2024 Edexcel IGCSE Further Pure Mathematics 模擬試題連答案詳解

To find the area of the region \(R\), we first find the vertices of the triangle that forms the boundary of \(R\).

1. Find the intersection of \(y = 2x + 4\) and \(y = 4 - x\):
\[2x + 4 = 4 - x \implies 3x = 0 \implies x = 0\]
Substituting \(x = 0\) into \(y = 4 - x\), we get \(y = 4\).
So, the first vertex is \(A(0, 4)\).

2. Find the intersection of \(y = 2x + 4\) and \(x = 3\):
Substituting \(x = 3\) into \(y = 2x + 4\), we get \(y = 2(3) + 4 = 10\).
So, the second vertex is \(B(3, 10)\).

3. Find the intersection of \(y = 4 - x\) and \(x = 3\):
Substituting \(x = 3\) into \(y = 4 - x\), we get \(y = 4 - 3 = 1\).
So, the third vertex is \(C(3, 1)\).

The region \(R\) is a triangle with vertices \(A(0, 4)\), \(B(3, 10)\), and \(C(3, 1)\).
Using the side along the line \(x = 3\) as the base:
The length of the base is the vertical distance between \(B(3, 10)\) and \(C(3, 1)\):
\[\text{Base} = 10 - 1 = 9\]
The height of the triangle is the horizontal distance from the vertex \(A(0, 4)\) to the line \(x = 3\):
\[\text{Height} = 3 - 0 = 3\]

The area of the region \(R\) is:
\[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 9 \times 3 = 13.5\]

M1: For finding the coordinates of the intersection of \(y = 2x + 4\) and \(y = 4 - x\) to get \((0, 4)\).
M1: For finding the coordinates of the other two vertices, \((3, 10)\) and \((3, 1)\).
M1: For an attempt to use a valid method to find the area of the triangle formed by these three points (e.g. \(\frac{1}{2} \times \text{base} \times \text{height}\) or the coordinate geometry formula).
A1: For identifying a correct base of \(9\) and height of \(3\).
A1: For the correct area of \(13.5\) (or \(\frac{27}{2}\)).

Let \(O\) be the centre of the square base \(ABCD\).
First, find the length of the diagonal \(AC\) of the square base:
\[AC = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}\text{ cm}\]
Since \(O\) is the centre of the base, the distance \(OA\) is half of \(AC\):
\[OA = \frac{1}{2} AC = 5\sqrt{2}\text{ cm}\]

The angle between the line \(VA\) and the base plane \(ABCD\) is the angle \(\angle VAO\).
In the right-angled triangle \(VOA\), the angle \(\theta = \angle VAO\) satisfies:
\[\tan \theta = \frac{VO}{OA} = \frac{12}{5\sqrt{2}}\]

Calculate the value of \(\theta\):
\[\tan \theta \approx \frac{12}{7.0711} \approx 1.6971\]
\[\theta = \arctan(1.6971) \approx 59.49^\circ\]

To 1 decimal place, the angle is \(59.5^\circ\).

M1: Use Pythagoras' Theorem to find the diagonal \(AC\) or half-diagonal \(OA\) of the square base.
A1: Correct length for \(OA = 5\sqrt{2}\) (or \(\sqrt{50}\) or approximately \(7.07\)).
M1: Use of a correct trigonometric ratio in the right-angled triangle \(VOA\) to find the angle \(\angle VAO\).
A1: Correct trigonometric expression, e.g., \(\tan \theta = \frac{12}{5\sqrt{2}}\) (or \(\cos \theta = \frac{5\sqrt{2}}{\sqrt{194}}\)).
A1: Correct angle of \(59.5^\circ\) (accept \(59.5\) or \(59.5^\circ\)).

Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\), substitute this into the given equation:
\[4(1 - \sin^2 \theta) + 5\sin \theta - 5 = 0\]
Expand and simplify:
\[4 - 4\sin^2 \theta + 5\sin \theta - 5 = 0\]
\[-4\sin^2 \theta + 5\sin \theta - 1 = 0\]
Multiply by \(-1\):
\[4\sin^2 \theta - 5\sin \theta + 1 = 0\]

This is a quadratic equation in \(\sin \theta\). Factorising it gives:
\[(4\sin \theta - 1)(\sin \theta - 1) = 0\]

This gives two possible cases:
1. \(\sin \theta = 1\)
Within the range \(0^\circ \le \theta \le 360^\circ\):
\[\theta = 90^\circ\]

2. \(\sin \theta = \frac{1}{4}\)
Within the range \(0^\circ \le \theta \le 360^\circ\):
\[\theta = \arcsin(0.25) \approx 14.4775^\circ \approx 14.5^\circ\]
The other solution in the range is:
\[\theta = 180^\circ - 14.4775^\circ = 165.5225^\circ \approx 165.5^\circ\]

Thus, the solutions in the given range are \(\theta = 14.5^\circ\), \(90^\circ\), and \(165.5^\circ\).

M1: For using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to rewrite the equation in terms of \(\sin \theta\).
A1: For obtaining a correct quadratic equation, e.g., \(4\sin^2 \theta - 5\sin \theta + 1 = 0\).
M1: For solving the quadratic equation to get values for \(\sin \theta\), i.e., \(\sin \theta = 1\) and \(\sin \theta = 0.25\).
A1: For \(\theta = 90^\circ\).
A1: For \(\theta = 14.5^\circ\) and \(\theta = 165.5^\circ\) (both required for this mark, rounded to 1 decimal place).

The plane \(BDE\) intersects the horizontal base plane \(ABCD\) along the line \(BD\).
Let \(P\) be the point on the line \(BD\) such that \(AP\) is perpendicular to \(BD\).
Since the vertical edge \(AE\) is perpendicular to the base plane \(ABCD\), the line \(EP\) is perpendicular to \(BD\) by the theorem of three perpendiculars.
Thus, the angle between the plane \(BDE\) and the base plane \(ABCD\) is \(\angle EPA\).

First, find the length of the diagonal \(BD\):
\[BD = \sqrt{AB^2 + AD^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\text{ cm}\]

Now, calculate the length of the perpendicular \(AP\):
The area of the right-angled triangle \(ABD\) can be expressed in two ways:
\[\text{Area} = \frac{1}{2} \times AB \times AD = \frac{1}{2} \times 8 \times 6 = 24\]
And also:
\[\text{Area} = \frac{1}{2} \times BD \times AP = \frac{1}{2} \times 10 \times AP = 5 \times AP\]
Therefore:
\[5 \times AP = 24 \implies AP = 4.8\text{ cm}\]

In the right-angled triangle \(EAP\), the angle \(\theta = \angle EPA\) is given by:
\[\tan \theta = \frac{AE}{AP} = \frac{5}{4.8} = \frac{25}{24}\]

Calculating the angle \(\theta\):
\[\theta = \arctan\left(\frac{25}{24}\right) \approx 46.169^\circ\]

To 1 decimal place, the angle is \(46.2^\circ\).

M1: For finding the length of the diagonal \(BD = 10\text{ cm}\) (or setting up a correct method to find it).
A1: For finding the correct length of \(AP = 4.8\text{ cm}\) (the perpendicular distance from \(A\) to \(BD\)).
M1: For identifying the required angle as \(\angle EPA\) (where \(P\) is on \(BD\) such that \(AP \perp BD\)).
A1: For setting up the correct trigonometric ratio, e.g., \(\tan(\angle EPA) = \frac{5}{4.8}\).
A1: For the correct angle of \(46.2^\circ\) (or to 1 decimal place, accept answers rounding to 46.2).

(a) When \( P \) is at rest, \( v = 0 \). This gives \( 3t^2 - 14t + 8 = 0 \), which factors to \( (3t - 2)(t - 4) = 0 \). Thus, \( t = \frac{2}{3} \) or \( t = 4 \). (b) Acceleration \( a = \frac{dv}{dt} = 6t - 14 \). When \( t = 3 \), \( a = 6(3) - 14 = 4 \text{ m/s}^2 \). (c) The particle changes direction at \( t = \frac{2}{3} \). Let \( s(t) = \int v \\, dt = t^3 - 7t^2 + 8t \). At \( t = 0 \), \( s(0) = 0 \). At \( t = \frac{2}{3} \), \( s(\frac{2}{3}) = \frac{68}{27} \). At \( t = 3 \), \( s(3) = 3^3 - 7(3)^2 + 8(3) = -12 \). Total distance = \( |s(\frac{2}{3}) - s(0)| + |s(3) - s(\frac{2}{3})| = \frac{68}{27} + | -12 - \frac{68}{27} | = \frac{68}{27} + \frac{392}{27} = \frac{460}{27} \text{ m} \).

(a) M1: Set \( v = 0 \) and attempt to solve the quadratic equation. A1: One correct value of \( t \). A1: Both correct values \( t = 2/3 \) and \( t = 4 \). (b) M1: Differentiate \( v \) to get \( a = 6t - 14 \). A1: Correct acceleration of \( 4 \text{ m/s}^2 \). (c) M1: Integrate \( v \) to get \( t^3 - 7t^2 + 8t \). A1: Correct integration. M1: Substitute \( t = 2/3 \) into their integrated expression. M1: Substitute \( t = 3 \) into their integrated expression. M1: Add the two absolute differences. A1: Correct total distance of \( \frac{460}{27} \text{ m} \).

(a) The total surface area is \( 2\pi r^2 + 2\pi rh = 150\pi \). Dividing by \( 2\pi \) gives \( r^2 + rh = 75 \), so \( h = \frac{75 - r^2}{r} \). Substituting \( h \) into the volume formula: \( V = \pi r^2 h = \pi r^2 \left(\frac{75 - r^2}{r}\right) = 75\pi r - \pi r^3 \). (b) Differentiating \( V \) with respect to \( r \): \( \frac{dV}{dr} = 75\pi - 3\pi r^2 \). Setting \( \frac{dV}{dr} = 0 \) gives \( 3\pi r^2 = 75\pi \implies r^2 = 25 \implies r = 5 \) (since \( r > 0 \)). Substituting \( r = 5 \) into the volume formula: \( V = 75\pi(5) - \pi(5)^3 = 375\pi - 125\pi = 250\pi \text{ cm}^3 \). (c) The second derivative is \( \frac{d^2V}{dr^2} = -6\pi r \). When \( r = 5 \), \( \frac{d^2V}{dr^2} = -30\pi \). Since \( -30\pi < 0 \), the volume is indeed a maximum.

(a) M1: State surface area formula and set equal to \( 150\pi \). M1: Express \( h \) in terms of \( r \). M1: Substitute their expression for \( h \) into the volume formula. A1: Show given volume expression with no errors. (b) M1: Differentiate to find \( \frac{dV}{dr} \). A1: Correct derivative. M1: Set derivative to 0 and solve for \( r \). A1: Obtain \( r = 5 \). A1: Correct maximum volume of \( 250\pi \). (c) M1: Find second derivative. A1: Evaluate and state that since it is negative, it is a maximum.

(a) \\text{Area} = \\int_{0}^{\\ln 3} (2 + e^{2x}) \\, dx = \\left[ 2x + \\frac{1}{2}e^{2x} \\right]_{0}^{\\ln 3} = \\left( 2\\ln 3 + \\frac{1}{2}e^{2\\ln 3} \\right) - \\left( 0 + \\frac{1}{2} \\right) = 2\\ln 3 + \\frac{9}{2} - \\frac{1}{2} = 4 + 2\\ln 3 \\. (b) \\text{Volume} = \\pi \\int_{0}^{\\ln 3} y^2 \\, dx = \\pi \\int_{0}^{\\ln 3} (2 + e^{2x})^2 \\, dx = \\pi \\int_{0}^{\\ln 3} (4 + 4e^{2x} + e^{4x}) \\, dx = \\pi \\left[ 4x + 2e^{2x} + \\frac{1}{4}e^{4x} \\right]_{0}^{\\ln 3} = \\pi \\left( \\left( 4\\ln 3 + 2(9) + \\frac{81}{4} \\right) - \\left( 0 + 2 + \\frac{1}{4} \\right) \\right) = \\pi \\left( 4\\ln 3 + 18 + \\frac{81}{4} - \\frac{9}{4} \\right) = \\pi \\left( 36 + 4\\ln 3 \\right) \\, so \( a = 36 \) and \( b = 4 \).

(a) M1: State integral with correct limits. A1: Correctly integrate terms. M1: Substitute limits correctly. A1: Correct area of \( 4 + 2\ln 3 \). (b) M1: Set up volume integral. M1: Expand bracket to obtain three terms. A1: Correct expansion. M1: Integrate terms. A1: Correct integration. M1: Substitute upper and lower limits. A1: Correct exact volume \( \pi (36 + 4\ln 3) \).

(a) When \( x = 2 \), \( y = \frac{8}{2^2} = 2 \), so \( P \) is \( (2,2) \). Differentiating, \( \frac{dy}{dx} = -16x^{-3} \). At \( x = 2 \), \( \frac{dy}{dx} = -2 \). Tangent line is \( y - 2 = -2(x - 2) \implies y = -2x + 6 \). (b) At \( y = 0 \), \( x = 3 \) so \( A \) is \( (3,0) \). At \( x = 0 \), \( y = 6 \) so \( B \) is \( (0,6) \). Area of triangle \( OAB = \frac{1}{2} \times 3 \times 6 = 9 \). (c) The curve lies above the tangent for \( x > 2 \). Area \( S = \\int_{2}^{4} \\left( \\frac{8}{x^2} - (-2x + 6) \\right) \\, dx = \\int_{2}^{4} (8x^{-2} + 2x - 6) \\, dx = \\left[ -8x^{-1} + x^2 - 6x \\right]_{2}^{4} = \\left( -\\frac{8}{4} + 16 - 24 \\right) - \\left( -\\frac{8}{2} + 4 - 12 \\right) = -10 - (-12) = 2 \\.

(a) M1: Find \( y \)-coordinate. M1: Differentiate to find gradient. M1: Form line equation. A1: Correct tangent equation. (b) M1: Find intercepts. A1: Correct coordinates. A1: Correct area of 9. (c) M1: Set up correct difference integral. M1: Integrate terms. A1: Correct integrated expression. A1: Correct exact area of 2.

(a) Using the sum formula \(S_n = \frac{n}{2}[2a + (n-1)d]\):
For \(S_4 = 44\):
\(\frac{4}{2}[2a + 3d] = 44 \implies 2a + 3d = 22\) (Equation 1)

For \(S_{10} = 230\):
\(\frac{10}{2}[2a + 9d] = 230 \implies 2a + 9d = 46\) (Equation 2)

Subtracting Equation 1 from Equation 2:
\(6d = 24 \implies d = 4\)

Substituting \(d = 4\) back into Equation 1:
\(2a + 3(4) = 22 \implies 2a = 10 \implies a = 5\)

(b) The \(r\)-th term is:
\(u_r = a + (r-1)d = 5 + (r-1)4 = 5 + 4r - 4 = 4r + 1\)

(c) For progression \(B\), the \(k\)-th term is given by:
\(v_k = b + (k-1)d_B\) where \(d_B = 3\)
\(b + 3(k-1) = 32\)
\(b + 3k - 3 = 32 \implies b = 35 - 3k\)

(d) Write expressions for \(S_k\) and \(T_{2k}\):
\(S_k = \frac{k}{2}[2a + (k-1)d] = \frac{k}{2}[10 + 4(k-1)] = \frac{k}{2}[4k + 6] = k(2k+3)\)

\(T_{2k} = \frac{2k}{2}[2b + (2k-1)3] = k[2b + 6k - 3]\)

Substitute \(b = 35 - 3k\) into the expression for \(T_{2k}\):
\(T_{2k} = k[2(35 - 3k) + 6k - 3]\)
\(T_{2k} = k[70 - 6k + 6k - 3] = 67k\)

Using the given relationship \(T_{2k} - S_k = 270\):
\(67k - k(2k + 3) = 270\)
\(67k - 2k^2 - 3k = 270\)
\(2k^2 - 64k + 270 = 0\)
\(k^2 - 32k + 135 = 0\)

Solving the quadratic equation:
\((k - 5)(k - 27) = 0\)
\(k = 5\) or \(k = 27\)

(a)
- M1: Uses the sum formula \(S_n\) correctly for at least one of \(S_4\) or \(S_{10}\).
- A1: Obtains two correct simultaneous equations, e.g., \(2a + 3d = 22\) and \(2a + 9d = 46\).
- M1: Solves the simultaneous equations to find a value for \(d\).
- A1: \(d = 4\)
- A1: \(a = 5\)

(b)
- M1: Uses their values of \(a\) and \(d\) in the general term formula \(u_r = a + (r-1)d\).
- A1*: Simplifies correctly to show \(u_r = 4r + 1\) (no errors allowed).

(c)
- M1: Uses the formula for the \(k\)-th term of an AP with \(d = 3\).
- A1: Correctly forms the equation \(b + 3(k-1) = 32\).
- A1: Obtains \(b = 35 - 3k\).

(d)
- M1: Writes down a correct expression for \(S_k\) in terms of \(k\).
- M1: Writes down a correct expression for \(T_{2k}\) in terms of \(b\) and \(k\).
- M1: Substitutes their expression for \(b\) in terms of \(k\) into \(T_{2k}\).
- A1: Formulates the simplified quadratic equation \(k^2 - 32k + 135 = 0\) (or equivalent).
- M1: Attempts to factor or solve their 3-term quadratic equation.
- A1: Both values \(k = 5\) and \(k = 27\).

(a) From \(2x^2 - 5x + 4 = 0\), we have:
\(\alpha + \beta = \frac{5}{2}\) and \(\alpha\beta = 2\)

Using the identity for the sum of squares:
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{5}{2}\right)^2 - 2(2) = \frac{25}{4} - 4 = \frac{9}{4}\)

Using the sum of cubes identity:
\(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\)
\(\alpha^3 + \beta^3 = \left(\frac{5}{2}\right)^3 - 3(2)\left(\frac{5}{2}\right) = \frac{125}{8} - 15 = \frac{125 - 120}{8} = \frac{5}{8}\)

(b) For roots \(\frac{1}{\alpha^2}\) and \(\frac{1}{\beta^2}\):
Sum of roots \(S = \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = \frac{9/4}{2^2} = \frac{9}{16}\)
Product of roots \(P = \frac{1}{\alpha^2\beta^2} = \frac{1}{(\alpha\beta)^2} = \frac{1}{4}\)

The equation is \(x^2 - Sx + P = 0\):
\(x^2 - \frac{9}{16}x + \frac{1}{4} = 0\)

Multiplying by 16 to get integer coefficients:
\(16x^2 - 9x + 4 = 0\)

(c) For roots \(\alpha^3 - \beta\) and \(\beta^3 - \alpha\):
Sum of roots \(S' = (\alpha^3 - \beta) + (\beta^3 - \alpha) = (\alpha^3 + \beta^3) - (\alpha + \beta) = \frac{5}{8} - \frac{5}{2} = -\frac{15}{8}\)

Product of roots \(P' = (\alpha^3 - \beta)(\beta^3 - \alpha) = \alpha^3\beta^3 - \alpha^4 - \beta^4 + \alpha\beta = (\alpha\beta)^3 + \alpha\beta - (\alpha^4 + \beta^4)\)

First find \(\alpha^4 + \beta^4\):
\(\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 = \left(\frac{9}{4}\right)^2 - 2(2)^2 = \frac{81}{16} - 8 = -\frac{47}{16}\)

Substitute this back to find \(P'\):
\(P' = 2^3 + 2 - \left(-\frac{47}{16}\right) = 10 + \frac{47}{16} = \frac{207}{16}\)

The equation is \(x^2 - S'x + P' = 0\):
\(x^2 - \left(-\frac{15}{8}\right)x + \frac{207}{16} = 0\)
\(x^2 + \frac{15}{8}x + \frac{207}{16} = 0\)

Multiplying by 16 to get integer coefficients:
\(16x^2 + 30x + 207 = 0\)

(a)
- B1: States correct sum and product: \(\alpha + \beta = \frac{5}{2}\) and \(\alpha\beta = 2\).
- M1: Employs the identity \(\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta\).
- A1: Correctly evaluates \(\alpha^2 + \beta^2 = \frac{9}{4}\).
- M1: Uses a valid algebraic identity for \(\alpha^3 + \beta^3\).
- A1*: Clearly establishes the given value of \(\frac{5}{8}\).

(b)
- M1: Expresses the new sum \(S\) in terms of \(\alpha^2 + \beta^2\) and \(\alpha^2\beta^2\).
- A1: Obtains \(S = \frac{9}{16}\).
- B1: Obtains product \(P = \frac{1}{4}\).
- A1: Provides the equation \(16x^2 - 9x + 4 = 0\) with integer coefficients (must include \(= 0\)).

(c)
- M1: Attempts to find the sum of the new roots in terms of \(\alpha^3+\beta^3\) and \(\alpha+\beta\).
- A1: Correctly evaluates \(S' = -\frac{15}{8}\).
- M1: Expands the product of the new roots to express it in terms of known algebraic components.
- M1: Employs the identity \(\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2\).
- A1: Finds \(\alpha^4 + \beta^4 = -\frac{47}{16}\) and consequently \(P' = \frac{207}{16}\).
- M1: Substitutes their \(S'\) and \(P'\) into \(x^2 - S'x + P' = 0\).
- A1: Obtains \(16x^2 + 30x + 207 = 0\) (or any integer multiple thereof, must include \(= 0\)).

Let \(\theta\) be the angle of the sector in radians. The perimeter of the sector is given by: \(P = 2r + r\theta\). Given that \(P = 40\), we have: \(2r + r\theta = 40 \implies r\theta = 40 - 2r\). The area of the sector is given by: \(A = \frac{1}{2}r^2\theta\). Given that \(A = 96\), we have: \( \frac{1}{2}r^2\theta = 96 \). Substituting \(r\theta = 40 - 2r\) into the area equation gives: \(\frac{1}{2}r(40 - 2r) = 96 \implies r(20 - r) = 96 \implies 20r - r^2 = 96\). Rearranging this into a standard quadratic equation: \(r^2 - 20r + 96 = 0\). Factoring the quadratic: \((r - 8)(r - 12) = 0\). This gives the two possible values for the radius: \(r = 8\) or \(r = 12\).

M1: Formulates the perimeter equation \(2r + r\theta = 40\) or equivalent. M1: Formulates the area equation \(\frac{1}{2}r^2\theta = 96\). M1: Eliminates \(\theta\) to form a quadratic equation in terms of \(r\) only. M1: Solves the resulting quadratic equation \(r^2 - 20r + 96 = 0\). A1: Correctly identifies both possible values: \(r = 8\) and \(r = 12\).

To find the gradient of the tangent, we find the derivative \(\frac{dy}{dx}\) using the product rule. Let \(u = 2x - 3 \implies \frac{du}{dx} = 2\) and \(v = e^{x^2 - 2} \implies \frac{dv}{dx} = 2x e^{x^2 - 2}\). Applying the product rule: \(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} = (2x - 3)(2x e^{x^2 - 2}) + 2 e^{x^2 - 2} = e^{x^2 - 2}[2x(2x - 3) + 2] = e^{x^2 - 2}[4x^2 - 6x + 2]\). Substituting \(x = \sqrt{2}\): at this point, \(e^{x^2 - 2} = e^{2 - 2} = e^0 = 1\). Thus, the gradient is \(1 \cdot [4(\sqrt{2})^2 - 6\sqrt{2} + 2] = 4(2) - 6\sqrt{2} + 2 = 10 - 6\sqrt{2}\).

M1: Applies the product rule to differentiate \(y\). A1: Correctly differentiates \(e^{x^2-2}\) using the chain rule. A1: Obtains the correct derivative expression \(\frac{dy}{dx} = e^{x^2-2}(4x^2 - 6x + 2)\). M1: Substitutes \(x = \sqrt{2}\) and simplifies \(e^0 = 1\). A1: Obtains the final correct exact value \(10 - 6\sqrt{2}\).

The volume of a cone is given by \(V = \frac{1}{3}\pi r^2 h\). Since the height is constant at \(h = 12\text{ cm}\), the volume becomes: \(V = \frac{1}{3}\pi r^2 (12) = 4\pi r^2\). Differentiating with respect to \(r\) yields: \(\frac{dV}{dr} = 8\pi r\). We are given that the rate of change of the radius is \(\frac{dr}{dt} = 0.5\text{ cm/s}\). By the chain rule: \(\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}\). Substituting \(r = 6\) and \(\frac{dr}{dt} = 0.5\): \(\frac{dV}{dt} = 8\pi(6) \times 0.5 = 48\pi \times 0.5 = 24\pi\text{ cm}^3/\text{s}\).

M1: Substitutes \(h = 12\) to express the volume as \(V = 4\pi r^2\). M1: Differentiates \(V\) correctly to get \(\frac{dV}{dr} = 8\pi r\). M1: Formulates the chain rule relation \(\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}\). M1: Substitutes the given values \(r = 6\) and \(\frac{dr}{dt} = 0.5\). A1: Obtains the correct final exact answer \(24\pi\).

First, find the \(y\)-coordinate of the point where \(x = 3\): \(y = \frac{3(3) + 1}{2(3) - 5} = \frac{10}{1} = 10\). So the point on the curve is \((3, 10)\). Next, differentiate \(y\) using the quotient rule: \(\frac{dy}{dx} = \frac{3(2x - 5) - 2(3x + 1)}{(2x - 5)^2} = \frac{6x - 15 - 6x - 2}{(2x - 5)^2} = \frac{-17}{(2x - 5)^2}\). Substituting \(x = 3\) to find the gradient of the tangent: \(m_{\text{tangent}} = \frac{-17}{(2(3) - 5)^2} = -17\). The gradient of the normal is the negative reciprocal of the tangent's gradient: \(m_{\text{normal}} = -\frac{1}{-17} = \frac{1}{17}\). The equation of the normal is: \(y - 10 = \frac{1}{17}(x - 3) \implies 17y - 170 = x - 3 \implies x - 17y + 167 = 0\).

M1: Finds the correct \(y\)-coordinate: \(y = 10\) at \(x = 3\). M1: Applies the quotient or product rule correctly to differentiate \(y\). A1: Obtains \(\frac{dy}{dx} = \frac{-17}{(2x-5)^2}\). M1: Calculates the normal gradient as \(\frac{1}{17}\) and writes down a linear equation through \((3, 10)\). A1: Expresses the normal equation correctly as \(x - 17y + 167 = 0\) (or any integer multiple).

(a) Using the change of base formula, \(\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2} = \frac{1}{2} \log_2 x\). (b) Substituting the result from (a), \(2 \log_4 (y + 2) = 2 \left(\frac{1}{2} \log_2 (y + 2)\right) = \log_2 (y + 2)\). The equation becomes \(\log_2 (y - 1) + \log_2 (y + 2) = 2\). Combining the logs: \(\log_2 ((y - 1)(y + 2)) = 2\). Converting to exponential form: \((y - 1)(y + 2) = 2^2 = 4\). Expanding gives \(y^2 + y - 2 = 4\), so \(y^2 + y - 6 = 0\). Factorizing, we get \((y + 3)(y - 2) = 0\), which gives \(y = -3\) or \(y = 2\). Since \(y - 1 > 0\), we must have \(y > 1\), so we reject \(y = -3\). Thus, \(y = 2\). (c) From \(2^{w + 2z} = 16\), we have \(w + 2z = 4\). From \(\log_3 (4w + z) = 2\), we have \(4w + z = 3^2 = 9\). Rearranging the second equation gives \(z = 9 - 4w\). Substituting this into the first equation: \(w + 2(9 - 4w) = 4\), which simplifies to \(w + 18 - 8w = 4\), leading to \(-7w = -14\), so \(w = 2\). Substituting \(w = 2\) back into \(z = 9 - 4w\) gives \(z = 9 - 8 = 1\).

(a) M1 for change of base formula, A1 for \(\frac{1}{2} \log_2 x\). (b) M1 for simplifying \(2 \log_4 (y + 2)\) to \(\log_2 (y + 2)\). M1 for combining logs to get \(\log_2 ((y - 1)(y + 2))\). M1 for removing logs to obtain a quadratic. M1 for solving the quadratic. A1 for \(y = 2\) and explicitly rejecting \(y = -3\). (c) M1 for finding \(w + 2z = 4\). M1 for finding \(4w + z = 9\). M1 for solving simultaneous equations. A1.25 for correct values \(w = 2, z = 1\).

(a)(i) \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}\). (ii) \(\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = 3\mathbf{b} - \frac{3}{4}\mathbf{a}\). (b) We expand both expressions for \(\overrightarrow{OR}\): First expression: \(\overrightarrow{OR} = (1 - m)\mathbf{a} + m\mathbf{b}\). Second expression: \(\overrightarrow{OR} = \left(\frac{3}{4} - \frac{3}{4}n\right)\mathbf{a} + 3n\mathbf{b}\). Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) respectively: \(1 - m = \frac{3}{4}(1 - n)\) and \(m = 3n\). Substituting \(m = 3n\) into the first coefficient equation: \(1 - 3n = \frac{3}{4} - \frac{3}{4}n\). Multiplying both sides by 4 gives \(4 - 12n = 3 - 3n\), which simplifies to \(9n = 1\), so \(n = \frac{1}{9}\). Then \(m = 3 \left(\frac{1}{9}\right) = \frac{1}{3}\). (c) Since \(\overrightarrow{OR} = \mathbf{a} + m(\mathbf{b} - \mathbf{a}) = \overrightarrow{OA} + m\overrightarrow{AB}\), we have \(\overrightarrow{AR} = m\overrightarrow{AB} = \frac{1}{3}\overrightarrow{AB}\). This shows that \(R\) divides \(AB\) in the ratio \(1 : 2\).

(a) B1 for \(\mathbf{b} - \mathbf{a}\). B1 for \(3\mathbf{b} - \frac{3}{4}\mathbf{a}\). (b) M1 for expressing \(\overrightarrow{OR}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\) for both expressions. M1 for setting up the equation for \(\mathbf{b}\) coefficients: \(m = 3n\). M1 for setting up the equation for \(\mathbf{a}\) coefficients: \(1 - m = \frac{3}{4}(1 - n)\). M1 for substituting and solving for \(n\). A1 for \(n = \frac{1}{9}\). A1 for \(m = \frac{1}{3}\). (c) M1 for identifying that \(\overrightarrow{AR} = \frac{1}{3}\overrightarrow{AB}\). M1 for writing the ratio equation. A1.25 for ratio \(1 : 2\).

(a) Equating the expressions for \(y\): \(6x - x^2 = x + 4\), which simplifies to \(x^2 - 5x + 4 = 0\). Factorizing gives \((x - 1)(x - 4) = 0\), so \(x = 1\) or \(x = 4\). When \(x = 1\), \(y = 1 + 4 = 5\). When \(x = 4\), \(y = 4 + 4 = 8\). The intersection points are \((1, 5)\) and \((4, 8)\). (b) Differentiating \(y = 6x - x^2\) with respect to \(x\) gives \(\frac{dy}{dx} = 6 - 2x\). At the maximum point, \(\frac{dy}{dx} = 0\), so \(6 - 2x = 0\), which gives \(x = 3\). When \(x = 3\), \(y = 6(3) - 3^2 = 18 - 9 = 9\). The coordinates of the maximum point are \((3, 9)\). (c) The area of the finite region is given by \(\int_{1}^{4} ((6x - x^2) - (x + 4)) dx = \int_{1}^{4} (5x - x^2 - 4) dx\). Integrating each term: \(\left[ \frac{5x^2}{2} - \frac{x^3}{3} - 4x \right]_{1}^{4}\). Evaluating at the upper limit \(4\): \(\frac{5(16)}{2} - \frac{64}{3} - 16 = 40 - \frac{64}{3} - 16 = \frac{8}{3}\). Evaluating at the lower limit \(1\): \(\frac{5(1)}{2} - \frac{1}{3} - 4 = \frac{5}{2} - \frac{1}{3} - 4 = -\frac{11}{6}\). The area is \(\frac{8}{3} - \left(-\frac{11}{6}\right) = \frac{16}{6} + \frac{11}{6} = \frac{27}{6} = \frac{9}{2} = 4.5\).

(a) M1 for equating the equations of the curve and line to form a quadratic. A1 for correct x-values \(1, 4\). A1 for correct y-values \(5, 8\). (b) M1 for differentiating \(y\) to get \(6 - 2x\). M1 for setting \(\frac{dy}{dx} = 0\) and solving for \(x\). A1 for the coordinates \((3, 9)\). (c) M1 for setting up the correct integral of \(y_C - y_L\) with correct limits. M1 for integrating correctly. M1 for substituting the limits \(4\) and \(1\). A1.25 for the correct exact area \(\frac{9}{2}\) (or \(4.5\)).

(a) Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), substitute \(\sin^2 \theta = 1 - \cos^2 \theta\) into the equation: \(6(1 - \cos^2 \theta) - \cos \theta - 4 = 0\). Expanding: \(6 - 6\cos^2 \theta - \cos \theta - 4 = 0\). Simplifying: \(-6\cos^2 \theta - \cos \theta + 2 = 0\). Multiplying by \(-1\) gives the required equation: \(6\cos^2 \theta + \cos \theta - 2 = 0\). (b) Factorizing the quadratic: \((3\cos\theta + 2)(2\cos\theta - 1) = 0\), which gives \(\cos\theta = -\frac{2}{3}\) or \(\cos\theta = \frac{1}{2}\). For \(\cos\theta = \frac{1}{2}\): \(\theta = 60^\circ\) or \(\theta = 360^\circ - 60^\circ = 300^\circ\). For \(\cos\theta = -\frac{2}{3}\): basic angle \(\alpha = \cos^{-1}\left(\frac{2}{3}\right) \approx 48.19^\circ\). In the second and third quadrants: \(\theta = 180^\circ - 48.19^\circ = 131.8^\circ\) and \(\theta = 180^\circ + 48.19^\circ = 228.2^\circ\). Thus, the solutions are \(60^\circ, 131.8^\circ, 228.2^\circ, 300^\circ\). (c) Let \(\theta = 2x\). Since \(0 \le x \le \pi\), we have \(0 \le 2x \le 2\pi\) (which corresponds to \(0^\circ \le \theta \le 360^\circ\) in degrees). Converting the solutions from (b) to radians: For \(2x = 60^\circ = \frac{\pi}{3}\), \(x = \frac{\pi}{6} \approx 0.52\) rad. For \(2x = 300^\circ = \frac{5\pi}{3}\), \(x = \frac{5\pi}{6} \approx 2.62\) rad. For \(2x = 131.81^\circ \approx 2.3005\) rad, \(x \approx 1.15\) rad. For \(2x = 228.19^\circ \approx 3.9827\) rad, \(x \approx 1.99\) rad. The solutions are \(x = 0.52, 1.15, 1.99, 2.62\).

(a) M1 for substituting \(1 - \cos^2 \theta\) for \(\sin^2 \theta\). M1 for expanding and rearranging. A1 for completing the show-that clearly. (b) M1 for factorizing the quadratic to get the two cosine roots. A1 for finding \(60^\circ, 300^\circ\). M1 for using the inverse cosine of \(2/3\) to find the principal angle. A1 for \(131.8^\circ\). A1 for \(228.2^\circ\). (c) M1 for transforming the limits and using their solutions from (b). M1 for converting degree angles to radians. A1.25 for all four correct values: \(0.52, 1.15, 1.99, 2.62\).

(a) The sum to infinity is given by \(S_{\infty} = \frac{a}{1-r} = 32\), which gives \(a = 32(1-r)\).
The second term is \(u_2 = ar\) and the third term is \(u_3 = ar^2\).
We are given that \(u_2 + u_3 = ar + ar^2 = ar(1+r) = 12\).
Substituting \(a = 32(1-r)\) into this equation:
\(32(1-r)r(1+r) = 12\)
\(32r(1-r^2) = 12\)
Divide by 4:
\(8r(1-r^2) = 3\)
\(8r - 8r^3 = 3\)
Rearranging gives:
\(8r^3 - 8r + 3 = 0\) (as required).

(b) Substitute \(r = \frac{1}{2}\) into \(8r^3 - 8r + 3\):
\(8\left(\frac{1}{2}\right)^3 - 8\left(\frac{1}{2}\right) + 3 = 8\left(\frac{1}{8}\right) - 4 + 3 = 1 - 4 + 3 = 0\).
Since the result is 0, \(r = \frac{1}{2}\) is a root of the equation.
Since \(r = \frac{1}{2}\) is a root, \((2r - 1)\) is a factor of \(8r^3 - 8r + 3\).
By polynomial division or inspection:
\(8r^3 - 8r + 3 = (2r - 1)(4r^2 + 2r - 3) = 0\).
To find the other two roots, solve \(4r^2 + 2r - 3 = 0\):
\(r = \frac{-2 \pm \sqrt{2^2 - 4(4)(-3)}}{2(4)}\)
\(r = \frac{-2 \pm \sqrt{4 + 48}}{8} = \frac{-2 \pm \sqrt{52}}{8} = \frac{-2 \pm 2\sqrt{13}}{8}\)
\(r = \frac{-1 \pm \sqrt{13}}{4}\).
So the other two roots are \(r = \frac{-1 + \sqrt{13}}{4}\) and \(r = \frac{-1 - \sqrt{13}}{4}\).

(c) For a geometric series to have a sum to infinity (converge), the common ratio must satisfy \(|r| < 1\).
Evaluating the roots:
\(\frac{-1 + \sqrt{13}}{4} \approx \frac{-1 + 3.606}{4} = 0.651\) (which satisfies \(|r| < 1\)).

\(\frac{-1 - \sqrt{13}}{4} \approx \frac{-1 - 3.606}{4} = -1.152\) (which does not satisfy \(|r| < 1\)).
Therefore, \(r = \frac{-1 - \sqrt{13}}{4}\) cannot be a possible value of \(r\) because its magnitude is greater than 1.

(d) For \(r = \frac{1}{2}\):
\(a = 32\left(1 - \frac{1}{2}\right) = 32\left(\frac{1}{2}\right) = 16\).

(e) We want \(S_n > 0.999 S_{\infty}\).
\(S_n = \frac{a(1-r^n)}{1-r}\) and \(S_{\infty} = 32\).
\(\frac{16(1 - (1/2)^n)}{1 - 1/2} > 0.999 \times 32\)
\(32(1 - (0.5)^n) > 31.968\)
\(1 - (0.5)^n > 0.999\)
\((0.5)^n < 0.001\)
Taking logarithms on both sides:
\(n \log(0.5) < \log(0.001)\)
Since \
\log(0.5)\) is negative, the inequality flips:
\(n > \frac{\log(0.001)}{\log(0.5)} \approx 9.966\).
Since \(n\) must be an integer, the least value of \(n\) is 10.

(a)
M1: Writes down correct expressions for \(S_{\infty}\) and \(u_2 + u_3\) in terms of \(a\) and \(r\).
M1: Substitutes \(a = 32(1-r)\) into the sum equation.
A1: Obtains a correct cubic expression in \(r\) (e.g., \(32r(1-r^2) = 12\)).
M1: Expands and simplifies terms towards the required equation.
A1*: Reaches the target equation \(8r^3 - 8r + 3 = 0\) with no errors.

(b)
B1: Shows clearly that substituting \(r = 1/2\) results in 0.
M1: Attempts to factorise the cubic equation, dividing by \((2r - 1)\) or equivalent.
A1: Obtains the correct quadratic factor \((4r^2 + 2r - 3)\).
M1: Applies the quadratic formula to solve \(4r^2 + 2r - 3 = 0\).
A1: Obtains the exact roots \(r = \frac{-1 \pm \sqrt{13}}{4}\).

(c)
B1: Correctly states the condition for convergence, \(|r| < 1\) (or \(-1 < r < 1\)).
B1: Explains that \(\frac{-1 - \sqrt{13}}{4} \approx -1.15\) lies outside this range, so it is invalid.

(d)
M1: Substitutes \(r = 1/2\) into \(a = 32(1-r)\).
A1: Correctly evaluates \(a = 16\).

(e)
M1: Sets up the inequality \(S_n > 0.999 S_{\infty}\) using their values.
A1: Simplifies to \((0.5)^n < 0.001\) (or equivalent).
A1: Identifies the correct least integer value of \(n = 10\).

(a) Using the quotient rule with \(u = 2x + 3\) and \(v = x - 1\):
\(\frac{du}{dx} = 2\) and \(\frac{dv}{dx} = 1\).
\(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} = \frac{2(x-1) - 1(2x+3)}{(x-1)^2}\)
\(\frac{dy}{dx} = \frac{2x - 2 - 2x - 3}{(x-1)^2} = -\frac{5}{(x-1)^2}\) (as required).

(b) First find the \(y\)-coordinate of \(P\) by substituting \(x = 2\) into the equation of \(C\):
\(y = \frac{2(2)+3}{2-1} = \frac{7}{1} = 7\).
So \(P\) is \((2, 7)\).
The gradient of the tangent at \(P\) is:
\(m_t = -\frac{5}{(2-1)^2} = -5\).
The gradient of the normal, \(m_n\), is the negative reciprocal of \(m_t\):
\(m_n = -\frac{1}{-5} = \frac{1}{5}\).
The equation of the normal is:
\(y - 7 = \frac{1}{5}(x - 2)\)
\(5(y - 7) = x - 2\)
\(5y - 35 = x - 2\)
\(x - 5y + 33 = 0\).

(c) To find \(Q\) (intersection with the \(x\)-axis, \(y=0\)):
\(x - 5(0) + 33 = 0 \implies x = -33\).
So \(Q\) is \((-33, 0)\).
To find \(R\) (intersection with the \(y\)-axis, \(x=0\)):
\(0 - 5y + 33 = 0 \implies 5y = 33 \implies y = \frac{33}{5} = 6.6\).
So \(R\) is \((0, 6.6)\) or \(\left(0, \frac{33}{5}\right)\).

(d) The triangle \(OQR\) has perpendicular base along the \(x\)-axis of length \(33\) and height along the \(y\)-axis of length \(6.6\).
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times |-33| \times \left|\frac{33}{5}\right| = \frac{1089}{10} = 108.9\).

(e) For a stationary point, we require \(\frac{dy}{dx} = 0\).
This gives \(-\frac{5}{(x-1)^2} = 0\).
Since the numerator is a non-zero constant (-5), this equation has no solutions. Hence, there are no stationary points on the curve \(C\).

(f) The line \(y = \frac{5}{4}x - 1\) has a gradient of \(\frac{5}{4}\).
The tangent to \(C\) is perpendicular to this line, so its gradient \(m_t\) must be the negative reciprocal:
\(m_t = -\frac{4}{5}\).
Set \(\frac{dy}{dx} = -\frac{4}{5}\):
\(-\frac{5}{(x-1)^2} = -\frac{4}{5}\)
\((x-1)^2 = \frac{25}{4}\)
\(x-1 = \pm \frac{5}{2}\).
Since \(x > 1\), we have:
\(x - 1 = 2.5 \implies x = 3.5 = \frac{7}{2}\).
Substitute \(x = 3.5\) back into the equation of \(C\):
\(y = \frac{2(3.5)+3}{3.5-1} = \frac{10}{2.5} = 4\).
So the coordinates of the point are \((3.5, 4)\) or \(\left(\frac{7}{2}, 4\right)\).

(a)
M1: Applies quotient rule (or product rule with chain rule) with correct derivatives of numerator and denominator.
M1: Obtains a correct algebraic fraction, simplifying the numerator.
A1*: Reaches \(-\frac{5}{(x-1)^2}\) with no errors.

(b)
B1: Finds correct coordinates of \(P\) as \((2, 7)\).
M1: Substitutes \(x=2\) into \(\frac{dy}{dx}\) to find gradient of tangent.
M1: Takes negative reciprocal to find gradient of normal (\(m_n = 1/5\)).
M1: Writes down equation of normal using their \(P\) and \(m_n\).
A1: Simplifies to correct integer form: \(x - 5y + 33 = 0\) (accept equivalent integer multiples).

(c)
B1: Finds \(Q = (-33, 0)\).
M1: Sets \(x = 0\) in their line equation to solve for \(y\).
A1: Finds \(R = (0, 6.6)\) or \((0, 33/5)\).

(d)
M1: Uses \(\frac{1}{2} \times \text{base} \times \text{height}\) with their magnitudes.
A1: Obtains \(108.9\) (or \(\frac{1089}{10}\)).

(e)
M1: States that stationary points occur when \(\frac{dy}{dx} = 0\).
A1: Explains clearly that \(-\frac{5}{(x-1)^2} = 0\) has no solutions because the numerator is non-zero.

(f)
M1: Finds perpendicular gradient \(m = -4/5\) and equates to \(\frac{dy}{dx}\).
A1: Solves for \(x\) (choosing positive branch because \(x > 1\)) and finds correct coordinates \((3.5, 4)\) (or \((7/2, 4)\)).