1. [Maximum mark: 4]
(a) Expand \( (1 + 3x)^6 \) in ascending powers of \( x \) up to, and including, the term in \( x^2 \).
(b) Hence find the coefficient of \( x^2 \) in the expansion of \( (1 – 7x + x^2)(1 + 3x)^6 \).
▶️Answer/Explanation
(a) Using the binomial expansion:
\( (1 + 3x)^6 = 1 + 18x + 135x^2 + \dots \)
(b) Multiply the expansion by \( (1 – 7x + x^2) \):
Coefficient of \( x^2 = 135 – 7 \times 18 + 1 = 10 \).
2. [Maximum mark: 4]
A line has equation \( y = 2cx + 3 \) and a curve has equation \( y = cx^2 + 3x – c \), where \( c \) is a constant. Determine which of the following statements is correct:
A) The line and curve intersect only for a particular set of values of \( c \).
B) The line and curve intersect for all values of \( c \).
C) The line and curve do not intersect for any values of \( c \).
▶️Answer/Explanation
Set the equations equal: \( cx^2 + (3 – 2c)x – (c + 3) = 0 \).
Discriminant: \( (3 – 2c)^2 + 4c(c + 3) = 8c^2 + 9 > 0 \) for all \( c \).
Thus, the correct statement is B.
3. [Maximum mark: 3]
The diagram shows a cubical closed container made of a thin elastic material which is filled with water and frozen. During the freezing process, the length, \( x \) cm, of each edge of the container increases at the constant rate of 0.01 cm per minute. The volume of the container at time \( t \) minutes is \( V \) cm\(^3\). Find the rate of increase of \( V \) when \( x = 20 \).
▶️Answer/Explanation
Volume \( V = x^3 \).
Rate of change: \( \frac{dV}{dt} = 3x^2 \cdot \frac{dx}{dt} \).
Substitute \( x = 20 \) and \( \frac{dx}{dt} = 0.01 \):
\( \frac{dV}{dt} = 3 \times 20^2 \times 0.01 = 12 \) cm\(^3\)/min.
4. [Maximum mark: 6]
The transformation R denotes a reflection in the \( x \)-axis, and the transformation T denotes a translation of \( \begin{pmatrix} 3 \\ -1 \end{pmatrix} \).
(a) Find the equation of the curve \( y = g(x) \) after transforming \( y = x^2 \) by R followed by T.
(b) Find the equation of the curve \( y = h(x) \) after transforming \( y = x^2 \) by T followed by R.
(c) State the transformation that maps \( y = g(x) \) onto \( y = h(x) \).
▶️Answer/Explanation
(a) After R: \( y = -x^2 \). After T: \( y = -(x – 3)^2 – 1 \).
(b) After T: \( y = (x – 3)^2 – 1 \). After R: \( y = -(x – 3)^2 + 1 \).
(c) The transformation is a translation \( \begin{pmatrix} 0 \\ 2 \end{pmatrix} \).
5. [Maximum mark: 6]
(a) Show that the equation \( 4 \sin x + \frac{5}{\tan x} + 2 \sin x = 0 \) can be expressed as \( a \cos^2 x + b \cos x + c = 0 \), where \( a \), \( b \), and \( c \) are integers.
(b) Solve the equation \( 4 \sin x + \frac{5}{\tan x} + 2 \sin x = 0 \) for \( 0^\circ \leq x \leq 360^\circ \).
▶️Answer/Explanation
(a) Multiply by \( \sin x \): \( 4 \sin^2 x + 5 \cos x + 2 \sin^2 x = 0 \).
Simplify: \( 4 \cos^2 x – 5 \cos x – 6 = 0 \).
(b) Solve the quadratic: \( \cos x = -\frac{3}{4} \) or \( \cos x = 2 \) (invalid).
Solutions: \( x = 138.6^\circ, 221.4^\circ \).
6. [Maximum mark: 7]
The diagram shows a motif formed by the major arc AB of a circle with radius \( r \) and centre O, and the minor arc AOB of a circle, also with radius \( r \) but with centre C. The point C lies on the circle with centre O.
(a) Given that angle \( ACB = k\pi \) radians, state the value of \( k \).
(b) State the perimeter of the shaded motif in terms of \( \pi \) and \( r \).
(c) Find the area of the shaded motif, giving your answer in terms of \( \pi \), \( r \), and \( \sqrt{3} \).
▶️Answer/Explanation
(a) \( k = \frac{2}{3} \).
(b) Perimeter \( = 2\pi r \).
(c) Area of major sector \( OAB = \frac{2}{3} \pi r^2 \).
Area of segments: \( 2 \left( \frac{\pi r^2}{6} – \frac{r^2 \sqrt{3}}{4} \right) \).
Shaded area \( = \frac{\pi r^2}{3} + \frac{r^2 \sqrt{3}}{2} \).
7. [Maximum mark: 7]
The sum of the first two terms of a geometric progression is 15, and the sum to infinity is \( \frac{125}{7} \). The common ratio is negative. Find the third term of the progression.
▶️Answer/Explanation
Let the first term be \( a \) and common ratio \( r \).
Given: \( a(1 + r) = 15 \) and \( \frac{a}{1 – r} = \frac{125}{7} \).
Solve for \( r \): \( r = -\frac{2}{5} \), \( a = 25 \).
Third term \( = ar^2 = 4 \).
8. [Maximum mark: 9]
The diagram shows the curves \( y = 2(2x – 3)^4 \) and \( y = (2x – 3)^2 + 1 \) meeting at points A and B.
(a) By using the substitution \( u = 2x – 3 \), find the coordinates of A and B.
(b) Find the exact area of the shaded region.
▶️Answer/Explanation
(a) Substitute \( u = 2x – 3 \): \( 2u^4 = u^2 + 1 \).
Solve: \( u = \pm 1 \), so \( x = 1 \) or \( 2 \).
Coordinates: \( (1, 2) \) and \( (2, 2) \).
(b) Integrate \( (2u^4 – u^2 – 1) \) from \( u = -1 \) to \( 1 \):
Area \( = \frac{14}{15} \).
9. [Maximum mark: 9]
(a) Express \( 4x^2 – 12x + 13 \) in the form \( (2x + a)^2 + b \), where \( a \) and \( b \) are constants.
(b) Given that the composite function \( gf \) can be formed, find the least possible value of \( p \) and the greatest possible value of \( q \).
(c) Find an expression for \( gf(x) \).
(d) Find an expression for \( h^{-1}(x) \), where \( h(x) = 4x^2 – 12x + 13 \) for \( x < 0 \).
▶️Answer/Explanation
(a) \( (2x – 3)^2 + 4 \).
(b) Solve \( (2x – 3)^2 < 4 \): \( \frac{1}{2} < x < \frac{5}{2} \).
Least \( p = \frac{1}{2} \), greatest \( q = \frac{5}{2} \).
(c) \( gf(x) = 12x^2 – 36x + 40 \).
(d) \( h^{-1}(x) = \frac{3}{2} – \frac{\sqrt{x – 4}}{2} \).
10. [Maximum mark: 11]
A curve has a stationary point at \( (2, -10) \) and satisfies \( \frac{d^2y}{dx^2} = 6x \).
(a) Find \( \frac{dy}{dx} \).
(b) Find the equation of the curve.
(c) Find the coordinates of the other stationary point and determine its nature.
(d) Find the equation of the tangent to the curve at the point where the curve crosses the y-axis.
▶️Answer/Explanation
(a) Integrate \( \frac{d^2y}{dx^2} \): \( \frac{dy}{dx} = 3x^2 – 12 \).
(b) Integrate \( \frac{dy}{dx} \): \( y = x^3 – 12x + 6 \).
(c) Other stationary point: \( (-2, 22) \), a maximum.
(d) Tangent at \( x = 0 \): \( y = -12x + 6 \).
11. [Maximum mark: 9]
The diagram shows the circle with equation \( (x – 4)^2 + (y + 1)^2 = 40 \). Parallel tangents, each with gradient 1, touch the circle at points A and B.
(a) Find the equation of the line AB, giving the answer in the form \( y = mx + c \).
(b) Find the coordinates of A, giving each coordinate in surd form.
(c) Find the equation of the tangent at A, giving the answer in the form \( y = mx + c \), where \( c \) is in surd form.
▶️Answer/Explanation
(a) Gradient of AB \( = -1 \). Centre \( (4, -1) \).
Equation: \( y = -x + 3 \).
(b) Substitute into circle equation: \( x = 4 \pm \sqrt{20} \).
Coordinates: \( (4 – \sqrt{20}, -1 + \sqrt{20}) \).
(c) Tangent at A: \( y = x – 5 + 4\sqrt{5} \).