1. [Maximum mark: 4]
Solve the inequality \( |2x – 1| < 3|x + 1| \).
▶️Answer/Explanation
Solution:
State or imply non-modular inequality \( (2x-1)^2 < 3^2 (x+1)^2 \), or corresponding quadratic equation.
Form and solve a 3-term quadratic in \( x \).
Obtain critical values \( x = -4 \) and \( x = -\frac{2}{5} \).
State final answer \( x < -4 \), \( x > -\frac{2}{5} \).
2. [Maximum mark: 4]
On a sketch of an Argand diagram, shade the region whose points represent complex numbers \( z \) satisfying the inequalities \( |z+1-\mathrm{i}| \leqslant 1 \) and \( \arg(z-1) \leqslant \frac{3}{4}\pi \).
▶️Answer/Explanation
Solution:
Show a circle with centre \(-1 + i\).
Show a circle with radius 1 and centre not at the origin (or relevant part thereof).
Show correct half line from 1 (or relevant part thereof).
Shade the correct region on a correct diagram.
3. [Maximum mark: 5]
The variables \( x \) and \( y \) satisfy the equation \( x = A(3^{-y}) \), where \( A \) is a constant.
(a) Explain why the graph of \( y \) against \(\ln x\) is a straight line and state the exact value of the gradient of the line.
It is given that the line intersects the \( y \)-axis at the point where \( y = 1.3 \).
(b) Calculate the value of \( A \), giving your answer correct to 2 decimal places.
▶️Answer/Explanation
(a) State or imply \( \ln x = \ln A – y \ln 3 \).
State that the graph of \( y \) against \( \ln x \) has an equation that is linear in \( y \) and \( \ln x \), or has an equation of the standard form \( y = mx + c \) and is thus a straight line.
State that the gradient is \( -\frac{1}{\ln 3} \).
(b) Substitute \( \ln x = 0, y = 1.3 \) and use correct method to solve for \( A \).
Obtain answer \( A = 4.17 \) only.
4. [Maximum mark: 5]
Using integration by parts, find the exact value of \( \int_{0}^{2} \tan^{-1}\left(\frac{1}{2}x\right) \, dx \).
▶️Answer/Explanation
Solution:
Commence integration and reach \( x \tan^{-1}\left(\frac{1}{2}x\right) – \int \frac{x}{4 + x^2} \, dx \).
Obtain \( x \tan^{-1}\left(\frac{1}{2}x\right) – \frac{1}{2} \ln(4 + x^2) \).
Substitute limits correctly in an expression of the form \( px \tan^{-1}x + q \ln(c + x^2) \).
Obtain final answer \( \frac{1}{2}\pi – \ln 2 \).
5. [Maximum mark: 5]
The complex number \( u \) is given by \( u = 10 – 4\sqrt{6}i \). Find the two square roots of \( u \), giving your answers in the form \( a + ib \), where \( a \) and \( b \) are real and exact.
▶️Answer/Explanation
Solution:
Square \( a + ib \), use \( i^2 = -1 \) and equate real and imaginary parts to 10 and \( -4\sqrt{6} \) respectively.
Obtain \( a^2 – b^2 = 10 \) and \( 2ab = -4\sqrt{6} \).
Eliminate one unknown and find an equation in the other.
Obtain \( a^4 – 10a^2 – 24 = 0 \), or equivalent.
Obtain final answers \( \pm \left( 2\sqrt{3} – \sqrt{2}i \right) \), or exact equivalents.
6. [Maximum mark: 7]
(a) Prove that \( \csc 2\theta – \cot 2\theta \equiv \tan \theta \).
(b) Hence show that \( \int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} (\csc 2\theta – \cot 2\theta) \, d\theta = \frac{1}{2} \ln 2 \).
▶️Answer/Explanation
(a) Express the LHS in terms of \( \cos 2\theta \) and \( \sin 2\theta \).
Use correct double angle formulae to express the LHS in terms of \( \cos \theta \) and \( \sin \theta \).
Obtain \( \tan \theta \) from correct working.
(b) State integral of the form \( \pm \ln \cos \theta \) or \( \pm \ln \sec \theta \).
Use correct limits correctly and insert exact values for the trigonometric ratios.
Obtain a correct expression, e.g., \( -\ln \frac{1}{2} + \ln \frac{1}{\sqrt{2}} \).
Obtain \( \frac{1}{2} \ln 2 \) from correct working.
7. [Maximum mark: 7]
A curve is such that the gradient at a general point with coordinates \( (x, y) \) is proportional to \( \frac{y}{\sqrt{x + 1}} \). The curve passes through the points with coordinates \( (0, 1) \) and \( (3, e) \). By setting up and solving a differential equation, find the equation of the curve, expressing \( y \) in terms of \( x \).
▶️Answer/Explanation
Solution:
State equation \( \frac{dy}{dx} = k \cdot \frac{y}{\sqrt{x + 1}} \).
Separate variables correctly and integrate at least one side.
Obtain \( \ln y \) and \( 2\sqrt{x + 1} \) terms.
Use \( (0, 1) \) and \( (3, e) \) to determine \( k \) and/or a constant of integration.
Obtain \( k = \frac{1}{2} \) and \( c = -1 \).
Obtain final answer \( y = \exp\left(\sqrt{x + 1} – 1\right) \).
8. [Maximum mark: 8]
The equation of a curve is \( y = e^{-5x} \tan 2x \) for \( -\frac{1}{2}\pi < x < \frac{1}{2}\pi \).
Find the \( x \)-coordinates of the stationary points of the curve. Give your answers correct to 3 decimal places where appropriate.
▶️Answer/Explanation
Solution:
Use correct product (or quotient) rule to find \( \frac{dy}{dx} \).
Obtain \( \frac{dy}{dx} = -5e^{-5x} \tan^2 x + 2e^{-5x} \tan x \sec^2 x \).
Equate derivative to zero, use \( \sec^2 x = 1 + \tan^2 x \), and obtain an equation in \( \tan x \).
Obtain \( 2\tan^2 x – 5\tan x + 2 = 0 \).
State answer \( x = 0 \).
Solve the quadratic in \( \tan x \) and obtain values of \( x \).
Obtain answers, e.g., \( 0.464 \) and \( 1.107 \), and no others in the given interval.
9. [Maximum mark: 10]
Let \( f(x) = \frac{14 – 3x + 2x^2}{(2 + x)(3 + x^2)} \).
(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) State or imply the form \( \frac{A}{2 + x} + \frac{B + Cx}{3 + x^2} \).
Use a correct method for finding constants \( A, B, \) and \( C \).
Obtain \( A = 4 \), \( B = 1 \), and \( C = -2 \).
(b) Use correct method to find the first two terms of the expansion of each partial fraction.
Obtain correct unsimplified expansions up to \( x^2 \).
Multiply out by \( B + Cx \) and combine terms.
Obtain final answer \( \frac{7}{3} – \frac{5}{3}x + \frac{7}{18}x^2 \).
10. [Maximum mark: 10]
The diagram shows a trapezium \( ABCD \) in which \( AD = BC = r \) and \( AB = 2r \). The acute angles \( BAD \) and \( ABC \) are both equal to \( x \) radians. Circular arcs of radius \( r \) with centres \( A \) and \( B \) meet at \( M \), the midpoint of \( AB \).
(a) Given that the sum of the areas of the shaded sectors is 90% of the area of the trapezium, show that \( x \) satisfies the equation \( x = 0.9(2 – \cos x) \sin x \).
(b) Verify by calculation that \( x \) lies between 0.5 and 0.7.
(c) Show that if a sequence of values in the interval \( 0 < x < \frac{1}{2}\pi \) given by the iterative formula \( x_{n+1} = \cos^{-1}\left(2 – \frac{x_n}{0.9 \sin x_n}\right) \) converges, then it converges to the root of the equation in part (a).
(d) Use this iterative formula to determine \( x \) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
▶️Answer/Explanation
(a) State or imply \( CD = 2r – 2r \cos x \).
Form an equation using areas of sectors and trapezium.
Obtain \( x = 0.9(2 – \cos x) \sin x \).
(b) Calculate values of the expression at \( x = 0.5 \) and \( x = 0.7 \).
Complete the argument with correct values.
(c) $\text{State a suitable equation, e.g. } \cos x = \left( 2 – \frac{x}{0.9 \sin x} \right)$
$\text{Rearrange this as } x = 0.9 \sin x (2 – \cos x)$
(d) Use the iterative process correctly at least once.
Obtain answer \( 0.62 \).
Show sufficient iterations to justify accuracy.
11. [Maximum mark: 10]
With respect to the origin \( O \), the points \( A \) and \( B \) have position vectors given by \( \overrightarrow{OA} = 2\mathbf{i} – \mathbf{j} \) and \( \overrightarrow{OB} = \mathbf{j} – 2\mathbf{k} \).
(a) Show that \( OA = OB \) and use a scalar product to calculate angle \( AOB \) in degrees.
(b) The midpoint of \( AB \) is \( M \). The point \( P \) on the line through \( O \) and \( M \) is such that \( PA : OA = \sqrt{7} : 1 \). Find the possible position vectors of \( P \).
▶️Answer/Explanation
(a) Show that \( OA = OB = \sqrt{5} \).
Evaluate the scalar product of the position vectors.
Divide by the product of the moduli and find the inverse cosine.
Obtain answer \( 101.5^\circ \).
(b) State or imply \( M \) has position vector \( \mathbf{i} – \mathbf{k} \).
Express \( AP = \sqrt{7} OA \) as an equation in \( \lambda \).
Solve the quadratic equation for \( \lambda \).
Obtain answers \( 5\mathbf{i} – 5\mathbf{k} \) and \( -3\mathbf{i} + 3\mathbf{k} \).