WMA11/01 Jun 2025, Question 9: Solving Equations with Surds
Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 10 marks
The question, in short
Using the curves y = √(2x) and y = 12 − √x meeting at P, describe transformations of y = √x and y = −√x, then solve for P's coordinates exactly.
Summarised in our own words. Use the official paper for the exact wording and any diagrams.
Part (ai) · 2 marks
Describe the single transformation mapping y = √x onto y = √(2x).
- 1√(2x) is √x with x replaced by 2x
- 2Replacing x by 2x is a stretch parallel to the x-axis, scale factor ½
- 3(Also √(2x) = √2 × √x: a stretch parallel to the y-axis, scale factor √2)
AnswerA stretch parallel to the x-axis with scale factor ½ (or equivalently a stretch parallel to the y-axis with scale factor √2)
Part (aii) · 2 marks
Describe the single transformation mapping y = −√x onto y = 12 − √x.
- 112 − √x = −√x + 12
- 2Adding 12 to y is a translation 12 units up, vector (0, 12)
AnswerA translation of 12 units up, by the vector (0, 12)
Part (bi) · 3 marks
Show that P's x coordinate satisfies √x = 12(√2 − 1).
- 1At P: √(2x) = 12 − √x
- 2√2 √x + √x = 12, so √x(√2 + 1) = 12
- 3√x = 12/(√2 + 1)
- 4Multiply top and bottom by (√2 − 1): √x = 12(√2 − 1)/(2 − 1) = 12(√2 − 1)
AnswerShown: √(2x) = 12 − √x gives √x(√2 + 1) = 12, so √x = 12/(√2 + 1) = 12(√2 − 1)
Part (bii) · 3 marks
Using the previous result, find the exact coordinates of P in simplest form.
- 1x = [12(√2 − 1)]² = 144(2 − 2√2 + 1)
- 2x = 144(3 − 2√2) = 432 − 288√2
- 3y = 12 − √x = 12 − 12(√2 − 1) = 24 − 12√2
- 4P(432 − 288√2, 24 − 12√2)
AnswerP(432 − 288√2, 24 − 12√2)
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