt{-3k}, and find their coordinates.

[4]

(f) Show that, as k varies, the turning points lie on a straight line through the origin.

[2]

(g) For k > 0, show that the area between the curve, its oblique asymptote and the lines x = 0 and x = a is \dfrac{k}{2}\ln\left(\dfrac{a^2 + k}{k}\right), and explain what happens to this area as a \to \infty.

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(h) Find the value of k > 0 for which the area in (g) with a = \sqrt{k} is equal to 1.

[2]

(i) Show that, for k > 0, f_k(x) = \sqrt{k}\,f_1\!\left(\dfrac{x}{\sqrt{k}}\right), and describe the transformation this represents.

[2]

Show solutionHide solution

(a) k=1: curve through O, increasing, with rotational symmetry about O; oblique asymptote y=x drawn (A1); k=-1: vertical asymptotes x=-1 and x=1, and oblique asymptote y=x, all drawn (A1); k=-1: middle branch decreasing through O; outer branches with a minimum for x>1 and a maximum for x<-1 (A1).

(b) x-\dfrac{kx}{x^2+k}=\dfrac{x(x^2+k)-kx}{x^2+k} (M1); =\dfrac{x^3+kx-kx}{x^2+k} (A1); =\dfrac{x^3}{x^2+k}=f_k(x) (AG); As x\to\pm\infty, \dfrac{kx}{x^2+k}\to0, so f_k(x)\to x: the oblique asymptote is y=x (A1).

(c) f_k'(x)=\dfrac{3x^2(x^2+k)-x^3\cdot2x}{(x^2+k)^2} (M1 A1); =\dfrac{3x^4+3kx^2-2x^4}{(x^2+k)^2}=\dfrac{x^4+3kx^2}{(x^2+k)^2} (A1); =\dfrac{x^2(x^2+3k)}{(x^2+k)^2} (AG); Alternative: differentiate the form in (b) to get 1-\dfrac{k(k-x^2)}{(x^2+k)^2}, then combine over (x^2+k)^2 (alternative method).

(d) Graph y=f_k'(x) for each value of k and count the zeros; or count the turning points of y=f_k(x) (A1 A1); k=-4: 3; k=-1: 3; k=1: 1; k=4: 1. Conjecture: 3 stationary points when k<0 and 1 when k>0 (A1).

(e) f_k'(x)=0\iff x^2=0 or x^2=-3k; for k<0 the second gives x=\pm\sqrt{-3k} (M1); For |x|<\sqrt{-3k}, x^2+3k<0; for |x|>\sqrt{-3k}, x^2+3k>0; \dfrac{x^2}{(x^2+k)^2}>0 throughout, so f_k' changes sign at each root (R1); At x^2=-3k: f_k(x)=\dfrac{x\cdot x^2}{x^2+k}=\dfrac{x(-3k)}{-2k}=\dfrac{3x}2 (M1); Turning points \left(\sqrt{-3k},\ \dfrac32\sqrt{-3k}\right) and \left(-\sqrt{-3k},\ -\dfrac32\sqrt{-3k}\right) (A1).

(f) For both turning points \dfrac yx=\dfrac{\frac32\sqrt{-3k}}{\sqrt{-3k}}=\dfrac32, which does not involve k (M1); All turning points satisfy y=\dfrac32x: a straight line through the origin of gradient \dfrac32 (A1).

(g) For x>0 and k>0: x-f_k(x)=\dfrac{kx}{x^2+k}>0, so \text{Area}=\displaystyle\int_0^a\left(x-f_k(x)\right)\mathrm dx (M1); \text{Area}=\displaystyle\int_0^a\dfrac{kx}{x^2+k}\,\mathrm dx (A1); =\left[\dfrac k2\ln\left(x^2+k\right)\right]_0^a (M1); =\dfrac k2\left(\ln(a^2+k)-\ln k\right) (A1); =\dfrac k2\ln\left(\dfrac{a^2+k}k\right) (AG); As a\to\infty, \ln\left(\dfrac{a^2+k}k\right)\to\infty: the area grows without bound, even though the curve approaches its asymptote (R1).

(h) a=\sqrt k: \dfrac k2\ln\left(\dfrac{2k}k\right)=\dfrac k2\ln2 (M1); \dfrac k2\ln2=1\Rightarrow k=\dfrac2{\ln2}=2.89 (A1).

(i) \sqrt k\,f_1\!\left(\dfrac x{\sqrt k}\right)=\sqrt k\cdot\dfrac{\left(x/\sqrt k\right)^3}{\left(x/\sqrt k\right)^2+1}=\sqrt k\cdot\dfrac{x^3k^{-3/2}}{(x^2+k)/k} (M1); =\dfrac{x^3}{x^2+k}=f_k(x) (AG); Enlargement, scale factor \sqrt k, centre the origin (both coordinates are multiplied by \sqrt k) (A1).

Final answer

Paper A1, Question 1. (a) sketches; (b) y=x; (c) proof; (d) 3,\,3,\,1,\,1: three when k<0, one when k>0; (e) \left(\pm\sqrt{-3k},\ \pm\tfrac32\sqrt{-3k}\right); (f) y=\tfrac32x; (g) the area grows without bound; (h) k=\dfrac2{\ln2}=2.89; (i) enlargement, scale factor \sqrt k, centre O.

Question 2

[28 marks]

Paper A1, Question 2: Sums that telescope [28 marks]

This question asks you to find closed forms for sums of reciprocals of products of consecutive integers.

Let S_n = \displaystyle\sum_{k=1}^{n} \frac{1}{k(k+1)}.

(a) Find S_1, S_2, S_3 and S_4, and suggest an expression for S_n.

[3]

(b) Prove your suggestion by mathematical induction.

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(c) By writing \dfrac{1}{k(k+1)} in partial fractions, give a second proof.

[3]

Let T_n = \displaystyle\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)}.

(d) Show that \dfrac{1}{k(k+1)(k+2)} = \dfrac{1}{2}\left(\dfrac{1}{k(k+1)} - \dfrac{1}{(k+1)(k+2)}\right).

[3]

(e) Hence show that T_n = \dfrac{1}{4} - \dfrac{1}{2(n+1)(n+2)}.

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(f) Write down the limit of T_n as n \to \infty.

[1]

(g) Find a similar identity for \dfrac{1}{k(k+1)(k+2)(k+3)}, and hence find \displaystyle\sum_{k=1}^{\infty} \frac{1}{k(k+1)(k+2)(k+3)}.

[5]

(h) Suggest a formula for \displaystyle\sum_{k=1}^{\infty} \frac{1}{k(k+1)\cdots(k+m)} in terms of m, and justify your suggestion.

[5]

Show solutionHide solution

(a) S_1=\dfrac12 and S_2=\dfrac12+\dfrac16=\dfrac23 (A1); S_3=\dfrac23+\dfrac1{12}=\dfrac34 and S_4=\dfrac34+\dfrac1{20}=\dfrac45 (A1); Conjecture: S_n=\dfrac n{n+1} (A1).

(b) n=1: LHS =\dfrac1{1\times2}=\dfrac12 and RHS =\dfrac12, so true for n=1 (A1); Assume true for n=m: \displaystyle\sum_{k=1}^{m}\dfrac1{k(k+1)}=\dfrac m{m+1} (M1); n=m+1: \displaystyle\sum_{k=1}^{m+1}\dfrac1{k(k+1)}=\dfrac m{m+1}+\dfrac1{(m+1)(m+2)} (M1); =\dfrac{m(m+2)+1}{(m+1)(m+2)}=\dfrac{(m+1)^2}{(m+1)(m+2)}=\dfrac{m+1}{m+2} (A1); True for n=1; true for n=m implies true for n=m+1; so true for all n\in\mathbb Z^+ (R1).

(c) \dfrac1{k(k+1)}=\dfrac Ak+\dfrac B{k+1}\Rightarrow A(k+1)+Bk=1 (M1); A=1, B=-1: \dfrac1{k(k+1)}=\dfrac1k-\dfrac1{k+1} (A1); S_n=\left(1-\dfrac12\right)+\left(\dfrac12-\dfrac13\right)+\dots+\left(\dfrac1n-\dfrac1{n+1}\right)=1-\dfrac1{n+1}=\dfrac n{n+1} (A1).

(d) \dfrac1{k(k+1)}-\dfrac1{(k+1)(k+2)}=\dfrac{(k+2)-k}{k(k+1)(k+2)} (M1); =\dfrac2{k(k+1)(k+2)} (A1); \dfrac12\left(\dfrac1{k(k+1)}-\dfrac1{(k+1)(k+2)}\right)=\dfrac12\times\dfrac2{k(k+1)(k+2)} (A1); =\dfrac1{k(k+1)(k+2)} (AG).

(e) T_n=\dfrac12\displaystyle\sum_{k=1}^{n}\left(\dfrac1{k(k+1)}-\dfrac1{(k+1)(k+2)}\right) (M1); \begin{aligned}&=\dfrac12\Big[\Big(\dfrac1{1\times2}-\dfrac1{2\times3}\Big)+\Big(\dfrac1{2\times3}-\dfrac1{3\times4}\Big)\\&\qquad+\dots+\Big(\dfrac1{n(n+1)}-\dfrac1{(n+1)(n+2)}\Big)\Big]\end{aligned} (A1); =\dfrac12\left[\dfrac12-\dfrac1{(n+1)(n+2)}\right] (A1); =\dfrac14-\dfrac1{2(n+1)(n+2)} (AG).

(f) \dfrac1{2(n+1)(n+2)}\to0, so T_n\to\dfrac14 (A1).

(g) \dfrac1{k(k+1)(k+2)}-\dfrac1{(k+1)(k+2)(k+3)}=\dfrac{(k+3)-k}{k(k+1)(k+2)(k+3)}=\dfrac3{k(k+1)(k+2)(k+3)} (M1 A1); \dfrac1{k(k+1)(k+2)(k+3)}=\dfrac13\left(\dfrac1{k(k+1)(k+2)}-\dfrac1{(k+1)(k+2)(k+3)}\right) (A1); \displaystyle\sum_{k=1}^{n}\dfrac1{k(k+1)(k+2)(k+3)}=\dfrac13\left[\dfrac1{1\times2\times3}-\dfrac1{(n+1)(n+2)(n+3)}\right] (M1); As n\to\infty the second term \to0: the sum to infinity is \dfrac13\times\dfrac16=\dfrac1{18} (A1).

(h) m=1,2,3: sums 1, \dfrac14, \dfrac1{18} (from (c), (f), (g)); conjecture \displaystyle\sum_{k=1}^{\infty}\dfrac1{k(k+1)\cdots(k+m)}=\dfrac1{m\cdot m!} (A1); \begin{aligned}\dfrac1{k(k+1)\cdots(k+m)}&=\dfrac1m\Big(\dfrac1{k(k+1)\cdots(k+m-1)}\\&\qquad-\dfrac1{(k+1)(k+2)\cdots(k+m)}\Big)\end{aligned} (M1); Check: the difference of the two fractions has numerator (k+m)-k=m (A1); \displaystyle\sum_{k=1}^{n}=\dfrac1m\left[\dfrac1{1\times2\times\cdots\times m}-\dfrac1{(n+1)(n+2)\cdots(n+m)}\right] (M1); As n\to\infty: \dfrac1m\times\dfrac1{m!}=\dfrac1{m\cdot m!} (A1).

Final answer

Paper A1, Question 2. (a) \tfrac12,\tfrac23,\tfrac34,\tfrac45; S_n=\dfrac n{n+1}; (b) induction; (c) \dfrac1k-\dfrac1{k+1} telescopes; (d), (e) proofs; (f) \tfrac14; (g) \tfrac13\left(\tfrac1{k(k+1)(k+2)}-\tfrac1{(k+1)(k+2)(k+3)}\right), sum \tfrac1{18}; (h) \dfrac1{m\cdot m!}.

Question 3

[26 marks]

Paper A2, Question 1: Letting in the light [26 marks]

This question asks you to design a window that lets in as much light as possible.

A window is a rectangle of width 2r metres and height h metres, with a semicircle of radius r on top. The total length of the frame around the outside of the window is 6 metres.

(a) Show that h = \dfrac{6 - (2 + \pi)r}{2}.

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(b) Show that the window has area A = 6r - \left(2 + \dfrac{\pi}{2}\right)r^2.

[2]

(c) Find the value of r that maximises A, justify that it is a maximum, and show that for this window h = r.

[5]

(d) Find the exact maximum area.

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The semicircle is made of tinted glass that lets through only half as much light per square metre as the clear rectangle.

(e) Show that the light let in is proportional to L = 6r - \left(2 + \dfrac{3\pi}{4}\right)r^2.

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(f) Find the dimensions of the window that maximise L.

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(g) If the semicircle lets through a fraction t of the light, where 0 \le t \le 1, show that the best window has \dfrac{h}{r} = 1 + \dfrac{\pi(1 - t)}{2}.

[5]

(h) Interpret your answer to (g) when t = 0 and when t = 1.

[3]

Show solutionHide solution

(a) Outside frame: two vertical sides, the base and the semicircular arc, so 2h+2r+\pi r=6 (M1); 2h=6-2r-\pi r (A1); h=\dfrac{6-(2+\pi)r}2 (AG).

(b) A=2rh+\dfrac12\pi r^2 (M1); A=r\left(6-(2+\pi)r\right)+\dfrac\pi2r^2=6r-2r^2-\pi r^2+\dfrac\pi2r^2 (A1); A=6r-\left(2+\dfrac\pi2\right)r^2 (AG).

(c) \dfrac{\mathrm dA}{\mathrm dr}=6-(4+\pi)r (A1); \dfrac{\mathrm dA}{\mathrm dr}=0\Rightarrow r=\dfrac6{4+\pi}=0.840\text{ m} (M1 A1); \dfrac{\mathrm d^2A}{\mathrm dr^2}=-(4+\pi)<0, so this is a maximum (R1); (4+\pi)r=6, so h=\dfrac{(4+\pi)r-(2+\pi)r}2=\dfrac{2r}2 (M1); h=r (AG).

(d) A=\left(4+\pi\right)r^2-\left(2+\dfrac\pi2\right)r^2=\left(2+\dfrac\pi2\right)r^2 using 6=(4+\pi)r (M1); A_{\max}=\dfrac{4+\pi}2\times\dfrac{36}{(4+\pi)^2}=\dfrac{18}{4+\pi}\text{ m}^2 (2.52\text{ m}^2) (A1).

(e) Light \propto L=2rh+\dfrac12\times\dfrac{\pi r^2}2 (M1); L=r\left(6-(2+\pi)r\right)+\dfrac{\pi r^2}4 (A1); L=6r-2r^2-\pi r^2+\dfrac\pi4r^2 (A1); L=6r-\left(2+\dfrac{3\pi}4\right)r^2 (AG).

(f) \dfrac{\mathrm dL}{\mathrm dr}=6-\left(4+\dfrac{3\pi}2\right)r=0 (M1); r=\dfrac6{4+\frac{3\pi}2}=0.689\text{ m} (A1); h=\dfrac{6-(2+\pi)(0.6887)}2 (M1); h=1.23\text{ m}: semicircle of radius 0.689 m on a rectangle 1.38 m wide and 1.23 m high (A1).

(g) L=2rh+t\times\dfrac{\pi r^2}2=r\left(6-(2+\pi)r\right)+\dfrac{t\pi r^2}2 (M1); L=6r-\left(2+\pi-\dfrac{t\pi}2\right)r^2 (A1); \dfrac{\mathrm dL}{\mathrm dr}=6-\left(4+2\pi-t\pi\right)r=0\Rightarrow r=\dfrac6{4+2\pi-t\pi} (M1 A1); \dfrac hr=\dfrac{6-(2+\pi)r}{2r}=\dfrac3r-\dfrac{2+\pi}2=\dfrac{4+2\pi-t\pi}2-\dfrac{2+\pi}2 (M1); \dfrac hr=\dfrac{2+\pi-t\pi}2=1+\dfrac{\pi(1-t)}2 (AG).

(h) t=1: \dfrac hr=1, agreeing with (c): the whole window is clear glass and the best window has h=r (A1); t=0: \dfrac hr=1+\dfrac\pi2=2.57 (A1); With t=0 the semicircle lets in no light but still uses frame, so the best window is taller and narrower; \dfrac hr rises as t falls (R1).

Final answer

Paper A2, Question 1. (a), (b) proofs; (c) r=\dfrac6{4+\pi}=0.840 m, a maximum, h=r; (d) \dfrac{18}{4+\pi} m^2; (e) proof; (f) r=0.689 m, h=1.23 m; (g) proof; (h) t=1: h/r=1; t=0: h/r=1+\tfrac\pi2.

Question 4

[29 marks]

Paper A2, Question 2: Roots on a vertical line [29 marks]

This question asks you to show that all the roots of a family of polynomial equations lie on one line in the Argand diagram.

For n \ge 2, consider the equation (z + 1)^n = z^n, z \in \mathbb{C}.

(a) Solve the equation when n = 2.

[1]

(b) Solve the equation when n = 3, giving your answers in the form a + bi.

[3]

(c) Explain why the equation has exactly n - 1 roots, and show that each root can be written as z = \dfrac{1}{\omega - 1}, where \omega^n = 1 and \omega \ne 1.

[4]

(d) Let \omega = e^{i\theta} with 0 < \theta < 2\pi. Show that \dfrac{1}{e^{i\theta} - 1} = -\dfrac{1}{2} - \dfrac{i}{2}\cot\dfrac{\theta}{2}.

[6]

(e) Hence state the real part of every root, and solve the equation when n = 4.

[3]

(f) Find the roots when n = 6 in exact form.

[4]

(g) By considering the coefficients of (z + 1)^n - z^n, find the sum of the roots, and show that this agrees with your answer to (e).

[4]

(h) Find the product of the roots, and verify your answer for n = 3.

[4]

Show solutionHide solution

(a) (z+1)^2-z^2=2z+1=0\Rightarrow z=-\dfrac12 (A1).

(b) (z+1)^3-z^3=3z^2+3z+1=0 (M1); z=\dfrac{-3\pm\sqrt{9-12}}6=\dfrac{-3\pm i\sqrt3}6 (A1); z=-\dfrac12+\dfrac{\sqrt3}6i and z=-\dfrac12-\dfrac{\sqrt3}6i (A1).

(c) (z+1)^n-z^n=nz^{n-1}+\dbinom n2z^{n-2}+\dots+1: the z^n terms cancel, leaving degree n-1 with leading coefficient n\ne0 (R1); z=0 is not a root, so divide by z^n: \left(\dfrac{z+1}z\right)^n=1 (M1); \dfrac{z+1}z=\omega, where \omega^n=1 (A1); \dfrac{z+1}z=1+\dfrac1z\ne1, so \omega\ne1; each of the n-1 values of \omega gives one root, so there are n-1 distinct roots (R1); 1+\dfrac1z=\omega\Rightarrow z=\dfrac1{\omega-1} (AG).

(d) \dfrac1{e^{i\theta}-1}=\dfrac{e^{-i\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}} (M1); e^{i\theta/2}-e^{-i\theta/2}=2i\sin\dfrac\theta2 (A1); e^{-i\theta/2}=\cos\dfrac\theta2-i\sin\dfrac\theta2 (A1); =\dfrac{\cos\frac\theta2}{2i\sin\frac\theta2}-\dfrac{i\sin\frac\theta2}{2i\sin\frac\theta2} (M1); \dfrac{\cos\frac\theta2}{2i\sin\frac\theta2}=-\dfrac i2\cot\dfrac\theta2 (A1); -\dfrac{i\sin\frac\theta2}{2i\sin\frac\theta2}=-\dfrac12 (A1); \dfrac1{e^{i\theta}-1}=-\dfrac12-\dfrac i2\cot\dfrac\theta2 (AG); Alternative: multiply by the conjugate, \dfrac{(\cos\theta-1)-i\sin\theta}{(\cos\theta-1)^2+\sin^2\theta}, then use double-angle identities (alternative method).

(e) Every root is \dfrac1{\omega-1} with \omega=e^{i\theta}, so every root has \operatorname{Re}z=-\dfrac12 (A1); n=4: \omega\in\{i,-1,-i\}, so \theta=\dfrac\pi2,\ \pi,\ \dfrac{3\pi}2 (M1); z=-\dfrac12-\dfrac i2,\quad-\dfrac12,\quad-\dfrac12+\dfrac i2 (A1).

(f) n=6: \theta=\dfrac\pi3,\dfrac{2\pi}3,\pi,\dfrac{4\pi}3,\dfrac{5\pi}3, so \dfrac\theta2=\dfrac\pi6,\dfrac\pi3,\dfrac\pi2,\dfrac{2\pi}3,\dfrac{5\pi}6 (M1); \cot\dfrac\pi6=\sqrt3 and \cot\dfrac\pi3=\dfrac1{\sqrt3} (A1); z=-\dfrac12-\dfrac{\sqrt3}2i and z=-\dfrac12-\dfrac{\sqrt3}6i (A1); z=-\dfrac12, \ -\dfrac12+\dfrac{\sqrt3}6i, \ -\dfrac12+\dfrac{\sqrt3}2i (A1).

(g) (z+1)^n-z^n=nz^{n-1}+\dbinom n2z^{n-2}+\dots (M1); Sum of the roots =-\dfrac{\binom n2}n (A1); =-\dfrac{n(n-1)}{2n}=-\dfrac{n-1}2 (A1); From (e): n-1 roots, each with real part -\dfrac12, and the imaginary parts cancel in conjugate pairs; the sum is -\dfrac{n-1}2 (R1).

(h) Product of the roots =(-1)^{n-1}\times\dfrac{\text{constant term}}{\text{leading coefficient}} (M1); Constant term =1, leading coefficient =n (A1); Product =\dfrac{(-1)^{n-1}}n (A1); n=3: \left(-\dfrac12+\dfrac{\sqrt3}6i\right)\left(-\dfrac12-\dfrac{\sqrt3}6i\right)=\dfrac14+\dfrac1{12}=\dfrac13=\dfrac{(-1)^2}3 (A1).

Final answer

Paper A2, Question 2. (a) -\tfrac12; (b) -\tfrac12\pm\tfrac{\sqrt3}6i; (c), (d) proofs; (e) \operatorname{Re}z=-\tfrac12; n=4: -\tfrac12,\ -\tfrac12\pm\tfrac12i; (f) -\tfrac12,\ -\tfrac12\pm\tfrac{\sqrt3}6i,\ -\tfrac12\pm\tfrac{\sqrt3}2i; (g) -\dfrac{n-1}2; (h) \dfrac{(-1)^{n-1}}n.

Question 5

[25 marks]

Paper A3, Question 1: Curves described by their tangents [25 marks]

This question asks you to find curves from a property of their tangents or normals, by setting up and solving differential equations.

Let P(x, y) be a point on a curve y = f(x) in the first quadrant, with f'(x) \ne 0.

(a) Show that the tangent at P meets the x-axis where X = x - \dfrac{y}{\frac{\mathrm{d}y}{\mathrm{d}x}}.

[3]

(b) A curve has the property that, for every point P, this tangent meets the x-axis at X = \dfrac{x}{2}. Show that \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2y}{x}.

[2]

(c) Solve this differential equation, and find the curve through (2, 12).

[4]

(d) A second curve has the property that the normal at every point P meets the x-axis exactly 2 units to the right of the point directly below P. Show that y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2, and find the curve through (1, 2).

[5]

(e) A third curve has the property that the tangent at every point P meets the y-axis at a height equal to the x-coordinate of P. Show that \dfrac{\mathrm{d}y}{\mathrm{d}x} - \dfrac{y}{x} = -1.

[3]

(f) Solve the equation in (e) using an integrating factor, and find the curve through (1, 2).

[5]

(g) Find the coordinates of the maximum point of the curve in (f).

[3]

Show solutionHide solution

(a) Tangent at P: Y-y=\dfrac{\mathrm dy}{\mathrm dx}(X-x) (M1); Meets the x-axis where Y=0: -y=\dfrac{\mathrm dy}{\mathrm dx}(X-x) (M1); X-x=-\dfrac y{\mathrm dy/\mathrm dx} (A1); X=x-\dfrac y{\mathrm dy/\mathrm dx} (AG).

(b) x-\dfrac y{\mathrm dy/\mathrm dx}=\dfrac x2 (M1); \dfrac y{\mathrm dy/\mathrm dx}=\dfrac x2 (A1); \dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2y}x (AG).

(c) \displaystyle\int\dfrac1y\,\mathrm dy=\int\dfrac2x\,\mathrm dx (M1); \ln y=2\ln x+c (A1); y=Cx^2 (A1); 12=4C\Rightarrow C=3: y=3x^2 (A1).

(d) Gradient of the normal =-\dfrac1{\mathrm dy/\mathrm dx}: Y-y=-\dfrac1{\mathrm dy/\mathrm dx}(X-x) (M1); Y=0: X=x+y\dfrac{\mathrm dy}{\mathrm dx} (A1); The point below P is (x,0) and X=x+2, so y\dfrac{\mathrm dy}{\mathrm dx}=2 (AG); \displaystyle\int y\,\mathrm dy=\int2\,\mathrm dx\Rightarrow\dfrac{y^2}2=2x+c (M1 A1); (1,2): 2=2+c\Rightarrow c=0, so y^2=4x (that is y=2\sqrt x in the first quadrant) (A1).

(e) Tangent: Y-y=\dfrac{\mathrm dy}{\mathrm dx}(X-x); at X=0: Y=y-x\dfrac{\mathrm dy}{\mathrm dx} (M1 A1); y-x\dfrac{\mathrm dy}{\mathrm dx}=x (A1); \dfrac{\mathrm dy}{\mathrm dx}-\dfrac yx=-1 (AG).

(f) Integrating factor =e^{\int-\frac1x\,\mathrm dx} (M1); =e^{-\ln x}=\dfrac1x (A1); \dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac y{x^2}=-\dfrac1x\Rightarrow\dfrac{\mathrm d}{\mathrm dx}\left(\dfrac yx\right)=-\dfrac1x (M1); \dfrac yx=-\ln x+C (A1); y=-x\ln x+Cx; (1,2): C=2, so y=2x-x\ln x (A1).

(g) \dfrac{\mathrm dy}{\mathrm dx}=2-(\ln x+1)=1-\ln x=0 (M1); x=e (A1); y=2e-e\ln e=e: maximum point (e,\,e) (A1).

Final answer

Paper A3, Question 1. (a), (b) proofs; (c) y=Cx^2, y=3x^2; (d) y^2=4x; (e) proof; (f) y=2x-x\ln x; (g) (e,\,e).

Question 6

[30 marks]

Paper A3, Question 2: Waiting for the first red [30 marks]

This question asks you to find expected waiting times in a discrete model and in a continuous model.

A bag contains 3 red balls and 7 blue balls. Balls are taken out at random, one at a time and without replacement, until the first red ball appears. Let X be the number of balls taken out.

(a) Find \mathrm{P}(X = 1), \mathrm{P}(X = 2) and \mathrm{P}(X = 3).

[3]

(b) Show that \mathrm{P}(X = k) = \dfrac{\binom{10-k}{2}}{\binom{10}{3}} for k = 1, 2, \dots, 8.

[4]

(c) Find \mathrm{E}(X).

[3]

(d) Find the expected number of draws if each ball is put back before the next draw, and comment on the difference.

[2]

(e) For a bag with r red balls and b blue balls, use your GDC or a table to find \mathrm{E}(X) for small cases, and suggest a formula for \mathrm{E}(X).

[3]

(f) Prove your formula when r = 1.

[3]

The time T minutes between buses at a stop is modelled by the probability density function f(t) = \lambda e^{-\lambda t} for t \ge 0, where \lambda > 0.

(g) Show that \mathrm{P}(T > t) = e^{-\lambda t} and that \mathrm{P}(T > s + t \mid T > s) = \mathrm{P}(T > t). Interpret this result.

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(h) Use integration by parts to show that \mathrm{E}(T) = \dfrac{1}{\lambda}.

[4]

(i) Find the median of T and compare it with the mean.

[2]

(j) Explain whether this model is realistic for buses that run to a timetable.

[2]

Show solutionHide solution

(a) \mathrm P(X=1)=\dfrac3{10} (A1); \mathrm P(X=2)=\dfrac7{10}\times\dfrac39=\dfrac7{30} (A1); \mathrm P(X=3)=\dfrac7{10}\times\dfrac69\times\dfrac38=\dfrac7{40} (A1).

(b) The positions of the 3 red balls form a random 3-element subset of \{1,\dots,10\}, all subsets equally likely (M1); Number of equally likely subsets =\dbinom{10}3 (A1); X=k: position k is red, positions 1,\dots,k-1 are blue, and the other two red balls lie among the 10-k later positions (M1); Number of favourable subsets =\dbinom{10-k}2 (A1); \mathrm P(X=k)=\dfrac{\binom{10-k}2}{\binom{10}3}, k=1,\dots,8 (AG).

(c) \mathrm E(X)=\displaystyle\sum_{k=1}^{8}k\,\mathrm P(X=k) (M1); \mathrm P(X=k)=\dfrac{36}{120},\dfrac{28}{120},\dfrac{21}{120},\dfrac{15}{120},\dfrac{10}{120},\dfrac6{120},\dfrac3{120},\dfrac1{120} (A1); \mathrm E(X)=\dfrac{330}{120}=\dfrac{11}4 (A1).

(d) With replacement, X is geometric with p=\dfrac3{10}: \mathrm E(X)=\dfrac1p=\dfrac{10}3 (3.33) (A1); Larger than \dfrac{11}4: without replacement each blue ball removed raises the chance of a red next time, so waits are shorter (and cannot exceed 8) (R1).

(e) Generalise (b): \mathrm P(X=k)=\dfrac{\binom{r+b-k}{r-1}}{\binom{r+b}r}; sum k\,\mathrm P(X=k) for small cases (M1); (r,b)=(1,1): \dfrac32; (1,2): 2; (2,2): \dfrac53; (2,3): 2; (3,7): \dfrac{11}4 (A1); Conjecture: \mathrm E(X)=\dfrac{r+b+1}{r+1} (A1).

(f) r=1: the red ball is equally likely to be in each of the b+1 positions, so \mathrm P(X=k)=\dfrac1{b+1}, k=1,\dots,b+1 (A1); \mathrm E(X)=\dfrac1{b+1}\displaystyle\sum_{k=1}^{b+1}k=\dfrac1{b+1}\times\dfrac{(b+1)(b+2)}2 (M1); =\dfrac{b+2}2=\dfrac{r+b+1}{r+1} when r=1 (A1).

(g) \mathrm P(T>t)=\displaystyle\int_t^\infty\lambda e^{-\lambda u}\,\mathrm du=\left[-e^{-\lambda u}\right]_t^\infty (M1); e^{-\lambda u}\to0 as u\to\infty, so \mathrm P(T>t)=e^{-\lambda t} (A1); \mathrm P(T>s+t\mid T>s)=\dfrac{\mathrm P(T>s+t)}{\mathrm P(T>s)} (M1); =\dfrac{e^{-\lambda(s+t)}}{e^{-\lambda s}}=e^{-\lambda t}=\mathrm P(T>t) (AG); The chance of waiting a further t minutes does not depend on how long you have already waited (memoryless) (R1).

(h) \mathrm E(T)=\displaystyle\int_0^\infty t\,\lambda e^{-\lambda t}\,\mathrm dt (M1); By parts with u=t, \dfrac{\mathrm dv}{\mathrm dt}=\lambda e^{-\lambda t}: \left[-te^{-\lambda t}\right]_0^\infty+\displaystyle\int_0^\infty e^{-\lambda t}\,\mathrm dt (M1 A1); te^{-\lambda t}\to0 as t\to\infty, so the bracket is 0 (A1); \mathrm E(T)=\left[-\dfrac1\lambda e^{-\lambda t}\right]_0^\infty=\dfrac1\lambda (AG).

(i) Median m: \mathrm P(T>m)=\dfrac12\Rightarrow e^{-\lambda m}=\dfrac12 (M1); m=\dfrac{\ln2}\lambda\approx\dfrac{0.693}\lambda, which is less than the mean \dfrac1\lambda: the distribution is skewed to the right (A1).

(j) Timetabled buses get more likely to arrive as the wait grows; the model says the wait ahead is the same however long you have waited (R1); The density is greatest at t=0 and has no upper limit, while timetabled gaps cluster near the scheduled interval and are bounded (R1).

Final answer

Paper A3, Question 2. (a) \tfrac3{10},\ \tfrac7{30},\ \tfrac7{40}; (b) proof; (c) \tfrac{11}4; (d) \tfrac{10}3, larger; (e) \dfrac{r+b+1}{r+1}; (f) proof; (g) proof, memoryless; (h) proof; (i) median \dfrac{\ln2}\lambda, below the mean; (j) not realistic: timetabled buses are not memoryless.

Question 7

[27 marks]

Paper A4, Question 1: How good is a small-angle approximation? [27 marks]

This question asks you to use Maclaurin series to evaluate limits, to approximate an integral and to bound the error.

(a) Write down the terms of the Maclaurin series of \sin x that have degree at most 5.

[2]

Let g(x) = \dfrac{x - \sin x}{x^3} for x \ne 0.

(b) Use your answer to (a) to show that g(x) = \dfrac16 - \dfrac{x^2}{120} + \dots, and write down \displaystyle\lim_{x \to 0} g(x).

[3]

(c) Confirm this limit by differentiating the numerator and the denominator of g(x) repeatedly.

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(d) Prove that \sin x > x - \dfrac{x^3}{6} for all x > 0.

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(e) Use the first three non-zero terms of the series for \dfrac{\sin x}{x} to find an approximation to \displaystyle\int_0^{1/2} \dfrac{\sin x}{x}\,\mathrm{d}x, giving your answer as a single fraction.

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(f) Explain why the approximation in (e) is within 10^{-6} of the true value of the integral.

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(g) Find the value of a for which \displaystyle\lim_{x \to 0} \dfrac{\sin x - x + ax^3}{x^5} exists, and find the value of the limit.

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(h) Use the approximation \cos x \approx 1 - \dfrac{x^2}{2} to find an approximate positive solution of \cos x = x, find the percentage error in your answer, and explain why the true solution is larger than your approximation.

[4]

Show solutionHide solution

(a) \sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dots (A1); series as far as x^5: x-\dfrac{x^3}{6}+\dfrac{x^5}{120} (A1).

(b) x-\sin x=\dfrac{x^3}{6}-\dfrac{x^5}{120}+\dots (M1 A1); g(x)=\dfrac16-\dfrac{x^2}{120}+\dots (AG); \displaystyle\lim_{x\to0}g(x)=\dfrac16 (A1).

(c) At x=0 numerator and denominator are both 0: \dfrac{1-\cos x}{3x^2} (M1); \dfrac{\sin x}{6x} (A1); \dfrac{\cos x}{6}\to\dfrac16 (A1).

(d) Let h(x)=\sin x-x+\dfrac{x^3}6, so h(0)=0; h'(x)=\cos x-1+\dfrac{x^2}2, h'(0)=0 (M1); h''(x)=x-\sin x (A1); Let p(x)=x-\sin x: p(0)=0 and p'(x)=1-\cos x\ge0, so p(x)>0 for x>0; hence h''>0 (R1); h''>0 and h'(0)=0 give h'>0; with h(0)=0 this gives h(x)>0, that is \sin x>x-\dfrac{x^3}6 (R1).

(e) \dfrac{\sin x}x\approx1-\dfrac{x^2}6+\dfrac{x^4}{120} (M1); \displaystyle\int_0^{1/2}\left(1-\dfrac{x^2}6+\dfrac{x^4}{120}\right)\mathrm dx=\left[x-\dfrac{x^3}{18}+\dfrac{x^5}{600}\right]_0^{1/2} (M1); =\dfrac12-\dfrac1{144}+\dfrac1{19\,200} (A1); =\dfrac{28\,800-400+3}{57\,600}=\dfrac{28\,403}{57\,600}\ (=0.493108) (A1).

(f) The series for \dfrac{\sin x}x alternates, and for 0\le x\le\dfrac12 its terms \dfrac{x^{2n}}{(2n+1)!} decrease, so the error is smaller than the first term omitted (R1); First omitted term: \displaystyle\int_0^{1/2}\dfrac{x^6}{5040}\,\mathrm dx=\dfrac{(1/2)^7}{7\times5040} (M1); =2.2\times10^{-7}<10^{-6} (A1).

(g) \sin x-x+ax^3=\left(a-\dfrac16\right)x^3+\dfrac{x^5}{120}-\dots (M1); Dividing by x^5: \dfrac{a-\frac16}{x^2}+\dfrac1{120}-\dots, which has no limit unless a-\dfrac16=0 (R1); a=\dfrac16 (A1); limit =\dfrac1{120} (A1).

(h) 1-\dfrac{x^2}2=x\Rightarrow x^2+2x-2=0 (M1); x=-1+\sqrt3=0.732 (A1); True solution (GDC) 0.739085: error =\dfrac{0.739085-0.732051}{0.739085}\times100=0.952\% (A1); For x>0, \cos x>1-\dfrac{x^2}2 (the inequality of (d) in the same style), so at x=0.732: \cos x-x>0; \cos x-x is decreasing because its derivative is -\sin x-1<0, so its root lies further to the right (R1).

Final answer

Paper A4, Question 1. (a) x-\tfrac{x^3}6+\tfrac{x^5}{120}; (b) proof; limit \tfrac16; (c) \tfrac16; (d) proof; (e) \dfrac{28\,403}{57\,600}; (f) first omitted term 2.2\times10^{-7}; (g) a=\tfrac16, limit \tfrac1{120}; (h) x=\sqrt3-1=0.732, error 0.952%, the true solution is larger.

Question 8

[28 marks]

Paper A4, Question 2: A drone on a spiral path [28 marks]

This question asks you to use vectors to describe the motion of a drone that climbs in a spiral.

The position of a drone, in metres, t seconds after take-off, is \mathbf{r}(t) = \begin{pmatrix} 3\cos t \\ 3\sin t \\ 2t \end{pmatrix} for t \ge 0, relative to an origin on the ground. The unit vector \mathbf{k} points vertically upwards.

(a) Find the velocity of the drone, and show that its speed is constant.

[3]

(b) Show that the acceleration of the drone is always perpendicular to its velocity.

[3]

(c) Find the angle between the velocity of the drone and \mathbf{k}, and show that it does not change.

[3]

(d) Find the distance the drone travels in its first complete turn, from t = 0 to t = 2\pi.

[2]

A straight cable L is fixed along the line \mathbf{r} = \begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix} + s\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}.

(e) Show that the path of the drone meets L at exactly one point, and find that point.

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(f) Find the acute angle between the path of the drone and L at this point.

[3]

(g) The drone crosses the plane x + y = z once. Find the time at which this happens, and the position of the drone at that moment.

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A second drone follows \mathbf{r}_k(t) = \begin{pmatrix} 3\cos t \\ 3\sin t \\ kt \end{pmatrix} for t \ge 0, where k > 0.

(h) Show that, for no value of k, does the path of this drone cross L at right angles.

[2]

(i) Find the value of k for which the path touches L at (3, 0, 0), that is, for which its velocity there is parallel to L.

[2]

(j) For this value of k, find the speed of the drone and the height it gains in one complete turn.

[2]

Show solutionHide solution

(a) \mathbf v=\dfrac{\mathrm d\mathbf r}{\mathrm dt}=\begin{pmatrix}-3\sin t\\3\cos t\\2\end{pmatrix} (M1 A1); |\mathbf v|=\sqrt{9\sin^2t+9\cos^2t+4}=\sqrt{13}\text{ m s}^{-1}, which does not depend on t (A1).

(b) \mathbf a=\dfrac{\mathrm d\mathbf v}{\mathrm dt}=\begin{pmatrix}-3\cos t\\-3\sin t\\0\end{pmatrix} (A1); \mathbf v\cdot\mathbf a=9\sin t\cos t-9\sin t\cos t+0=0 (M1); The scalar product is zero for all t and neither vector is the zero vector, so \mathbf a\perp\mathbf v (R1).

(c) \cos\varphi=\dfrac{\mathbf v\cdot\mathbf k}{|\mathbf v||\mathbf k|}=\dfrac2{\sqrt{13}} (M1 A1); \varphi=56.3^\circ, the same at every time (A1).

(d) Distance =\displaystyle\int_0^{2\pi}|\mathbf v|\,\mathrm dt=\int_0^{2\pi}\sqrt{13}\,\mathrm dt (M1); =2\pi\sqrt{13}=22.7\text{ m} (A1).

(e) A point on both: 3\cos t=3, 3\sin t=s, 2t=s (M1); \cos t=1\Rightarrow t=2n\pi, n\in\mathbb Z^{+}\cup\{0\} (A1); Then \sin t=0, so s=0; and 2t=s=0 gives t=0 (M1); the only common point is (3,0,0), at t=0 (A1).

(f) At t=0: \mathbf v=\begin{pmatrix}0\\3\\2\end{pmatrix} and the direction of L is \mathbf d=\begin{pmatrix}0\\1\\1\end{pmatrix} (M1); \cos\theta=\dfrac{|\mathbf v\cdot\mathbf d|}{|\mathbf v||\mathbf d|}=\dfrac5{\sqrt{13}\sqrt2}=\dfrac5{\sqrt{26}} (A1); \theta=11.3^\circ (A1).

(g) On the plane: 3\cos t+3\sin t=2t (M1); Solve with the GDC: t=1.54 s (A1 A1); \mathbf r(1.54)=\begin{pmatrix}0.0857\\3.00\\3.08\end{pmatrix} m (A1).

(h) \mathbf v_k(0)=\begin{pmatrix}0\\3\\k\end{pmatrix} at (3,0,0), so \mathbf v_k(0)\cdot\mathbf d=3+k (M1); 3+k>0 when k>0, so the scalar product is never 0: the path never crosses L at right angles (R1).

(i) \begin{pmatrix}0\\3\\k\end{pmatrix}=\lambda\begin{pmatrix}0\\1\\1\end{pmatrix} (M1); \lambda=3, so k=3 (A1).

(j) k=3: speed =\sqrt{9+9}=3\sqrt2=4.24\text{ m s}^{-1} (A1); Height gained =3\times2\pi=6\pi=18.8\text{ m} (A1).

Final answer

Paper A4, Question 2. (a) \mathbf v=(-3\sin t,\,3\cos t,\,2), speed \sqrt{13}; (b) \mathbf v\cdot\mathbf a=0; (c) 56.3^\circ; (d) 2\pi\sqrt{13}=22.7 m; (e) (3,0,0); (f) 11.3^\circ; (g) t=1.54 s, (0.0857,\,3.00,\,3.08); (h) \mathbf v\cdot\mathbf d=3+k>0; (i) k=3; (j) 3\sqrt2=4.24 m s^{-1}, 6\pi=18.8 m.

From the book

Photon Guidebook: Mathematics HL Paper 3 for the IB Diploma

Chapter 16 teaches practice papers: analysis and approaches hl: how the IB builds this kind of Paper 3 question, the techniques, marked worked investigations, and practice investigations like these.

About the book