WMA11/01 Jan 2025, Question 9: Harder Inequalities and Curved Regions

Part (a) · 3 marks

Write 6x − 27/4 − x² as a + b(x + c)², finding constants a, b and c.

1. 16x − 27/4 − x² = −(x² − 6x) − 27/4
2. 2= −[(x − 3)² − 9] − 27/4
3. 3= 9 − 27/4 − (x − 3)²
4. 4= 9/4 − (x − 3)², so a = 9/4, b = −1, c = −3

Answer9/4 − (x − 3)² (a = 9/4, b = −1, c = −3)

Part (b) · 2 marks

Given P is the maximum point of C₁, state its coordinates.

1. 1y = 9/4 − (x − 3)² is greatest when x − 3 = 0
2. 2Maximum point P = (3, 9/4)

AnswerP(3, 9/4)

Part (ci) · 1 mark

State the x coordinate of point B where C₁ and C₂ meet on the x-axis.

1. 1C₁ meets the x-axis where (x − 3)² = 9/4, so x − 3 = ±3/2
2. 2x = 3/2 (A) or x = 9/2 (B), so B has x = 9/2

Answerx = 9/2

Part (cii) · 1 mark

State the value of the constant k in y = cos(kx).

1. 1A (x = 3/2) is the first zero of cos(kx) for x > 0, so k × 3/2 = π/2
2. 2k = π/3 (check: at B, k × 9/2 = 3π/2, also a zero)

Answerk = π/3

Part (ciii) · 1 mark

State the period of curve C₂.

1. 1Period of cos(kx) is 2π/k
2. 22π ÷ (π/3) = 6

Answer6

Part (d) · 5 marks

Use inequalities to define the shaded region R bounded by line segment L, C₁ and C₂.

1. 1Gradient of PB = (0 − 9/4)/(9/2 − 3) = −3/2
2. 2Line L: y − 0 = −(3/2)(x − 9/2), so y = −(3/2)x + 27/4
3. 3R is above C₂: y > cos(πx/3)
4. 4R is below C₁ and below L: y < 6x − 27/4 − x² and y < −(3/2)x + 27/4

Answery > cos(πx/3), y < 6x − 27/4 − x², y < −(3/2)x + 27/4

More past-paper questions on harder inequalities and curved regions

• WMA11/01 Jun 2024 Q6 · 7 marks
• WMA11/01 Jan 2023 Q10 · 10 marks
• WMA11/01 Oct 2022 Q7 · 8 marks
• WMA11/01 Oct 2022 Q9 · 14 marks
• WMA11/01 Jan 2021 Q4 · 8 marks
• WMA11/01 Oct 2021 Q3 · 6 marks