(a) Using substitution, find ∫ x³√(1 + x²) dx.(b) At any point (x, y) on the curve Γ, the slope of the tangent is 15x³√(1 + x²), and the y-intercept of Γ is 2. Find the equation of Γ.
A tangent slides along Γ. Its slope 15x³√(1 + x²) is negative on the left, zero at the y-intercept (0, 2) and positive on the right, growing very quickly as x moves away from 0. Try it yourself →
(a) Let u = 1 + x², du = 2x dx, x² = u − 1: ∫ x³√(1 + x²) dx = ½∫ (u − 1)√u du
= ½[(2/5)u^(5/2) − (2/3)u^(3/2)] + C = (1 + x²)^(5/2)/5 − (1 + x²)^(3/2)/3 + C
(b) y = 15 × (the result of (a)) = 3(1 + x²)^(5/2) − 5(1 + x²)^(3/2) + C
2018 Q8 — Extreme value, asymptotes and inflexion of y = A/(x² − 4x + 7)
Let f(x) = A/(x² − 4x + 7), where A is a constant, and the extreme value of f(x) is 4.(a) Find f′(x).(b) Someone claims that the graph of y = f(x) has at least two asymptotes. Do you agree?(c) Find the point(s) of inflexion of the graph.
The tangent rolls over the bell-shaped curve y = 12/(x² − 4x + 7). It is horizontal at the maximum (2, 4) and steepest at the two points of inflexion (1, 3) and (3, 3); the curve only ever approaches the x-axis. Try it yourself →
x² − 4x + 7 = (x − 2)² + 3 has least value 3 at x = 2, so the extreme value is A/3 = 4 and A = 12
f″ changes sign at x = 1 and x = 3, and f(1) = f(3) = 12/4 = 3
Answer:(a) −24(x − 2)/(x² − 4x + 7)²; (b) no, only y = 0; (c) (1, 3) and (3, 3)
2018 Q9 — Greatest area of a triangle formed by a normal to y = ln √x
P is a moving point on the curve C: y = ln √x (x > 1) with x-coordinate r. The normal to C at P cuts the x-axis at Q, and the vertical line through P cuts the x-axis at R.(a) Show that the x-coordinate of Q is (4r² + ln r)/(4r).(b) Find the greatest area of △PQR.
The triangle itself is far too thin to see, so the video plots its area (ln r)²/(16r) against r. The tangent to this area curve is horizontal at r = e² ≈ 7.39, where the area peaks at 1/(4e²) ≈ 0.034. Try it yourself →
(a) y = ½ ln x, so the tangent at P has slope 1/(2r) and the normal has slope −2r
Normal: y − ½ ln r = −2r(x − r). At y = 0: x = r + (ln r)/(4r) = (4r² + ln r)/(4r)
(b) QR = (ln r)/(4r) and PR = ½ ln r, with a right angle at R: area S = (ln r)²/(16r)
For r > 1, dS/dr changes from + to − at ln r = 2, i.e. r = e², so S is greatest there
S = 2²/(16e²) = 1/(4e²)
Answer:Greatest area = 1/(4e²)
2018 Q10(c) — Volume of revolution of y = √x sin²x
Earlier parts give ∫₀^π sin⁴x dx = 3π/8 and ∫₀^π x sin⁴x dx = 3π²/16. Consider the curve G: y = √x sin²x for π ≤ x ≤ 2π. The region bounded by G and the x-axis is revolved about the x-axis. Find the volume of the solid.
The single arch of y = √x sin²x between π and 2π (red) turns about the x-axis into a lemon-shaped solid. Its cross-sections are discs of area πy² = πx sin⁴x. Try it yourself →
Disc method: V = π∫_π^{2π} y² dx = π∫_π^{2π} x sin⁴x dx
Let x = u + π; since sin⁴(u + π) = sin⁴u: ∫_π^{2π} x sin⁴x dx = ∫₀^π (u + π) sin⁴u du
= 3π²/16 + π · 3π/8 = 9π²/16
V = π · 9π²/16
Answer:V = 9π³/16
2018 Q12 — Tetrahedron volume, projection on a plane and the angle between two planes
The position vectors of A, B, C, D are 4i − 3j + k, −i + 3j − 3k, 7i − j + 5k and 3i − 2j − 5k. Π is the plane through A, B, C, and E is the projection of D on Π.(a) Find AB × AC, the volume of tetrahedron ABCD, and DE.(b)F lies on BC with DF ⊥ BC. Find DF, and decide whether BC is perpendicular to EF.(c) Find the angle between △BCD and Π.
The camera circles the two faces that meet along BC: plane Π (blue, ABC) and triangle BCD (orange). DE (red) drops straight onto Π and DF (green) meets BC at right angles, so angle DFE, about 33.7°, is the angle between the two planes. Try it yourself →
(a) AB = −5i + 6j − 4k, AC = 3i + 2j + 4k: AB × AC = 32i + 8j − 28k = 4(8i + 2j − 7k)
AD = −i + j − 6k, so AD · (AB × AC) = −32 + 8 + 168 = 144 and the volume is 144/6 = 24
With n = 8i + 2j − 7k (|n|² = 117, AD · n = 36): DE = −(36/117)n = −(4/13)(8i + 2j − 7k)
(b) BC = 8i − 4j + 8k, BD = 4i − 5j − 2k, so BF = (BD · BC/|BC|²)BC = (36/144)BC = 2i − j + 2k. Then F = (1, 2, −1) and DF = −2i + 4j + 4k
EF = DF − DE. BC · DF = 0, and BC · DE = 0 because DE is normal to Π, which contains BC. So BC · EF = 0: yes, BC ⊥ EF
(c) Both DF and EF are perpendicular to BC, so the angle is ∠DFE: sin ∠DFE = |DE|/|DF| = (12/√13)/6 = 2/√13, about 33.7°
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 2018 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.
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