1. [Maximum mark: 4]
The coefficient of \( x^3 \) in the expansion of \( (3 + 2ax)^5 \) is six times the coefficient of \( x^2 \) in the expansion of \( (2 + ax)^6 \). Find the value of the constant \( a \).
▶️Answer/Explanation
Coefficient of \( x^3 \) from \( (3 + 2ax)^5 = \binom{5}{3} \cdot 3^{2} \cdot (2a)^3 = 10 \times 9 \times 8a^3 = 720a^3 \).
Coefficient of \( x^2 \) from \( (2 + ax)^6 = \binom{6}{2} \cdot 2^{4} \cdot a^2 = 15 \times 16 \times a^2 = 240a^2 \).
Given \( 720a^3 = 6 \times 240a^2 \):
\( 720a^3 = 1440a^2 \).
Divide both sides by \( 720a^2 \): \( a = 2 \).
Final Answer: \( a = 2 \).
2. [Maximum mark: 2]
Find the exact solution of the equation:
\( \frac{1}{6}\pi + \tan^{-1}(4x) = -\cos^{-1}\left(\frac{1}{2}\sqrt{3}\right) \).
▶️Answer/Explanation
Evaluate \( -\cos^{-1}\left(\frac{1}{2}\sqrt{3}\right) \):
\( \cos^{-1}\left(\frac{1}{2}\sqrt{3}\right) = \frac{\pi}{6} \), so \( -\cos^{-1}\left(\frac{1}{2}\sqrt{3}\right) = -\frac{\pi}{6} \).
Substitute back into the original equation:
\( \frac{1}{6}\pi + \tan^{-1}(4x) = -\frac{\pi}{6} \).
Subtract \( \frac{1}{6}\pi \) from both sides:
\( \tan^{-1}(4x) = -\frac{\pi}{3} \).
Take the tangent of both sides:
\( 4x = \tan\left(-\frac{\pi}{3}\right) = -\sqrt{3} \).
Divide by 4:
\( x = -\frac{\sqrt{3}}{4} \).
Final Answer: \( x = -\frac{\sqrt{3}}{4} \).
3. [Maximum mark: 6]
The equation of a curve is such that \( \frac{dy}{dx} = \frac{1}{2x} + \frac{72}{x^4} \). The curve passes through \( P(2, 8) \).
(a) Find the equation of the normal to the curve at \( P \). [2]
(b) Find the equation of the curve. [4]
▶️Answer/Explanation
(a) Gradient at \( P(2, 8) \):
\( \frac{dy}{dx} = \frac{1}{4} + \frac{72}{16} = \frac{19}{4} \).
Normal gradient: \( -\frac{4}{19} \).
Equation: \( y – 8 = -\frac{4}{19}(x – 2) \) → \( 4x + 19y = 160 \).
(b) Integrate \( \frac{dy}{dx} \):
\( y = \frac{1}{2} \ln|x| – \frac{24}{x^3} + C \).
Substitute \( P(2, 8) \): \( C = 11 – \frac{1}{2} \ln 2 \).
Final Answer: \( y = \frac{1}{2} \ln|x| – \frac{24}{x^3} + 11 – \frac{1}{2} \ln 2 \).
4. [Maximum mark: 6]
The diagram shows a coin with three circular arcs centered at the vertices of an equilateral triangle \( ABC \) (side length 2 cm).
(a) Find the perimeter of the coin. [2]
(b) Find the area of the coin’s face in terms of \( \pi \) and \( \sqrt{3} \). [4]
▶️Answer/Explanation
(a) Each arc length: \( \frac{60^\circ}{360^\circ} \times 2\pi \times 2 = \frac{2\pi}{3} \).
Perimeter: \( 3 \times \frac{2\pi}{3} = 2\pi \) cm.
(b) Area of one sector: \( \frac{60^\circ}{360^\circ} \times \pi \times 2^2 = \frac{2\pi}{3} \).
Area of triangle \( ABC \): \( \sqrt{3} \).
Total area: \( 3 \times \frac{2\pi}{3} – 2 \times \sqrt{3} = 2\pi – 2\sqrt{3} \).
Final Answer: \( 2(\pi – \sqrt{3}) \) cm².
5. [Maximum mark: 6]
A geometric progression has terms \( \sin \theta \), \( \cos \theta \), and \( 2 – \sin \theta \) (where \( \theta \) is acute).
(a) Find \( \theta \). [3]
(b) Find the sum of the first 10 terms in the form \( \frac{b}{\sqrt{c} – 1} \). [3]
▶️Answer/Explanation
(a) GP property: \( \cos^2 \theta = \sin \theta (2 – \sin \theta) \).
Simplify: \( 1 = 2 \sin \theta \) → \( \theta = \frac{\pi}{6} \).
(b) Common ratio \( r = \sqrt{3} \).
Sum: \( S_{10} = \frac{1}{2} \cdot \frac{243 – 1}{\sqrt{3} – 1} = \frac{121}{\sqrt{3} – 1} \).
Final Answer: \( \frac{121}{\sqrt{3} – 1} \).
6. [Maximum mark: 8]
The equation of a curve is y = x² – 8x + 5.
(a) Find the coordinates of the minimum point of the curve.
(b) The curve is stretched by a factor of 2 parallel to the y-axis and then translated by (4, 1). Find the coordinates of the minimum point of the transformed curve.
(c) Find the equation of the transformed curve in the form y = ax² + bx + c.
▶️Answer/Explanation
(a) Complete the square: y = (x – 4)² – 11.
Minimum point: (4, -11).
(b) After transformations: x-coordinate becomes 4 + 4 = 8.
y-coordinate becomes 2 × (-11) + 1 = -21.
New minimum: (8, -21).
(c) Stretched equation: y = 2(x² – 8x + 5).
Translated equation: y = 2[(x – 4)² – 11] + 1.
Expand: y = 2x² – 32x + 107.
Final Answers:
(a) (4, -11)
(b) (8, -21)
(c) y = 2x² – 32x + 107
7. [Maximum mark: 9]
(a) Verify that (2x – 1)(4x² + 2x – 1) ≡ 8x³ – 4x + 1.
(b) Prove the identity (tan²θ + 1)/(tan²θ – 1) ≡ 1/(1 – 2cos²θ).
(c) Solve (tan²θ + 1)/(tan²θ – 1) = 4cosθ for 0° ≤ θ ≤ 180°.
▶️Answer/Explanation
(a) Expand LHS: 8x³ + 4x² – 2x – 4x² – 2x + 1 = 8x³ – 4x + 1 ≡ RHS.
(b) Replace tanθ with sinθ/cosθ:
LHS = (sin²θ/cos²θ + 1)/(sin²θ/cos²θ – 1) = (1/cos²θ)/(-cos2θ/cos²θ) = -sec²θ/cos2θ.
Simplify to 1/(1 – 2cos²θ).
(c) Using (a) and (b): 1/(1 – 2cos²θ) = 4cosθ.
Let x = cosθ: 1 = 4x(1 – 2x²) → 8x³ – 4x + 1 = 0.
Factorize using (a): (2x – 1)(4x² + 2x – 1) = 0.
Solutions: θ = 60°, 72°, 144°.
Final Answers:
(a) Verified
(b) Proven
(c) θ = 60°, 72°, 144°
8. [Maximum mark: 8]
Functions f and g are defined by:
f(x) = (x + a)² – a for x ≤ -a
g(x) = 2x – 1 for x ∈ ℝ
(a) Find f⁻¹(x). [3]
(b) (i) State the domain of f⁻¹. [1]
(ii) State the range of f⁻¹. [1]
(c) Given a = 7/2, solve gf(x) = 0. [3]
▶️Answer/Explanation
(a) Let y = (x + a)² – a.
Swap x and y: x = (y + a)² – a → y = -√(x + a) – a.
f⁻¹(x) = -√(x + a) – a.
(b) (i) Domain: x ≥ -a.
(ii) Range: f⁻¹(x) ≤ -a.
(c) gf(x) = 2[(x + 7/2)² – 7/2] – 1 = 0.
Solve: (x + 7/2)² = 4 → x = -7/2 ± 2.
Only valid solution: x = -11/2.
Final Answers:
(a) f⁻¹(x) = -√(x + a) – a
(b)(i) x ≥ -a (ii) f⁻¹(x) ≤ -a
(c) x = -11/2
9. [Maximum mark: 9]
Two curves intersect at points A and B:
y = 2x¹ᐟ² + 13x⁻¹ᐟ² and y = 3x⁻¹ᐟ² + 12.
(a) Find the coordinates of A and B. [4]
(b) Find the area between the curves. [5]
▶️Answer/Explanation
(a) Set equations equal: 2x¹ᐟ² + 13x⁻¹ᐟ² = 3x⁻¹ᐟ² + 12.
Multiply by x¹ᐟ²: 2x + 13 = 3 + 12x¹ᐟ² → 2x – 12x¹ᐟ² + 10 = 0.
Let u = x¹ᐟ²: 2u² – 12u + 10 = 0 → u = 1 or 5.
Solutions: x = 1 (y = 15), x = 25 (y = 12.6).
(b) Area = ∫[3x⁻¹ᐟ² + 12 – (2x¹ᐟ² + 13x⁻¹ᐟ²)]dx from 1 to 25.
= ∫[-2x¹ᐟ² – 10x⁻¹ᐟ² + 12]dx.
Integrate: [-4/3 x³ᐟ² – 20x¹ᐟ² + 12x] from 1 to 25.
Evaluate: 128/3.
Final Answers:
(a) A(1,15), B(25,12.6)
(b) 128/3
10. [Maximum mark: 7]
The curve has equation y = (4x – 3)⁵ᐟ³ – (20/3)x.
(a) Find the x-coordinates of the stationary points and determine their nature. [6]
(b) State where the function is increasing. [1]
▶️Answer/Explanation
(a) dy/dx = (20/3)(4x – 3)²ᐟ³ – 20/3.
Set dy/dx = 0: (4x – 3)²ᐟ³ = 1 → x = 1/2, 1.
Second derivative test:
At x = 1/2: d²y/dx² < 0 (maximum).
At x = 1: d²y/dx² > 0 (minimum).
(b) Increasing when x < 1/2 or x > 1.
Final Answers:
(a) Max at x=1/2, Min at x=1
(b) x < 1/2 or x > 1
11. [Maximum mark: 10]
Points A(6,4), B(p,7), C(14,18) lie on a circle with AB perpendicular to BC.
(a) Find p (p < 10). [4]
(b) Find the circle’s equation. [3]
(c) Find the tangent at C in the form dx + ey + f = 0. [3]
▶️Answer/Explanation
(a) Gradient AB × Gradient BC = -1:
(7-4)/(p-6) × (18-7)/(14-p) = -1 → p² – 20p + 51 = 0.
Solve: p = 3 (since p < 10).
(b) Midpoint of AC is center (10,11).
Radius² = (14-10)² + (18-11)² = 65.
Equation: (x-10)² + (y-11)² = 65.
(c) Gradient at C: (18-11)/(14-10) = 7/4.
Tangent gradient: -4/7.
Equation: 4x + 7y – 182 = 0.
Final Answers:
(a) p=3
(b) (x-10)² + (y-11)²=65
(c) 4x + 7y – 182 = 0