Free worked solutions · 2026 Edition

IB Maths AA SL Chapter 12: Exam strategy worked solutions

Worked solutions to the timed Section A questions in Chapter 12 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches SL (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Timed Section A

Six Section A questions, 35 marks: about 39 minutes under exam timing.

Question 1

[5 marks]
An arithmetic sequence has u_6=25 and u_3+u_8=46. (a) Calculate its first term and common difference.
[3]
(b) Identify the first term of the sequence that exceeds 60, giving its position.
[2]
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a+5d=25 and 2a+9d=46 (M1: translate the term positions). Subtracting the second equation from twice the first gives d=4 (A1), then a=5 (A1). 5+4(n-1)>60, so n>59/4 (M1: solve the strict inequality). The first position is n=15, with u_{15}=61 (A1: check the integer threshold).

Final answer

(a) u_1=5, d=4. (b) u_{15}=61, position 15.

Question 2

[5 marks]
A proposed solution to \ln(x-1)+\ln(x+3)=\ln12 is x=-5. (a) Explain why this value cannot be used in the equation.
[1]
(b) Solve the equation, showing how you select a valid root.
[4]
Show solutionHide solution
The arguments of both logarithms must be positive; together they require x>1, so -5 is excluded (R1). \ln((x-1)(x+3))=\ln12 (M1: combine logarithms), so (x-1)(x+3)=12 (A1). x^2+2x-15=(x-3)(x+5)=0 (M1: factorise). The domain leaves x=3 (A1).

Final answer

(a) Both log arguments must be positive; x>1. (b) x=3.

Question 3

[7 marks]
For x\ne3, a function is defined by g(x)=2+\dfrac5{x-3}. (a) Find the coordinates of the two points where its graph meets the coordinate axes.
[2]
(b) State both asymptotes.
[2]
(c) Find a formula for the inverse function and give its domain.
[3]
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g(0)=1/3 and g(x)=0 gives x=1/2 (A1 each: substitute zero for the appropriate coordinate). The asymptotes are x=3 and y=2 (A1 each: excluded denominator and limiting height). Write y-2=5/(x-3) (M1), then x=3+5/(y-2) (A1: rearrange). Thus g^{-1}(x)=3+5/(x-2), with domain x\ne2 (A1: use the range of g).

Final answer

(a) (0,\frac13), (\frac12,0). (b) x=3, y=2. (c) g^{-1}(x)=3+\frac5{x-2}, x\ne2.

Question 4

[5 marks]
A token machine awards X points per play. The possible awards are -2, 1 and 4, with probabilities \frac14, p and \frac34-p, respectively. The mean award is 1 point. (a) Determine p.
[2]
(b) Find the probability of a positive award.
[1]
(c) A player pays c points per play. Calculate c for a fair game and explain your choice.
[2]
Show solutionHide solution
(-2)/4+p+4(3/4-p)=1 (M1: a weighted mean), so 5/2-3p=1 and p=1/2 (A1). P(X>0)=1/2+1/4=3/4 (A1: add the two positive-award cases). A fair game has mean net gain zero (R1). The mean award is 1, so the charge is c=1 point (A1).

Final answer

(a) p=\frac12. (b) \frac34. (c) c=1: mean net gain zero.

Question 5

[6 marks]
A circular badge has radius 6 cm. A straight chord removes a cap whose shorter curved edge has length 4\pi cm. (a) Find the angle at the centre subtended by this curved edge, in radians.
[2]
(b) Calculate the area of the sector containing the cap.
[2]
(c) Find the exact area of the cap.
[2]
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6\theta=4\pi (M1: arc length), giving \theta=2\pi/3 (A1). Sector area =\frac12(6)^2(2\pi/3) (M1), =12\pi square centimetres (A1). Subtract the triangle: 12\pi-\frac12(6)^2\sin(2\pi/3) (M1). The cap area is 12\pi-9\sqrt3 square centimetres (A1: exact sine value).

Final answer

(a) \frac{2\pi}{3}. (b) 12\pi\ \mathrm{cm}^2. (c) (12\pi-9\sqrt3)\ \mathrm{cm}^2.

Question 6

[7 marks]
Let f(x)=x^3-6x^2+9x+2. (a) Show that f'(x)=3(x-1)(x-3).
[2]
(b) Find the coordinates of the local maximum and justify its nature.
[3]
(c) Find the exact value of \displaystyle\int_0^1 f(x)\,dx.
[2]
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f'(x)=3x^2-12x+9 (M1: differentiate), =3(x-1)(x-3) (A1, AG: reach the target). f'(x)=0 at x=1,3 (M1). f''(1)=6-12=-6<0, so x=1 gives a local maximum (R1). The point is (1,6) (A1). \int_0^1f(x)dx=[x^4/4-2x^3+9x^2/2+2x]_0^1 (M1: antiderivative and limits), =19/4 (A1).

Final answer

(b) (1,6); f''(1)=-6<0. (c) \frac{19}{4}.

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.

JobTI-84 Plus CE keystrokes
Solve an equationMATH, 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
IntersectionsGraph screen: Y1 and Y2, 2nd TRACE (CALC), 5:intersect, three ENTERs per crossing
Normal distribution2nd VARS (DISTR), 2:normalcdf(lower, upper, mean, sd); 3:invNorm(area, mean, sd)
Binomial2nd VARS (DISTR), binomcdf(n, p, k) then 1 minus binomcdf(n, p, k-1) for P(Y≥k)
Gradient at a pointMATH, 8:nDeriv(expression, X, value)
Definite integralMATH, 9:fnInt(expression, X, lower, upper)
Polynomial rootsAPPS, PlySmlt2 (or graph and 2nd TRACE, 2:zero, for real roots)
Linear system2nd x⁻¹ (MATRIX), enter the augmented matrix, then MATH, rref(
Radian modeMODE, Radian, ENTER

From the book

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

Chapter 12 teaches Exam strategy 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

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