2022 Q4 — Derivatives and points of inflexion of y = (7x − 2x²)e⁻ˣ
Let y = (7x − 2x²)e⁻ˣ.(a) Find dy/dx and d²y/dx².(b) Someone claims that the graph has two points of inflexion. Do you agree?
The tangent rolls along the curve. It crosses from one side of the curve to the other at the two points of inflexion: x = 3/2 on the way down, and x = 6 where the shallow dip bends back towards the x-axis. Try it yourself →
It is negative for x < 3/2, positive for 3/2 < x < 6 and negative for x > 6: the sign changes twice
Answer:(a) (2x² − 11x + 7)e⁻ˣ and (−2x² + 15x − 18)e⁻ˣ; (b) agree: at x = 3/2 and x = 6
2022 Q7 — Area between y = ln(x + 2) and its tangent, and how fast it changes
P moves on the curve Γ: y = ln(x + 2) (x > 0) with x-coordinate h. L is the tangent to Γ at P, and A is the area of the region bounded by Γ, L and the y-axis.(a) Prove that A = (h² + 4h)/(2h + 4) − 2 ln(h + 2) + 2 ln 2.(b) If h = 3^(−t), where t is the time in seconds, find the rate of change of A when t = 1.
As P moves along the curve, the orange sliver between the tangent (red) and the curve (blue), from the y-axis to P, grows with h. Its area and the rate dA/dh = h²/(2(h + 2)²) are shown as h changes. Try it yourself →
(a) Tangent at P: y = ln(h + 2) + (x − h)/(h + 2); the curve is concave, so the tangent is above it
At t = 1: h = 1/3, dh/dt = −3^(−t) ln 3 = −(ln 3)/3, and dA/dh = (1/9)/(2 · 49/9) = 1/98, so dA/dt = −(ln 3)/294
Answer:dA/dt = −(ln 3)/294 square units per second
2022 Q9 — Asymptotes, turning points and a volume about the line y = 10
Let f(x) = (x² + 3x)/(x − 1) for x ≠ 1, with graph H.(a) Find the asymptotes of H.(b) Find the maximum and minimum points of H.(c) Sketch H.(d)R is the region bounded by H and the line y = 10. Find the volume of the solid generated by revolving R about the line y = 10.
The solid formed when the thin region between y = 10 and the curve (from x = 2 to x = 5) turns about the line y = 10, which is drawn as the horizontal axis. The radius 10 − f(x) is greatest, 1 unit, at the minimum point x = 3. Try it yourself →
(a) Divide: x² + 3x = (x − 1)(x + 4) + 4, so f(x) = x + 4 + 4/(x − 1). Asymptotes: x = 1 and y = x + 4
(b) f′(x) = 1 − 4/(x − 1)² = (x + 1)(x − 3)/(x − 1)²: + to − at x = −1, − to + at x = 3. Maximum point (−1, 1), minimum point (3, 9)
(c) The graph has two branches, one each side of x = 1, through (0, 0) and (−3, 0), approaching y = x + 4 at both ends
(d) f(x) = 10: x² − 7x + 10 = 0, so x = 2 or 5. Radius = 10 − f(x) = 6 − x − 4/(x − 1); with u = x − 1: V = π∫₁⁴ (5 − u − 4/u)² du
(5 − u − 4/u)² = u² − 10u + 33 − 40/u + 16/u²
V = π[u³/3 − 5u² + 33u − 40 ln u − 16/u]₁⁴ = π(57 − 40 ln 4)
Answer:(a) x = 1, y = x + 4; (b) max (−1, 1), min (3, 9); (d) π(57 − 80 ln 2)
2022 Q12(b) — Incentre and inradius of a triangle in space
An earlier part shows that, with a = BC, b = AC and c = AB, the angle bisectors of △ABC meet at the point J with OJ = (a·OA + b·OB + c·OC)/(a + b + c). Now let OA = 35i + 9j + k, OB = 40i − 3j + k and OC = −3j + k, and let I be the incentre of △ABC. Find OI and the radius of the inscribed circle.
All three vertices have z = 1, so the triangle is drawn flat in that plane. A circle centred at the incentre I grows until, at radius 16/3, it touches all three sides at once; the dashed lines are the angle bisectors. Try it yourself →
Side lengths: a = |BC| = 40, b = |AC| = √(35² + 12²) = 37, c = |AB| = √(5² + 12²) = 13
OI = (40·OA + 37·OB + 13·OC)/90 = (2880i + 210j + 90k)/90 = 32i + (7/3)j + k
The radius is the distance from I to side AB: r = |AI × AB|/|AB|
AI = −3i − (20/3)j, AB = 5i − 12j: AI × AB = (36 + 100/3)k = (208/3)k
r = (208/3)/13 = 16/3
Check: BC lies on y = −3, and I is 7/3 + 3 = 16/3 above it ✓
Answer:OI = 32i + (7/3)j + k; inradius 16/3
How to use these 2022 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 2022 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.