2025 Edexcel IGCSE Further Pure Mathematics 模擬試題連答案詳解

For the arithmetic series:
Using the sum formula \(S_n = \frac{n}{2}[2a + (n-1)d]\):
\(S_4 = \frac{4}{2}[2a + 3d] = 56\)
Given \(d = 4\):
\(2(2a + 12) = 56\)
\(2a + 12 = 28\)
\(2a = 16 \implies a = 8\)

For the geometric series:
First term is \(a = 8\).
Using the sum to infinity formula \(S_{\infty} = \frac{a}{1-r}\):
\(\frac{8}{1-r} = 32\)
\(1-r = \frac{8}{32} = \frac{1}{4}\)
\(r = \frac{3}{4}\)

The third term of the geometric series is given by:
\(u_3 = ar^2 = 8 \times \left(\frac{3}{4}\right)^2 = 8 \times \frac{9}{16} = 4.5\)

M1: Uses the arithmetic series sum formula with \(n=4\) and \(d=4\) to set up an equation in terms of \(a\).
A1: Correctly solves to find \(a = 8\).
M1: Sets up the sum to infinity equation using their value of \(a\).
A1: Correctly solves to find \(r = \frac{3}{4}\) (or 0.75).
M1: Uses \(ar^2\) to calculate the third term.
A0.5: Correct value of \(4.5\) (or \(\frac{9}{2}\)).

For the equation \(2x^2 - 5x + 4 = 0\):
\(\alpha + \beta = \frac{5}{2}\)
\(\alpha\beta = 2\)

Let the new roots be \(p = \frac{\alpha}{\beta} + 1 = \frac{\alpha+\beta}{\beta}\) and \(q = \frac{\beta}{\alpha} + 1 = \frac{\alpha+\beta}{\alpha}\).

Find the sum of the new roots:
\(p + q = \frac{\alpha+\beta}{\beta} + \frac{\alpha+\beta}{\alpha} = (\alpha+\beta)\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) = \frac{(\alpha+\beta)^2}{\alpha\beta}\)

Substituting the known values:
\(p + q = \frac{\left(\frac{5}{2}\right)^2}{2} = \frac{\frac{25}{4}}{2} = \frac{25}{8}\)

Find the product of the new roots:
\(p \times q = \left(\frac{\alpha+\beta}{\beta}\right)\left(\frac{\alpha+\beta}{\alpha}\right) = \frac{(\alpha+\beta)^2}{\alpha\beta} = \frac{25}{8}\)

The quadratic equation is of the form:
\(x^2 - (p+q)x + pq = 0\)
\(x^2 - \frac{25}{8}x + \frac{25}{8} = 0\)

Multiplying by 8 to obtain integer coefficients:
\(8x^2 - 25x + 25 = 0\)

B1: Identifies \(\alpha+\beta = 2.5\) and \(\alpha\beta = 2\).
M1: Algebraically simplifies the sum or product of the new roots in terms of \(\alpha+\beta\) and \(\alpha\beta\).
A1: Obtains the correct sum of the new roots as \(\frac{25}{8}\).
A1: Obtains the correct product of the new roots as \(\frac{25}{8}\).
M1: Applies the formula \(x^2 - (\text{Sum})x + (\text{Product}) = 0\).
A0.5: Gives the final equation in the correct integer coefficient form: \(8x^2 - 25x + 25 = 0\).

The binomial expansion of \((1 + ax)^n\) is:
\((1 + ax)^n = 1 + n(ax) + \frac{n(n-1)}{2}(ax)^2 + \frac{n(n-1)(n-2)}{6}(ax)^3 + \dots\)

Given the coefficients:
1) Coefficient of \(x\): \(na = -12 \implies a = -\frac{12}{n}\)
2) Coefficient of \(x^2\): \(\frac{n(n-1)}{2}a^2 = 60\)

Substitute \(a = -\frac{12}{n}\) into the second equation:
\(\frac{n(n-1)}{2} \left(-\frac{12}{n}\right)^2 = 60\)
\(\frac{n(n-1)}{2} \times \frac{144}{n^2} = 60\)
\(\frac{72(n-1)}{n} = 60\)
\(72n - 72 = 60n\)
\(12n = 72 \implies n = 6\)

Substitute \(n = 6\) back to find \(a\):
\(a = -\frac{12}{6} = -2\)

The coefficient of \(x^3\) is:
\(\frac{n(n-1)(n-2)}{6} a^3 = \frac{6 \times 5 \times 4}{6} (-2)^3 = 20 \times (-8) = -160\)

M1: Sets up equations for the coefficients of \(x\) and \(x^2\) in terms of \(n\) and \(a\).
M1: Eliminates \(a\) (or \(n\)) to obtain a single variable equation.
A1: Obtains \(n = 6\).
A1: Obtains \(a = -2\).
M1: Uses the correct third term coefficient term formula \(\frac{n(n-1)(n-2)}{6} a^3\) with their values.
A0.5: Obtains the correct coefficient of \(-160\).

Using the properties of logarithms:
\(2\log_3 x - \log_3(x - 2) = 2\)

Apply the power law \(2\log_3 x = \log_3 (x^2)\):
\(\log_3(x^2) - \log_3(x - 2) = 2\)

Apply the division law of logarithms:
\(\log_3\left(\frac{x^2}{x - 2}\right) = 2\)

Convert from logarithmic to exponential form:
\(\frac{x^2}{x - 2} = 3^2\)
\(\frac{x^2}{x - 2} = 9\)

Multiply through by \((x - 2)\):
\(x^2 = 9(x - 2)\)
\(x^2 = 9x - 18\)
\(x^2 - 9x + 18 = 0\)

Factorise the quadratic equation:
\((x - 3)(x - 6) = 0\)

This gives:
\(x = 3\) or \(x = 6\)
Both values are valid as they satisfy the domain conditions \(x > 0\) and \(x > 2\).

M1: Applies the power law of logarithms to get \(\log_3(x^2)\).
M1: Applies the division law of logarithms to combine into a single log expression.
M1: Converts the logarithmic equation into an exponential/index equation.
A1: Obtains the correct quadratic equation \(x^2 - 9x + 18 = 0\).
M1: Solves the quadratic equation by factoring or using the quadratic formula.
A0.5: Identifies both solutions \(x = 3\) and \(x = 6\) and confirms their validity.

(a) First, find the gradient of the tangent at P.
\( y = x^2 - 4x + 5 \)
\( \frac{dy}{dx} = 2x - 4 \)

At \( x = 3 \):
\( \frac{dy}{dx} = 2(3) - 4 = 2 \)

The gradient of the tangent at P is 2, so the gradient of the normal is \( m_n = -\frac{1}{2} \).

The equation of the normal is:
\( y - 2 = -\frac{1}{2}(x - 3) \)
\( 2y - 4 = -x + 3 \)
\( x + 2y - 7 = 0 \)

(b) To find where the normal intersects C again, solve the equations simultaneously:
\( x + 2y - 7 = 0 \implies x = 7 - 2y \)
Substitute \( y = x^2 - 4x + 5 \):
\( x + 2(x^2 - 4x + 5) - 7 = 0 \)
\( x + 2x^2 - 8x + 10 - 7 = 0 \)
\( 2x^2 - 7x + 3 = 0 \)

Factorising the quadratic equation:
\( (2x - 1)(x - 3) = 0 \)

Since \( x = 3 \) is the x-coordinate of P, the x-coordinate of Q is:
\( x = \frac{1}{2} = 0.5 \)

Substitute \( x = 0.5 \) back into the curve equation to find the y-coordinate:
\( y = (0.5)^2 - 4(0.5) + 5 = 0.25 - 2 + 5 = 3.25 \)

Thus, the coordinates of Q are \( (0.5, 3.25) \).

(a)
- M1: Differentiates the curve equation to find \( \frac{dy}{dx} \) (at least one term correct).
- M1: Evaluates gradient of normal as the negative reciprocal of \( \frac{dy}{dx} \) at \( x = 3 \).
- M1: Uses their normal gradient and point \( (3,2) \) to set up the straight line equation.
- A1: Correct linear equation in the form \( ax + by + c = 0 \) (e.g., \( x + 2y - 7 = 0 \) or any integer multiple).

(b)
- M1: Equates the normal equation with the curve equation to form a quadratic in x (or y).
- M1: Solves the quadratic equation by factorisation or formula.
- A1: Identifies the correct x-coordinate of Q as \( x = 0.5 \) (or \( 1/2 \)).
- A1: Finds the correct y-coordinate as \( y = 3.25 \) (or \( 13/4 \)).
- A1: Clearly states the final coordinates as \( (0.5, 3.25) \) or equivalent fractions.

(a) To find the stationary points, differentiate y with respect to x:
\( \frac{dy}{dx} = 6x^2 - 18x - 24 \)

Set \( \frac{dy}{dx} = 0 \) to find the stationary points:
\( 6x^2 - 18x - 24 = 0 \)
\( x^2 - 3x - 4 = 0 \)
\( (x - 4)(x + 1) = 0 \)

This gives \( x = 4 \) and \( x = -1 \).

For \( x = 4 \):
\( y = 2(4)^3 - 9(4)^2 - 24(4) + 15 \)
\( y = 2(64) - 9(16) - 96 + 15 = 128 - 144 - 96 + 15 = -97 \)

For \( x = -1 \):
\( y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 15 \)
\( y = -2 - 9 + 24 + 15 = 28 \)

The stationary points are \( (4, -97) \) and \( (-1, 28) \).

(b) Differentiate again to find the second derivative:
\( \frac{d^2y}{dx^2} = 12x - 18 \)

Test \( x = 4 \):
\( \frac{d^2y}{dx^2} = 12(4) - 18 = 48 - 18 = 30 \)
Since \( 30 > 0 \), the stationary point \( (4, -97) \) is a local minimum.

Test \( x = -1 \):
\( \frac{d^2y}{dx^2} = 12(-1) - 18 = -12 - 18 = -30 \)
Since \( -30 < 0 \), the stationary point \( (-1, 28) \) is a local maximum.

(a)
- M1: Differentiates to find \( \frac{dy}{dx} \) (at least two terms correct).
- M1: Equates their derivative to 0 and attempts to solve the quadratic equation.
- A1: Finds both correct x-values: \( x = 4 \) and \( x = -1 \).
- A1: Evaluates one y-coordinate correctly (either \( y = 28 \) or \( y = -97 \)).
- A1: Evaluates both coordinates correctly: \( (-1, 28) \) and \( (4, -97) \).

(b)
- M1: Differentiates to find \( \frac{d^2y}{dx^2} \).
- M1: Evaluates the sign of \( \frac{d^2y}{dx^2} \) at \( x = 4 \) or \( x = -1 \).
- A1: Correctly identifies \( (4, -97) \) as a local minimum with working.
- A1: Correctly identifies \( (-1, 28) \) as a local maximum with working.

(a) To find the points of intersection, set the equation of the curve equal to the equation of the line:
\( 8x - x^2 = 2x \)
\( x^2 - 6x = 0 \)
\( x(x - 6) = 0 \)

So, \( x = 0 \) or \( x = 6 \).

When \( x = 0 \), \( y = 2(0) = 0 \).
When \( x = 6 \), \( y = 2(6) = 12 \).

The points of intersection are \( (0, 0) \) and \( (6, 12) \).

(b) The area \( A \) of the bounded region is given by the integral:
\( A = \int_{0}^{6} (y_{\text{curve}} - y_{\text{line}}) \, dx \)
\( A = \int_{0}^{6} ((8x - x^2) - 2x) \, dx \)
\( A = \int_{0}^{6} (6x - x^2) \, dx \)

Integrate each term:
\( A = \left[ 3x^2 - \frac{x^3}{3} \right]_{0}^{6} \)

Substitute the upper limit \( x = 6 \):
\( A = \left( 3(6)^2 - \frac{6^3}{3} \right) - (0) \)
\( A = \left( 3(36) - \frac{216}{3} \right) \)
\( A = 108 - 72 = 36 \)

The area of the bounded region is 36.

(a)
- M1: Equates curve and line equations to form a quadratic equation.
- A1: Finds the correct x-coordinates \( x = 0 \) and \( x = 6 \).
- A1: States both intersection points \( (0, 0) \) and \( (6, 12) \).

(b)
- M1: Sets up the definite integral for the area with correct integrand \( (8x - x^2) - 2x \).
- M1: Uses their limits from part (a) (e.g., 0 to 6) in the definite integral.
- A1: Correctly simplifies the integrand to \( 6x - x^2 \).
- M1: Integrates to find \( 3x^2 - \frac{x^3}{3} \) (allow one minor error).
- A1: Evaluates the definite integral substituting the limits \( 6 \) and \( 0 \).
- A1: Obtains the correct final area of 36.

(a) Complete the square for both x and y:
\( (x - 3)^2 - 9 + (y - 4)^2 - 16 + 5 = 0 \)
\( (x - 3)^2 + (y - 4)^2 - 20 = 0 \)
\( (x - 3)^2 + (y - 4)^2 = 20 \)

The centre of C is \( (3, 4) \).
The radius is \( \sqrt{20} = 2\sqrt{5} \).

(b) Method 1: Using the distance from the centre of the circle to the line.
The equation of the line L is \( 2x - y + c = 0 \).
The distance from the centre \( (3, 4) \) to this line must be equal to the radius \( 2\sqrt{5} \).
Using the perpendicular distance formula:
\( d = \frac{|2(3) - 4 + c|}{\sqrt{2^2 + (-1)^2}} = 2\sqrt{5} \)
\( \frac{|6 - 4 + c|}{\sqrt{5}} = 2\sqrt{5} \)
\( |2 + c| = 2(5) \)
\( |2 + c| = 10 \)

This gives two equations:
\( 2 + c = 10 \implies c = 8 \)
\( 2 + c = -10 \implies c = -12 \)

Method 2: Using the discriminant.
Substitute \( y = 2x + c \) into \( (x - 3)^2 + (y - 4)^2 = 20 \):
\( (x - 3)^2 + (2x + c - 4)^2 = 20 \)
\( x^2 - 6x + 9 + 4x^2 + 4(c - 4)x + (c - 4)^2 - 20 = 0 \)
\( 5x^2 + (4c - 22)x + (c^2 - 8c - 11) = 0 \)

For L to be a tangent, the discriminant \( B^2 - 4AC = 0 \):
\( (4c - 22)^2 - 4(5)(c^2 - 8c - 11) = 0 \)
\( 16c^2 - 176c + 484 - 20c^2 + 160c + 220 = 0 \)
\( -4c^2 - 16c + 704 = 0 \)
\( c^2 + 4c - 176 = 0 \) is incorrect in expansion; let's check:
\( (c-4)^2 - 11 = c^2 - 8c + 16 - 11 = c^2 - 8c + 5 \).
So: \( -4(5)(c^2 - 8c + 5) = -20c^2 + 160c - 100 \).
Correct equation:
\( 16c^2 - 176c + 484 - 20c^2 + 160c - 100 = 0 \)
\( -4c^2 - 16c + 384 = 0 \)
\( c^2 + 4c - 96 = 0 \)
\( (c + 12)(c - 8) = 0 \)
This yields \( c = 8 \) or \( c = -12 \).

(a)
- M1: Attempts to complete the square for both x and y.
- A1: Finds the correct centre \( (3, 4) \).
- A1: Finds the correct radius \( 2\sqrt{5} \).

(b)
- M1: Sets up an equation for tangency using either the perpendicular distance formula or substituting into the circle equation.
- M1: (If distance method) Substitutes center and radius into the formula: \( \frac{|2(3) - 4 + c|}{\sqrt{5}} = 2\sqrt{5} \).
- M1: (If discriminant method) Obtains a quadratic in x: \( 5x^2 + (4c-22)x + (c^2-8c+5) = 0 \).
- A1: Correctly simplifies the condition to \( |c + 2| = 10 \) or \( c^2 + 4c - 96 = 0 \).
- A1: Finds one correct value of c (either 8 or -12).
- A1: Finds both correct values of c: \( c = 8 \) and \( c = -12 \).

(a) Let r be the radius of the water surface when the depth is h.
Since the semi-vertical angle is \( 30^\circ \), we have:
\( \tan(30^\circ) = \frac{r}{h} \)
\( \frac{1}{\sqrt{3}} = \frac{r}{h} \implies r = \frac{h}{\sqrt{3}} \)

The volume of water in the cone is:
\( V = \frac{1}{3}\pi r^2 h \)
Substitute \( r = \frac{h}{\sqrt{3}} \):
\( V = \frac{1}{3}\pi \left(\frac{h}{\sqrt{3}}\right)^2 h \)
\( V = \frac{1}{3}\pi \left(\frac{h^2}{3}\right) h = \frac{1}{9}\pi h^3 \) (as required)

(b) We are given that \( \frac{dV}{dt} = 6\pi \).
We want to find \( \frac{dh}{dt} \) when \( h = 4 \).

Using the chain rule:
\( \frac{dh}{dt} = \frac{dV}{dt} \times \frac{dh}{dV} \)

First, differentiate V with respect to h:
\( \frac{dV}{dh} = \frac{d}{dh}\left(\frac{1}{9}\pi h^3\right) = \frac{3}{9}\pi h^2 = \frac{1}{3}\pi h^2 \)

Therefore:
\( \frac{dh}{dt} = 6\pi \times \frac{3}{\pi h^2} = \frac{18}{h^2} \)

When \( h = 4 \):
\( \frac{dh}{dt} = \frac{18}{4^2} = \frac{18}{16} = 1.125 \text{ cm s}^{-1} \)

Rounding to 3 significant figures:
\( \frac{dh}{dt} \approx 1.13 \text{ cm s}^{-1} \).

(a)
- M1: Uses trigonometry to relate r and h, i.e., \( r = h \tan 30^\circ \).
- M1: Expresses r in terms of h using the exact value \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).
- A1: Substitutes into the volume of a cone formula and correctly simplifies to show the given formula.

(b)
- B1: Identifies the rate of change of volume as \( \frac{dV}{dt} = 6\pi \).
- M1: Differentiates the volume formula to find \( \frac{dV}{dh} \).
- M1: Formulates a correct chain rule relation connecting \( \frac{dh}{dt} \), \( \frac{dV}{dt} \), and \( \frac{dV}{dh} \).
- A1: Expresses \( \frac{dh}{dt} \) in terms of h as \( \frac{18}{h^2} \).
- M1: Substitutes \( h = 4 \) into their expression for \( \frac{dh}{dt} \).
- A1: Correctly calculates \( 1.13 \) (accept \( 1.125 \) or exact fraction \( 9/8 \)).

(a) Find the midpoint \(M\) of \(AB\):
\(M = \left(\frac{3+11}{2}, \frac{4+10}{2}\right) = (7, 7)\)

Find the gradient of \(AB\), denoted \(m_{AB}\):
\(m_{AB} = \frac{10 - 4}{11 - 3} = \frac{6}{8} = \frac{3}{4}\)

The gradient of the perpendicular bisector \(L\) is:
\(m_L = -\frac{1}{m_{AB}} = -\frac{4}{3}\)

Using the point-slope form with \(M(7, 7)\):
\(y - 7 = -\frac{4}{3}(x - 7)\)
\(3y - 21 = -4x + 28\)
\(4x + 3y = 49\) (as required)

(b) Since \(C\) passes through \(A\) and \(B\), its center must lie on the perpendicular bisector of \(AB\), which is \(L\). The center also lies on the line \(y = 3x - 1\).
Substitute \(y = 3x - 1\) into \(4x + 3y = 49\):
\(4x + 3(3x - 1) = 49\)
\(4x + 9x - 3 = 49\)
\(13x = 52 \implies x = 4\)

Find the \(y\)-coordinate of the center:
\(y = 3(4) - 1 = 11\)

So the center of \(C\) is \((4, 11)\).

(c) The radius \(r\) is the distance from the center \((4, 11)\) to point \(A(3, 4)\):
\(r = \sqrt{(4 - 3)^2 + (11 - 4)^2} = \sqrt{1^2 + 7^2} = \sqrt{50} = 5\sqrt{2}\)

Here, \(a = 5\) and \(b = 2\) (where 2 is prime).

(d) The tangent at \(A\) is perpendicular to the radius \(OA\), where \(O(4, 11)\) is the center.
Gradient of radius \(OA\):
\(m_{OA} = \frac{11 - 4}{4 - 3} = 7\)

Gradient of the tangent line \(m_T\):
\(m_T = -\frac{1}{7}\)

Using the point \(A(3, 4)\) to find the tangent equation:
\(y - 4 = -\frac{1}{7}(x - 3)\)
\(7y - 28 = -x + 3\)
\(x + 7y = 31\) (as required)

(a)
- M1: Correct method to find the midpoint of \(AB\).
- M1: Correct method to find the gradient of \(AB\).
- M1: Correct use of negative reciprocal gradient in a straight line equation through their midpoint.
- A1: Fully correct derivation showing \(4x + 3y = 49\).

(b)
- M1: Realises that the center is the intersection of \(L\) and the given line, attempting to solve them simultaneously.
- M1: Eliminates one variable to find a linear equation in \(x\) or \(y\).
- A1: Finds \(x = 4\) correctly.
- A1: Finds \(y = 11\) correctly.

(c)
- M1: Correct application of the distance formula between their center and \(A\) (or \(B\)).
- A1: Simplifies \(r = \sqrt{50}\) to \(5\sqrt{2}\).

(d)
- M1: Correct method to find the gradient of the radius \(OA\).
- M1: Finds the negative reciprocal of their radial gradient and attempts to write the equation of a line passing through \(A(3, 4)\).
- A1: Correctly simplifies to show \(x + 7y = 31\).

(a) We express \(\cos 3\theta\) using compound angle formulas:
\(\cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta \cos \theta - \sin 2\theta \sin \theta\)

Substitute the double angle formulas \(\cos 2\theta = 2\cos^2 \theta - 1\) and \(\sin 2\theta = 2\sin \theta \cos \theta\):
\(\cos 3\theta = (2\cos^2 \theta - 1)\cos \theta - (2\sin \theta \cos \theta)\sin \theta\)
\(\cos 3\theta = 2\cos^3 \theta - \cos \theta - 2\sin^2 \theta \cos \theta\)

Using identity \(\sin^2 \theta = 1 - \cos^2 \theta\):
\(\cos 3\theta = 2\cos^3 \theta - \cos \theta - 2(1 - \cos^2 \theta)\cos \theta\)
\(\cos 3\theta = 2\cos^3 \theta - \cos \theta - 2\cos \theta + 2\cos^3 \theta\)
\(\cos 3\theta = 4\cos^3 \theta - 3\cos \theta\) (as required)

(b) Substitute \(\cos 3x = 4\cos^3 x - 3\cos x\) and \(\sin 2x = 2\sin x \cos x\) into the given equation:
\((4\cos^3 x - 3\cos x) + 3\cos x = 2(2\sin x \cos x)\)
\(4\cos^3 x = 4\sin x \cos x\)
\(\cos^3 x - \sin x \cos x = 0\)
\(\cos x (\cos^2 x - \sin x) = 0\)

This yields two possible cases:

Case 1: \(\cos x = 0\)
In the interval \(0 \le x < 2\pi\):
\(x = \frac{\pi}{2}\) (or \(1.57\)) and \(x = \frac{3\pi}{2}\) (or \(4.71\))

Case 2: \(\cos^2 x - \sin x = 0\)
Substitute \(\cos^2 x = 1 - \sin^2 x\):
\(1 - \sin^2 x - \sin x = 0 \implies \sin^2 x + \sin x - 1 = 0\)

Solving using the quadratic formula:
\(\sin x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}\)

Since \(\sin x\) must lie in the range \([-1, 1]\), we discard \(\frac{-1 - \sqrt{5}}{2} \approx -1.618\).
Thus, \(\sin x = \frac{\sqrt{5} - 1}{2} \approx 0.61803\)

Finding the solutions for \(x\):
\(x_1 = \arcsin(0.61803) \approx 0.666\) rad
\(x_2 = \pi - 0.6662 \approx 2.48\) rad

All solutions in \(0 \le x < 2\pi\) are \(x = \frac{\pi}{2}, \frac{3\pi}{2}, 0.666, 2.48\).

(c) Prove \(\frac{1 - \cos 2\theta + \sin 2\theta}{1 + \cos 2\theta + \sin 2\theta} = \tan \theta\):

Using \(\cos 2\theta = 1 - 2\sin^2 \theta\) in the numerator:
\(1 - \cos 2\theta + \sin 2\theta = 1 - (1 - 2\sin^2 \theta) + 2\sin\theta\cos\theta = 2\sin^2\theta + 2\sin\theta\cos\theta = 2\sin\theta(\sin\theta + \cos\theta)\)

Using \(\cos 2\theta = 2\cos^2 \theta - 1\) in the denominator:
\(1 + \cos 2\theta + \sin 2\theta = 1 + (2\cos^2 \theta - 1) + 2\sin\theta\cos\theta = 2\cos^2\theta + 2\sin\theta\cos\theta = 2\cos\theta(\cos\theta + \sin\theta)\)

Substituting these back into the expression:
\(\text{LHS} = \frac{2\sin\theta(\sin\theta + \cos\theta)}{2\cos\theta(\cos\theta + \sin\theta)} = \frac{\sin\theta}{\cos\theta} = \tan\theta = \text{RHS}\)

(a)
- M1: Applies the compound angle formula to expand \(\cos(2\theta + \theta)\).
- M1: Correctly substitutes double angle formulas for both \(\cos 2\theta\) and \(\sin 2\theta\).
- M1: Uses \(\sin^2 \theta = 1 - \cos^2 \theta\) to form an expression entirely in terms of \(\cos \theta\).
- A1: Fully correct simplified identity reached without errors.

(b)
- M1: Substitutes \(\cos 3x\) expansion from part (a) and writes \(\sin 2x = 2\sin x \cos x\).
- M1: Factorizes to get \(\cos x(\cos^2 x - \sin x) = 0\) (or equivalent form).
- A1: Finds \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\).
- M1: Sets up and attempts to solve the quadratic equation \(\sin^2 x + \sin x - 1 = 0\).
- A1: Identifies \(\sin x = \frac{\sqrt{5} - 1}{2}\) and finds one angle (\(x \approx 0.666\)).
- A1: Finds the second angle (\(x \approx 2.48\)).

(c)
- M1: Chooses and substitutes the correct version of \(\cos 2\theta\) for either the numerator or the denominator.
- A1: Simplifies the numerator successfully to \(2\sin\theta(\sin\theta + \cos\theta)\).
- M1: Simplifies the denominator successfully to \(2\cos\theta(\cos\theta + \sin\theta)\).
- A1: Shows that common factors cancel to yield \(\tan\theta\) with no logical errors.

Using the double angle identity \(\cos 2\theta = 1 - 2\sin^2 \theta\), we substitute this into the equation: \(3(1 - 2\sin^2 \theta) - \sin \theta - 1 = 0\). Simplifying gives: \(3 - 6\sin^2 \theta - \sin \theta - 1 = 0\), which rearranges to: \(6\sin^2 \theta + \sin \theta - 2 = 0\). Factoring the quadratic expression, we get: \((2\sin \theta - 1)(3\sin \theta + 2) = 0\). This gives two possible cases. Case 1: \(2\sin \theta - 1 = 0 \implies \sin \theta = \frac{1}{2}\). For the interval \(0^\circ \le \theta \le 360^\circ\), the solutions are \(\theta = 30^\circ, 150^\circ\). Case 2: \(3\sin \theta + 2 = 0 \implies \sin \theta = -\frac{2}{3}\). The reference angle is \(\sin^{-1}(2/3) \approx 41.81^\circ\). Since the sine value is negative, the angle lies in the third and fourth quadrants: \(\theta = 180^\circ + 41.81^\circ = 221.8^\circ\) and \(\theta = 360^\circ - 41.81^\circ = 318.2^\circ\). Combining all solutions, we get \(\theta = 30^\circ, 150^\circ, 221.8^\circ, 318.2^\circ\).

M1: Substitute \(\cos 2\theta = 1 - 2\sin^2 \theta\) to form a quadratic equation in terms of \(\sin \theta\). A1: Correctly simplify to obtain \(6\sin^2 \theta + \sin \theta - 2 = 0\). M1: Solve the quadratic equation to find \(\sin \theta = \frac{1}{2}\) and \(\sin \theta = -\frac{2}{3}\). A1: Identify \(\theta = 30^\circ, 150^\circ\). A1: Identify \(\theta = 221.8^\circ, 318.2^\circ\) (accept answers rounding to 222 and 318 if 1 decimal place is not specified, but penalise 1 mark if incorrect rounding is consistently applied).

To solve the inequality \(\frac{2x - 5}{x + 3} < 1\), we multiply both sides by \((x + 3)^2\) to ensure the multiplier is positive (noting that \(x \neq -3\)): \((2x - 5)(x + 3) < (x + 3)^2\). Expanding and moving all terms to one side: \((2x - 5)(x + 3) - (x + 3)^2 < 0\). Factor out the common term \((x + 3)\): \((x + 3)[(2x - 5) - (x + 3)] < 0\), which simplifies to \((x + 3)(x - 8) < 0\). The critical values for this inequality are \(x = -3\) and \(x = 8\). For the product to be strictly negative, \(x\) must lie between these critical values: \(-3 < x < 8\). Since the denominator cannot be zero, \(x \neq -3\), which is already excluded from our range. Thus, the solution set is \(-3 < x < 8\).

M1: Multiply both sides by \((x + 3)^2\) or rearrange the terms to get \(\frac{2x - 5}{x + 3} - 1 < 0\) and combine into a single fraction. A1: Correctly identify critical values \(x = -3\) and \(x = 8\). M1: Establish the inequality \((x + 3)(x - 8) < 0\) or use a sign table to analyze the intervals. A1: Correctly identify the interval between the roots. A1: State the final solution set as \(-3 < x < 8\) (or in set builder notation \(\{x : -3 < x < 8\}\)).

Using the trigonometric identity \(\sec^2 \theta = 1 + \tan^2 \theta\), substitute this into the equation: \(2(1 + \tan^2 \theta) + \tan \theta = 5\). Expanding this yields: \(2 + 2\tan^2 \theta + \tan \theta = 5\). Rearranging terms gives a quadratic equation in \(\tan \theta\): \(2\tan^2 \theta + \tan \theta - 3 = 0\). Factoring the quadratic equation gives: \((2\tan \theta + 3)(\tan \theta - 1) = 0\). This results in two cases. Case 1: \(\tan \theta = 1\). In the range \(0^\circ \le \theta \le 360^\circ\), the solutions are \(\theta = 45^\circ, 225^\circ\). Case 2: \(\tan \theta = -1.5\). The reference angle is \(\tan^{-1}(1.5) \approx 56.31^\circ\). Since tangent is negative, the angles lie in the second and fourth quadrants: \(\theta = 180^\circ - 56.31^\circ = 123.7^\circ\) and \(\theta = 360^\circ - 56.31^\circ = 303.7^\circ\). Thus, the complete solution set is \(\theta = 45^\circ, 123.7^\circ, 225^\circ, 303.7^\circ\).

M1: Substitute \(\sec^2 \theta = 1 + \tan^2 \theta\) to form an equation in terms of \(\tan \theta\). A1: Correctly write the quadratic equation as \(2\tan^2 \theta + \tan \theta - 3 = 0\). M1: Solve the quadratic equation to find \(\tan \theta = 1\) and \(\tan \theta = -1.5\). A1: Find the solutions \(\theta = 45^\circ, 225^\circ\). A1: Find the solutions \(\theta = 123.7^\circ, 303.7^\circ\) (accept answers rounding to 124 and 304 if 1 decimal place is not specified, but penalise 1 mark if incorrect rounding is consistently applied).

(a) The common difference \(d\) is given by:
\[d = u_2 - u_1 = \log_p 8 - \log_p 2\]
Using the laws of logarithms:
\[d = \log_p \left(\frac{8}{2}\right) = \log_p 4\]
Hence, \(k = 4\).

(b) The sum of the first \(n\) terms of an arithmetic series is:
\[S_n = \frac{n}{2}[2u_1 + (n-1)d]\]
Substitute \(u_1 = \log_p 2\) and \(d = \log_p 4 = 2\log_p 2\):
\[S_n = \frac{n}{2}[2\log_p 2 + (n-1)(2\log_p 2)]\]
Factor out \(2\log_p 2\):
\[S_n = \frac{n}{2} \cdot 2\log_p 2 [1 + (n-1)]\]
\[S_n = n \log_p 2 [n] = n^2 \log_p 2\] (as required).

(c) Given \(S_{12} = 36\):
\[12^2 \log_p 2 = 36\]
\[144 \log_p 2 = 36\]
\[\log_p 2 = \frac{36}{144} = \frac{1}{4}\]
Converting to exponential form:
\[p^{\frac{1}{4}} = 2\]
\[p = 2^4 = 16\]

(a)
M1: Attempting to find the common difference by subtracting \(u_1\) from \(u_2\).
A1: Correctly simplifying to obtain \(d = \log_p 4\) (or stating \(k = 4\)).

(b)
M1: Using the sum of an arithmetic series formula with \(u_1\) and \(d\).
M1: Substituting \(d = 2\log_p 2\) or expressing all terms in terms of \(\log_p 2\).
M1: Factoring out \(\log_p 2\) and simplifying the algebraic expression inside the bracket.
A1: Correctly completing the algebraic proof to show \(S_n = n^2 \log_p 2\).

(c)
M1: Setting \(12^2 \log_p 2 = 36\) and attempting to evaluate \(12^2\).
M1: Converting \(\log_p 2 = \frac{1}{4}\) to index form \(p^{\frac{1}{4}} = 2\).
A1: Finding the exact value \(p = 16\).

(a) The vertical asymptote occurs where the denominator is zero, so \(x = 1\).
As \(x \to \pm\infty\), \(y \to 1\), so the horizontal asymptote is \(y = 1\).

(b) For the y-intercept, set \(x = 0\):
\[y = \frac{0+3}{0-1} = -3 \implies (0, -3)\]
For the x-intercept, set \(y = 0\):
\[0 = \frac{x+3}{x-1} \implies x + 3 = 0 \implies x = -3 \implies (-3, 0)\]

(c) The sketch should show a rectangular hyperbola with branches in the region where \(x > 1, y > 1\) and \(x < 1, y < 1\), crossing the axes at \((-3, 0)\) and \((0, -3)\), with asymptotes clearly marked as dashed lines at \(x = 1\) and \(y = 1\).

(d) To solve \(\frac{x+3}{x-1} > x\):
Subtract \(x\) from both sides:
\[\frac{x+3}{x-1} - x > 0\]
\[\frac{x+3 - x(x-1)}{x-1} > 0\]
\[\frac{-x^2 + 2x + 3}{x-1} > 0\]
Multiply both sides by \((x-1)^2\) (since \((x-1)^2 > 0\) for \(x \neq 1\)) to clear the denominator:
\[(-x^2 + 2x + 3)(x-1) > 0\]
\[-(x-3)(x+1)(x-1) > 0\]
\[(x-3)(x-1)(x+1) < 0\]
The critical values are \(x = -1\), \(x = 1\), and \(x = 3\).
Testing the regions:
- For \(x < -1\), the expression is negative (True).
- For \(-1 < x < 1\), the expression is positive (False).
- For \(1 < x < 3\), the expression is negative (True).
- For \(x > 3\), the expression is positive (False).
Thus, the solution set is \(x < -1\) or \(1 < x < 3\).

(a)
B1: For \(x = 1\).
B1: For \(y = 1\).

(b)
B1: For \((0, -3)\) or \(y = -3\) when \(x = 0\).
B1: For \((-3, 0)\) or \(x = -3\) when \(y = 0\).

(c)
B1: Two correctly positioned branches with curves approaching \(x = 1\) and \(y = 1\).
B1: All intercepts and asymptotes correctly labeled.

(d)
M1: Expressing as a single fraction and finding the quadratic numerator \(-x^2 + 2x + 3\) or equivalent.
M1: Finding all three critical values: \(x = -1\), \(x = 1\), and \(x = 3\).
A1: Correctly specifying the ranges \(x < -1\) or \(1 < x < 3\) (allow set notation).

(a) The sum to infinity is given by:
\[S_{\infty} = \frac{a}{1-r} = 16\]
The sum of the first three terms is:
\[S_3 = \frac{a(1-r^3)}{1-r} = 14\]
We can rewrite the expression for \(S_3\) as:
\[S_3 = \left(\frac{a}{1-r}\right)(1-r^3) = 14\]
Substitute \(S_{\infty} = 16\) into the equation:
\[16(1-r^3) = 14\]
\[1-r^3 = \frac{14}{16} = \frac{7}{8}\]
\[r^3 = 1 - \frac{7}{8} = \frac{1}{8}\] (as required).

(b) From part (a):
\[r = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}\]
Substitute \(r = \frac{1}{2}\) into the sum to infinity equation:
\[\frac{a}{1 - 1/2} = 16 \implies \frac{a}{1/2} = 16 \implies a = 8\]

(c) We require \(S_n > 15.8\):
\[S_n = \frac{8(1 - (0.5)^n)}{1 - 0.5} = 16(1 - (0.5)^n)\]
Set up the inequality:
\[16(1 - (0.5)^n) > 15.8\]
\[1 - (0.5)^n > \frac{15.8}{16} = 0.9875\]
\[(0.5)^n < 0.0125\]
Taking logarithms on both sides:
\[\log(0.5^n) < \log(0.0125)\]
\[n \log(0.5) < \log(0.0125)\]
Since \(\log(0.5) < 0\), dividing by it reverses the inequality sign:
\[n > \frac{\log(0.0125)}{\log(0.5)} \approx 6.32\]
Since \(n\) must be an integer, the least value is \(n = 7\).

(a)
M1: Recalling correct formula for \(S_{\infty}\) or \(S_3\).
M1: Attempting to link the two equations by substitution or division.
A1: Reaching \(16(1-r^3) = 14\) or equivalent.
A1: Showing the step-by-step reduction to \(r^3 = \frac{1}{8}\) clearly.

(b)
B1: Deducing \(r = 0.5\).
M1: Correctly substituting their \(r\) into \(\frac{a}{1-r} = 16\).
A1: Finding \(a = 8\).

(c)
M1: Setting up \(16(1 - 0.5^n) > 15.8\) (or with \(=\)).
M1: Converting to the form \(0.5^n < 0.0125\) and applying logarithms or systematic trial.
A1: Correctly concluding that the least integer is \(n = 7\).

(a) Rewrite the term \(3^{2x+1}\) as \(3 \cdot (3^x)^2\).
Let \(u = 3^x\). The equation becomes:
\[3u^2 - 10u + 3 = 0\]
Factorising the quadratic:
\[(3u - 1)(u - 3) = 0\]
Thus, \(u = \frac{1}{3}\) or \(u = 3\).
Now solve for \(x\):
- If \(3^x = \frac{1}{3} = 3^{-1} \implies x = -1\)
- If \(3^x = 3 = 3^1 \implies x = 1\)
So, \(x = -1\) or \(x = 1\).

(b) From the first equation \(\log_y x = 2\):
\[x = y^2\]
where \(y > 0\) and \(y \neq 1\).
From the second equation \(\log_x (y+6) = 1\):
\[y+6 = x^1 = x\]
where \(x > 0\) and \(x \neq 1\).
Substitute \(x = y^2\) into the second equation:
\[y^2 = y+6\]
\[y^2 - y - 6 = 0\]
Factorising the quadratic:
\[(y-3)(y+2) = 0\]
So, \(y = 3\) or \(y = -2\).
Since \(y\) is the base of a logarithm, \(y > 0\), so we reject \(y = -2\). Therefore, \(y = 3\).
Using \(x = y^2\):
\[x = 3^2 = 9\]
Checking conditions: \(x = 9\) and \(y = 3\) are both valid bases. Thus, the solution is \(x = 9, y = 3\).

(a)
M1: Recognising \(3^{2x+1} = 3 \cdot (3^x)^2\) and substituting \(u = 3^x\) to form a quadratic.
M1: Solving the quadratic equation to obtain \(u = 1/3\) and \(u = 3\).
M1: Substituting back \(u = 3^x\) and attempting to solve for \(x\).
A1: Both correct values: \(x = -1\) and \(x = 1\).

(b)
M1: Converting \(\log_y x = 2\) to exponential form: \(x = y^2\).
M1: Converting \(\log_x (y+6) = 1\) to exponential form: \(x = y+6\).
M1: Equating the two expressions to form the quadratic equation \(y^2 - y - 6 = 0\) and attempting to solve it.
A1: Getting \(y = 3\) and showing rejection of \(y = -2\).
A1: Finding \(x = 9\).

(a) The \(n\)-th term of an arithmetic series is \(u_n = a + (n-1)d\).
For the 5th term:
\[a + 4d = 18\] (Equation 1)
The sum of the first \(n\) terms is \(S_n = \frac{n}{2}[2a + (n-1)d]\).
For the sum of the first 8 terms:
\[S_8 = \frac{8}{2}[2a + 7d] = 136\]
\[4[2a + 7d] = 136\]
\[2a + 7d = 34\] (Equation 2)
From Equation 1, multiply by 2:
\[2a + 8d = 36\] (Equation 3)
Subtract Equation 2 from Equation 3:
\[(2a + 8d) - (2a + 7d) = 36 - 34\]
\[d = 2\]
Substitute \(d = 2\) into Equation 1:
\[a + 4(2) = 18 \implies a = 10\]

(b) We are given \(\sum_{r=1}^{k} u_r = S_k = 580\).
Using \(a = 10\) and \(d = 2\):
\[S_k = \frac{k}{2}[2(10) + (k-1)2] = 580\]
\[\frac{k}{2}[20 + 2k - 2] = 580\]
\[\frac{k}{2}[2k + 18] = 580\]
\[k(k + 9) = 580\]
\[k^2 + 9k - 580 = 0\]
Factorising this quadratic equation:
\[(k - 20)(k + 29) = 0\]
Since \(k\) must be a positive integer, we reject \(k = -29\).
Thus, \(k = 20\).

(a)
M1: Setting up an equation for \(u_5\) in terms of \(a\) and \(d\): \(a + 4d = 18\).
M1: Setting up an equation for \(S_8\) in terms of \(a\) and \(d\): \(4(2a+7d) = 136\).
M1: Attempting to solve the two simultaneous equations for \(a\) or \(d\).
A1: Finding \(a = 10\) and \(d = 2\).

(b)
M1: Substituting their values of \(a\) and \(d\) into the arithmetic sum formula.
M1: Equating their sum to 580 and expanding to form a quadratic equation.
M1: Simplifying to \(k^2 + 9k - 580 = 0\) (or equivalent).
M1: Attempting to solve the quadratic equation (by factorisation or formula).
A1: Correctly choosing \(k = 20\) and rejecting the negative value.

(a)
We have \(\vec{OP} = \frac{2}{3}\mathbf{a}\) and \(\vec{OQ} = \frac{1}{4}\mathbf{b}\).
\(\vec{AQ} = \vec{AO} + \vec{OQ} = -\mathbf{a} + \frac{1}{4}\mathbf{b}\).
Since \(\vec{AX} = \lambda \vec{AQ}\), we get:
\(\vec{OX} = \vec{OA} + \vec{AX} = \mathbf{a} + \lambda\left(-\mathbf{a} + \frac{1}{4}\mathbf{b}\right) = (1-\lambda)\mathbf{a} + \frac{1}{4}\lambda \mathbf{b}\).

(b)
\(\vec{PB} = \vec{PO} + \vec{OB} = -\frac{2}{3}\mathbf{a} + \mathbf{b}\).
Since \(\vec{PX} = \mu \vec{PB}\), we get:
\(\vec{OX} = \vec{OP} + \vec{PX} = \frac{2}{3}\mathbf{a} + \mu\left(-\frac{2}{3}\mathbf{a} + \mathbf{b}\right) = \frac{2}{3}(1-\mu)\mathbf{a} + \mu \mathbf{b}\).

(c)
Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) from parts (a) and (b):
From the \(\mathbf{b}\) term: \(\frac{1}{4}\lambda = \mu\)
From the \(\mathbf{a}\) term: \(1-\lambda = \frac{2}{3}(1-\mu)\)
Substitute \(\mu = \frac{1}{4}\lambda\) into the second equation:
\(1-\lambda = \frac{2}{3}\left(1-\frac{1}{4}\lambda\right)\)
\(3(1-\lambda) = 2 - \frac{1}{2}\lambda\)
\(3 - 3\lambda = 2 - \frac{1}{2}\lambda\)
\(1 = \frac{5}{2}\lambda \implies \lambda = \frac{2}{5}\)
Then, \(\mu = \frac{1}{4}\left(\frac{2}{5}\right) = \frac{1}{10}\).

(d)
The area of triangle \(OAQ\) is \(\frac{1}{4}\) of the area of triangle \(OAB\) since they share the vertex \(A\) and \(OQ = \frac{1}{4}OB\).
Since \(X\) lies on \(AQ\) and \(\vec{AX} = \frac{2}{5}\vec{AQ}\), the area of triangle \(OAX\) is \(\frac{2}{5}\) of the area of triangle \(OAQ\).
Therefore, \(\text{Area}(OAX) = \frac{2}{5} \times \text{Area}(OAQ) = \frac{2}{5} \times \left(\frac{1}{4} \times \text{Area}(OAB)\right) = \frac{1}{10}\text{Area}(OAB)\).
Thus, \(k = \frac{1}{10}\).

(a)
M1: Writes down a correct expression for \(\vec{AQ}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
M1: Writes down an expression for \(\vec{OX}\) using \(\vec{OA} + \lambda \vec{AQ}\).
A1: Correctly simplifies to \((1-\lambda)\mathbf{a} + \frac{1}{4}\lambda \mathbf{b}\).

(b)
M1: Writes down a correct expression for \(\vec{PB}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
M1: Writes down an expression for \(\vec{OX}\) using \(\vec{OP} + \mu \vec{PB}\).
A1: Correctly simplifies to \(\frac{2}{3}(1-\mu)\mathbf{a} + \mu \mathbf{b}\).

(c)
M1: Equates the coefficients of \(\mathbf{b}\) to find a relation between \(\lambda\) and \(\mu\).
M1: Equates the coefficients of \(\mathbf{a}\) to set up a second simultaneous equation.
A1: Obtains \(\lambda = \frac{2}{5}\) (accept decimal equivalent).
A1: Obtains \(\mu = \frac{1}{10}\) (accept decimal equivalent).

(d)
M1: Realises that \(\text{Area}(OAQ) = \frac{1}{4}\text{Area}(OAB)\) or equivalent ratio logic.
A1: Correctly shows that \(k = \frac{1}{10}\) (accept 0.1).

(a)
Let \(O\) be the center of the square base \(ABCD\). The vertical height is \(VO = h\).
In the right-angled triangle \(VAO\), \(\angle VAO = \theta\) and the hypotenuse \(VA = 13\).
Given \(\tan \theta = \frac{12}{5}\), we have \(\sin \theta = \frac{12}{\sqrt{12^2 + 5^2}} = \frac{12}{13}\) and \(\cos \theta = \frac{5}{13}\).
Therefore, \(VO = h = VA \sin \theta = 13 \times \frac{12}{13} = 12\) cm (as required).
Also, \(AO = VA \cos \theta = 13 \times \frac{5}{13} = 5\) cm.
The diagonal \(AC\) of the square base is \(2 \times AO = 10\) cm.
Since \(ABCD\) is a square of side length \(x\):
\(x^2 + x^2 = AC^2 = 100 \implies 2x^2 = 100 \implies x^2 = 50 \implies x = 5\sqrt{2}\) cm.

(b)
Let \(M\) be the midpoint of \(AB\). Since \(VAB\) is an isosceles triangle, \(VM \perp AB\).
Also, \(OM \perp AB\) where \(OM = \frac{1}{2}x = \frac{5\sqrt{2}}{2} = 2.5\sqrt{2}\) cm.
The angle between the face \(VAB\) and the base \(ABCD\) is \(\angle VMO\).
In the right-angled triangle \(VMO\):
\(\tan(\angle VMO) = \frac{VO}{OM} = \frac{12}{2.5\sqrt{2}} = 2.4\sqrt{2} \approx 3.3941\).
Thus, \(\angle VMO = \arctan(2.4\sqrt{2}) \approx 73.58^\circ \approx 73.6^\circ\).

(c)
Let \(AP\) be the perpendicular from \(A\) to \(VB\) in triangle \(VAB\). By symmetry, \(CP\) is also perpendicular to \(VB\).
The angle between the face \(VAB\) and face \(VBC\) is \(\angle APC\).
First, find the slant height \(VM\):
\(VM = \sqrt{VO^2 + OM^2} = \sqrt{12^2 + 12.5} = \sqrt{156.5}\) cm.
The area of triangle \(VAB\) is:
\(\text{Area} = \frac{1}{2} \times AB \times VM = \frac{1}{2} \times 5\sqrt{2} \times \sqrt{156.5} = \frac{5\sqrt{313}}{2}\) cm\(^2\).
We also have \(\text{Area} = \frac{1}{2} \times VB \times AP = \frac{13}{2} AP\).
Therefore, \(\frac{13}{2} AP = \frac{5\sqrt{313}}{2} \implies AP = \frac{5\sqrt{313}}{13}\) cm.
Now, in triangle \(APC\), \(AP = CP = \frac{5\sqrt{313}}{13}\) and \(AC = 10\).
Using the cosine rule in triangle \(APC\):
\(\cos(\angle APC) = \frac{AP^2 + CP^2 - AC^2}{2 \times AP \times CP} = \frac{\frac{7825}{169} + \frac{7825}{169} - 100}{2 \times \frac{7825}{169}} = \frac{15650 - 16900}{15650} = -\frac{1250}{15650} = -\frac{25}{313}\).
Therefore, \(\angle APC = \arccos\left(-\frac{25}{313}\right) \approx 94.58^\circ \approx 94.6^\circ\).

(a)
M1: Sets up the trigonometric ratios using \(\tan \theta\) to find \(\sin \theta\) or \(\cos \theta\).
A1: Shows clearly that \(h = VO = 12\).
M1: Uses Pythagoras' theorem on the base diagonal or coordinates to write \(2x^2 = AC^2\).
A1: Finds the correct exact value of \(x = 5\sqrt{2}\) (or equivalent simplified surd).

(b)
M1: Identifies the required angle as \(\angle VMO\) where \(M\) is the midpoint of \(AB\).
M1: Calculates the length \(OM = 2.5\sqrt{2}\) or equivalent decimal \(\approx 3.536\).
M1: Uses a correct trigonometric ratio (e.g. \(\tan\)) for \(\angle VMO\).
A1: Obtains \(73.6^\circ\) (must be rounded to 1 decimal place).

(c)
M1: Identifies that the angle between the two planes is represented by \(\angle APC\) where \(AP \perp VB\) and \(CP \perp VB\).
M1: Uses the area of triangle \(VAB\) to calculate the height \(AP\).
M1: Applies the cosine rule on triangle \(APC\) to express \(\cos(\angle APC)\).
A1: Obtains \(94.6^\circ\) (must be rounded to 1 decimal place).

(a)
Let \(M\) be the midpoint of \(BC\). Since \(ABC\) is isosceles with \(AB=AC\), \(AM\) is perpendicular to \(BC\).
Thus, \(BM = MC = 6\) cm.
In the right-angled triangle \(ABM\):
\(AM = \sqrt{AB^2 - BM^2} = \sqrt{10^2 - 6^2} = 8\) cm.
The area of the base triangle \(ABC\) is:
\(\text{Area} = \frac{1}{2} \times BC \times AM = \frac{1}{2} \times 12 \times 8 = 48\text{ cm}^2\).

(b)
The vertical edge \(BE\) is perpendicular to the base plane \(ABC\), so \(BE \perp AB\).
In the right-angled triangle \(ABE\):
\(AE = \sqrt{AB^2 + BE^2} = \sqrt{10^2 + 15^2} = \sqrt{325} = 5\sqrt{13} \approx 18.0\) cm (to 1 d.p.).

(c)
The line \(AM\) is horizontal and perpendicular to \(BC\). Since the face \(BCFD\) is vertical and contains \(BC\), \(AM\) is perpendicular to the entire face \(BCFD\).
Thus, the projection of the line \(AE\) onto the face \(BCFD\) is the line \(ME\).
The angle between \(AE\) and the face \(BCFD\) is \(\angle AEM\).
In the right-angled triangle \(AME\) (where \(\angle AME = 90^\circ\)):
\(AM = 8\) cm, and \(AE = \sqrt{325}\) cm.
\(\sin(\angle AEM) = \frac{AM}{AE} = \frac{8}{\sqrt{325}} \approx 0.44376\).
Therefore, \(\angle AEM = \arcsin(0.44376) \approx 26.34^\circ \approx 26.3^\circ\) (to 1 d.p.).

(d)
The vertical plane \(AED\) contains the vertical edge \(AD\). Since \(BE\) is parallel to \(AD\), this plane is the vertical plane \(ABED\).
The face \(BCFD\) is also a vertical plane.
The line of intersection of these two vertical planes is the vertical edge \(BE\).
The angle between the two planes is the angle between their horizontal cross-sections (traces on the base plane), which is the angle \(\angle ABC\).
In the right-angled triangle \(ABM\):
\(\cos(\angle ABC) = \frac{BM}{AB} = \frac{6}{10} = 0.6\).
Therefore, \(\angle ABC = \arccos(0.6) \approx 53.13^\circ \approx 53.1^\circ\) (to 1 d.p.).

(a)
M1: Uses Pythagoras' theorem on the half-base to find the perpendicular height \(AM\) of the base triangle.
A1: Correct height \(AM = 8\) cm.
A1: Correct area of \(48\text{ cm}^2\).

(b)
M1: Identifies the right-angled triangle \(ABE\) and applies Pythagoras' theorem.
A1: Obtains the exact expression \(\sqrt{325}\) or \(5\sqrt{13}\).
A1: Correctly rounds to \(18.0\) cm (must have 1 decimal place).

(c)
M1: Identifies the projection of \(AE\) onto the plane \(BCFD\) as the line \(ME\) or identifies the angle as \(\angle AEM\).
M1: Sets up a correct trigonometric ratio (sine, cosine, or tangent) for the angle \(\angle AEM\).
A1: Obtains \(26.3^\circ\) (must be rounded to 1 decimal place).

(d)
M1: Identifies the dihedral angle as the angle \(\angle ABC\) (or equivalent angle in the horizontal plane).
M1: Applies basic trigonometry in triangle \(ABM\) or the cosine rule in triangle \(ABC\).
A1: Obtains \(53.1^\circ\) (must be rounded to 1 decimal place).