1. [Maximum mark: 3]
Solve the equation \( |5x − 2| = |4x + 9| \).
▶️Answer/Explanation
Solution:
Solve \(5x – 2 = 4x + 9\) to obtain \(x = 11\).
Attempt solution of linear equation where signs of \(5x\) and \(4x\) are different.
Obtain final value \(x = -\frac{7}{9}\).
Alternative method:
State or imply non-modulus equation \((5x – 2)^2 = (4x + 9)^2\).
Attempt solution of 3-term quadratic equation.
Obtain \(x = -\frac{7}{9}\) and \(x = 11\).
2. [Maximum mark: 5]
A curve has equation \( y = 7 + 4 \ln(2x + 5) \). Find the equation of the tangent to the curve at the point \((-2, 7)\), giving your answer in the form \( y = mx + c \).
▶️Answer/Explanation
Solution:
Differentiate to obtain form \(\frac{k}{2x + 5}\).
Obtain correct \(\frac{8}{2x + 5}\).
Substitute \(x = -2\) to obtain gradient \(8\).
Attempt equation of tangent through \((-2, 7)\) with numerical gradient.
Obtain \(y = 8x + 23\).
3. [Maximum mark: 5]
The variables \( x \) and \( y \) satisfy the equation \( y = 3^{2a} a^x \), where \( a \) is a constant. The graph of \(\ln y\) against \( x \) is a straight line with gradient 0.239.
(a) Find the value of \( a \) correct to 3 significant figures.
(b) Hence find the value of \( x \) when \( y = 36 \). Give your answer correct to 3 significant figures.
▶️Answer/Explanation
(a) Solution:
State or imply equation is \(\ln y = \ln 3^{2a} + x \ln a\).
Equate gradient of line involving \(a\) to 0.239.
Obtain \(\ln a = 0.239\) and hence \(a = 1.27\).
(b) Solution:
Substitute \(y = 36\) in \(\ln y = …\) equation and solve for \(x\).
Obtain \(3.32\).
4. [Maximum mark: 7]
(a) Show that \(\sin 2\theta \cot \theta – \cos 2\theta \equiv 1\).
(b) Hence find the exact value of \(\sin \frac{1}{6}\pi \cot \frac{1}{12}\pi\).
(c) Find the smallest positive value of \(\theta\) (in radians) satisfying the equation \(\sin 2\theta \cot \theta – 3 \cos 2\theta = 1\).
▶️Answer/Explanation
(a) Solution:
Use at least two of \(\sin 2\theta = 2\sin\theta \cos\theta\), \(\cos 2\theta = \cos^2\theta – \sin^2\theta\), \(\cot\theta = \frac{\cos\theta}{\sin\theta}\).
Express LHS in terms of \(\sin\theta\) and \(\cos\theta\) only and attempt valid simplification.
Obtain \(\cos^2\theta + \sin^2\theta\) or equivalent and hence 1.
(b) Solution:
Substitute \(\theta = \frac{1}{2}\pi\) and show or imply \(\sin\frac{1}{2}\pi \cot\frac{1}{2}\pi = 1 + \cos\frac{1}{2}\pi\).
Obtain \(1 + \frac{1}{2}\sqrt{3}\) or exact equivalent.
(c) Solution:
Use the identity from part (a) to obtain \(-2\cos 2\theta = 0\) or equivalent.
Obtain \(\theta = \frac{1}{4}\pi\).
5. [Maximum mark: 8]
(a) Given that \( y = \tan^2 x \), show that \(\frac{dy}{dx} = 2 \tan x + 2 \tan^3 x\).
(b) Find the exact value of \(\int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} (\tan x + \tan^2 x + \tan^3 x) dx\).
▶️Answer/Explanation
(a) Solution:
Differentiate to obtain \(2\tan x \sec^2 x\).
Use \(\sec^2 x = 1 + \tan^2 x\) to confirm \(2\tan x + 2\tan^3 x\).
(b) Solution:
Attempt to use part (a) result to integrate \(\tan x + \tan^3 x\).
Obtain \(\frac{1}{2} \tan^2 x\).
Use relevant identity to integrate \(\tan^2 x\).
Obtain \(\sec^2 x – 1\) and hence \(\tan x – x\).
Use limits correctly for integrand of form \(k_1 \tan^2 x + k_2 \tan x + k_3 x\).
Obtain \(\sqrt{3} – \frac{1}{12}\pi\) or exact equivalent.
6. [Maximum mark: 10]
The polynomial \( p(x) \) is defined by \( p(x) = 4x^3 + 16x^2 + 9x – 15 \).
(a) Find the quotient when \( p(x) \) is divided by \( (2x + 3) \), and show that the remainder is \(-6\).
(b) Find \( \int \frac{p(x)}{2x + 3} dx \).
(c) Factorise \( p(x) + 6 \) completely and hence solve the equation \( p(\csc 2\theta) + 6 = 0 \) for \( 0^\circ < \theta < 135^\circ \).
▶️Answer/Explanation
(a) Solution:
Carry out division at least as far as \(2x^2 + kx\).
Obtain quotient \(2x^2 + 5x – 3\).
Confirm remainder is \(-6\).
(b) Solution:
Integrate to obtain at least \(k_ix^3\) and \(k_2 \ln(2x+3)\) terms.
Obtain \(\frac{4}{3}x^3 + \frac{4}{2}x^2 – 3x – 3\ln(2x+3)\).
(c) Solution:
State or imply \(p(x) + 6 = (2x + 3)(2x^2 + 5x – 3)\).
Conclude \((2x + 3)(2x – 1)(x + 3)\).
State or imply \(\sin 2\theta = -\frac{2}{3}\) or \(\sin 2\theta = -\frac{1}{3}\) or both.
Carry out correct process to find \(\theta\) in at least one case.
Obtain 99.7 and 110.9 or greater accuracy and no others between \(0^\circ\) and \(135^\circ\).
7. [Maximum mark: 12]
A curve has equation \( e^{2x}y – e^y = 100 \).
(a) Show that \( \frac{dy}{dx} = \frac{2e^{2x}y}{e^y – e^{2x}} \).
(b) Show that the curve has no stationary points.
(c) It is required to find the x-coordinate of P, the point on the curve at which the tangent is parallel to the y-axis. Show that the x-coordinate of P satisfies the equation \( x = \ln 10 – \frac{1}{2} \ln(2x – 1) \).
(d) Use an iterative formula, based on the equation in part (c), to find the x-coordinate of P correct to 3 significant figures. Use an initial value of 2 and give the result of each iteration to 5 significant figures.
▶️Answer/Explanation
(a) Solution:
Use product rule to differentiate \( e^{2x}y \).
Obtain \( 2e^{2x}y + e^{2x} \frac{dy}{dx} \).
Obtain \( 2e^{2x}y + e^{2x} \frac{dy}{dx} – e^y \frac{dy}{dx} = 0 \) and rearrange to confirm given result.
(b) Solution:
Consider \( e^{2x}y = 0 \) and either state \( e^{2x} \neq 0 \) or substitute \( y = 0 \) in equation of curve.
Complete argument with \( e^{2x} \neq 0 \) or \( e^{2x} > 0 \) and substitution to show \( y \) cannot be zero.
(c) Solution:
State or imply \( e^y – e^{2x} = 0 \) and hence \( y = 2x \).
Substitute for \( y \) in equation of curve and attempt rearrangement as far as \( e^{2x} = … \).
Use relevant logarithm properties.
Confirm given equation.
(d) Solution:
Use iteration process correctly at least once.
Obtain final answer 1.82.
Show sufficient iterations to 5 sf to justify answer or show sign change in interval [1.815, 1.825].