2019 Q3 — Rate of change of the depth of liquid in a vessel
A vessel contains 580 cm³ of a liquid at the start of a 24-hour experiment. During the experiment dV/dt = −2t, where V cm³ is the volume after t hours.(a) Is there still liquid in the vessel at the end?(b) Given V = h² + 24h, where h cm is the depth, find dh/dt when t = 18.
Depth h plotted against time t. The curve falls faster and faster; the red tangent shows the rate of change, which is −0.9 cm per hour at t = 18, when the depth is 8 cm. At t = 24 a little liquid is left. Try it yourself →
(a) Integrate: V = −t² + C, and V = 580 at t = 0, so V = 580 − t²
At t = 24: V = 580 − 576 = 4 > 0, so some liquid remains
(b) At t = 18: V = 580 − 324 = 256, so h² + 24h − 256 = 0, (h − 8)(h + 32) = 0, h = 8
2019 Q8 — An increasing function and its point of inflexion
A function h(x) is defined for x > 0 with h′(x) = (2x² − 7x + 8)/x.(a) Is h(x) an increasing function?(b) The curve H: y = h(x) passes through (1, 3). Find the equation of H and its point(s) of inflexion.
The tangent to H always slopes upwards, so the function is increasing. The slope is smallest at the point of inflexion (2, 8 ln 2 − 1), where the curve changes from concave down to concave up. Try it yourself →
(a) For 2x² − 7x + 8, Δ = 49 − 64 < 0, so it is always positive; with x > 0, h′(x) > 0. Yes, h(x) is increasing
(b) h′(x) = 2x − 7 + 8/x, so h(x) = x² − 7x + 8 ln x + C
Through (1, 3): 1 − 7 + 0 + C = 3, so C = 9
h″(x) = 2 − 8/x² = 2(x − 2)(x + 2)/x² changes from − to + at x = 2
h(2) = 4 − 14 + 8 ln 2 + 9 = 8 ln 2 − 1
Answer:(a) Yes; (b) y = x² − 7x + 8 ln x + 9, point of inflexion (2, 8 ln 2 − 1)
2019 Q9 — A line tangent to two curves, and the area between them
Let Γ be the curve y = (1/3)√(12 − x²) (0 < x < 2√3) and L the tangent to Γ at x = 3.(a) Find the equation of L.(b)C is the curve y = √(4 − x²) (0 < x < 2), and L is also a tangent to C. Find(i) the point of contact of L and C,(ii) the point of intersection of C and Γ,(iii) the area of the region bounded by L, C and Γ.
The red line touches the green circle arc at (1, √3) and the blue ellipse arc at (3, √3/3). The orange region under the line fills from one point of contact to the other, switching from the circle to the ellipse where they cross at (√3, 1). Try it yourself →
(a) At x = 3: y = √3/3 and dy/dx = −x/(3√(12 − x²)) = −1/√3. So L: y − √3/3 = −(x − 3)/√3, i.e. x + √3y − 4 = 0
(b)(i) C is part of the circle x² + y² = 4. The radius to the point of contact is perpendicular to L, so it has slope √3: y = √3x, giving (1, √3)
(ii) 4 − x² = (12 − x²)/9 gives x² = 3: the point is (√3, 1)
The first integral is 4√3/3. With x = 2 sin θ the second is [2θ + sin 2θ] from π/6 to π/3= π/3; with x = 2√3 sin θ the third is [2θ + sin 2θ] from π/6 to π/3= π/3
4√3/3 − π/3 − π/3 = 4√3/3 − 2π/3
Answer:(a) x + √3y − 4 = 0; (b)(i) (1, √3); (ii) (√3, 1); (iii) 4√3/3 − 2π/3
2019 Q12 — Pyramid volume and the projection of the origin on a plane
Let OA = i − 4j + 2k, OB = −5i − 4j + 8k and OC = −5i − 12j + tk, with |AC| = |BC|.(a) Find t.(b) Find AB × AC.(c) Find the volume of the pyramid OABC.(d)Π is the plane through A, B, C; it meets the axes at P, Q, R with OP = pi, OQ = qj, OR = rk, and D is the projection of O on Π. Find OD, and describe the relationship between D, O and the point E with OE = (1/p)i + (1/q)j + (1/r)k.
The camera circles plane Π, drawn as the grey triangle PQR through its axis intercepts, with triangle ABC (blue) lying in it. The red segment OD is the perpendicular from the origin to the plane. Try it yourself →
(d) Π: 4x − 3y + 4z = 24 (substitute A). D is on the normal through O: OD = λ(4i − 3j + 4k) with 41λ = 24
Intercepts: p = 6, q = −8, r = 6, so OE = (1/6)i − (1/8)j + (1/6)k = (1/24)(4i − 3j + 4k), which is parallel to OD
Answer:(a) t = 2; (b) 48i − 36j + 48k; (c) 48; (d) OD = (24/41)(4i − 3j + 4k); O, E, D are collinear
How to use these 2019 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 2019 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.