Free worked solutions · 2026 Edition

Chapter 19: Exam strategy worked solutions

Worked solutions to the timed Section A questions in Chapter 19 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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5 timed Section A questions
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Timed Section A

Five Section A questions, 30 marks: about 33 minutes under exam timing.

Question 1

[6 marks]

A sector of a circle has radius r cm and angle \theta radians, where 0<\theta<2\pi. The perimeter of the sector is 20 cm and its area is 16\ \text{cm}^2. Find the value of \theta.

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2r+r\theta=20 M1; \tfrac12r^2\theta=16 M1. From the first, r\theta=20-2r, so \tfrac12 r(20-2r)=16 M1, giving r^2-10r+16=0, (r-2)(r-8)=0 A1. r=2 gives \theta=8>2\pi, rejected R1; r=8 gives \theta=\tfrac{4}{8}=0.5 A1. Answer: \theta=0.5.

Final answer

\theta=0.5

Question 2

[5 marks]

Solve the equation \log_4(x+6)-\log_2 x=\dfrac12.

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\log_4(x+6)=\dfrac{\log_2(x+6)}{2} M1; \log_2(x+6)-2\log_2x=1, so \log_2\dfrac{x+6}{x^2}=1 M1; x+6=2x^2, 2x^2-x-6=0 A1; (2x+3)(x-2)=0 A1; x=-\tfrac32 is rejected since \log_2x needs x>0, so x=2 R1. Check: \log_48-\log_22=\tfrac32-1=\tfrac12.

Final answer

x=2

Question 3

[6 marks]

The cubic equation z^3+az^2+bz+10=0 has real coefficients a and b, and one of its roots is 1+2\mathrm{i}. Determine a and b.

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Real coefficients, so 1-2\mathrm{i} is also a root A1. Quadratic factor (z-1)^2+4=z^2-2z+5 M1A1. Product of the roots =-10, so 5c=-10, c=-2 M1A1 (or by division). (z+2)(z^2-2z+5)=z^3+z+10, so a=0, b=1 A1.

Final answer

a=0,\ b=1

Question 4

[7 marks]

The discrete random variable X has probability distribution \mathrm{P}(X=x)=k(x^2+1) for x=0,1,2,3.

(a) Find the value of k.

[2]

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

[2]

(c) Find \mathrm{Var}(3X-1).

[3]

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(a) k(1+2+5+10)=1 M1, k=\tfrac1{18} A1. (b) \mathrm{E}(X)=\tfrac{0+2+10+30}{18} M1 =\tfrac73 A1. (c) \mathrm{E}(X^2)=\tfrac{0+2+20+90}{18}=\tfrac{56}{9} M1; \mathrm{Var}(X)=\tfrac{56}9-\tfrac{49}9=\tfrac79 A1; \mathrm{Var}(3X-1)=9\times\tfrac79=7 A1.

Final answer

(a) k=\frac1{18}
(b) \mathrm{E}(X)=\frac73
(c) \operatorname{Var}(3X-1)=7

Question 5

[6 marks]

Let f(x)=e^{x}(x^2-3), for x\in\mathbb{R}.

(a) Show that f'(x)=e^{x}(x+3)(x-1).

[2]

(b) Find the exact coordinates of the local maximum point of the graph of f, justifying that it is a maximum.

[4]

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(a) Product rule: f'(x)=e^x(x^2-3)+2xe^x M1 =e^x(x^2+2x-3)=e^x(x+3)(x-1) A1AG. (b) f'(x)=0 at x=-3 and x=1, since e^x>0 A1. f'(x)>0 for x<-3 and f'(x)<0 for -3<x<1, so the sign of f' goes from + to - as x passes -3, which gives a maximum R1. f(-3)=e^{-3}(9-3)=6e^{-3} M1. Maximum point (-3,\,6e^{-3}) A1.

Final answer

(b) \left(-3,\,6e^{-3}\right)

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.

Job	TI-84 Plus CE keystrokes
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`
Intersections	`Graph screen: Y1 and Y2, 2nd TRACE (CALC), 5:intersect, three ENTERs per crossing`
Normal distribution	`2nd VARS (DISTR), 2:normalcdf(lower, upper, mean, sd); 3:invNorm(area, mean, sd)`
Binomial	`2nd VARS (DISTR), binomcdf(n, p, k) then 1 minus binomcdf(n, p, k-1) for P(Y≥k)`
Gradient at a point	`MATH, 8:nDeriv(expression, X, value)`
Definite integral	`MATH, 9:fnInt(expression, X, lower, upper)`
Polynomial roots	`APPS, PlySmlt2 (or graph and 2nd TRACE, 2:zero, for real roots)`
Linear system	`2nd x⁻¹ (MATRIX), enter the augmented matrix, then MATH, rref(`
Radian mode	`MODE, Radian, ENTER`

From the book

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

Chapter 19 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