2017 DSE M2 Past Paper: Solutions with Video Explanations

We are featuring 3 questions from the 2017 DSE M2 paper, each explained with a video. Take a look.

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2017 Q6 — Wet curved surface of an inverted cone

A container is an inverted right circular cone of height 20 cm and base radius 15 cm, held vertically. Water is poured in.(a) Let A cm² be the wet curved surface area and h cm the depth. Prove that A = 15πh²/16.(b) The depth increases at a constant rate of 3/π cm/s. Find the rate of change of the wet curved surface area when the volume of water is 96π cm³.

Side view of the cone as it fills. The water forms a smaller similar cone of radius 3h/4; the red slant edge, of length 5h/4, sweeps out the wet surface. At a depth of 8 cm the wet area is growing at 45 cm²/s. Try it yourself →
  1. (a) By similar triangles the water surface has radius r = (15/20)h = 3h/4, and the slant length is √(h² + r²) = 5h/4
  2. A = π r l = π(3h/4)(5h/4) = 15πh²/16
  3. (b) Volume V = (1/3)πr²h = 3πh³/16. Set 3πh³/16 = 96π: h³ = 512, so h = 8
  4. dA/dt = (15πh/8)(dh/dt) = (15π · 8/8)(3/π)

Answer: dA/dt = 45 cm²/s

2017 Q8 — Tangent, equation and inflexion of a curve with f′(x) = (1/x) ln x²

The curve Γ: y = f(x) is defined for x > 0, passes through P(e³, 7), and has f′(x) = (1/x) ln x². Find(a) the equation of the tangent to Γ at P,(b) the equation of Γ,(c) the point(s) of inflexion of Γ.

A tangent slides along Γ: y = (ln x)² − 2. The curve falls to its lowest point at x = 1, changes from concave up to concave down at (e, −1), and at P(e³, 7) the tangent is y = 6x/e³ + 1. Try it yourself →
  1. For x > 0, ln x² = 2 ln x, so f′(x) = (2 ln x)/x
  2. (a) Slope at P: 2 · 3/e³ = 6/e³. Tangent: y − 7 = (6/e³)(x − e³), i.e. y = 6x/e³ + 1
  3. (b) Let u = ln x: y = ∫ 2u du = (ln x)² + C. Through P: 9 + C = 7, so y = (ln x)² − 2
  4. (c) f″(x) = (2 − 2 ln x)/x², which is 0 at x = e and changes from + to − there
  5. f(e) = 1 − 2 = −1

Answer: (a) y = 6x/e³ + 1; (b) y = (ln x)² − 2; (c) (e, −1)

2017 Q9 — Asymptotes, turning points and a volume for y = (x² − 5x)/(x + 4)

Let f(x) = (x² − 5x)/(x + 4) for x ≠ −4, with graph G.(a) Find the asymptotes of G.(b) Find f′(x).(c) Find the maximum and minimum points of G.(d) R is the region bounded by G and the x-axis. Find the volume of the solid generated by revolving R about the x-axis.

Between its x-intercepts 0 and 5 the curve (red) dips just below the x-axis, to the minimum (2, −1). Turning that shallow region about the x-axis gives the spindle-shaped solid, of volume about 8.28. Try it yourself →
  1. (a) Divide: x² − 5x = (x + 4)(x − 9) + 36, so f(x) = x − 9 + 36/(x + 4). Asymptotes: x = −4 and y = x − 9
  2. (b) f′(x) = 1 − 36/(x + 4)² = (x + 10)(x − 2)/(x + 4)²
  3. (c) f′ changes + to − at x = −10: maximum point (−10, −25). It changes − to + at x = 2: minimum point (2, −1)
  4. (d) G cuts the x-axis at x = 0 and x = 5. With u = x + 4: V = π∫₄⁹ (u − 13 + 36/u)² du
  5. (u − 13 + 36/u)² = u² − 26u + 241 − 936/u + 1296/u²
  6. V = π[u³/3 − 13u² + 241u − 936 ln u − 1296/u]₄⁹ = π(2285/3 − 936 ln(9/4))

Answer: (a) x = −4, y = x − 9; (c) max (−10, −25), min (2, −1); (d) π(2285/3 − 1872 ln(3/2)) ≈ 8.28

How to use these 2017 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 2017 HKDSE Mathematics Extended Part Module 2 paper, and the solutions and animations are written by SketchMath.

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