1. [Maximum mark: 4]
Solve the inequality \( |2x + 3| > 3|x + 2| \).
▶️Answer/Explanation
Step 1: Square both sides to eliminate the absolute values:
\( (2x + 3)^2 > 9(x + 2)^2 \)
Step 2: Expand and simplify:
\( 4x^2 + 12x + 9 > 9x^2 + 36x + 36 \)
\( -5x^2 – 24x – 27 > 0 \)
Step 3: Solve the quadratic inequality:
Critical points: \( x = -3 \) and \( x = -\frac{9}{5} \)
Final Answer: \( -3 < x < -\frac{9}{5} \)
2. [Maximum mark: 4]
On a sketch of an Argand diagram, shade the region whose points represent complex numbers \( z \) satisfying the inequalities \( |z + 2 – 3i| \leq 2 \) and \( \arg z \leq \frac{3}{4}\pi \).
▶️Answer/Explanation
Step 1: The inequality \( |z + 2 – 3i| \leq 2 \) represents a circle centered at \( -2 + 3i \) with radius 2.
Step 2: The inequality \( \arg z \leq \frac{3}{4}\pi \) represents all complex numbers with an argument less than or equal to \( \frac{3}{4}\pi \).
Step 3: Shade the region inside the circle and within the angle \( \frac{3}{4}\pi \) from the positive real axis.
3. [Maximum mark: 5]
The variables \( x \) and \( y \) satisfy the equation \( x^n y^2 = C \), where \( n \) and \( C \) are constants. The graph of \( \ln y \) against \( \ln x \) is a straight line passing through the points \( (0.31, 1.21) \) and \( (1.06, 0.91) \). Find the value of \( n \) and the value of \( C \) correct to 2 decimal places.
▶️Answer/Explanation
Step 1: Take the natural logarithm of both sides of the equation:
\( \ln(x^n y^2) = \ln C \)
\( n \ln x + 2 \ln y = \ln C \)
Step 2: Rearrange to express \( \ln y \) in terms of \( \ln x \):
\( \ln y = -\frac{n}{2} \ln x + \frac{\ln C}{2} \)
Step 3: Use the given points to solve for \( n \) and \( C \):
Slope \( = -\frac{n}{2} = \frac{0.91 – 1.21}{1.06 – 0.31} \)
\( n = 0.8 \)
Intercept \( = \frac{\ln C}{2} = 1.21 + \frac{0.8}{2} \times 0.31 \)
\( C \approx 14.41 \)
4. [Maximum mark: 5]
The parametric equations of a curve are \( x = 1 – \cos \theta \), \( y = \cos \theta – \frac{1}{4} \cos 2\theta \). Show that \( \frac{dy}{dx} = -2 \sin^2 \left( \frac{1}{2} \theta \right) \).
▶️Answer/Explanation
Step 1: Compute \( \frac{dx}{d\theta} = \sin \theta \).
Step 2: Compute \( \frac{dy}{d\theta} = -\sin \theta + \frac{1}{2} \sin 2\theta \).
Step 3: Use the chain rule:
\( \frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta} = \frac{-\sin \theta + \frac{1}{2} \sin 2\theta}{\sin \theta} \)
Step 4: Simplify using trigonometric identities:
\( \frac{dy}{dx} = -1 + \cos \theta = -2 \sin^2 \left( \frac{1}{2} \theta \right) \)
5. [Maximum mark: 6]
The angles \( \alpha \) and \( \beta \) lie between \( 0^\circ \) and \( 180^\circ \) and are such that \( \tan(\alpha + \beta) = 2 \) and \( \tan \alpha = 3 \tan \beta \). Find the possible values of \( \alpha \) and \( \beta \).
▶️Answer/Explanation
Step 1: Use the tangent addition formula:
\( \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 – \tan \alpha \tan \beta} = 2 \)
Step 2: Substitute \( \tan \alpha = 3 \tan \beta \):
\( \frac{3 \tan \beta + \tan \beta}{1 – 3 \tan^2 \beta} = 2 \)
\( 4 \tan \beta = 2 – 6 \tan^2 \beta \)
Step 3: Solve the quadratic equation:
\( 6 \tan^2 \beta + 4 \tan \beta – 2 = 0 \)
Solutions: \( \tan \beta = \frac{1}{3} \) or \( \tan \beta = -1 \)
Step 4: Find corresponding \( \alpha \) and \( \beta \):
Case 1: \( \beta = 18.4^\circ \), \( \alpha = 45^\circ \)
Case 2: \( \beta = 135^\circ \), \( \alpha = 108.4^\circ \)
6. [Maximum mark: 6]
Find the complex numbers \( w \) which satisfy the equation \( w^2 + 2iw^* = 1 \) and are such that \( \text{Re } w \leq 0 \). Give your answers in the form \( x + iy \), where \( x \) and \( y \) are real.
▶️Answer/Explanation
Step 1: Let \( w = x + iy \), then \( w^* = x – iy \).
Substitute into the equation:
\( (x + iy)^2 + 2i(x – iy) = 1 \)
Step 2: Expand and equate real and imaginary parts:
Real: \( x^2 – y^2 – 2y = 1 \)
Imaginary: \( 2xy + 2x = 0 \)
Step 3: Solve the system of equations:
From the imaginary part: \( x(y + 1) = 0 \) ⇒ \( x = 0 \) or \( y = -1 \)
Case 1: \( x = 0 \) ⇒ \( -y^2 – 2y = 1 \) ⇒ \( y = -1 \) ⇒ \( w = -i \)
Case 2: \( y = –
7. [Maximum mark: 7]
(a) By sketching a suitable pair of graphs, show that the equation \( 4 – x^2 = \sec \left( \frac{1}{2}x \right) \) has exactly one root in the interval \( 0 \leq x < \pi \).
(b) Verify by calculation that this root lies between 1 and 2.
(c) Use the iterative formula \( x_{n+1} = \sqrt{4 – \sec \left( \frac{1}{2}x_n \right)} \) to determine the root correct to 2 decimal places.
▶️Answer/Explanation
(a) Sketch the graphs of \( y = 4 – x^2 \) and \( y = \sec \left( \frac{1}{2}x \right) \). The graphs intersect once in \( [0, \pi) \).
(b) Evaluate at \( x = 1 \): \( 4 – 1 = 3 > \sec(0.5) \approx 1.139 \)
Evaluate at \( x = 2 \): \( 4 – 4 = 0 < \sec(1) \approx 1.851 \). Root lies in \( (1, 2) \).
(c) Iterations:
\( x_1 = 1.5 \) ⇒ \( x_2 \approx 1.60 \) ⇒ \( x_3 \approx 1.60 \) (converged)
Final Answer: \( x \approx 1.60 \)
8. [Maximum mark: 8]
(a) Find the quotient and remainder when \( 8x^3 + 4x^2 + 2x + 7 \) is divided by \( 4x^2 + 1 \).
(b) Hence find the exact value of \( \int_0^{\frac{1}{2}} \frac{8x^3 + 4x^2 + 2x + 7}{4x^2 + 1} \, dx \).
▶️Answer/Explanation
(a) Perform polynomial division:
Quotient: \( 2x + 1 \)
Remainder: \( 6 \)
(b) Rewrite the integrand using the division result:
\( \frac{8x^3 + 4x^2 + 2x + 7}{4x^2 + 1} = 2x + 1 + \frac{6}{4x^2 + 1} \)
Integrate term by term:
\( \int (2x + 1) \, dx = x^2 + x \)
\( \int \frac{6}{4x^2 + 1} \, dx = 3 \tan^{-1}(2x) \)
Evaluate from 0 to \( \frac{1}{2} \):
\( \left[ x^2 + x + 3 \tan^{-1}(2x) \right]_0^{\frac{1}{2}} = \frac{3}{4} + \frac{3\pi}{4} \)
Final Answer: \( \frac{3}{4}(1 + \pi) \)
9. [Maximum mark: 9]
The variables \( x \) and \( y \) satisfy the differential equation \( (x + 1)(3x + 1) \frac{dy}{dx} = y \), and it is given that \( y = 1 \) when \( x = 1 \). Solve the differential equation and find the exact value of \( y \) when \( x = 3 \), giving your answer in a simplified form.
▶️Answer/Explanation
Step 1: Separate variables:
\( \frac{dy}{y} = \frac{dx}{(x + 1)(3x + 1)} \)
Step 2: Use partial fractions:
\( \frac{1}{(x + 1)(3x + 1)} = \frac{A}{x + 1} + \frac{B}{3x + 1} \)
Solve for \( A \) and \( B \): \( A = -\frac{1}{2} \), \( B = \frac{3}{2} \)
Step 3: Integrate both sides:
\( \ln y = -\frac{1}{2} \ln(x + 1) + \frac{1}{2} \ln(3x + 1) + C \)
Step 4: Use initial condition \( y(1) = 1 \) to find \( C \):
\( C = \frac{1}{2} \ln 2 \)
Step 5: Solve for \( y \) and evaluate at \( x = 3 \):
\( y = \frac{\sqrt{3x + 1}}{\sqrt{2(x + 1)}} \) ⇒ \( y(3) = \frac{\sqrt{10}}{\sqrt{8}} = \frac{\sqrt{5}}{2} \)
10. [Maximum mark: 10]
The points \( A \) and \( B \) have position vectors \( 2i + j + k \) and \( i – 2j + 2k \) respectively. The line \( l \) has vector equation \( r = i + 2j – 3k + \mu(i – 3j – 2k) \).
(a) Find a vector equation for the line through \( A \) and \( B \).
(b) Find the acute angle between the directions of \( AB \) and \( l \), giving your answer in degrees.
(c) Show that the line through \( A \) and \( B \) does not intersect the line \( l \).
▶️Answer/Explanation
(a) Direction vector \( \vec{AB} = \vec{B} – \vec{A} = -i – 3j + k \)
Equation: \( r = 2i + j + k + \lambda(-i – 3j + k) \)
(b) Use the dot product formula:
\( \cos \theta = \frac{(-1)(1) + (-3)(-3) + (1)(-2)}{\sqrt{1 + 9 + 1} \sqrt{1 + 9 + 4}} = \frac{6}{\sqrt{11} \sqrt{14}} \)
\( \theta \approx 61.1^\circ \)
(c) Set up parametric equations for both lines and solve for \( \lambda \) and \( \mu \):
No solution exists, hence the lines do not intersect.
11. [Maximum mark: 11]
The diagram shows the curve \( y = \sin x \cos 2x \) for \( 0 \leq x \leq \frac{1}{2}\pi \), and its maximum point \( M \).
(a) Find the \( x \)-coordinate of \( M \), giving your answer correct to 3 significant figures.
(b) Using the substitution \( u = \cos x \), find the area of the shaded region enclosed by the curve and the \( x \)-axis in the first quadrant, giving your answer in a simplified exact form.
▶️Answer/Explanation
(a) Differentiate \( y = \sin x \cos 2x \):
\( \frac{dy}{dx} = \cos x \cos 2x – 2 \sin x \sin 2x \)
Set derivative to zero and solve:
\( \cos x \cos 2x = 2 \sin x \sin 2x \)
Use trigonometric identities to simplify:
\( x \approx 0.421 \) radians
(b) Substitute \( u = \cos x \), \( du = -\sin x \, dx \):
\( \int \sin x \cos 2x \, dx = -\int (2u^2 – 1) \, du = – \left( \frac{2}{3}u^3 – u \right) \)
Evaluate from \( x = 0 \) to \( x = \frac{\pi}{2} \):
Area \( = \frac{1}{3} \left( \sqrt{2} – 1 \right) \)