1. [Maximum mark: 3]
Find the quotient and remainder when \( 2x^4 + 1 \) is divided by \( x^2 – x + 2 \).
▶️Answer/Explanation
Solution:
Commence division and reach partial quotient of the form \( 2x^2 + kx \).
Obtain quotient \( 2x^2 + 2x – 2 \).
Obtain remainder \( -6x + 5 \).
2. [Maximum mark: 4]
(a) Sketch the graph of \( y = |2x – 3| \).
(b) Solve the inequality \( |2x – 3| < 3x + 2 \).
▶️Answer/Explanation
(a) Show a recognizable sketch graph of \( y = |2x – 3| \).
(b) Find \( x \)-coordinate of intersection with \( y = 3x + 2 \).
Obtain \( x = \frac{1}{5} \).
State final answer \( x > \frac{1}{5} \) only.
3. [Maximum mark: 4]
Solve the equation \( 4^{x-2} = 4^x – 42 \), giving your answer correct to 3 decimal places.
▶️Answer/Explanation
Solution:
Use laws of indices correctly and solve for \( 4^x \).
Obtain correct solution in any form, e.g., \( 4^x = \frac{256}{15} \).
Use a correct method for solving an equation of the form \( 4^x = a \), where \( a > 0 \).
Obtain answer \( 2.047 \).
4. [Maximum mark: 5]
Find the exact value of \( \int_{\pi}^{\frac{1}{3}\pi} x \sin\left(\frac{1}{2}x\right) \, dx \).
▶️Answer/Explanation
Solution:
Commence integration and reach \( \frac{1}{2}x + b \cos\left(\frac{1}{2}x\right) \).
Obtain \( -2x \cos\left(\frac{1}{2}x\right) + 2 \int \cos\left(\frac{1}{2}x\right) \, dx \).
Complete integration obtaining \( -2x \cos\left(\frac{1}{2}x\right) + 4 \sin\left(\frac{1}{2}x\right) \).
Use limits correctly, having integrated twice.
Obtain answer \( 2 + \frac{\sqrt{3}}{3}\pi \), or exact equivalent.
5. [Maximum mark: 5]
Solve the equation \( \sin \theta = 3 \cos 2\theta + 2 \), for \( 0^\circ \leq \theta \leq 360^\circ \).
▶️Answer/Explanation
Solution:
Use double angle formula and obtain an equation in \( \sin \theta \).
Reduce to \( 6\sin^2\theta + \sin\theta – 5 = 0 \), or 3-term equivalent.
Solve a 3-term quadratic in \( \sin\theta \) and calculate \( \theta \).
Obtain answer, e.g., \( 56.4^\circ \).
Obtain second and third answers, e.g., \( 123.6^\circ \) and \( 270^\circ \), and no others in the given interval.
6. [Maximum mark: 7]
(a) By first expanding \( \cos(x – 60^\circ) \), show that the expression \( 2 \cos(x – 60^\circ) + \cos x \) can be written in the form \( R \cos(x – \alpha) \), where \( R > 0 \) and \( 0^\circ < \alpha < 90^\circ \). Give the exact value of \( R \) and the value of \( \alpha \) correct to 2 decimal places.
(b) Hence find the value of \( x \) in the interval \( 0^\circ < x < 360^\circ \) for which \( 2 \cos(x – 60^\circ) + \cos x \) takes its least possible value.
▶️Answer/Explanation
(a) Use \( \cos(A – B) \) formula and obtain an expression in terms of \( \sin x \) and \( \cos x \).
Collect terms and reach \( 2 \cos x + \sqrt{3} \sin x \).
State \( R = \sqrt{7} \).
Use trig formula to find \( \alpha \).
Obtain \( \alpha = 40.89^\circ \).
(b) Use correct method to find \( x \).
Obtain answer \( x = 220.9^\circ \).
7. [Maximum mark: 7]
The equation of a curve is \( \ln(x + y) = x – 2y \).
(a) Show that \( \frac{dy}{dx} = \frac{x + y – 1}{2(x + y) + 1} \).
(b) Find the coordinates of the point on the curve where the tangent is parallel to the \( x \)-axis.
▶️Answer/Explanation
(a) Use chain rule to differentiate LHS.
Obtain \( \frac{1}{x+y} \left( 1 + \frac{dy}{dx} \right) \).
Equate derivative of LHS to \( 1 – 2 \frac{dy}{dx} \) and solve for \( \frac{dy}{dx} \).
Obtain the given answer correctly.
(b) State \( x + y = 1 \).
Obtain or imply \( x – 2y = 0 \).
Obtain coordinates \( x = \frac{2}{3} \) and \( y = \frac{1}{3} \).
8. [Maximum mark: 9]
In the diagram, OABCD is a pyramid with vertex D. The horizontal base OABC is a square of side 4 units. The edge OD is vertical and OD = 4 units. The unit vectors i, j and k are parallel to OA, OC and OD respectively. The midpoint of AB is M and the point N on CD is such that DN = 3NC.
(a) Find a vector equation for the line through M and N.
(b) Show that the length of the perpendicular from O to MN is \( \frac{1}{3}\sqrt{82} \).
▶️Answer/Explanation
(a) State \( \overrightarrow{OM} = 4i + 2j \).
Use a correct method to find \( \overrightarrow{ON} \).
Obtain answer \( 3j + k \).
Use a correct method to find a line equation for MN.
Obtain answer \( r = 3j + k + \lambda(4i – j – k) \), or equivalent.
(b) Taking a general point P on MN, form an equation in \( \lambda \) by equating a relevant scalar product to zero.
Obtain \( \lambda = \frac{2}{9} \).
Use correct method to find OP.
Obtain the given answer correctly.
9. [Maximum mark: 10]
Let \( f(x) = \frac{1}{\sqrt{(9 – x)x}} \).
(a) Find the x-coordinate of the stationary point of the curve with equation y = f(x).
(b) Using the substitution \( u = \sqrt{x} \), show that \( \int_{0}^{4} f(x)dx = \frac{1}{3} \ln 5 \).
▶️Answer/Explanation
(a) Use quotient or product rule.
Obtain correct derivative in any form.
Equate derivative to zero and solve for x.
Obtain answer \( x = 3 \).
(b) State \( \frac{du}{dx} = \frac{1}{2\sqrt{x}} \), or equivalent.
Substitute and obtain integrand \( \frac{2}{9 – u^2} \).
Use given formula for the integral or integrate relevant partial fractions.
Obtain integral \( \frac{1}{3} \ln \left( \frac{3 + u}{3 – u} \right) \).
Use limits \( u = 0 \) and \( u = 2 \) correctly.
Obtain the given answer correctly.
10. [Maximum mark: 11]
A large plantation of area 20 km² is becoming infected with a plant disease. At time t years the area infected is x km² and the rate of increase of x is proportional to the ratio of the area infected to the area not yet infected. When t = 0, x = 1 and \( \frac{dx}{dt} = 1 \).
(a) Show that x and t satisfy the differential equation \( \frac{dx}{dt} = \frac{19x}{20 – x} \).
(b) Solve the differential equation and show that when t = 1 the value of x satisfies the equation \( x = e^{0.9 + 0.05x} \).
(c) Use an iterative formula based on the equation in part (b), with an initial value of 2, to determine x correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
(d) Calculate the value of t at which the entire plantation becomes infected.
▶️Answer/Explanation
(a) State or imply equation of the form \( \frac{dx}{dt} = k \frac{x}{20 – x} \).
Obtain \( k = 19 \).
(b) Separate variables and integrate at least one side.
Obtain terms \( 20 \ln x – x \) and \( 19t \), or equivalent.
Evaluate a constant or use \( t = 0 \) and \( x = 1 \) as limits.
Substitute \( t = 1 \) and rearrange the equation in the given form.
(c) Use \( x_{n+1} = e^{0.9 + 0.05x_n} \) correctly at least once.
Obtain final answer \( x = 2.83 \).
Show sufficient iterations to justify 2.83 to 2 d.p.
(d) Set \( x = 20 \) and obtain answer \( t = 2.15 \).
11. [Maximum mark: 10]
The complex number \( -\sqrt{3} + i \) is denoted by u.
(a) Express u in the form \( re^{i\theta} \), where \( r > 0 \) and \( -\pi < \theta \leq \pi \), giving the exact values of r and θ.
(b) Hence show that \( u^6 \) is real and state its value.
(c) (i) On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying the inequalities \( 0 \leq \arg(z – u) \leq \frac{1}{4}\pi \) and \( \text{Re } z \leq 2 \).
(ii) Find the greatest value of \( |z| \) for points in the shaded region. Give your answer correct to 3 significant figures.
▶️Answer/Explanation
(a) State or imply \( r = 2 \).
State or imply \( \theta = \frac{5}{6}\pi \).
(b) Use a correct method for finding the modulus or argument of \( u^6 \).
Show correctly that \( u^6 \) is real and has value -64.
(c)(i) Show half lines from the point representing \( -\sqrt{3} + i \).
Show correct half lines.
Show the line \( x = 2 \) in the first quadrant.
Shade the correct region.
(c)(ii) Carry out a correct method to find the greatest value of \( |z| \).
Obtain answer 5.14.