1. [Maximum mark: 4]
Solve the inequality \( 2|3x – 1| < |x + 1| \).
▶️Answer/Explanation
State or imply non-modular inequality \( (3x – 1)^2 < (x + 1)^2 \), or corresponding quadratic equation, or pair of linear equations.
Form and solve a 3-term quadratic, e.g., \( 35x^2 – 26x + 3 = 0 \).
Obtain critical values \( x = \frac{3}{5} \) and \( x = \frac{1}{7} \).
Final Answer: \( \frac{1}{7} < x < \frac{3}{5} \).
2. [Maximum mark: 5]
Find the real root of the equation \( \frac{2e^x + e^{-x}}{2 + e^x} = 3 \), giving your answer correct to 3 decimal places.
Your working should show clearly that the equation has only one real root
▶️Answer/Explanation
Multiply through by the denominator to get \( 2e^x + e^{-x} = 6 + 3e^x \). Let \( u = e^x \) to form the quadratic \( u^2 + 6u – 1 = 0 \). The positive solution \( u = \sqrt{10} – 3 \) gives \( x = \ln(\sqrt{10} – 3) \approx -1.818 \). Since \( e^x > 0 \), this is the only real root.
3. [Maximum mark: 6]
(a) Given that \( \cos(x – 30^\circ) = 2 \sin(x + 30^\circ) \), show that \( \tan x = \frac{2 – \sqrt{3}}{1 – 2\sqrt{3}} \).
(b) Hence solve the equation
$\cos(x – 30^\circ) = 2 \sin(x + 30^\circ),$
for \( 0^\circ < x < 360^\circ \).
▶️Answer/Explanation
(a) Expand both sides using compound angle formulas and collect terms to isolate \( \tan x \). The exact form simplifies to \( \frac{2 – \sqrt{3}}{1 – 2\sqrt{3}} \).
(b) Solving the tangent equation gives \( x = 173.8^\circ \) and \( x = 353.8^\circ \) within the specified range.
4. [Maximum mark: 6]
(a) Prove that \( \frac{1 – \cos 2\theta}{1 + \cos 2\theta} = \tan^2 \theta \).
(b) Hence find the exact value of \( \int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \frac{1 – \cos 2\theta}{1 + \cos 2\theta} \, d\theta \).
▶️Answer/Explanation
(a) Substitute double-angle identities \( \cos 2\theta = 1 – 2\sin^2 \theta = 2\cos^2 \theta – 1 \) and simplify both sides to match.
(b) Replace the integrand with \( \tan^2 \theta \) and integrate to get \( \tan \theta – \theta \). Evaluating between limits yields \( \frac{2}{3}\sqrt{3} – \frac{1}{6}\pi \).
5. [Maximum mark: 7]
(a) Solve \( z^2 – 2\pi z – q = 0 \) for real constants \( p, q \).
In an Argand diagram with origin O, the roots of this equation are represented by the distinct points A and B.
(b) Find a relation between \( p \) and \( q \) when roots lie on the imaginary axis.
(c) Express \( q \) in terms of \( p \) when triangle \( OAB \) is equilateral.
▶️Answer/Explanation
(a) Applying the quadratic formula gives \( z = \pi \pm \sqrt{\pi^2 + q} \).
(b) For purely imaginary roots, the discriminant must be negative: \( q < -\pi^2 \).
(c) Using geometric properties of equilateral triangles, the relation simplifies to \( q = \frac{4}{3}\pi^2 \).
6. [Maximum mark: 8]
The parametric equations of a curve are \( x = \ln(2 + 3t) \), \( y = \frac{t}{2 + 3t} \).
(a) Show that the gradient is always positive.
(b) Find the tangent equation at the y-intercept.
▶️Answer/Explanation
(a) Differentiating both parametric equations and computing \( \frac{dy}{dx} \) yields \( \frac{2}{3(2 + 3t)} \), which is always positive since \( 2 + 3t > 0 \).
$\text{Obtain } \frac{1 – \cos 2\theta}{1 + \cos 2\theta} = \tan^2 \theta \text{ from correct working}$
(b) At the y-intercept (\( x = 0 \)), \( t = -\frac{1}{3} \). The gradient here is \( \frac{2}{3} \), giving the tangent equation \( 3y = 2x – 1 \).
7. [Maximum mark: 9]
The curve \( y = \frac{\tan^{-1}x}{\sqrt{x}} \) has a maximum point at \( x = a \).
(a) Show that a satisfies the equation
\( a = \tan\left(\frac{2a}{1 + a^2}\right) \).
(b) Verify by calculation that a lies between l.3 and 1.5.
(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) Differentiating using the quotient rule and setting \( y’ = 0 \) leads to the given transcendental equation after simplification.
(b) Substituting \( a = 1.3 \) and \( a = 1.5 \) shows a sign change in \( a – \tan(\frac{2a}{1 + a^2}) \).
(c) Iterating with \( a_{n+1} = \tan(\frac{2a_n}{1 + a_n^2}) \) converges to \( a \approx 1.39 \).
8. [Maximum mark: 9]
$\text{With respect to the origin } O, \text{ the points } A \text{ and } B \text{ have position vectors given by }$ $\vec{OA} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} \text{ and } \vec{OB} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}.$ $\text{ The line } l \text{ has equation } \mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}.$
(a) Find the acute angle between \( AB \) and \( l \).
(b) Find point \( P \) on \( l \) where \( AP = BP \).
▶️Answer/Explanation
(a) The vector \( \mathbf{AB} = (2, -1, -3) \). The angle \( \theta \) with \( l \)’s direction vector satisfies \( \cos \theta = \frac{1}{\sqrt{14}\sqrt{6}} \), giving \( \theta \approx 83.7^\circ \).
(b) Parameterizing \( P \) as \( (2 + \lambda, 3 – 2\lambda, 1 + \lambda) \) and equating distances gives \( \lambda = 6 \), so \( P \) is \( (8, -9, 7) \).
9. [Maximum mark: 10]
The equation of a curve is \( y = x^{-2/3} \ln x \) for x > 0. The curve has one stationary point.
(a) Find its exact coordinates of the stationary point.
(b) Show \( \int_1^8 y \, dx = 18 \ln 2 – 9 \).
▶️Answer/Explanation
(a) Setting the derivative \( y’ = -\frac{2}{3}x^{-5/3}\ln x + x^{-2/3}/x = 0 \) leads to \( \ln x = \frac{3}{2} \). Thus, the stationary point is at \( \left(e^{3/2}, \frac{3}{2e}\right) \).
(b) Integrate by parts with \( u = \ln x \), \( dv = x^{-2/3} dx \) to obtain \( 3x^{1/3}(\ln x – 3) \). Evaluating from 1 to 8 yields \( 18 \ln 2 – 9 \).
10. [Maximum mark: 11]
The variables x and t satisfy the differential equation \( \frac{dx}{dt} = x^2(1 + 2x) \), and \( x = 1 \) when t = 0.
Using partial fractions, solve the differential equation, obtaining an expression for t in terms of x.
▶️Answer/Explanation
Separate variables and decompose \( \frac{1}{x^2(1 + 2x)} \) into \( \frac{1}{x^2} – \frac{2}{x} + \frac{4}{1 + 2x} \). Integrating both sides gives: \[ t = -\frac{1}{x} – 2\ln|x| + 2\ln|1 + 2x| + C \] Applying \( x = 1 \) at \( t = 0 \) determines \( C = 1 + 2\ln 3 \). The final solution simplifies to: \[ t = \frac{1}{x} + 2\ln\left(\frac{1 + 2x}{3x}\right) + 1 \]