2020 Q6 — Related rates for a point moving on y = 2^(x − 1)
P(u, v) moves on the curve C₁: y = 2^(x − 1) (x > 0) so that the area of the circle with OP as a diameter increases at 5π square units per second.(a) Let S = u² + v². Does S increase at a constant rate?(b) The vertical line through P cuts C₂: y = 2ˣ at Q. Find the rate of change of the area of △OPQ when u = 2.
P moves along the blue curve and Q stays directly above it on the green one, so PQ = 2^(u − 1) is the base of the red triangle and u its height. At u = 2 the area is 2 and growing at 5 square units per second. Try it yourself →
(a) The circle has radius OP/2, so its area is π(u² + v²)/4 = πS/4. Then (π/4)(dS/dt) = 5π gives dS/dt = 20: yes, constant
(b) PQ = 2ᵘ − 2^(u − 1) = 2^(u − 1), and the distance from O to the line PQ is u: area A = ½u · 2^(u − 1) = u · 2^(u − 2)
With v = 2^(u − 1): dS/dt = (2u + 2v · v ln 2)(du/dt). At u = 2, v = 2: (4 + 8 ln 2)(du/dt) = 20, so du/dt = 5/(1 + 2 ln 2)
Answer:(a) Yes, dS/dt = 20; (b) 5 square units per second
2020 Q7 — Tangent from an external point and the normal at the point of contact
The curve Γ: y = f(x) passes through (1, 2) and f′(x) = −2x + 8.(a) Find the equation of Γ.(b)L is a tangent to Γ that passes through (5, 14) and has negative slope; it touches Γ at P. Find the coordinates of P and the equation of the normal to Γ at P.
The red tangent slides along the parabola. It passes through the outside point (5, 14) twice: at x = 3 with a positive slope and at x = 7 with a negative slope. The green segment is that second tangent and the dashed line is the normal at P(7, 2). Try it yourself →
(a) f(x) = −x² + 8x + C; through (1, 2): −1 + 8 + C = 2, so y = −x² + 8x − 5
(b) Let P have x-coordinate p. The tangent there has slope −2p + 8 and passes through (5, 14): 14 − (−p² + 8p − 5) = (−2p + 8)(5 − p)
Simplify: p² − 10p + 21 = 0, so p = 3 (slope 2) or p = 7 (slope −6)
The slope must be negative: p = 7, f(7) = −49 + 56 − 5 = 2, so P = (7, 2)
Normal slope = 1/6: y − 2 = (x − 7)/6
Answer:(a) y = −x² + 8x − 5; (b) P(7, 2), normal x − 6y + 5 = 0
2020 Q9 — Asymptotes, turning point, inflexion and area for y = (x + 4)³/(x − 4)²
Let f(x) = (x + 4)³/(x − 4)² for x ≠ 4, with graph H.(a) Find the asymptotes of H.(b) Find f″(x).(c) Someone claims that H has two turning points. Do you agree?(d) Find the point(s) of inflexion.(e) Find the area of the region bounded by H, the x-axis and the y-axis.
The curve only touches the x-axis at (−4, 0): it flattens there and carries on rising, so that point is an inflexion, not a turning point. The shaded region between the curve and the axes, from x = −4 to 0, has area 136 − 192 ln 2 ≈ 2.92. Try it yourself →
(a) Divide: (x + 4)³ = (x + 20)(x − 4)² + 192x − 256, so f(x) = x + 20 + 192/(x − 4) + 512/(x − 4)². Asymptotes: x = 4 and y = x + 20
Answer:(a) x = 4, y = x + 20; (b) 384(x + 4)/(x − 4)⁴; (c) no, only one; (d) (−4, 0); (e) 136 − 192 ln 2
2020 Q12 — Cross product, a parallelogram and the angle between two planes
Let OP = i + j + 4k and OQ = 5i − 7j − 4k; R lies on PQ with PR : RQ = 1 : 3.(a) Find OP × OR.(b) With OS = OP + OR, find the area of quadrilateral OPSR.(c)N satisfies ON = λ(OP × OR).(i) Is NR perpendicular to PQ?(ii) If NQ is parallel to 11i + μj − 10k, find λ and μ, and find tan θ, where θ is the angle between △OPQ and △NPQ.
The camera circles the two triangles hinged along PQ: OPQ (blue) and NPQ (orange). The green segments OR and NR both meet PQ at right angles, and the red segment ON is perpendicular to the blue triangle, so tan θ = ON/OR = 2/3. Try it yourself →
(a) OR = (3·OP + OQ)/4 = 2i − j + 2k, so OP × OR = 6i + 6j − 3k
(b) OPSR is a parallelogram: area = |OP × OR| = √(36 + 36 + 9) = 9
(c)(i) PQ = 4i − 8j − 8k. OR · PQ = 8 + 8 − 16 = 0, and ON · PQ = 0 because ON is normal to plane OPQ. So NR · PQ = 0: yes
(ii) NQ = (5 − 6λ)i + (−7 − 6λ)j + (−4 + 3λ)k. Comparing the i and k parts: −10(5 − 6λ) = 11(−4 + 3λ), so λ = 2/9
Then NQ = (1/3)(11i − 25j − 10k), so μ = −25
OR and NR are both perpendicular to PQ, so θ = ∠ORN; ON ⊥ OR, hence tan θ = |ON|/|OR| = (2/9 × 9)/3 = 2/3
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 2020 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.