1. [Maximum mark: 3]
Solve the equation \(3x + 2 = \frac{2}{x – 1}\).
▶️Answer/Explanation
Steps:
1. Multiply both sides by \((x – 1)\): \[ (3x + 2)(x – 1) = 2 \]
2. Expand and simplify: \[ 3x^2 – x – 4 = 0 \]
3. Factorize: \[ (3x – 4)(x + 1) = 0 \]
4. Solutions: \[ x = \frac{4}{3} \quad \text{or} \quad x = -1 \]
2. [Maximum mark: 6]
The equation of a curve is such that \(\frac{dy}{dx} = 12\left(\frac{1}{2}x – 1\right)^{-4}\). The curve passes through \(P(6, 4)\).
(a) Find the equation of the tangent to the curve at \(P\).
(b) Find the equation of the curve.
▶️Answer/Explanation
(a)
1. Substitute \(x = 6\) into \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = 12(2)^{-4} = \frac{3}{4} \]
2. Equation of tangent: \[ y – 4 = \frac{3}{4}(x – 6) \quad \Rightarrow \quad y = \frac{3}{4}x – \frac{1}{2} \] (b) Integrate d y d x dx dy : = \int 12\left(\frac{1}{2}x – 1\right)^{-4} dx = -8\left(\frac{1}{2}x – 1\right)^{-3} + C ] 2. Substitute P ( 6 , 4 ) P(6,4) to find C C: 4 = − 8 ( 2 ) − 3 + C ⇒ C = 5 4=−8(2) −3 +C⇒C=5
3. [Maximum mark: 5]
A curve has equation \(y = ax^{\frac{1}{2}} – 2x\) (\(x > 0\)) with a stationary point at \(P(9, y)\). Find the y-coordinate of \(P\).
▶️Answer/Explanation
1. Find \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{a}{2}x^{-\frac{1}{2}} – 2 \]
2. Set \(\frac{dy}{dx} = 0\) at \(x = 9\): \[ \frac{a}{2}(9)^{-\frac{1}{2}} = 2 \quad \Rightarrow \quad a = 12 \]
3. Substitute \(a = 12\) and \(x = 9\) into \(y\): \[ y = 12(9)^{\frac{1}{2}} – 2(9) = 18 \]
4. [Maximum mark: 6]
The coefficient of \(x^2\) in the expansion of \(\left(1 + \frac{2}{p}x\right)^5 + \left(1 + px\right)^6\) is 70. Find the possible values of the constant \(p\).
▶️Answer/Explanation
(a) Find coefficients of \(x^2\):
1. For \(\left(1 + \frac{2}{p}x\right)^5\): \[ \text{Coefficient} = \binom{5}{2} \left(\frac{2}{p}\right)^2 = \frac{40}{p^2} \]
2. For \(\left(1 + px\right)^6\): \[ \text{Coefficient} = \binom{6}{2} p^2 = 15p^2 \]
(b) Solve for p p: Total coefficient equation: 40 p 2 + 15 p 2 = 70 p 2 40 +15p 2 =70 Multiply by p 2 p 2 : 15 p 4 − 70 p 2 + 40 = 0 15p 4 −70p 2 +40=0 Substitute y = p 2 y=p 2 : Final solutions: p = ± 2 or p = ± 2 3 p=±2orp=± 3 2
5. [Maximum mark: 6]
A sector OAB has arc length 8 cm and perimeter 20 cm.
(a) Find the perimeter of the shaded segment.
(b) Find the area of the shaded segment.
▶️Answer/Explanation
(a)
1. Radius \(r = 6\) cm (from \(2r + 8 = 20\)).
2. Angle \(\theta = \frac{8}{6} = \frac{4}{3}\) radians.
3. Chord length \(AB = 2r \sin\left(\frac{\theta}{2}\right) \approx 7.42\) cm.
4. Perimeter of segment: \(8 + 7.42 \approx 15.4\) cm. (b) Sector area: 1 2 r 2 θ = 24 2 1 r 2 θ=24 cm². Triangle area: 1 2 r 2 sin θ ≈ 17.49 2 1 r 2 sinθ≈17.49 cm². Shaded area: 24 − 17.49 ≈ 6.51 24−17.49≈6.51 cm².
6. [Maximum mark: 6]
(a) Show that \(\frac{1}{\sin\theta + \cos\theta} + \frac{1}{\sin\theta – \cos\theta} = 1\) simplifies to \(2\sin^2\theta – 2\sin\theta – 1 = 0\).
(b) Solve the equation for \(0^\circ \leq \theta \leq 360^\circ\).
▶️Answer/Explanation
(a)
1. Combine fractions: \[ \frac{2\sin\theta}{\sin^2\theta – \cos^2\theta} = 1 \]
2. Substitute \(\cos^2\theta = 1 – \sin^2\theta\): \[ 2\sin\theta = \sin^2\theta – (1 – \sin^2\theta) \quad \Rightarrow \quad 2\sin^2\theta – 2\sin\theta – 1 = 0 \] (b) Solve quadratic: sin\theta = \frac{1 \pm \sqrt{3}}{2} ] 2. Solutions: θ ≈ 201. 5 ∘ , 338. 5 ∘ θ≈201.5 ∘ ,338.5 ∘
7. [Maximum mark: 7]
A post-rammer’s impacts follow a geometric progression: 50 mm, 40 mm, 32 mm, …
(a) Show the 9th impact is the first <10 mm.
(b) Find total depth after 20 impacts.
(c) Find theoretical maximum depth.
▶️Answer/Explanation
(a)
1. Common ratio \(r = 0.8\).
2. 9th term: \(50(0.8)^8 \approx 8.39\) mm (first <10 mm). (b) Sum of GP: _{20} = \frac{50(1 – 0.8^{20})}{1 – 0.8} \approx 247 \text{ mm} ] (c) Sum to infinity: _\infty = \frac{50}{1 – 0.8} = 250 \text{ mm} ]
8. [Maximum mark: 8]
For \(f(x) = 2 – \frac{3}{4x – p}\) (\(x > \frac{p}{4}\)):
(a) Find \(f'(x)\) and determine if \(f\) is increasing/decreasing.
(b) Find \(f^{-1}(x)\).
(c) Find \(p\) if \(f^{-1}(x) \equiv f(x)\).
▶️Answer/Explanation
(a)
1. Derivative: \[ f'(x) = \frac{12}{(4x – p)^2} > 0 \quad \Rightarrow \quad \text{Increasing} \] (b) Let y = 2 − 3 4 x − p y=2− 4x−p 3 . Solve for x x: = \frac{p}{4} – \frac{3}{4(y – 2)} \quad \Rightarrow \quad f^{-1}(x) = \frac{p}{4} – \frac{3}{4(x – 2)} ] (c) Set f − 1 ( x ) = f ( x ) f −1 (x)=f(x): frac{p}{4} = 2 \quad \Rightarrow \quad p = 8 ]
9. [Maximum mark: 8]
For \(f(x) = x^2 – 4x + 9\) and \(g(x) = 2x^2 + 4x + 12\):
(a) Write \(f(x)\) in completed square form.
(b) Write \(g(x)\) as \(2[(x + c)^2 + d]\).
(c) Express \(g(x)\) as \(kf(x + h)\).
(d) Describe transformations from \(f(x)\) to \(g(x)\).
▶️Answer/Explanation
(a) \[ f(x) = (x – 2)^2 + 5 \] (b) g ( x ) = 2 [ ( x + 1 ) 2 + 5 ] g(x)=2[(x+1) 2 +5] (c) g ( x ) = 2 f ( x + 3 ) g(x)=2f(x+3) (d) Horizontal translation 3 units left. Vertical stretch by factor 2.
10. [Maximum mark: 10]
Curves \(y = 2x^{\frac{1}{2}} + 1\) and \(y = \frac{1}{2}x^2 – x + 1\) intersect at \(A(0, 1)\) and \(B(4, 5)\).
(a) Find the area between the curves.
(b) Find the acute angle between tangents at \(B\).
▶️Answer/Explanation
(a)
1. Integrate difference from \(0\) to \(4\): \[ \int_0^4 \left(2x^{\frac{1}{2}} + 1 – \left(\frac{1}{2}x^2 – x + 1\right)\right) dx \]
2. Result: \[ \left[\frac{4}{3}x^{\frac{3}{2}} – \frac{1}{6}x^3 + \frac{1}{2}x^2\right]_0^4 = 8 \text{ units}^2 \] (b) Gradients at B B: _1 = \frac{1}{\sqrt{4}} = \frac{1}{2}, \quad m_2 = 4 – 1 = 3 ] 2. Angle: tan − 1 ( 3 ) − tan − 1 ( 1 2 ) ≈ 4 5 ∘ tan −1 (3)−tan −1 ( 2 1 )≈45 ∘
11. [Maximum mark: 10]
The circle \(x^2 + y^2 = 20\) has tangents from \(A(0, 10)\) touching at \(B\) and \(C\).
(a) Find slopes \(m\) of tangents.
(b) Find coordinates of \(B\) and \(C\).
(c) Find angle \(BDC\) where \(D\) is on the positive x-axis.
▶️Answer/Explanation
(a)
1. Substitute \(y = mx + 10\) into circle: \[ x^2(1 + m^2) + 20mx + 80 = 0 \]
2. Discriminant = 0: \[ 400m^2 – 320(1 + m^2) = 0 \quad \Rightarrow \quad m = \pm 2 \] (b) For m = 2 m=2: Solve 5 x 2 + 40 x + 80 = 0 5x 2 +40x+80=0 → x = − 4 x=−4, y = 2 y=2. Points: B ( − 4 , 2 ) B(−4,2), C ( 4 , 2 ) C(4,2). (c) Use cosine rule in triangle B D C BDC: cos \theta = \frac{\sqrt{5}}{5} \quad \Rightarrow \quad \theta \approx 63.4^\circ ]