1. [Maximum mark: 4]
Solve the equation \( \ln(2x – 1) = 2 \ln(x + 1) – \ln(x) \). Give your answer correct to 3 decimal places.
▶Answer/Explanation
To solve the equation, we start by using the properties of logarithms:
1. Rewrite the equation:
\( \ln(2x – 1) = \ln((x + 1)^2) – \ln(x) \)
\( \ln(2x – 1) = \ln\left(\frac{(x + 1)^2}{x}\right) \)
2. Exponentiate both sides to eliminate the logarithm:
\( 2x – 1 = \frac{(x + 1)^2}{x} \)
3. Multiply both sides by \( x \) to clear the fraction:
\( x(2x – 1) = (x + 1)^2 \)
4. Expand both sides:
\( 2x^2 – x = x^2 + 2x + 1 \)
5. Rearrange to form a quadratic equation:
\( 2x^2 – x – x^2 – 2x – 1 = 0 \)
\( x^2 – 3x – 1 = 0 \)
6. Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \):
\( x = \frac{3 \pm \sqrt{(-3)^2 – 4(1)(-1)}}{2(1)} \)
\( x = \frac{3 \pm \sqrt{9 + 4}}{2} \)
\( x = \frac{3 \pm \sqrt{13}}{2} \)
7. Calculate the roots:
\( x_1 = \frac{3 + \sqrt{13}}{2} \approx 3.302 \)
\( x_2 = \frac{3 – \sqrt{13}}{2} \approx -0.302 \) (not valid since \( x \) must be positive)
Thus, the solution is \( x \approx 3.302 \) (to 3 decimal places).
2. [Maximum mark: 5]
Expand \(\sqrt{\frac{1 + 2x}{1 – 2x}}\) in ascending powers of \(x\), up to and including the term in \(x^2\), simplifying the coefficients.
▶Answer/Explanation
Solution:
State a correct unsimplified term in \(x\) or \(x^2\) of the expansion of either \((1+2x)^{\frac{1}{2}}\) or \((1-2x)^{\frac{1}{2}}\).
State correct unsimplified expansion of \((1+2x)^{\frac{1}{2}}\) up to the term in \(x^2\).
State correct unsimplified expansion of \((1-2x)^{\frac{1}{2}}\) up to the term in \(x^2\).
Obtain sufficient terms of the product of the expansions.
Obtain final answer \(1 + 2x + 2x^2\).
3. [Maximum mark: 5]
Find the exact value of \(\int_{0}^{\frac{1}{4}\pi} x\sec^{2}x\,dx\).
▶Answer/Explanation
Solution:
Commence integration by parts and reach \(x \tan x \pm \int \tan x \,dx\).
Use a correct method to integrate \(\tan x\).
Obtain integral \(x \tan x – \ln \sec x\), or equivalent.
Use limits correctly, having integrated twice.
Obtain answer \(\frac{1}{4}\pi – \frac{1}{2}\ln 2\), or exact equivalent.
4. [Maximum mark: 5]
The parametric equations of a curve are
\(x = 2t – \tan t\),
\(y = \ln(\sin 2t)\),
for \(0 < t < \frac{1}{2}\pi\).
Show that \(\frac{dy}{dx} = \cot t\).
▶Answer/Explanation
Solution:
State or imply \(\frac{dx}{dt} = 2 – \sec^2 t\) or \(\frac{dy}{dt} = 2\frac{\cos 2t}{\sin 2t}\).
Use \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\).
Obtain correct answer in any form.
Use double angle formula to express derivative in terms of \(\cos x\) and \(\sin x\).
Obtain the given answer correctly.
5. [Maximum mark: 6]
(a) On a sketch of an Argand diagram, shade the region whose points represent complex numbers \(z\) satisfying the inequalities \(|z + 2| \leq 2\) and \(\text{Im } z \geq 1\). [4]
(b) Find the greatest value of \(\arg z\) for points in the shaded region. [2]
▶Answer/Explanation
Solution:
(a)
– Show a circle with center \(-2\)
– Show a circle with radius 2 and center not the origin
– Show the line \(y = 1\)
– Shade the correct region
(b)
Identify the correct point and carry out a correct method to find the argument.
Obtain answer \(\frac{11}{12}\pi\) (2.88 radians or 165°).
6. [Maximum mark: 6]
Solve the quadratic equation \((1 – 3i)z^2 – (2 + i)z + i = 0\), giving your answers in the form \(x + iy\), where \(x\) and \(y\) are real.
▶Answer/Explanation
Solution:
Use quadratic formula to solve for \(z\).
Use \(i^2 = -1\) throughout.
Obtain correct answer in any form.
Multiply numerator and denominator by \((1 + 3i)\), or equivalent.
Obtain final answer, e.g. \(-\frac{1}{2} + \frac{1}{2}i\).
Obtain second final answer, e.g. \(\frac{2}{5} + \frac{1}{5}i\).
7. [Maximum mark: 8]
(a) Show that the equation \(\sqrt{5}\sec x + \tan x = 4\) can be expressed as \(R\cos(x + \alpha) = \sqrt{5}\), where \(R > 0\) and \(0° < \alpha < 90°\). Give the exact value of \(R\) and the value of \(\alpha\) correct to 2 decimal places. [4]
(b) Hence solve the equation \(\sqrt{5}\sec 2x + \tan 2x = 4\), for \(0° < x < 180°\). [4]
▶Answer/Explanation
Solution:
(a)
Rearrange and obtain \(4\cos x – \sin x = \sqrt{5}\).
State \(R = \sqrt{17}\).
Use trig formulae to find \(\alpha\).
Obtain \(\alpha = 14.04°\).
(b)
Evaluate \(\cos^{-1}\left(\frac{\sqrt{5}}{\sqrt{17}}\right)\).
Carry out a correct method to find a value of \(x\) in the given interval.
Obtain answer, e.g. \(21.6°\).
Obtain a second answer, e.g. \(144.4°\) and no other in the interval.
8. [Maximum mark: 8]
The curve with equation \(y = \frac{x^3}{e^x – 1}\) has a stationary point at \(x = p\), where \(p > 0\).
(a) Show that \(p = 3(1 – e^{-p})\). [3]
(b) Verify by calculation that \(p\) lies between 2.5 and 3. [2]
(c) Use an iterative formula based on the equation in part (a) to determine \(p\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
▶Answer/Explanation
Solution:
(a)
Use quotient or product rule.
Obtain correct derivative in any form.
Equate derivative at \(x = p\) to zero and obtain the given equation.
(b)
Evaluate a relevant expression or pair of relevant expressions at \(p = 2.5\) and \(p = 3\).
Complete the argument with correct calculated values.
(c)
Use the iterative formula \(p_{n+1} = 3(1 – e^{-p_n})\) correctly at least once.
Obtain final answer \(p = 2.82\).
Show sufficient iterations to 4 d.p. to justify 2.82 to 2 d.p., or show there is a sign change in the interval (2.815, 2.825).
9. [Maximum mark: 9]
With respect to the origin O, the position vectors of the points A, B and C are given by \(\overrightarrow{OA} = \begin{pmatrix} 0 \\ 5 \\ 2 \end{pmatrix}\), \(\overrightarrow{OB} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\) and \(\overrightarrow{OC} = \begin{pmatrix} 4 \\ -3 \\ -2 \end{pmatrix}\). The midpoint of AC is M and the point N lies on BC, between B and C, and is such that BN = 2NC.
(a) Find the position vectors of M and N. [3]
(b) Find a vector equation for the line through M and N. [2]
(c) Find the position vector of the point Q where the line through M and N intersects the line through A and B. [4]
▶Answer/Explanation
Solution:
(a)
State \(\overrightarrow{OM} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}\).
Use a correct method to find \(\overrightarrow{ON}\).
Obtain answer \(\overrightarrow{ON} = \begin{pmatrix} 3 \\ -2 \\ -1 \end{pmatrix}\).
(b)
Carry out a correct method to form a vector equation for MN.
Obtain a correct equation in any form, e.g. \(\mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -3 \\ -1 \end{pmatrix}\).
(c)
1. Equation for line AB:
– Direction vector: \(\overrightarrow{AB} = \overrightarrow{OB} – \overrightarrow{OA} = \begin{pmatrix} 1 \\ -5 \\ -1 \end{pmatrix}\)
– Parametric form: \(\mathbf{r} = \begin{pmatrix} 0 \\ 5 \\ 2 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -5 \\ -1 \end{pmatrix}\)
2. Equation for line MN:
– Direction vector: \(\overrightarrow{MN} = \overrightarrow{ON} – \overrightarrow{OM} = \begin{pmatrix} 1 \\ -3 \\ -1 \end{pmatrix}\)
– Parametric form: \(\mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -3 \\ -1 \end{pmatrix}\)
3. Intersection condition:
Set the parametric equations equal:
\[ \begin{cases} 0 + \mu = 2 + \lambda \quad (x) \\ 5 – 5\mu = 1 – 3\lambda \quad (y) \\ 2 – \mu = 0 – \lambda \quad (z) \end{cases} \]
Solve the system:
– From \(z\): \(\mu = 2 + \lambda\)
– Substitute into \(x\): \(\mu = 2 + \lambda\) (consistent)
– Substitute into \(y\): \(5 – 5(2 + \lambda) = 1 – 3\lambda \Rightarrow -5 – 5\lambda = 1 – 3\lambda \Rightarrow \lambda = -3\)
– Then \(\mu = 2 + (-3) = -1\)
4. Find Q:
Substitute \(\mu = -1\) into AB’s equation:
\[ \overrightarrow{OQ} = \begin{pmatrix} 0 \\ 5 \\ 2 \end{pmatrix} + (-1)\begin{pmatrix} 1 \\ -5 \\ -1 \end{pmatrix} = \begin{pmatrix} -1 \\ 10 \\ 3 \end{pmatrix} \]
Final answer: \(\boxed{\begin{pmatrix} -1 \\ 10 \\ 3 \end{pmatrix}}\)
10. [Maximum mark: 9]
A gardener is filling an ornamental pool with water, using a hose that delivers 30 litres of water per minute. Initially the pool is empty. At time \(t\) minutes after filling begins the volume of water in the pool is \(V\) litres. The pool has a small leak and loses water at a rate of 0.01\(V\) litres per minute.
The differential equation satisfied by \(V\) and \(t\) is of the form \(\frac{dV}{dt} = a – bV\).
(a) Write down the values of the constants \(a\) and \(b\). [1]
(b) Solve the differential equation and find the value of \(t\) when \(V = 1000\). [6]
(c) Obtain an expression for \(V\) in terms of \(t\) and hence state what happens to \(V\) as \(t\) becomes large. [2]
▶Answer/Explanation
Solution:
(a)
\(a = 30\) and \(b = 0.01\).
(b)
Separate variables and integrate one side.
Obtain terms \(-100\ln(30-0.01V)\) and \(t\), or equivalent.
Evaluate a constant, or use \(t = 0\), \(V = 0\) as limits.
Obtain solution \(100\ln(30) – 100\ln(30-0.01V) = t\), or equivalent.
Substitute \(V = 1000\) and obtain answer \(t = 40.5\).
(c)
Obtain \(V = 3000(1 – e^{-0.01t})\).
State that \(V\) approaches 3000.
11. [Maximum mark: 10]
Let \(f(x) = \frac{5 – x + 6x^2}{(3 – x)(1 + 3x^2)}\).
(a) Express \(f(x)\) in partial fractions. [5]
(b) Find the exact value of \(\int_{0}^{1} f(x)\,dx\), simplifying your answer. [5]
▶Answer/Explanation
Solution:
(a)
State or imply the form \(\frac{A}{3-x} + \frac{Bx+C}{1+3x^2}\).
Use a correct method to find a constant.
Obtain one of \(A = 2\), \(B = 0\) and \(C = 1\).
Obtain a second value.
Obtain the third value.
(b)
Integrate and obtain term \(-2\ln(3-x)\).
Obtain term of the form \(b\tan^{-1}(\sqrt{3}x)\).
Obtain term \(\frac{1}{\sqrt{3}}\tan^{-1}(\sqrt{3}x)\).
Substitute limits correctly in an integral with terms \(a\ln(3-x)\) and \(b\tan^{-1}(\sqrt{3}x)\).
Obtain answer \(2\ln\frac{3}{2} + \frac{1}{3\sqrt{3}}\pi\), or equivalent.