1. [Maximum mark: 4]
Find, in terms of \( a \), the set of values of \( x \) satisfying the inequality \[ 2|3x + a| < |2x + 3a|, \] where \( a \) is a positive constant.
▶️Answer/Explanation
Step 1: Square both sides to eliminate the absolute values: \[ 2^2 (3x + a)^2 < (2x + 3a)^2 \] \[ 4(9x^2 + 6ax + a^2) < 4x^2 + 12ax + 9a^2 \] \[ 36x^2 + 24ax + 4a^2 < 4x^2 + 12ax + 9a^2 \] Step 2: Rearrange to form a quadratic inequality: \[ 32x^2 + 12ax – 5a^2 < 0 \] Step 3: Solve the quadratic equation \( 32x^2 + 12ax – 5a^2 = 0 \): Using the quadratic formula: \[ x = \frac{-12a \pm \sqrt{(12a)^2 – 4 \cdot 32 \cdot (-5a^2)}}{2 \cdot 32} \] \[ x = \frac{-12a \pm \sqrt{144a^2 + 640a^2}}{64} \] \[ x = \frac{-12a \pm \sqrt{784a^2}}{64} \] \[ x = \frac{-12a \pm 28a}{64} \] Step 4: Find critical values: \[ x = \frac{16a}{64} = \frac{1}{4}a \quad \text{or} \quad x = \frac{-40a}{64} = -\frac{5}{8}a \] Final Answer: The solution to the inequality is: \[ -\frac{5}{8}a < x < \frac{1}{4}a \]
2. [Maximum mark: 5]
Solve the equation \( \cos(\theta – 60^\circ) = 3 \sin \theta \), for \( 0^\circ \leq \theta \leq 360^\circ \).
▶️Answer/Explanation
Step 1: Expand \( \cos(\theta – 60^\circ) \) using the cosine subtraction formula: \[ \cos \theta \cos 60^\circ + \sin \theta \sin 60^\circ = 3 \sin \theta \] \[ \frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta = 3 \sin \theta \] Step 2: Rearrange to group terms: \[ \frac{1}{2} \cos \theta = 3 \sin \theta – \frac{\sqrt{3}}{2} \sin \theta \] \[ \frac{1}{2} \cos \theta = \left(3 – \frac{\sqrt{3}}{2}\right) \sin \theta \] Step 3: Divide both sides by \( \cos \theta \) to express in terms of \( \tan \theta \): \[ \frac{1}{2} = \left(3 – \frac{\sqrt{3}}{2}\right) \tan \theta \] \[ \tan \theta = \frac{1}{2 \left(3 – \frac{\sqrt{3}}{2}\right)} = \frac{1}{6 – \sqrt{3}} \] Step 4: Rationalize the denominator: \[ \tan \theta = \frac{6 + \sqrt{3}}{33} \approx 0.2343 \] Step 5: Find \( \theta \): \[ \theta = \tan^{-1}(0.2343) \approx 13.2^\circ \] Second solution in the interval: \[ \theta \approx 180^\circ + 13.2^\circ = 193.2^\circ \] Final Answer: The solutions are \( \theta = 13.2^\circ \) and \( \theta = 193.2^\circ \).
3. [Maximum mark: 5]
(a) Show that the equation \( \log_3 (2x + 1) = 1 + 2 \log_3 (x – 1) \) can be written as a quadratic equation in \( x \).
(b) Hence solve the equation \( \log_3 (4y + 1) = 1 + 2 \log_3 (2y – 1) \), giving your answer correct to 2 decimal places.
▶️Answer/Explanation
(a) Step 1: Use logarithm power rule on the right-hand side: \[ 1 + 2 \log_3 (x – 1) = \log_3 3 + \log_3 (x – 1)^2 \] \[ = \log_3 [3(x – 1)^2] \] Step 2: Set the arguments equal: \[ 2x + 1 = 3(x – 1)^2 \] \[ 2x + 1 = 3(x^2 – 2x + 1) \] \[ 2x + 1 = 3x^2 – 6x + 3 \] Step 3: Rearrange to form a quadratic equation: \[ 3x^2 – 8x + 2 = 0 \] (b) Step 1: Substitute \( x = 2y \) into the quadratic equation from part (a): \[ 3(2y)^2 – 8(2y) + 2 = 0 \] \[ 12y^2 – 16y + 2 = 0 \] Step 2: Solve the quadratic equation: \[ y = \frac{16 \pm \sqrt{(-16)^2 – 4 \cdot 12 \cdot 2}}{2 \cdot 12} \] \[ y = \frac{16 \pm \sqrt{256 – 96}}{24} \] \[ y = \frac{16 \pm \sqrt{160}}{24} \] \[ y \approx \frac{16 \pm 12.649}{24} \] Final Answer: The valid solution is \( y \approx 1.19 \) (to 2 decimal places).
4. [Maximum mark: 7]
The curve \( y = e^{-4x} \tan x \) has two stationary points in the interval \( 0 \leq x < \frac{1}{2}\pi \).
(a) Obtain an expression for \( \frac{dy}{dx} \) and show it can be written in the form \( \sec^2 x (a + b \sin 2x) e^{-4x} \), where \( a \) and \( b \) are constants.
(b) Hence find the exact \( x \)-coordinates of the two stationary points.
▶️Answer/Explanation
(a) Step 1: Differentiate using the product rule: \[ \frac{dy}{dx} = -4e^{-4x} \tan x + e^{-4x} \sec^2 x \] \[ = e^{-4x} (\sec^2 x – 4 \tan x) \] Step 2: Express \( \tan x \) in terms of \( \sin \) and \( \cos \): \[ \sec^2 x – 4 \tan x = \sec^2 x – 4 \frac{\sin x}{\cos x} \] \[ = \sec^2 x (1 – 4 \sin x \cos x) \] \[ = \sec^2 x (1 – 2 \sin 2x) \] Final Form: \[ \frac{dy}{dx} = \sec^2 x (1 – 2 \sin 2x) e^{-4x} \] (b) Step 1: Set \( \frac{dy}{dx} = 0 \): \[ \sec^2 x (1 – 2 \sin 2x) e^{-4x} = 0 \] Since \( \sec^2 x \neq 0 \) and \( e^{-4x} \neq 0 \), we have: \[ 1 – 2 \sin 2x = 0 \] \[ \sin 2x = \frac{1}{2} \] Step 2: Solve for \( x \) in the given interval: \[ 2x = \frac{\pi}{6} \quad \text{or} \quad 2x = \frac{5\pi}{6} \] \[ x = \frac{\pi}{12} \quad \text{or} \quad x = \frac{5\pi}{12} \] Final Answer: The exact \( x \)-coordinates are \( \frac{\pi}{12} \) and \( \frac{5\pi}{12} \).
5. [Maximum mark: 8]
The complex number \( 3 – i \) is denoted by \( u \).
(a) Show, on an Argand diagram with origin \( O \), the points \( A \), \( B \) and \( C \) representing the complex numbers \( u \), \( u^* \) and \( u^* – u \) respectively. State the type of quadrilateral formed by the points \( O \), \( A \), \( B \) and \( C \).
(b) Express \( \frac{u^*}{u} \) in the form \( x + iy \), where \( x \) and \( y \) are real.
(c) By considering the argument of \( \frac{u^*}{u} \), or otherwise, prove that \( \tan^{-1}\left(\frac{3}{4}\right) = 2 \tan^{-1}\left(\frac{1}{3}\right) \).
▶️Answer/Explanation
(a) Step 1: Plot the points: – \( A \) represents \( u = 3 – i \). – \( B \) represents \( u^* = 3 + i \). – \( C \) represents \( u^* – u = 2i \). Step 2: The quadrilateral \( OABC \) is a parallelogram. (b) Step 1: Compute \( \frac{u^*}{u} \): \[ \frac{u^*}{u} = \frac{3 + i}{3 – i} \] Step 2: Multiply numerator and denominator by the conjugate of the denominator: \[ \frac{(3 + i)(3 + i)}{(3 – i)(3 + i)} = \frac{9 + 6i + i^2}{9 – i^2} \] \[ = \frac{9 + 6i – 1}{9 + 1} = \frac{8 + 6i}{10} \] \[ = \frac{4}{5} + \frac{3}{5}i \] (c) Step 1: Find the argument of \( \frac{u^*}{u} \): \[ \arg\left(\frac{u^*}{u}\right) = \tan^{-1}\left(\frac{\frac{3}{5}}{\frac{4}{5}}\right) = \tan^{-1}\left(\frac{3}{4}\right) \] Step 2: Find the argument of \( u^* \) and \( u \): \[ \arg(u^*) = \tan^{-1}\left(\frac{1}{3}\right) \] \[ \arg(u) = -\tan^{-1}\left(\frac{1}{3}\right) \] Step 3: Use the property of arguments: \[ \arg\left(\frac{u^*}{u}\right) = \arg(u^*) – \arg(u) = 2 \tan^{-1}\left(\frac{1}{3}\right) \] Final Result: \[ \tan^{-1}\left(\frac{3}{4}\right) = 2 \tan^{-1}\left(\frac{1}{3}\right) \]
6. [Maximum mark: 8]
The parametric equations of a curve are \( x = \frac{1}{\cos t} \), \( y = \ln \tan t \), where \( 0 < t < \frac{1}{2}\pi \).
(a) Show that \( \frac{dy}{dx} = \frac{\cos t}{\sin^2 t} \).
(b) Find the equation of the tangent to the curve at the point where \( y = 0 \).
▶️Answer/Explanation
(a) Step 1: Find \( \frac{dx}{dt} \): \[ x = \sec t \Rightarrow \frac{dx}{dt} = \sec t \tan t \] Step 2: Find \( \frac{dy}{dt} \): \[ y = \ln \tan t \Rightarrow \frac{dy}{dt} = \frac{\sec^2 t}{\tan t} = \frac{1}{\sin t \cos t} \] Step 3: Compute \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{1}{\sin t \cos t}}{\sec t \tan t} = \frac{\cos t}{\sin^2 t} \] (b) Step 1: Find \( t \) when \( y = 0 \): \[ \ln \tan t = 0 \Rightarrow \tan t = 1 \Rightarrow t = \frac{\pi}{4} \] Step 2: Find the gradient at \( t = \frac{\pi}{4} \): \[ \frac{dy}{dx} = \frac{\cos \frac{\pi}{4}}{\sin^2 \frac{\pi}{4}} = \frac{\frac{\sqrt{2}}{2}}{(\frac{\sqrt{2}}{2})^2} = \sqrt{2} \] Step 3: Find the point and equation of tangent: \[ x = \sec \frac{\pi}{4} = \sqrt{2} \] \[ y = 0 \] \[ y – 0 = \sqrt{2}(x – \sqrt{2}) \] \[ y = \sqrt{2}x – 2 \] Final Answer: The tangent equation is \( y = \sqrt{2}x – 2 \).
7. [Maximum mark: 10]
Let \( f(x) = \frac{5x^2 + 8x – 3}{(x – 2)(2x^2 + 3)} \).
(a) Express \( f(x) \) in partial fractions.
(b) Hence obtain the expansion of \( f(x) \) in ascending powers of \( x \), up to and including the term in \( x^2 \).
▶️Answer/Explanation
(a) Step 1: Set up partial fractions: \[ \frac{5x^2 + 8x – 3}{(x – 2)(2x^2 + 3)} = \frac{A}{x – 2} + \frac{Bx + C}{2x^2 + 3} \] Step 2: Multiply through and equate coefficients: \[ 5x^2 + 8x – 3 = A(2x^2 + 3) + (Bx + C)(x – 2) \] Solving gives \( A = 3 \), \( B = -1 \), \( C = 6 \). Final Form: \[ \frac{3}{x – 2} + \frac{-x + 6}{2x^2 + 3} \] (b) Step 1: Expand each partial fraction: \[ \frac{3}{x – 2} = -\frac{3}{2} \left(1 – \frac{x}{2}\right)^{-1} = -\frac{3}{2} \left(1 + \frac{x}{2} + \frac{x^2}{4}\right) \] \[ \frac{-x + 6}{2x^2 + 3} = \frac{-x + 6}{3} \left(1 + \frac{2x^2}{3}\right)^{-1} \approx \frac{-x + 6}{3} \left(1 – \frac{2x^2}{3}\right) \] Step 2: Combine expansions: \[ f(x) \approx -\frac{3}{2} – \frac{3x}{4} – \frac{3x^2}{8} + 2 – \frac{x}{3} – \frac{4x^2}{3} + 2x^2 \] Final Expansion: \[ f(x) \approx \frac{1}{2} – \frac{13}{12}x – \frac{41}{24}x^2 \]
[‘8. [Maximum mark: 9]
At time \( t \) days after the start of observations, the number of insects in a population is \( N \). The variation in the number of insects is modelled by a differential equation of the form \( \frac{dN}{dt} = kN^{3/2} \cos 0.02t \), where \( k \) is a constant and \( N \) is a continuous variable. It is given that when \( t = 0 \), \( N = 100 \).
(a) Solve the differential equation, obtaining a relation between \( N \), \( k \) and \( t \).
(b) Given also that \( N = 625 \) when \( t = 50 \), find the value of \( k \).
(c) Obtain an expression for \( N \) in terms of \( t \), and find the greatest value of \( N \) predicted by this model.
▶️Answer/Explanation
(a) Step 1: Separate variables: \[ \frac{dN}{N^{3/2}} = k \cos 0.02t \, dt \] Step 2: Integrate both sides: \[ -2N^{-1/2} = \frac{k}{0.02} \sin 0.02t + C \] Step 3: Apply initial condition \( t = 0 \), \( N = 100 \): \[ -2(100)^{-1/2} = 50k \sin 0 + C \] \[ -0.2 = C \] Final Relation: \[ -\frac{2}{\sqrt{N}} = 50k \sin 0.02t – 0.2 \] (b) Step 1: Substitute \( t = 50 \), \( N = 625 \): \[ -\frac{2}{25} = 50k \sin 1 – 0.2 \] Solving gives \( k \approx 0.00285 \). (c) Step 1: Rearrange to find \( N \): \[ N = \left( \frac{10}{0.7125 \sin 0.02t – 1} \right)^2 \] Greatest Value: Occurs when \( \sin 0.02t = 1 \): \[ N_{\text{max}} \approx 1210 \]
9. [Maximum mark: 9]
With respect to the origin \( O \), the point \( A \) has position vector given by \( \overrightarrow{OA} = \mathbf{i} + 5\mathbf{j} + 6\mathbf{k} \). The line \( l \) has vector equation \( \mathbf{r} = 4\mathbf{i} + \mathbf{k} + \lambda(-\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) \).
(a) Find in degrees the acute angle between the directions of \( \overrightarrow{OA} \) and \( l \).
(b) Find the position vector of the foot of the perpendicular from \( A \) to \( l \).
(c) Hence find the position vector of the reflection of \( A \) in \( l \).
▶️Answer/Explanation
(a) Step 1: Find direction vectors: \[ \overrightarrow{OA} = (1, 5, 6) \] \[ \text{Line direction} = (-1, 2, 3) \] Step 2: Compute scalar product and moduli: \[ \overrightarrow{OA} \cdot \mathbf{d} = -1 + 10 + 18 = 27 \] \[ |\overrightarrow{OA}| = \sqrt{62}, \quad |\mathbf{d}| = \sqrt{14} \] Step 3: Find angle: \[ \cos \theta = \frac{27}{\sqrt{62} \sqrt{14}} \Rightarrow \theta \approx 23.6^\circ \] (b) Step 1: General point on \( l \): \[ \mathbf{P} = (4 – \lambda, 2\lambda, 1 + 3\lambda) \] Step 2: Find \( \lambda \) for perpendicularity: \[ \overrightarrow{AP} \cdot \mathbf{d} = 0 \] Solving gives \( \lambda = 2 \). Foot of Perpendicular: \[ \mathbf{P} = (2, 4, 7) \] (c) Reflection: \[ \mathbf{A’} = 2\mathbf{P} – \mathbf{A} = (3, 3, 8) \]
10. [Maximum mark: 10]
The constant \( a \) is such that \( \int_1^a x^2 \ln x \, dx = 4 \).
(a) Show that \( a = \left( \frac{35}{3 \ln a – 1} \right)^{1/3} \).
(b) Verify by calculation that \( a \) lies between 2.4 and 2.8.
(c) Use an iterative formula based on the equation in part (a) to determine \( a \) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
▶️Answer/Explanation
(a) Step 1: Integrate by parts: \[ \int x^2 \ln x \, dx = \frac{1}{3}x^3 \ln x – \frac{1}{9}x^3 + C \] Step 2: Evaluate definite integral: \[ \left[ \frac{1}{3}x^3 \ln x – \frac{1}{9}x^3 \right]_1^a = 4 \] Step 3: Rearrange to given form. (b) Verification: At \( a = 2.4 \): LHS ≈ 3.78 < 4 At \( a = 2.8 \): LHS ≈ 4.67 > 4 (c) Iteration: \[ a_{n+1} = \left( \frac{35}{3 \ln a_n – 1} \right)^{1/3} \] Converges to \( a \approx 2.64 \).