2016 Q3 — Rate of change of the area of a triangle under y = 2eˣ
P is a point on the curve C: y = 2eˣ (x > 0) with x-coordinate u. The horizontal line through P cuts the y-axis at Q, and O is the origin.(a) Express the area of △OPQ in terms of u.(b) If OQ increases at a constant rate of 6 units per second, find the rate of change of the area when u = 4.
P climbs the curve y = 2eˣ and the right-angled triangle OPQ (red) grows with it. Its base PQ = u grows slowly while its height OQ = 2eᵘ shoots up; at u = 4 the area is growing at 15 square units per second. Try it yourself →
(a) PQ = u and OQ = 2eᵘ, with a right angle at Q: area A = ½ · u · 2eᵘ = u eᵘ
2016 Q7(b) — Volume of revolution about the y-axis
Consider the curve Γ: y = 4x² − 4x for 1 ≤ x ≤ 4. R is the region bounded by Γ, the line y = 48 and the two axes. Find the volume of the solid generated by revolving R about the y-axis.
The red arc of y = 4x² − 4x turns about the vertical axis to form a bowl-shaped solid, 1 unit wide at the bottom and 4 units wide at the top, where y = 48. Each horizontal slice is a disc of radius x. Try it yourself →
The curve runs from (1, 0) to (4, 48), so R lies between the y-axis and Γ for 0 ≤ y ≤ 48
Disc method about the y-axis: V = π∫₀⁴⁸ x² dy
Change the variable to x: dy = (8x − 4) dx, and x goes from 1 to 4
V = π∫₁⁴ x²(8x − 4) dx = π[2x⁴ − 4x³/3]₁⁴
= π[(512 − 256/3) − (2 − 4/3)] = π(510 − 84)
Answer:V = 426π
2016 Q9 — A cubic: turning points, inflexion and the area cut off by a tangent
Let f(x) = x³ + ax² + bx + 5, and let C be the curve y = f(x). P(−1, 10) is a turning point of C.(a) Find a and b.(b) Is P a maximum point?(c) Find the minimum value of f(x).(d) Find the point of inflexion.(e)L is the tangent to C at P; find the area of the region bounded by C and L.
The tangent at the maximum point P is the horizontal line y = 10 (blue). It meets the cubic (red) again at x = 5, and the region between them fills up to an area of 108. Try it yourself →
(a) f(−1) = 10: a − b = 6. f′(−1) = 0: 3 − 2a + b = 0. Solving: a = −3, b = −9
(b) f′(x) = 3x² − 6x − 9 = 3(x + 1)(x − 3) changes from + to − at x = −1, so P is a maximum point
(c) f′ changes from − to + at x = 3: minimum value f(3) = 27 − 27 − 27 + 5 = −22
(d) f″(x) = 6x − 6 changes sign at x = 1, and f(1) = −6: point of inflexion (1, −6)
(e) L is y = 10. f(x) = 10 gives x³ − 3x² − 9x − 5 = (x + 1)²(x − 5) = 0, so they meet again at x = 5
Answer:(a) a = −3, b = −9; (b) yes; (c) −22; (d) (1, −6); (e) 108
2016 Q12 — Equidistant point, unit normal and the angle between a line and a plane
Let OA = 2j + 2k, OB = 4i + j + k and OP = i + tj, where P is equidistant from A and B.(a) Find t.(b) Let OC = 2i − j + 4k and OD = 3i + 2j + 5k, and let Π be the plane through A, B, C. Find(i) a unit vector perpendicular to Π,(ii) the angle between CD and Π.
The camera circles plane Π (blue triangle ABC). The green segment DE is the perpendicular from D to the plane; the angle between the red segment CD and its shadow CE on the plane is about 64.8°. Try it yourself →
(b)(i) AB = 4i − j − k, AC = 2i − 3j + 2k, so AB × AC = −5i − 10j − 10k = −5(i + 2j + 2k)
Unit normal: (i + 2j + 2k)/3 (or its negative)
(ii) CD = i + 3j + k, |CD| = √11. If θ is the angle between CD and Π: sin θ = |CD · n̂| = (1 + 6 + 2)/(3√11) = 3/√11
θ = sin⁻¹(3/√11) ≈ 64.8°
Answer:(a) t = −1; (b)(i) ±(i + 2j + 2k)/3; (ii) sin⁻¹(3/√11) ≈ 64.8°
How to use these 2016 DSE M2 solutions
Try each question on paper first, then watch its animation: every video is a real GeoGebra construction of that question, so you can see why the answer is what it is before reading the steps. Click a video to open the same graph in SketchMath and drag or rotate it yourself. For the theory behind a question, see the DSE M2 notes; for other years, see all M2 past paper solutions.
Each question is described in our own words with the same numbers as the 2016 HKDSE Mathematics Extended Part Module 2 paper, and the solutions and animations are written by SketchMath.
Frequently Asked Questions
Featured questions on curve sketching, tangents, areas, volumes of revolution, rates of change and 3D vectors. Each one has a step-by-step solution and a short GeoGebra animation.
No. They are SketchMath’s own worked solutions. The final answers have been checked numerically, but the layout of the working is ours, not the marking scheme’s.
Yes. Click any video to open that construction in SketchMath Chat, then ask for a variation — a different curve, limit or vector — and the graph is redrawn with the new working.