Which 2014 DSE M2 questions are explained here?

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.

Are these the official HKEAA answers?

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.

Can I change the numbers and see the graph update?

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.

2014 DSE M2 Past Paper: Solutions with Video Explanations

We are featuring 5 questions from the 2014 DSE M2 paper, each explained with a video. Take a look.

Ask AI about a 2014 DSE M2 question

2014 Q3 — Tangent to the implicit curve x ln y + y = 2

Find the equation of the tangent to the curve x ln y + y = 2 at the point where the curve cuts the y-axis.

A tangent slides along the curve x ln y + y = 2. Where the curve crosses the y-axis, at (0, 2), its slope is −ln 2 ≈ −0.69. Try it yourself →

On the y-axis x = 0, so y = 2: the point is (0, 2)
Differentiate implicitly: ln y + (x/y)(dy/dx) + dy/dx = 0
At (0, 2): ln 2 + 0 + dy/dx = 0, so dy/dx = −ln 2
Tangent: y − 2 = −(ln 2)(x − 0)

Answer: y = −(ln 2)x + 2

2014 Q6 — Area between y = xe⁻ˣ and y = x/e

(a) Find ∫ x e⁻ˣ dx.(b) Find the area of the region bounded by the curve y = x e⁻ˣ and the straight line y = x/e.

The shaded region between the curve y = xe⁻ˣ (blue) and the line y = x/e (red) grows from x = 0 to the second intersection at x = 1, reaching 1 − 5/(2e) ≈ 0.08. Try it yourself →

(a) By parts with u = x, dv = e⁻ˣ dx: ∫ x e⁻ˣ dx = −x e⁻ˣ + ∫ e⁻ˣ dx = −x e⁻ˣ − e⁻ˣ + C
(b) Intersections: x e⁻ˣ = x/e gives x = 0 or e⁻ˣ = e⁻¹, i.e. x = 1
Between them the curve is above the line, so area = ∫₀¹ (x e⁻ˣ − x/e) dx
= [−x e⁻ˣ − e⁻ˣ]₀¹ − [x²/(2e)]₀¹ = (−2/e + 1) − 1/(2e)

Answer: (a) −x e⁻ˣ − e⁻ˣ + C; (b) 1 − 5/(2e)

2014 Q8 — Volume of a tetrahedron and the angle between a line and a plane

Let OP = −i + 2j + 2k, OQ = i − j + 2k and OR = 2i − 3j + 6k.(a) Find OP × OQ and hence the volume of tetrahedron OPQR.(b) Find the acute angle between the plane OPQ and the line OR, correct to the nearest 0.1°.

The camera circles tetrahedron OPQR. The red arrow is the normal to the blue face OPQ; OR lies almost in that face, which is why the angle between them is only about 6.8°. Try it yourself →

OP × OQ = (2·2 − 2·(−1))i − ((−1)·2 − 2·1)j + ((−1)(−1) − 2·1)k = 6i + 4j − k
(a) Volume = (1/6)|OR · (OP × OQ)| = (1/6)|12 − 12 − 6| = 1
(b) Let θ be the angle between OR and the plane. With the normal n = 6i + 4j − k: sin θ = |OR · n| / (|OR| |n|)
|OR| = √(4 + 9 + 36) = 7, |n| = √(36 + 16 + 1) = √53
sin θ = 6/(7√53), θ ≈ 6.8°

Answer: (a) OP × OQ = 6i + 4j − k, volume = 1; (b) 6.8°

2014 Q10 — The shortest ladder over a bookcase

A bookcase stands against a wall; its side view is a rectangle 24 cm wide and 192 cm high. A ladder HK touches the wall at H, the top outer edge of the bookcase, and the floor at K, making an angle θ with the floor.(a) Express HK in terms of θ.(b) Prove that the shortest such ladder is 120√5 cm.

The ladder pivots on the top corner of the bookcase as θ changes. Its length is long when it is nearly flat or nearly upright and shortest, 120√5 ≈ 268.3 cm, when tan θ = 2 (θ ≈ 63.4°). Try it yourself →

(a) Split the ladder at the corner B of the bookcase: the upper part spans the 24 cm width, the lower part the 192 cm height. HK = 24/cos θ + 192/sin θ
d(HK)/dθ = 24 sin θ/cos² θ − 192 cos θ/sin² θ
Set it to 0: 24 sin³ θ = 192 cos³ θ, so tan³ θ = 8 and tan θ = 2
The derivative changes from − to + there (it is negative for small θ, positive near 90°), so HK is a minimum
With tan θ = 2: cos θ = 1/√5, sin θ = 2/√5, so HK = 24√5 + 96√5 = 120√5

Answer: (a) HK = 24/cos θ + 192/sin θ cm; (b) shortest length 120√5 cm

2014 Q13(d) — Volume about the y-axis using a symmetry result

Earlier parts show that ∫₀^{nπ} cos²x sin⁴x dx = nπ/16, and that if f(k − x) = f(x) then ∫₀ᵏ x f(x) dx = (k/2)∫₀ᵏ f(x) dx. The region bounded by y = cos²x sin⁴x and the x-axis for π ≤ x ≤ 2π is revolved about the y-axis. Find the volume of the solid.

The two humps of y = cos²x sin⁴x between π and 2π turn about the y-axis into two ring-shaped ridges. (The height is drawn 12 times taller so the ridges can be seen.) Try it yourself →

Shell method about the y-axis: V = 2π∫_π^{2π} x f(x) dx, where f(x) = cos²x sin⁴x
f(2π − x) = f(x), so with k = 2π: ∫₀^{2π} x f(x) dx = π · (2π/16) = π²/8
f(π − x) = f(x), so with k = π: ∫₀^π x f(x) dx = (π/2)(π/16) = π²/32
∫_π^{2π} x f(x) dx = π²/8 − π²/32 = 3π²/32
V = 2π · 3π²/32

Answer: V = 3π³/16

How to use these 2014 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 2014 HKDSE Mathematics Extended Part Module 2 paper, and the solutions and animations are written by SketchMath.