The slope of a curve at any point (x, y) is dy/dx = eˣ − 1, and the curve passes through (1, e). Find(a) the equation of the curve,(b) the equation of the tangent at the point where the curve cuts the y-axis.
A tangent slides along y = eˣ − x + 1. Its slope eˣ − 1 is negative for x < 0 and positive for x > 0, so it is horizontal exactly where the curve crosses the y-axis, at (0, 2). Try it yourself →
(a) Integrate: y = ∫(eˣ − 1) dx = eˣ − x + C
Through (1, e): e = e − 1 + C, so C = 1 and y = eˣ − x + 1
(b) On the y-axis x = 0: y = 1 − 0 + 1 = 2
Slope there = e⁰ − 1 = 0, so the tangent is horizontal
Answer:(a) y = eˣ − x + 1; (b) y = 2
2013 Q5 — Turning point, inflexion and asymptote from a sign table
Let f(x) = (3 − 3x²)/(3 + x²). A table gives the signs of f′(x) and f″(x): f′ > 0 for x < 0, f′(0) = 0, f′ < 0 for x > 0; f″ > 0 for |x| > 1, f″(±1) = 0, f″ < 0 for |x| < 1. Find the maximum/minimum points, the points of inflexion and the asymptote of y = f(x), and sketch the graph.
The tangent rises to the maximum (0, 1) and falls after it. The curve bends from concave up to concave down at (−1, 0) and back at (1, 0), and flattens towards y = −3 on both sides. Try it yourself →
f′ changes from + to − at x = 0, and f(0) = 3/3 = 1: maximum point (0, 1); there is no minimum point
f″ changes sign at x = −1 and x = 1, and f(±1) = 0/4 = 0: points of inflexion (−1, 0) and (1, 0)
Divide top and bottom by x²: f(x) = (3/x² − 3)/(3/x² + 1) → −3 as x → ±∞
The denominator 3 + x² is never zero, so there is no vertical asymptote
Answer:Maximum (0, 1); inflexion (−1, 0) and (1, 0); horizontal asymptote y = −3
2013 Q6 — Area between a parabola and y = 4, and the volume about y = 4
The curve C: y = −x²/2 + 2x + 4 meets the line L₁: y = 4 at (0, 4) and (4, 4); L₂ is the line x = 5. The shaded region lies between C and L₁ from x = 0 to x = 5. Find(a) its area,(b) the volume when it is revolved about L₁.
The region fills from left to right. From x = 0 to 4 the curve is above y = 4 (blue, 16/3); from 4 to 5 it is below (green, 7/6). The two pieces must be added, not subtracted. Try it yourself →
Height of the region above L₁: y − 4 = −x²/2 + 2x, positive on (0, 4) and negative on (4, 5)
2013 Q10 — Division of a segment and concyclic points
Let OA = 2i and OB = i + 2j. M is the mid-point of OA, and N lies on AB with BN : NA = k : 1. BM meets ON at P.(a) Express ON in terms of k.(b) If A, N, P, M are concyclic, find k.
N slides along AB as k changes; the red circle always passes through A, N and M. P falls on that circle only when ON is perpendicular to AB, which happens at k = 3/2. Try it yourself →
(b) M = (1, 0) and B = (1, 2), so BM is vertical and ∠PMA = 90°
Opposite angles of a cyclic quadrilateral add up to 180°, so ∠PNA = 90°, i.e. ON ⊥ AB
AB = OB − OA = −i + 2j
ON · AB = 0: −(1 + 2k) + 4 = 0
Answer:(a) ON = [(1 + 2k)i + 2j]/(k + 1); (b) k = 3/2
How to use these 2013 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 2013 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.