1. [Maximum mark: 4]
(a) Sketch the graph of \( y = |2x + 1| \)
(b) Solve the inequality \( 3x + 5 < |2x + 1| \)
▶Answer/Explanation
(a) The graph of \( y = |2x + 1| \) is a V-shaped graph with its vertex at \( x = -\frac{1}{2} \).
(b) Solve the inequality by considering two cases:
Case 1: \( 2x + 1 \geq 0 \) (i.e., \( x \geq -\frac{1}{2} \))
\( 3x + 5 < 2x + 1 \) → \( x < -4 \) (no solution in this interval).
Case 2: \( 2x + 1 < 0 \) (i.e., \( x < -\frac{1}{2} \))
\( 3x + 5 < -(2x + 1) \) → \( 5x < -6 \) → \( x < -\frac{6}{5} \).
Final solution: \( x < -\frac{6}{5} \).
2. [Maximum mark: 4]
On a sketch of an Argand diagram, shade the region whose points represent complex numbers \( z \) satisfying the inequalities \( |z| \leq 3 \), \( \text{Re } z \geq -2 \), and \( \frac{1}{4}\pi \leq \arg z \leq \pi \).
▶Answer/Explanation
The shaded region includes all points within the circle of radius 3 centered at the origin, to the right of the vertical line \( \text{Re } z = -2 \), and within the angular sector from \( \frac{\pi}{4} \) to \( \pi \). 
3. [Maximum mark: 4]
Solve the equation \( 2^{3x-1} = 5(3^{-x}) \). Give your answer in the form \( \frac{\ln a}{\ln b} \), where \( a \) and \( b \) are integers.
▶Answer/Explanation
Take the natural logarithm of both sides:
\( (3x – 1)\ln 2 = \ln 5 – x \ln 3 \).
Rearrange to solve for \( x \):
\( x(3\ln 2 + \ln 3) = \ln 5 + \ln 2 \).
\( x = \frac{\ln 10}{\ln 24} \).
4. [Maximum mark: 5]
Solve the equation \( \tan(x + 45^\circ) = 2 \cot x \) for \( 0^\circ < x < 180^\circ \).
▶Answer/Explanation
Use the tangent addition formula:
\( \frac{\tan x + 1}{1 – \tan x} = \frac{2}{\tan x} \).
Multiply through by \( \tan x(1 – \tan x) \):
\( \tan^2 x + \tan x = 2 – 2\tan x \).
Rearrange to form a quadratic equation:
\( \tan^2 x + 3\tan x – 2 = 0 \).
Solve for \( \tan x \):
\( \tan x = \frac{-3 \pm \sqrt{17}}{2} \).
Solutions: \( x \approx 29.3^\circ \) and \( x \approx 105.7^\circ \).
5. [Maximum mark: 4]
The complex numbers u and w are defined by u = 2e¼πi and w = 3e⅓πi.
(a) Find u²/w, giving your answer in the form reiθ, where r > 0 and -π < θ ≤ π. Give the exact values of r and θ.
(b) State the least positive integer n such that both Im(wn) = 0 and Re(wn) > 0.
▶Answer/Explanation
(a) Finding u²/w:
1. Calculate u²: u² = (2e¼πi)² = 4e½πi
2. Divide by w: u²/w = (4e½πi)/(3e⅓πi) = (4/3)e(½π – ⅓π)i
3. Simplify exponent: = (4/3)e(⅙π)i
Final answer: \(\frac{4}{3}e^{\frac{1}{6}\pi i\)
(r = 4/3, θ = π/6) (b) Finding least positive integer n:
1. Consider wn = 3nenπi/3
2. Im(wn) = 0 when nπ/3 is multiple of π ⇒ n must be multiple of 3
3. Re(wn) > 0 when cos(nπ/3) > 0
4. Smallest such n is 6 (as n=3 gives Re(w³)=-27)
Final answer: 6
6. [Maximum mark: 7]
(a) Prove the identity \( \cos 4\theta + 4 \cos 2\theta + 3 \equiv 8 \cos^4 \theta \).
(b) Hence solve the equation \( \cos 4\theta + 4 \cos 2\theta = 4 \) for \( 0^\circ \leq \theta \leq 180^\circ \).
▶Answer/Explanation
(a) Use double-angle identities:
\( \cos 4\theta = 2\cos^2 2\theta – 1 \).
\( \cos 2\theta = 2\cos^2 \theta – 1 \).
Substitute and simplify to show \( \cos 4\theta + 4\cos 2\theta + 3 = 8\cos^4 \theta \).
(b) Substitute into the equation:
\( 8\cos^4 \theta – 3 = 4 \).
Solve for \( \cos \theta \):
\( \cos \theta = \pm \left(\frac{7}{8}\right)^{1/4} \).
Solutions: \( \theta \approx 14.7^\circ \) and \( \theta \approx 165.3^\circ \).
7. [Maximum mark: 8]
The equation of a curve is \( y = \frac{x}{\cos^2 x} \), for \( 0 \leq x < \frac{\pi}{2} \). At the point where \( x = a \), the tangent to the curve has gradient equal to 12.
(a) Show that \( a = \cos^{-1}\left( \sqrt{\frac{\cos a + 2a \sin a}{12}} \right). \)
(b) Verify by calculation that \( a \) lies between 0.9 and 1.
(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) Showing the relationship for \( a \):
1. Differentiate \( y = x \sec^2 x \) using the product rule:
\( \frac{dy}{dx} = \sec^2 x + 2x \sec^2 x \tan x \)
2. Set gradient = 12:
\( \sec^2 a (1 + 2a \tan a) = 12 \)
3. Convert to cosine form:
\( \frac{\cos a + 2a \sin a}{\cos^3 a} = 12 \)
4. Rearrange:
\( \cos a = \sqrt{\frac{\cos a + 2a \sin a}{12}} \)
5. Final form:
\( a = \cos^{-1}\left( \sqrt{\frac{\cos a + 2a \sin a}{12}} \right) \) (b) Verification that \( a \in (0.9, 1) \):
• \( f(0.9) \approx 0.085 > 0 \)
• \( f(1) \approx -0.036 < 0 \)
∴ Solution exists in \( (0.9, 1) \) (c) Iterative solution:
1. Formula: \( a_{n+1} = \cos^{-1}\left( \sqrt{\frac{\cos a_n + 2a_n \sin a_n}{12}} \right) \)
2. Iterations:
\( 0.9500 \rightarrow 0.9743 \rightarrow 0.9694 \rightarrow 0.9704 \rightarrow 0.9702 \)
3. Final answer: \( a = 0.97 \) (2 d.p.)
8. [Maximum mark: 8]
In a certain chemical reaction the amount, x grams, of a substance is increasing. The differential equation satisfied by x and t, the time in seconds since the reaction began, is:
\(\frac{dx}{dt} = kxe^{-0.1t}\),
where k is a positive constant. It is given that x = 20 at the start of the reaction.
(a) Solve the differential equation, obtaining a relation between x, t and k.
(b) Given that x = 40 when t = 10, find the value of k and find the value approached by x as t becomes large.
▶Answer/Explanation
(a) Solving the differential equation:
1. Separate variables: \(\frac{1}{x}dx = ke^{-0.1t}dt\)
2. Integrate both sides: \(\int \frac{1}{x}dx = \int ke^{-0.1t}dt\)
\(\ln x = -10ke^{-0.1t} + C\)
3. Use initial condition (t=0, x=20): \(\ln 20 = -10k + C\) → \(C = \ln 20 + 10k\)
4. Final relation: \(\ln x = 10k(1 – e^{-0.1t}) + \ln 20\)
or equivalently: \(x = 20e^{10k(1-e^{-0.1t})}\) (b) Finding k and limiting value:
1. Substitute t=10, x=40: \(\ln 40 = 10k(1-e^{-1}) + \ln 20\)
\(\ln 2 = 10k(1-\frac{1}{e})\)
\(k = \frac{\ln 2}{10(1-\frac{1}{e})} ≈ 0.109654\) (exact form preferred)
2. As t→∞, \(e^{-0.1t}\)→0, so: \(x→20e^{10k} ≈ 59.9\) (exact form: \(20e^{\frac{\ln 2}{1-e^{-1}}}\))
9. [Maximum mark: 10]
The diagram shows part of the curve \( y = (3 – x)e^{-\frac{1}{3}x} \) for \( x \geq 0 \), and its minimum point \( M \).
(a) Find the exact coordinates of \( M \).
(b) Find the area of the shaded region bounded by the curve and the axes, giving your answer in terms of \( e \).
▶Answer/Explanation
(a) To find the minimum point \( M \):
1. Differentiate \( y = (3 – x)e^{-\frac{1}{3}x} \) using the product rule:
\( \frac{dy}{dx} = -e^{-\frac{1}{3}x} + (3 – x)(-\frac{1}{3})e^{-\frac{1}{3}x} \).
2. Set derivative equal to zero for stationary points:
\( -e^{-\frac{1}{3}x} – \frac{1}{3}(3 – x)e^{-\frac{1}{3}x} = 0 \).
3. Simplify and solve for \( x \):
\( -1 – \frac{3 – x}{3} = 0 \) → \( x = 6 \).
4. Find \( y \)-coordinate by substituting \( x = 6 \):
\( y = (3 – 6)e^{-2} = -3e^{-2} \).
Exact coordinates of \( M \): \( (6, -3e^{-2}) \). (b) To find the area bounded by the curve and the axes (from \( x = 0 \) to \( x = 3 \)):
1. Integrate \( y = (3 – x)e^{-\frac{1}{3}x} \) with respect to \( x \):
Use integration by parts with \( u = 3 – x \), \( dv = e^{-\frac{1}{3}x}dx \).
2. First integration gives:
\( \int (3 – x)e^{-\frac{1}{3}x} dx = -3(3 – x)e^{-\frac{1}{3}x} – 9e^{-\frac{1}{3}x} + C \).
3. Evaluate definite integral from 0 to 3:
\( \left[ -3(3 – x)e^{-\frac{1}{3}x} – 9e^{-\frac{1}{3}x} \right]_0^3 = 9e^{-1} – 9e^{0} = \frac{9}{e} – 9 \).
4. Take absolute value for area (since curve is below x-axis for \( x > 3 \)):
Area = \( \frac{9}{e} \) (exact form).
10. [Maximum mark: 10]
Let \( f(x) = \frac{2x^2 + 7x + 8}{(1 + x)(2 + x)^2} \).
(a) Express \( f(x) \) in partial fractions. [5]
(b) Hence obtain the expansion of \( f(x) \) in ascending powers of \( x \), up to and including the term in \( x^2 \). [5]
▶Answer/Explanation
(a) Partial fractions:
1. Set up the form:
\( f(x) = \frac{A}{1 + x} + \frac{B}{2 + x} + \frac{C}{(2 + x)^2} \)
2. Multiply through by the denominator:
\( 2x^2 + 7x + 8 = A(2 + x)^2 + B(1 + x)(2 + x) + C(1 + x) \)
3. Solve for coefficients:
– Let \( x = -1 \): \( 2(-1)^2 + 7(-1) + 8 = A(1)^2 \) ⇒ \( A = 3 \)
– Let \( x = -2 \): \( 2(-2)^2 + 7(-2) + 8 = C(-1) \) ⇒ \( C = -2 \)
– Compare \( x^2 \) terms: \( 2 = A + B \) ⇒ \( B = -1 \)
4. Final partial fractions:
\( f(x) = \frac{3}{1 + x} – \frac{1}{2 + x} – \frac{2}{(2 + x)^2} \)
(b) Binomial expansion:
1. Expand each term up to \( x^2 \):
– \( \frac{3}{1 + x} = 3(1 – x + x^2 + \ldots) \)
– \( -\frac{1}{2 + x} = -\frac{1}{2}\left(1 – \frac{x}{2} + \frac{x^2}{4} + \ldots\right) \)
– \( -\frac{2}{(2 + x)^2} = -\frac{1}{2}\left(1 – x + \frac{3x^2}{4} + \ldots\right) \)
2. Combine expansions:
\( = [3 – 3x + 3x^2] + \left[-\frac{1}{2} + \frac{x}{4} – \frac{x^2}{8}\right] + \left[-\frac{1}{2} + x – \frac{3x^2}{8}\right] \)
3. Collect like terms:
Constant: \( 3 – \frac{1}{2} – \frac{1}{2} = 2 \)
\( x \) term: \( -3 + \frac{1}{4} + 1 = -\frac{9}{4} \)
\( x^2 \) term: \( 3 – \frac{1}{8} – \frac{3}{8} = \frac{5}{2} \)
4. Final expansion:
\( f(x) = 2 – \frac{9}{4}x + \frac{5}{2}x^2 + \ldots \)
11. [Maximum mark: 11]
In the diagram, \( OABCD \) is a solid figure where \( OA = OB = 4 \) units and \( OD = 3 \) units. The edge \( OD \) is vertical, \( DC \) is parallel to \( OB \) and \( DC = 1 \) unit. The base \( OAB \) is horizontal and angle \( AOB = 90^\circ \). Unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are parallel to \( OA, OB, OD \) respectively. The midpoint of \( AB \) is \( M \) and the point \( N \) on \( BC \) is such that \( CN = 2NB \).
(a) Express vectors \( \overline{MD} \) and \( \overline{ON} \) in terms of \( \mathbf{i}, \mathbf{j}, \mathbf{k} \). [4]
(b) Calculate the angle in degrees between the directions of \( \overline{MD} \) and \( \overline{ON} \). [3]
(c) Show that the length of the perpendicular from \( M \) to \( ON \) is \( \sqrt{\frac{22}{5}} \). [4]
▶Answer/Explanation
(a) Finding vectors:
1. Coordinates setup:
– Let \( O \) be at the origin \( (0,0,0) \)
– \( OA = 4\mathbf{i}, OB = 4\mathbf{j}, OD = 3\mathbf{k} \)
2. Find \( M \) (midpoint of \( AB \)):
\( M = \frac{OA + OB}{2} = \frac{4\mathbf{i} + 4\mathbf{j}}{2} = 2\mathbf{i} + 2\mathbf{j} \)
3. Find \( D \):
\( D = OD = 3\mathbf{k} \)
4. Vector \( \overline{MD} \):
\( \overline{MD} = D – M = 3\mathbf{k} – (2\mathbf{i} + 2\mathbf{j}) = -2\mathbf{i} – 2\mathbf{j} + 3\mathbf{k} \)
5. Find \( N \) (divides \( BC \) in ratio \( CN:NB = 2:1 \)):
– \( C = OA + DC = 4\mathbf{i} + 1\mathbf{j} \) (since \( DC \parallel OB \) and \( DC=1 \))
– \( B = 4\mathbf{j} \)
– \( N = B + \frac{2}{3}(C-B) = 4\mathbf{j} + \frac{2}{3}(4\mathbf{i} – 3\mathbf{j}) = \frac{8}{3}\mathbf{i} + 2\mathbf{j} \)
6. Vector \( \overline{ON} \):
\( \overline{ON} = \frac{8}{3}\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \)
(b) Calculating angle:
1. Dot product:
\( \overline{MD} \cdot \overline{ON} = (-2)\left(\frac{8}{3}\right) + (-2)(2) + (3)(3) = -\frac{16}{3} – 4 + 9 = -\frac{7}{3} \)
2. Magnitudes:
\( |\overline{MD}| = \sqrt{(-2)^2 + (-2)^2 + 3^2} = \sqrt{17} \)
\( |\overline{ON}| = \sqrt{\left(\frac{8}{3}\right)^2 + 2^2 + 3^2} = \sqrt{\frac{181}{9}} \)
3. Angle calculation:
\( \cos \theta = \frac{-\frac{7}{3}}{\sqrt{17} \cdot \sqrt{\frac{181}{9}}} \approx -0.229 \)
\( \theta \approx 103.3^\circ \)
(c) Perpendicular distance:
1. Parametric point \( P \) on \( ON \): \( P = \lambda\left(\frac{8}{3} \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\right) \)
2. Vector \( MP = P – M = \left(\frac{8\lambda}{3} – 2\right)\mathbf{i} + (2\lambda – 2)\mathbf{j} + 3\lambda\mathbf{k} \)
3. \( MP \) must be perpendicular to \( ON \):
\( MP \cdot \overline{ON} = 0 \) ⇒ \( \lambda = \frac{3}{5} \)
4. Calculate \( MP \) at \( \lambda = \frac{3}{5} \):
\( MP = \left(-\frac{6}{5}\right)\mathbf{i} + \left(-\frac{4}{5}\right)\mathbf{j} + \left(\frac{9}{5}\right)\mathbf{k} \)
5. Distance \( |MP| \):
\( \sqrt{\left(-\frac{6}{5}\right)^2 + \left(-\frac{4}{5}\right)^2 + \left(\frac{9}{5}\right)^2} = \sqrt{\frac{133}{25}} = \sqrt{\frac{22}{5}} \)