2021 Q5 — Asymptotes, derivative and inflexion of a rational function
Let r(x) = (x³ − x² − 2x + 3)/(x − 1)² for x ≠ 1.(a) Find the asymptotes of the graph of y = r(x).(b) Find r′(x).(c) Someone claims that the graph has only one point of inflexion. Do you agree?
A tangent slides along the right-hand branch. The curve is concave up until x = 4 and concave down after it, slipping from above the asymptote y = x + 1 to just below it — the single point of inflexion is (4, 43/9). Try it yourself →
(a) Divide: x³ − x² − 2x + 3 = (x + 1)(x − 1)² − (x − 2), so r(x) = x + 1 − 1/(x − 1) + 1/(x − 1)². Asymptotes: x = 1 and y = x + 1
r″(x) = 0 only at x = 4, where it changes from + to −; r(4) = 5 − 1/3 + 1/9 = 43/9
Answer:(a) x = 1, y = x + 1; (b) 1 + 1/(x − 1)² − 2/(x − 1)³; (c) agree: the only one is (4, 43/9)
2021 Q6 — Area bounded by y = e^(2x − 6), its normal and a vertical line
Let L be the normal to the curve Γ: y = e^(2x − 6) at the point (3, 1), and let c be the x-intercept of L. Find(a)c,(b) the area of the region bounded by L, Γ and the line x = c.
The region between the steeply rising curve (blue) and its normal at (3, 1) (red) fills from x = 3 to the normal’s x-intercept, x = 5. Almost all of the area, (e⁴ − 3)/2 ≈ 25.8, comes from the last half unit. Try it yourself →
(a) dy/dx = 2e^(2x − 6) = 2 at x = 3, so the normal has slope −1/2: y − 1 = −(x − 3)/2
At y = 0: x − 3 = 2, so c = 5
(b) For 3 ≤ x ≤ 5 the curve is above L: area = ∫₃⁵ [e^(2x − 6) − 1 + (x − 3)/2] dx
2021 Q7 — Volume of revolution of y = √x ln(x² + 1)
(a) Using integration by parts, find ∫ (ln x)² dx.(b) Consider the curve C: y = √x ln(x² + 1) for x ≥ 0. The region bounded by C, the line x = 1 and the x-axis is revolved about the x-axis. Find the volume of the solid.
The red curve y = √x ln(x² + 1) from x = 0 to 1 turns about the x-axis into a trumpet-shaped solid that opens to radius ln 2 ≈ 0.69 at x = 1. Its volume is π(1 − ln 2)² ≈ 0.296. Try it yourself →
(a) ∫ (ln x)² dx = x(ln x)² − ∫ 2 ln x dx = x(ln x)² − 2x ln x + 2x + C
(b) V = π∫₀¹ y² dx = π∫₀¹ x[ln(x² + 1)]² dx
Let u = x² + 1, du = 2x dx: V = (π/2)∫₁² (ln u)² du
Answer:(a) x(ln x)² − 2x ln x + 2x + C; (b) V = π(1 − ln 2)²
2021 Q10 — Shortest vertical distance between two curves, and a rate of change
For 0 < x < 20, let F be the graph of y = √(x² + 36) and G the graph of y = −√((20 − x)² + 16). P moves on F with x-coordinate u, and the vertical line through P cuts G at Q. PQ is shortest when u = a.(a) Find a.(b) Horizontal lines through P and Q cut the y-axis at R and S.(i) Is the area of rectangle PQSR least when u = a?(ii) If OP increases at 28 units per minute, find the rate of change of the perimeter of PQSR when u = a.
The red segment PQ slides between the two curves. It is shortest, 10√5 ≈ 22.36, at u = 12, but the rectangle PQSR is still getting wider there, so its area keeps increasing and is not at a minimum. Try it yourself →
Set it to 0 and square: 16u² = 36(20 − u)², so 4u = 6(20 − u) and u = 12. The derivative changes from − to + there, so a = 12
(b)(i) Area = u · PQ, so d(Area)/du = PQ + u · d(PQ)/du = PQ + 0 > 0 at u = 12. The area is still increasing: disagree
(ii) OP = √(u² + u² + 36) = √(2u² + 36), so d(OP)/dt = 2u(du/dt)/√(2u² + 36). At u = 12: 24(du/dt)/18 = 28, du/dt = 21
Perimeter = 2(u + PQ), so its rate is 2(1 + d(PQ)/du)(du/dt) = 2(1 + 0)(21) = 42
Answer:(a) a = 12; (b)(i) no; (ii) 42 units per minute
2021 Q12 — Area, volume, distance to a plane and a circumcentre test
The position vectors of A, B, C, D are ti + 14j + sk, 12i − sj − 2k, (s + 2)i − 16j + 10k and −ti + (s + 2)j + 14k, and AB is parallel to 5i − 4j − 2k. Π is the plane through A, B, C.(a) Find s and t, the area of △ABC, the volume of tetrahedron ABCD, and the shortest distance from D to Π.(b)E is the projection of D on Π. Is E the circumcentre of △ABC?
The camera circles tetrahedron ABCD. The red segment DE, of length 24, is the perpendicular from D to the blue triangle ABC; its foot E is clearly much closer to B and C than to A, so E is not the circumcentre. Try it yourself →
(a) AB = (12 − t)i − (s + 14)j − (s + 2)k. From the j and k parts: (s + 14)/4 = (s + 2)/2, so s = 10; then AB = 6(5i − 4j − 2k) and 12 − t = 30, t = −18
So A(−18, 14, 10), B(12, −10, −2), C(12, −16, 10), D(18, 12, 14); AB × AC = −180(2i + 2j + k), and area = ½ · 180 · 3 = 270
EA ≠ EB, so E is not equidistant from the vertices
Answer:(a) s = 10, t = −18; area 270; volume 2160; distance 24; (b) no
How to use these 2021 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 2021 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.