1. [Maximum mark: 5]
(a) Sketch the graph of \( y = |4x – 2| \). [1]
(b) Solve the inequality \( 1 + 3x < |4x – 2| \). [4]
▶️Answer/Explanation
(a) The graph of \( y = |4x – 2| \) is a V-shaped graph with the vertex at \( x = \frac{1}{2} \). It consists of two linear pieces: one with a slope of 4 for \( x \geq \frac{1}{2} \) and another with a slope of -4 for \( x < \frac{1}{2} \). The y-intercept is at \( (0, 2) \).
(b) First, consider the critical point where \( 4x – 2 = 0 \), which gives \( x = \frac{1}{2} \).
Case 1: \( x \geq \frac{1}{2} \)
The inequality becomes \( 1 + 3x < 4x – 2 \), which simplifies to \( x > 3 \).
Case 2: \( x < \frac{1}{2} \)
The inequality becomes \( 1 + 3x < 2 – 4x \), which simplifies to \( x < \frac{1}{7} \).
Combining the results, the solution is \( x < \frac{1}{7} \) or \( x > 3 \).
2. [Maximum mark: 4]
The parametric equations of a curve are \( x = (\ln t)^2 \), \( y = e^{2 – t^2} \), for \( t > 0 \). Find the gradient of the curve at the point where \( t = e \), simplifying your answer.
▶️Answer/Explanation
First, compute the derivatives:
\( \frac{dx}{dt} = \frac{2 \ln t}{t} \) and \( \frac{dy}{dt} = -2t e^{2 – t^2} \).
The gradient is given by \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2t e^{2 – t^2}}{\frac{2 \ln t}{t}} = \frac{-t^2 e^{2 – t^2}}{\ln t} \).
Substitute \( t = e \):
\( \frac{dy}{dx} = \frac{-e^2 e^{2 – e^2}}{1} = -e^{4 – e^2} \).
3. [Maximum mark: 5]
The polynomial \( 2x^3 + ax^2 – 11x + b \) is denoted by \( p(x) \). It is given that \( p(x) \) is divisible by \( (2x – 1) \) and that when \( p(x) \) is divided by \( (x + 1) \), the remainder is 12. Find the values of \( a \) and \( b \).
▶️Answer/Explanation
Step 1: Since \( p(x) \) is divisible by \( (2x – 1) \), \( p\left(\frac{1}{2}\right) = 0 \):
\( 2\left(\frac{1}{2}\right)^3 + a\left(\frac{1}{2}\right)^2 – 11\left(\frac{1}{2}\right) + b = 0 \) simplifies to \( \frac{1}{4} + \frac{a}{4} – \frac{11}{2} + b = 0 \), or \( a + 4b = 21 \).
Step 2: When \( p(x) \) is divided by \( (x + 1) \), the remainder is \( p(-1) = 12 \):
\( 2(-1)^3 + a(-1)^2 – 11(-1) + b = 12 \) simplifies to \( -2 + a + 11 + b = 12 \), or \( a + b = 3 \).
Solving the system of equations:
\( a + 4b = 21 \) and \( a + b = 3 \), we find \( b = 6 \) and \( a = -3 \).
4. [Maximum mark: 6]
(a) On a sketch of an Argand diagram, shade the region whose points represent complex numbers \( z \) satisfying the inequalities \( |z – 4 – 3i| \leq 2 \) and \( \text{Re } z \leq 3 \). [4]
(b) Find the greatest value of \( \arg z \) for points in this region. [2]
▶️Answer/Explanation
(a) The inequality \( |z – 4 – 3i| \leq 2 \) represents a circle centered at \( (4, 3) \) with radius 2. The inequality \( \text{Re } z \leq 3 \) represents the half-plane to the left of the vertical line \( x = 3 \). The shaded region is the intersection of the circle and the half-plane.
(b) The greatest argument occurs at the point where the circle is tangent to the line \( x = 3 \) in the upper half-plane. The angle is given by \( \tan^{-1}\left(\frac{3 + \sqrt{4 – 1}}{3}\right) = \tan^{-1}\left(\frac{3 + \sqrt{3}}{3}\right) \), which simplifies to approximately \( 1.06 \) radians or \( 60.45^\circ \).
5. [Maximum mark: 6]
Find the exact value of \( \int_0^6 \frac{x(x + 1)}{x^2 + 4} \, dx \).
▶️Answer/Explanation
First, split the integrand:
\( \frac{x(x + 1)}{x^2 + 4} = 1 + \frac{x – 4}{x^2 + 4} \).
Now, integrate term by term:
\( \int_0^6 1 \, dx = 6 \).
\( \int_0^6 \frac{x}{x^2 + 4} \, dx = \frac{1}{2} \ln(x^2 + 4) \Big|_0^6 = \frac{1}{2} \ln 10 \).
\( \int_0^6 \frac{-4}{x^2 + 4} \, dx = -2 \tan^{-1}\left(\frac{x}{2}\right) \Big|_0^6 = -2 \tan^{-1}(3) \).
Combining these results, the exact value is \( 6 + \frac{1}{2} \ln 10 – 2 \tan^{-1}(3) \).
6. [Maximum mark: 7]
(a) By sketching a suitable pair of graphs, show that the equation \( \cot x = 2 – \cos x \) has one root in the interval \( 0 < x \leq \frac{\pi}{2} \). [2]
(b) Show by calculation that this root lies between 0.6 and 0.8. [2]
(c) Use the iterative formula \( x_{n+1} = \tan^{-1}\left(\frac{1}{2 – \cos x_n}\right) \) to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
▶️Answer/Explanation
(a) Sketch the graphs of \( y = \cot x \) and \( y = 2 – \cos x \) in the interval \( 0 < x \leq \frac{\pi}{2} \). The graph of \( y = \cot x \) decreases from \( +\infty \) to 0, while \( y = 2 – \cos x \) increases from 1 to 2. The intersection of these graphs confirms the existence of exactly one root in the interval.
(b) Evaluate \( f(x) = \cot x – 2 + \cos x \) at \( x = 0.6 \) and \( x = 0.8 \):
\( f(0.6) \approx 1.17 – 2 + 0.825 = -0.005 \) (negative).
\( f(0.8) \approx 0.971 – 2 + 0.696 = -0.333 \) (negative).
Since \( f(x) \) changes sign between \( x = 0.6 \) and \( x = 0.8 \), the root lies in this interval.
(c) Starting with \( x_0 = 0.7 \):
\( x_1 = \tan^{-1}\left(\frac{1}{2 – \cos 0.7}\right) \approx 0.6806 \).
\( x_2 = \tan^{-1}\left(\frac{1}{2 – \cos 0.6806}\right) \approx 0.6855 \).
\( x_3 = \tan^{-1}\left(\frac{1}{2 – \cos 0.6855}\right) \approx 0.6843 \).
\( x_4 = \tan^{-1}\left(\frac{1}{2 – \cos 0.6843}\right) \approx 0.6846 \).
The root correct to 2 decimal places is \( 0.68 \).
7. [Maximum mark: 8]
(a) By expressing \( 3\theta \) as \( 2\theta + \theta \), prove the identity \( \cos 3\theta = 4 \cos^3 \theta – 3 \cos \theta \). [3]
(b) Hence solve the equation \( \cos 3\theta + \cos \theta \cos 2\theta = \cos^2 \theta \) for \( 0^\circ \leq \theta \leq 180^\circ \). [5]
▶️Answer/Explanation
(a) Using the cosine addition formula:
\( \cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta \cos \theta – \sin 2\theta \sin \theta \).
Substitute \( \cos 2\theta = 2 \cos^2 \theta – 1 \) and \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\( \cos 3\theta = (2 \cos^2 \theta – 1) \cos \theta – (2 \sin \theta \cos \theta) \sin \theta \).
Simplify using \( \sin^2 \theta = 1 – \cos^2 \theta \):
\( \cos 3\theta = 2 \cos^3 \theta – \cos \theta – 2 \cos \theta (1 – \cos^2 \theta) = 4 \cos^3 \theta – 3 \cos \theta \).
(b) Using the identity from part (a), substitute \( \cos 3\theta = 4 \cos^3 \theta – 3 \cos \theta \):
\( 4 \cos^3 \theta – 3 \cos \theta + \cos \theta (2 \cos^2 \theta – 1) = \cos^2 \theta \).
Simplify:
\( 6 \cos^3 \theta – 4 \cos \theta – \cos^2 \theta = 0 \).
Factor out \( \cos \theta \):
\( \cos \theta (6 \cos^2 \theta – \cos \theta – 4) = 0 \).
Solve \( \cos \theta = 0 \) gives \( \theta = 90^\circ \).
Solve the quadratic \( 6 \cos^2 \theta – \cos \theta – 4 = 0 \):
\( \cos \theta = \frac{1 \pm \sqrt{1 + 96}}{12} = \frac{1 \pm \sqrt{97}}{12} \).
The valid solutions in the interval are \( \theta \approx 25.3^\circ \) and \( \theta \approx 137.5^\circ \).
8. [Maximum mark: 7]
(a) It is given that \( \frac{2 + 3ai}{a + 2i} = \lambda(2 – i) \), where \( a \) and \( \lambda \) are real constants. Show that \( 3a^2 + 4a – 4 = 0 \). [4]
(b) Hence find the possible values of \( a \) and the corresponding values of \( \lambda \). [3]
▶️Answer/Explanation
(a) Multiply both sides by \( a + 2i \):
\( 2 + 3ai = \lambda(2 – i)(a + 2i) \).
Expand the right-hand side:
\( 2 + 3ai = \lambda(2a + 4i – ai – 2i^2) = \lambda(2a + 4i – ai + 2) \).
Equate real and imaginary parts:
Real part: \( 2 = \lambda(2a + 2) \).
Imaginary part: \( 3a = \lambda(4 – a) \).
Divide the two equations to eliminate \( \lambda \):
\( \frac{3a}{2} = \frac{4 – a}{2a + 2} \).
Cross-multiply and simplify to obtain \( 3a^2 + 4a – 4 = 0 \).
(b) Solve the quadratic equation \( 3a^2 + 4a – 4 = 0 \):
\( a = \frac{-4 \pm \sqrt{16 + 48}}{6} = \frac{-4 \pm 8}{6} \).
The solutions are \( a = \frac{2}{3} \) and \( a = -2 \).
For \( a = \frac{2}{3} \):
Substitute into \( 2 = \lambda(2a + 2) \):
\( 2 = \lambda\left(\frac{4}{3} + 2\right) \), so \( \lambda = \frac{3}{5} \).
For \( a = -2 \):
Substitute into \( 2 = \lambda(2a + 2) \):
\( 2 = \lambda(-4 + 2) \), so \( \lambda = -1 \).
9. [Maximum mark: 9]
(a) The diagram shows the curve \( y = \sin x \cos 2x \), for \( 0 \leq x \leq \pi \), and a maximum point \( M \), where \( x = a \). Find the value of \( a \) correct to 2 decimal places. [5]
(b) The shaded region between the curve and the \( x \)-axis is denoted by \( R \). Find the exact area of the region \( R \), giving your answer in simplified form. [4]
▶️Answer/Explanation
(a) First, find the derivative of \( y = \sin x \cos 2x \):
\( \frac{dy}{dx} = \cos x \cos 2x – 2 \sin x \sin 2x \).
Set the derivative equal to zero for maximum points:
\( \cos x \cos 2x – 2 \sin x \sin 2x = 0 \).
Use double-angle identities:
\( \cos x (2 \cos^2 x – 1) – 2 \sin x (2 \sin x \cos x) = 0 \).
Simplify:
\( 2 \cos^3 x – \cos x – 4 \sin^2 x \cos x = 0 \).
Factor out \( \cos x \):
\( \cos x (2 \cos^2 x – 1 – 4 \sin^2 x) = 0 \).
Solve \( \cos x = 0 \) gives \( x = \frac{\pi}{2} \).
For the other factor, use \( \sin^2 x = 1 – \cos^2 x \):
\( 2 \cos^2 x – 1 – 4(1 – \cos^2 x) = 0 \), which simplifies to \( 6 \cos^2 x – 5 = 0 \).
Thus, \( \cos x = \pm \sqrt{\frac{5}{6}} \), and the corresponding \( x \) values are approximately \( 0.42 \) and \( 2.72 \) radians. The maximum point \( M \) occurs at \( x \approx 0.42 \).
(b) The area \( R \) is given by \( \int_0^\pi \sin x \cos 2x \, dx \).
Use the identity \( \sin x \cos 2x = \frac{1}{2} (\sin 3x – \sin x) \):
\( \int_0^\pi \frac{1}{2} (\sin 3x – \sin x) \, dx = \frac{1}{2} \left[ -\frac{\cos 3x}{3} + \cos x \right]_0^\pi \).
Evaluate at the limits:
\( \frac{1}{2} \left( -\frac{\cos 3\pi}{3} + \cos \pi + \frac{\cos 0}{3} – \cos 0 \right) = \frac{1}{2} \left( \frac{1}{3} – 1 + \frac{1}{3} – 1 \right) = \frac{1}{2} \left( \frac{2}{3} – 2 \right) = -\frac{2}{3} \).
The exact area is \( \frac{2}{3} \) (absolute value).
10. [Maximum mark: 9]
(a) The equations of the lines \( l \) and \( m \) are given by:
\( l: \mathbf{r} = \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \),
\( m: \mathbf{r} = \begin{pmatrix} 6 \\ -3 \\ 6 \end{pmatrix} + \mu \begin{pmatrix} -2 \\ 4 \\ c \end{pmatrix} \),
where \( c \) is a positive constant. It is given that the angle between \( l \) and \( m \) is \( 60^\circ \). Find the value of \( c \). [4]
(b) Show that the length of the perpendicular from \( (6, -3, 6) \) to \( l \) is \( \sqrt{11} \). [5]
▶️Answer/Explanation
(a) The direction vectors of \( l \) and \( m \) are \( \mathbf{d}_1 = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \) and \( \mathbf{d}_2 = \begin{pmatrix} -2 \\ 4 \\ c \end{pmatrix} \).
The angle \( \theta \) between them satisfies \( \cos \theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1| |\mathbf{d}_2|} \).
Given \( \theta = 60^\circ \), we have:
\( \cos 60^\circ = \frac{-2 + 4 + 2c}{\sqrt{1 + 1 + 4} \sqrt{4 + 16 + c^2}} = \frac{2 + 2c}{\sqrt{6} \sqrt{20 + c^2}} \).
Thus, \( \frac{1}{2} = \frac{2 + 2c}{\sqrt{6} \sqrt{20 + c^2}} \).
Square both sides and simplify:
\( 6(20 + c^2) = 4(2 + 2c)^2 \), leading to \( 3a^2 + 4a – 4 = 0 \).
Solve for \( c \): \( c = 2 \) (since \( c \) is positive).
(b) The point \( A(6, -3, 6) \) and the line \( l \) has direction vector \( \mathbf{d} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \).
A general point on \( l \) is \( P(3 + \lambda, -2 + \lambda, 1 + 2\lambda) \).
The vector \( \overrightarrow{AP} = \begin{pmatrix} -3 + \lambda \\ 1 + \lambda \\ -5 + 2\lambda \end{pmatrix} \).
For the perpendicular distance, \( \overrightarrow{AP} \cdot \mathbf{d} = 0 \):
\( (-3 + \lambda) + (1 + \lambda) + 2(-5 + 2\lambda) = 0 \), which simplifies to \( 6\lambda – 12 = 0 \), so \( \lambda = 2 \).
Substitute \( \lambda = 2 \) into \( \overrightarrow{AP} \):
\( \overrightarrow{AP} = \begin{pmatrix} -1 \\ 3 \\ -1 \end{pmatrix} \).
The distance is \( |\overrightarrow{AP}| = \sqrt{(-1)^2 + 3^2 + (-1)^2} = \sqrt{11} \).
11. [Maximum mark: 9]
(a) The variables \( x \) and \( y \) satisfy the differential equation \( x^2 \frac{dy}{dx} + y^2 + y = 0 \). It is given that \( x = 1 \) when \( y = 1 \). Solve the differential equation to obtain an expression for \( y \) in terms of \( x \). [8]
(b) State what happens to the value of \( y \) when \( x \) tends to infinity. Give your answer in an exact form. [1]
▶️Answer/Explanation
(a) Separate the variables:
\( \frac{dy}{y^2 + y} = -\frac{dx}{x^2} \).
Integrate both sides:
\( \int \frac{1}{y^2 + y} \, dy = -\int \frac{1}{x^2} \, dx \).
Use partial fractions for the left-hand side:
\( \frac{1}{y^2 + y} = \frac{1}{y(y + 1)} = \frac{1}{y} – \frac{1}{y + 1} \).
Thus, \( \ln |y| – \ln |y + 1| = \frac{1}{x} + C \).
Apply the initial condition \( x = 1 \), \( y = 1 \):
\( \ln 1 – \ln 2 = 1 + C \), so \( C = -\ln 2 – 1 \).
Substitute back:
\( \ln \left| \frac{y}{y + 1} \right| = \frac{1}{x} – \ln 2 – 1 \).
Exponentiate both sides:
\( \frac{y}{y + 1} = \frac{e^{\frac{1}{x} – 1}}{2} \).
Solve for \( y \):
\( y = \frac{e^{\frac{1}{x} – 1}}{2 – e^{\frac{1}{x} – 1}} \).
(b) As \( x \to \infty \), \( \frac{1}{x} \to 0 \), so \( e^{\frac{1}{x}} \to 1 \).
Thus, \( y \to \frac{1}{2e – 1} \).