plumiLocating roots by change of sign: AQA A‑Level Maths questions
AQA A-Level Maths past-paper questions on I.1 Locating roots by change of sign, 2018–2025, with mark schemes and examiner’s reports where available. Add MadAsMaths questions to your practice, or browse them separately by topic and difficulty.
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All 9 questions on Locating roots by change of sign, as a list
- June 2025, Paper 3, Q9 (5 marks): (a) Describe a sequence of two transformations which maps the graph with equation y = 1 x onto the graph with equation y = 3 x – 4 [2 marks] 9 (b) State the equ
- June 2024, Paper 1, Q14 (10 marks): (a) The equation x3 = e6 – 2x has a single solution, x = α By considering a suitable change of sign, show that α lies between 0 and 4 [2 marks] 14 (b) Show that
- June 2022, Paper 1, Q10 (12 marks): The diagram shows a sector of a circle OAB. O θ C A B The point C lies on OB such that AC is perpendicular to OB. Angle AOB is y radians. 10 (a) Given the area
- November 2021, Paper 1, Q7 (7 marks): The equation x2 = x3 + x − 3 has a single solution, x = a 7 (a) By considering a suitable change of sign, show that a lies between 1.5 and 1.6 [2 marks] 7 (b) S
- November 2021, Paper 2, Q10 (11 marks): The function h is defined by h(x) = x p x − 3 where h has its maximum possible domain. 10 (a) Find the domain of h. Give your answer using set notation. [3 mark
- November 2020, Paper 1, Q2 (1 mark): A student is searching for a solution to the equation f (x) = 0 He correctly evaluates f (×1) = ×1 and f (1) = 1 and concludes that there must be a root between
- November 2020, Paper 1, Q14 (9 marks): The function f is defined by f (x) = 3x x p × 1 where x ≥ 0 14 (a) f (x) = 0 has a single solution at the point x = a By considering a suitable change of sign,
- June 2019, Paper 1, Q7 (11 marks): (a) By sketching the graphs of y = 1 x and y = sec 2x on the axes below, show that the equation 1 x = sec 2x has exactly one solution for x > 0 [3 marks] x ϖ 2ϖ
- June 2018, Paper 1, Q11 (10 marks): The daily world production of oil can be modelled using V = 10 + 100 t 30 3 −50 t 30 4 where V is volume of oil in millions of barrels, and t is time in years s
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