2015 Q9 — Extreme values, asymptotes and an area for y = (x² + 12)/(x − 2)
Let f(x) = (x² + 12)/(x − 2) for x ≠ 2.(a) Find f′(x).(b) Show that the maximum and minimum values of f(x) are −4 and 12.(c) Find the asymptotes of y = f(x).(d) Find the area of the region bounded by the graph and the line y = 14.
The region between the line y = 14 (blue) and the right-hand branch of the curve (red) fills from x = 4 to x = 10. The curve dips to its minimum (6, 12), so the region is a thin crescent of area 30 − 32 ln 2 ≈ 7.82. Try it yourself →
Answer:(a) (x − 6)(x + 2)/(x − 2)²; (c) x = 2 and y = x + 2; (d) 30 − 32 ln 2
2015 Q10 — Ratios on a triangle, then an area and a volume in 3D
In triangle OAB, P is the mid-point of OA, Q lies on AB with AQ : QB = 1 : 2, and R lies on OB with OR : RB = 3 : 1. PR and OQ meet at C.(a) With PC : CR = t : (1 − t), find t and CQ : OQ.(b) Given OA = 20i − 6j − 12k, OB = 16i − 16j and OD = i + 3j − 6k, find the area of △OAB and the volume of tetrahedron ABCD.
The camera circles the figure. Triangle OAB (grey) has area 176; C lies inside it on OQ, and the tetrahedron ABCD is built on the blue triangle ABC with apex D, giving volume 42. Try it yourself →
(a) Write a = OA, b = OB. Then OQ = (2a + b)/3, OP = a/2, OR = 3b/4, and OC = (1 − t)a/2 + 3t b/4
C is on OQ, so the coefficients of a and b are in the ratio 2 : 1: (1 − t)/2 = 2 · (3t/4), giving t = 1/4
Then OC = (3/8)a + (3/16)b = (9/16)·OQ, so CQ : OQ = 7 : 16
(b) OA × OB = −192i − 192j − 224k = −32(6i + 6j + 7k), so area of △OAB= ½ · 32 · √(36 + 36 + 49) = 16 · 11 = 176
C lies in plane OAB and CQ = (7/16)OQ, so area of △ABC= (7/16) · 176 = 77
Height from D to plane OAB (normal 6i + 6j + 7k, through O): |6 + 18 − 42|/11 = 18/11. Volume = (1/3) · 77 · 18/11 = 42
Answer:(a) t = 1/4, CQ : OQ = 7 : 16; (b) area 176, volume 42
2015 Q12 — A cup formed by revolution and the rate the water rises
A curve consists of the arc AB of x² − 4y + 8 = 0 and the segments BC (on 3x + y − 9 = 0) and CO, where B is in the first quadrant and C is on the x-axis.(a) Find B, and show that when A has y-coordinate h > 3 the cup formed by revolving the curve about the y-axis has capacity π(2h² − 8h + 25).(b) A cup of this shape has base radius 3 cm and lip radius 6 cm. Find its capacity, and the rate at which the depth rises when water is poured in at 24π cm³/s and the volume of water is 35π cm³.
The camera circles the cup. Its lower part is a cone frustum narrowing from radius 3 to 2, and above 3 cm it opens out along the parabola. The blue disc is the water surface at a depth of 5 cm, where the volume is 35π cm³. Try it yourself →
(a) Substitute y = 9 − 3x: x² + 12x − 28 = 0, so x = 2 (first quadrant) and B = (2, 3); C is (3, 0)
Lower part (0 ≤ y ≤ 3, x = (9 − y)/3): π∫₀³ (9 − y)²/9 dy = (π/27)(9³ − 6³) = 19π
Upper part (3 ≤ y ≤ h, x² = 4y − 8): π∫₃ʰ (4y − 8) dy = π(2h² − 8h + 6). Total = π(2h² − 8h + 25)
(b) Lip radius 6: 36 = 4h − 8, so h = 11 and capacity = π(242 − 88 + 25) = 179π cm³
35π > 19π, so the surface is in the upper part: 2y² − 8y + 25 = 35, giving depth y = 5
dV/dy = πx² = π(4y − 8) = 12π at y = 5, so dy/dt = 24π/12π = 2
Answer:(a) B = (2, 3); (b) capacity 179π cm³, the depth rises at 2 cm/s
How to use these 2015 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 2015 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.