2023 Edexcel IGCSE Further Pure Mathematics Practice Paper with Answers

Let \( y = 2x + 15^\circ \). Since \( 0^\circ \le x \le 180^\circ \), the interval for \( y \) is \( 15^\circ \le y \le 375^\circ \). The given equation is: \( 6\sin^2 y - \sin y - 2 = 0 \). Factorising this quadratic equation in terms of \( \sin y \): \( (3\sin y - 2)(2\sin y + 1) = 0 \). This gives: \( \sin y = \frac{2}{3} \) or \( \sin y = -\frac{1}{2} \). Case 1: \( \sin y = \frac{2}{3} \). The solutions for \( y \) in the interval \( 15^\circ \le y \le 375^\circ \) are: \( y = \sin^{-1}\left(\frac{2}{3}\right) \approx 41.81^\circ \) and \( y = 180^\circ - 41.81^\circ \approx 138.19^\circ \). Case 2: \( \sin y = -\frac{1}{2} \). The solutions for \( y \) in the interval \( 15^\circ \le y \le 375^\circ \) are: \( y = 180^\circ - (-30^\circ) = 210^\circ \) and \( y = 360^\circ + (-30^\circ) = 330^\circ \). Now, solve for \( x \) using \( x = \frac{y - 15^\circ}{2} \): For \( y = 41.81^\circ \): \( x = \frac{41.81^\circ - 15^\circ}{2} \approx 13.4^\circ \). For \( y = 138.19^\circ \): \( x = \frac{138.19^\circ - 15^\circ}{2} \approx 61.6^\circ \). For \( y = 210^\circ \): \( x = \frac{210^\circ - 15^\circ}{2} = 97.5^\circ \). For \( y = 330^\circ \): \( x = \frac{330^\circ - 15^\circ}{2} = 157.5^\circ \). Thus, the values of \( x \) are \( 13.4^\circ \), \( 61.6^\circ \), \( 97.5^\circ \), and \( 157.5^\circ \).

M1: Attempt to solve the quadratic equation to find values for \( \sin(2x + 15^\circ) \) (e.g., by factorisation, formula, or calculator).
A1: Obtain \( \sin(2x + 15^\circ) = \frac{2}{3} \) and \( \sin(2x + 15^\circ) = -\frac{1}{2} \).
B1: Identify the correct range of interest for the substituted angle: \( 15^\circ \le 2x + 15^\circ \le 375^\circ \) (implied by correct values of \( y \)).
M1: Find at least two correct values for the angle \( 2x + 15^\circ \).
A1: Find all four correct values for the angle: \( 41.8^\circ \), \( 138.2^\circ \), \( 210^\circ \), and \( 330^\circ \) (accept answers to 1 d.p. or better).
M1: Complete process of solving for \( x \) by subtracting \( 15^\circ \) and dividing by 2.
A1: All four correct values of \( x \) written to 1 d.p.: \( x = 13.4^\circ, 61.6^\circ, 97.5^\circ, 157.5^\circ \). Deduct 1 mark if extra solutions are given within the range.

Let \( a \) be the first term and \( r \) be the common ratio of the geometric series. The sum to infinity is given by: \( S_\infty = \frac{a}{1-r} = 32 \) (Equation 1). The sum of the first three terms is given by: \( S_3 = \frac{a(1-r^3)}{1-r} = 28 \) (Equation 2). Substituting Equation 1 into Equation 2: \( 32(1-r^3) = 28 \). Solving for \( r^3 \): \( 1-r^3 = \frac{28}{32} = \frac{7}{8} \implies r^3 = \frac{1}{8} \). Taking the cube root gives: \( r = \frac{1}{2} \). Substitute \( r = \frac{1}{2} \) back into Equation 1 to find \( a \): \( \frac{a}{1 - 1/2} = 32 \implies 2a = 32 \implies a = 16 \). The \( n \)-th term of a geometric series is given by \( u_n = a r^{n-1} \). To find the 5th term: \( u_5 = 16 \left(\frac{1}{2}\right)^4 = 16 \times \frac{1}{16} = 1 \).

M1: Use the correct formulas for \( S_3 \) and \( S_\infty \) to set up equations \( \frac{a(1-r^3)}{1-r} = 28 \) and \( \frac{a}{1-r} = 32 \).
M1: Eliminate \( a \) to obtain a single equation in terms of \( r \): \( 32(1-r^3) = 28 \) or equivalent.
A1: Solve to find the common ratio \( r = \frac{1}{2} \) (or 0.5).
M1: Substitute their \( r \) into \( S_\infty = 32 \) to solve for \( a \).
A1: Obtain the first term \( a = 16 \).
M1: Use \( u_n = a r^{n-1} \) with \( n=5 \) and their values of \( a \) and \( r \).
A1: Obtain the correct 5th term \( u_5 = 1 \).

Part (a): The closed box has a square base of side \( x \) cm and height \( h \) cm. The surface area \( A \) is composed of the top and bottom squares, and four vertical rectangular faces: \( A = 2x^2 + 4xh \). Since \( A = 2400 \text{ cm}^2 \): \( 2x^2 + 4xh = 2400 \implies x^2 + 2xh = 1200 \). Expressing \( h \) in terms of \( x \) gives: \( 2xh = 1200 - x^2 \implies h = \frac{1200 - x^2}{2x} = \frac{600}{x} - \frac{x}{2} \). The volume \( V \) of the box is: \( V = x^2 h = x^2 \left(\frac{600}{x} - \frac{x}{2}\right) = 600x - \frac{1}{2}x^3 \) (as required).

Part (b): Differentiate \( V \) with respect to \( x \) to find stationary points: \( \frac{dV}{dx} = 600 - \frac{3}{2}x^2 \). Set \( \frac{dV}{dx} = 0 \implies 600 - \frac{3}{2}x^2 = 0 \implies x^2 = 400 \). Since length must be positive, \( x = 20 \) cm. To justify that this is a maximum, find the second derivative: \( \frac{d^2V}{dx^2} = -3x \). Substituting \( x = 20 \) gives \( \frac{d^2V}{dx^2} = -60 \). Since \( \frac{d^2V}{dx^2} < 0 \), the volume is maximum at \( x = 20 \). The maximum volume is: \( V = 600(20) - \frac{1}{2}(20)^3 = 12000 - 4000 = 8000\text{ cm}^3 \).

M1: Write a correct expression for the total surface area of the box: \( 2x^2 + 4xh = 2400 \).
M1: Rearrange this expression to make \( h \) the subject: \( h = \frac{1200 - x^2}{2x} \) or equivalent.
A1: Substitute this expression into \( V = x^2 h \) and simplify correctly to obtain \( V = 600x - \frac{1}{2}x^3 \) (cso).
M1: Differentiate \( V = 600x - \frac{1}{2}x^3 \) with respect to \( x \) to find \( \frac{dV}{dx} = 600 - \frac{3}{2}x^2 \).
M1: Equate their derivative to zero and solve for \( x \) to find \( x = 20 \).
A1: Calculate the maximum volume \( V = 8000\text{ cm}^3 \).
B1: Find the second derivative \( \frac{d^2V}{dx^2} = -3x \) and show that it is negative at \( x = 20 \) to justify that it is a maximum.

Subtract the right-hand side from both sides of the inequality:
\(\frac{2x + 4}{x + 3} - \frac{x + 2}{x - 1} < 0\).

Combine into a single fraction:
\(\frac{(2x + 4)(x - 1) - (x + 2)(x + 3)}{(x + 3)(x - 1)} < 0\).

Expand the numerator:
\((2x + 4)(x - 1) = 2x^2 + 2x - 4\)
\((x + 2)(x + 3) = x^2 + 5x + 6\).

Subtracting these:
\((2x^2 + 2x - 4) - (x^2 + 5x + 6) = x^2 - 3x - 10\).

So the inequality is:
\(\frac{x^2 - 3x - 10}{(x + 3)(x - 1)} < 0\).

Factor the numerator:
\(\frac{(x - 5)(x + 2)}{(x + 3)(x - 1)} < 0\).

The critical values are \(x = -3, -2, 1, 5\).

Test the intervals between the critical values:
1) For \(x < -3\), \(f(x) > 0\).
2) For \(-3 < x < -2\), \(f(x) < 0\) (Solution region).
3) For \(-2 < x < 1\), \(f(x) > 0\).
4) For \(1 < x < 5\), \(f(x) < 0\) (Solution region).
5) For \(x > 5\), \(f(x) > 0\).

Thus, the solution is \(-3 < x < -2\) or \(1 < x < 5\).

M1: For rearranging the inequality to compare with zero, e.g., \(\frac{2x + 4}{x + 3} - \frac{x + 2}{x - 1} < 0\).
M1: For finding a common denominator and expressing as a single fraction.
M1: For expanding the numerator: \((2x + 4)(x - 1) - (x + 2)(x + 3)\).
A1: For simplifying the numerator to \(x^2 - 3x - 10\).
M1: For factoring the numerator to get \((x - 5)(x + 2)\).
B1: For identifying all four critical values: \(x = -3, -2, 1, 5\).
M1: For testing intervals or using a sign diagram/graph to determine regions where the fraction is negative.
A1: For the interval \(-3 < x < -2\).
A1: For the interval \(1 < x < 5\).
A1: For expressing the complete and correct final answer as \(-3 < x < -2\) or \(1 < x < 5\).

(a) LHS = \(\frac{\sin 2\theta}{1 + \cos 2\theta}\). Use the double angle identities: \(\sin 2\theta = 2\sin\theta\cos\theta\) and \(\cos 2\theta = 2\cos^2\theta - 1\). Substituting these gives \(\frac{2\sin\theta\cos\theta}{1 + (2\cos^2\theta - 1)} = \frac{2\sin\theta\cos\theta}{2\cos^2\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta\) (QED).

(b) The equation is \(\frac{4\sin 2\theta}{1 + \cos 2\theta} = 3 \sec^2 \theta - 5\). Using part (a), this becomes \(4\tan\theta = 3\sec^2\theta - 5\). Since \(\sec^2\theta = 1 + \tan^2\theta\), we have \(4\tan\theta = 3(1 + \tan^2\theta) - 5\) which simplifies to \(3\tan^2\theta - 4\tan\theta - 2 = 0\). Let \(t = \tan\theta\), so \(3t^2 - 4t - 2 = 0\). Solving using the quadratic formula gives \(t = \frac{4 \pm \sqrt{16 - 4(3)(-2)}}{6} = \frac{2 \pm \sqrt{10}}{3}\). Case 1: \(\tan\theta = \frac{2 + \sqrt{10}}{3} \approx 1.7208 \implies \theta \approx 59.8^\circ\) or \(\theta \approx 180^\circ + 59.8^\circ = 239.8^\circ\). Case 2: \(\tan\theta = \frac{2 - \sqrt{10}}{3} \approx -0.3874 \implies \theta \approx 180^\circ - 21.2^\circ = 158.8^\circ\) or \(\theta \approx 360^\circ - 21.2^\circ = 338.8^\circ\). Thus, the solutions are \(\theta = 59.8^\circ, 158.8^\circ, 239.8^\circ, 338.8^\circ\).

Part (a):
M1: For using \(\sin 2\theta = 2\sin\theta\cos\theta\) and \(\cos 2\theta = 2\cos^2\theta - 1\) (or equivalent standard identities).
M1: For substituting these into the LHS expression to get \(\frac{2\sin\theta\cos\theta}{2\cos^2\theta}\).
A1: For simplifying to \(\tan\theta\) with complete and correct working shown.

Part (b):
M1: For substituting the result of (a) to obtain \(4\tan\theta = 3\sec^2\theta - 5\).
M1: For using \(\sec^2\theta = 1 + \tan^2\theta\) to obtain a quadratic equation in \(\tan\theta\).
A1: For the correct quadratic equation \(3\tan^2\theta - 4\tan\theta - 2 = 0\).
M1: For solving their quadratic equation to find two values for \(\tan\theta\) (decimal values are \(\approx 1.72\) and \(\approx -0.39\)).
M1: For finding at least one correct value of \(\theta\) to 1 d.p. from their tangent values.
A1: For obtaining any two of the correct angles (e.g., \(59.8^\circ\) and \(239.8^\circ\)).
A1: For finding all four correct angles: \(59.8^\circ, 158.8^\circ, 239.8^\circ, 338.8^\circ\) (to 1 d.p.), with no extra incorrect angles in the range.

(a) The total surface area of a closed cylinder is \(S = 2\pi r^2 + 2\pi r h = 450\pi\). Dividing by \(2\pi\) gives \(r^2 + rh = 225 \implies rh = 225 - r^2 \implies h = \frac{225 - r^2}{r}\).
The volume of the cylinder is \(V = \pi r^2 h\). Substituting \(h\) gives \(V = \pi r^2 \left(\frac{225 - r^2}{r}\right) = \pi r (225 - r^2) = 225\pi r - \pi r^3\) (QED).

(b) Differentiating \(V\) with respect to \(r\) gives \(\frac{\mathrm{d}V}{\mathrm{d}r} = 225\pi - 3\pi r^2\).
For a maximum value, set \(\frac{\mathrm{d}V}{\mathrm{d}r} = 0 \implies 225\pi - 3\pi r^2 = 0 \implies r^2 = 75\). Since \(r > 0\), we have \(r = \sqrt{75} = 5\sqrt{3}\).
Substituting this value back into the expression for \(V\): \(V = 225\pi(5\sqrt{3}) - \pi(5\sqrt{3})^3 = 1125\pi\sqrt{3} - 375\pi\sqrt{3} = 750\pi\sqrt{3}\).

(c) Differentiating again: \(\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -6\pi r\).
At \(r = 5\sqrt{3}\), we have \(\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -30\pi\sqrt{3} < 0\). Since the second derivative is negative, the value is a maximum.

Part (a):
M1: For writing the correct surface area formula \(S = 2\pi r^2 + 2\pi r h = 450\pi\).
M1: For expressing \(h\) in terms of \(r\): \(h = \frac{225 - r^2}{r}\).
M1: For substituting their \(h\) into the volume formula \(V = \pi r^2 h\).
A1: For obtaining \(V = 225\pi r - \pi r^3\) with correct working.

Part (b):
M1: For differentiating \(V\) to obtain \(\frac{\mathrm{d}V}{\mathrm{d}r} = 225\pi - 3\pi r^2\).
M1: For setting \(\frac{\mathrm{d}V}{\mathrm{d}r} = 0\) and solving for \(r\).
A1: For finding \(r = 5\sqrt{3}\) (accept \(\sqrt{75}\)).
A1: For substituting \(r = 5\sqrt{3}\) to find the maximum volume \(750\pi\sqrt{3}\).

Part (c):
M1: For finding \(\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -6\pi r\).
A1: For substituting \(r = 5\sqrt{3}\), evaluating it as negative, and concluding that it is a maximum.

(a) The 2nd, 5th, and 14th terms of the arithmetic series are \(t_2 = a + d\), \(t_5 = a + 4d\), and \(t_{14} = a + 13d\). Since these are the first three terms of a geometric series, \(\frac{a+4d}{a+d} = \frac{a+13d}{a+4d} \implies (a+4d)^2 = (a+d)(a+13d)\). Expanding both sides: \(a^2 + 8ad + 16d^2 = a^2 + 14ad + 13d^2\). Simplifying, we get \(3d^2 - 6ad = 0 \implies 3d(d - 2a) = 0\). Since \(r \ne 1\), \(d\) cannot be 0 (otherwise all terms of the geometric series would be equal, giving \(r = 1\)). Hence, \(d = 2a\) (QED).

(b) The sum of the first 4 terms of the arithmetic series is \(S_4 = \frac{4}{2}(2a + 3d) = 2(2a + 3d) = 32 \implies 2a + 3d = 16\). Substituting \(d = 2a\) into this equation: \(2a + 3(2a) = 16 \implies 8a = 16 \implies a = 2\). Then, \(d = 2(2) = 4\).

(c) The first term of the geometric series is \(A = t_2 = a + d = 2 + 4 = 6\). The common ratio is \(r = \frac{a+4d}{a+d} = \frac{2+16}{6} = 3\). The sum of the first 10 terms is \(S_{10} = \frac{A(r^{10} - 1)}{r - 1} = \frac{6(3^{10} - 1)}{3 - 1} = 3(59049 - 1) = 177144\).

Part (a):
M1: For writing expressions for \(t_2\), \(t_5\) and \(t_{14}\) in terms of \(a\) and \(d\).
M1: For using the geometric progression property \((a+4d)^2 = (a+d)(a+13d)\).
M1: For expanding both sides correctly to get \(a^2 + 8ad + 16d^2 = a^2 + 14ad + 13d^2\).
M1: For simplifying to get \(3d^2 - 6ad = 0\).
A1: For concluding \(d = 2a\) and clearly stating why \(d \ne 0\) (i.e., because \(r \ne 1\)).

Part (b):
M1: For using \(S_4 = 32\) to obtain \(4a + 6d = 32\) and substituting \(d = 2a\).
A1: For finding \(a = 2\) and \(d = 4\).

Part (c):
M1: For finding the first term of the geometric series \(A = 6\) and common ratio \(r = 3\).
M1: For substituting \(A = 6\) and \(r = 3\) into the sum formula \(S_{10} = \frac{A(r^{10}-1)}{r-1}\).
A1: For obtaining \(177144\).

(a) Since \(OP : PA = 2 : 1\), we have \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\).
Since \(OQ : QB = 1 : 3\), we have \(\overrightarrow{OQ} = \frac{1}{4}\mathbf{b}\).
\(\overrightarrow{AQ} = \overrightarrow{AO} + \overrightarrow{OQ} = -\mathbf{a} + \frac{1}{4}\mathbf{b}\).
\(\overrightarrow{BP} = \overrightarrow{BO} + \overrightarrow{OP} = -\mathbf{b} + \frac{2}{3}\mathbf{a} = \frac{2}{3}\mathbf{a} - \mathbf{b}\).

(b) First expression: \(\overrightarrow{OX} = \overrightarrow{OA} + \overrightarrow{AX} = \mathbf{a} + \lambda\overrightarrow{AQ} = \mathbf{a} + \lambda\left(-\mathbf{a} + \frac{1}{4}\mathbf{b}\right) = (1 - \lambda)\mathbf{a} + \frac{1}{4}\lambda\mathbf{b}\).
Second expression: \(\overrightarrow{OX} = \overrightarrow{OB} + \overrightarrow{BX} = \mathbf{b} + \mu\overrightarrow{BP} = \mathbf{b} + \mu\left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2}{3}\mu\mathbf{a} + (1 - \mu)\mathbf{b}\).
Equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\):
1) \(1 - \lambda = \frac{2}{3}\mu\)
2) \(\frac{1}{4}\lambda = 1 - \mu \implies \mu = 1 - \frac{1}{4}\lambda\).
Substitute (2) into (1): \(1 - \lambda = \frac{2}{3}\left(1 - \frac{1}{4}\lambda\right) \implies 1 - \lambda = \frac{2}{3} - \frac{1}{6}\lambda \implies 6 - 6\lambda = 4 - \lambda \implies 5\lambda = 2 \implies \lambda = \frac{2}{5} = 0.4\).
Substituting \(\lambda\) back into (2): \(\mu = 1 - \frac{1}{4}(0.4) = 0.9\).

(c) Since \(\overrightarrow{BX} = \mu\overrightarrow{BP} = 0.9\overrightarrow{BP} = \frac{9}{10}\overrightarrow{BP}\), the point \(X\) lies on the segment \(BP\) such that \(BX\) is \(\frac{9}{10}\) of the length of \(BP\). Thus, \(XP = \frac{1}{10}BP\). Therefore, the ratio of lengths is \(PX : XB = 1 : 9\).

Part (a):
B1: For \(\overrightarrow{AQ} = -\mathbf{a} + \frac{1}{4}\mathbf{b}\).
B1: For \(\overrightarrow{BP} = \frac{2}{3}\mathbf{a} - \mathbf{b}\).

Part (b):
M1: For writing an expression for \(\overrightarrow{OX}\) via \(A\): \(\overrightarrow{OX} = \mathbf{a} + \lambda \overrightarrow{AQ}\).
M1: For writing an expression for \(\overrightarrow{OX}\) via \(B\): \(\overrightarrow{OX} = \mathbf{b} + \mu \overrightarrow{BP}\).
A1: Both expressions correct in terms of \(\mathbf{a}\), \(\mathbf{b}\), \(\lambda\), and \(\mu\).
M1: For equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to form two simultaneous equations.
A1: For solving to find \(\lambda = 0.4\).
A1: For solving to find \(\mu = 0.9\).

Part (c):
M1: For writing \(\overrightarrow{BX} = \frac{9}{10}\overrightarrow{BP}\) or \(\overrightarrow{XP} = \frac{1}{10}\overrightarrow{BP}\) (or equivalent vector statement using their value of \(\mu\)).
A1: For finding the correct ratio \(PX : XB = 1 : 9\).

(a) Using the product rule for differentiation with \(u = x - 2\) and \(v = e^{2x}\):
\(\frac{\text{d}u}{\text{d}x} = 1\)
\(\frac{\text{d}v}{\text{d}x} = 2e^{2x}\)

\(\frac{\text{d}y}{\text{d}x} = u\frac{\text{d}v}{\text{d}x} + v\frac{\text{d}u}{\text{d}x}\)
\(\frac{\text{d}y}{\text{d}x} = (x - 2)(2e^{2x}) + e^{2x}(1) = e^{2x}(2x - 4 + 1) = (2x - 3)e^{2x}\)

(b) At the stationary point, \(\frac{\text{d}y}{\text{d}x} = 0\).
Since \(e^{2x} \neq 0\) for all real \(x\):
\(2x - 3 = 0 \Rightarrow x = \frac{3}{2}\)

Substitute \(x = \frac{3}{2}\) into the equation of \(C\) to find the \(y\)-coordinate:
\(y = \left(\frac{3}{2} - 2\right)e^{2(3/2)} = -\frac{1}{2}e^3\)

So the coordinates of the stationary point are \(\left(\frac{3}{2}, -\frac{1}{2}e^3\right)\).

To determine its nature, find the second derivative \(\frac{\text{d}^2y}{\text{d}x^2}\):
\(\frac{\text{d}^2y}{\text{d}x^2} = \frac{\text{d}}{\text{d}x}\left((2x - 3)e^{2x}\right) = 2e^{2x} + (2x - 3)(2e^{2x}) = e^{2x}(2 + 4x - 6) = (4x - 4)e^{2x}\)

At \(x = \frac{3}{2}\):
\(\frac{\text{d}^2y}{\text{d}x^2} = \left(4\left(\frac{3}{2}\right) - 4\right)e^3 = (6 - 4)e^3 = 2e^3\)

Since \(2e^3 > 0\), the stationary point is a local minimum.

(c) The curve crosses the \(y\)-axis where \(x = 0\).
When \(x = 0\), \(y = (0 - 2)e^0 = -2\).

The gradient of the tangent at \(x = 0\) is:
\(m_t = (2(0) - 3)e^0 = -3\)

The gradient of the normal, \(m_n\), is:
\(m_n = -\frac{1}{m_t} = \frac{1}{3}\)

The equation of the normal at \((0, -2)\) is:
\(y - (-2) = \frac{1}{3}(x - 0)\)
\(y + 2 = \frac{1}{3}x\)
\(3y + 6 = x\)
\(x - 3y - 6 = 0\)

(d) For a point of inflection, \(\frac{\text{d}^2y}{\text{d}x^2} = 0\).
\((4x - 4)e^{2x} = 0\)

Since \(e^{2x} \neq 0\):
\(4x - 4 = 0 \Rightarrow x = 1\)

Since \(\frac{\text{d}^2y}{\text{d}x^2} < 0\) for \(x < 1\) and \(\frac{\text{d}^2y}{\text{d}x^2} > 0\) for \(x > 1\), there is a change of sign in the second derivative, confirming that the point at \(x = 1\) is indeed a point of inflection.

(a)
M1: Correctly applies the product rule to find \(\frac{\text{d}y}{\text{d}x}\) with at least one term correct.
A1: Correct unsimplified derivative, e.g., \(2(x-2)e^{2x} + e^{2x}\).
A1: Correct simplified derivative \((2x - 3)e^{2x}\).

(b)
M1: Sets \(\frac{\text{d}y}{\text{d}x} = 0\) and solves for \(x\).
A1: Correct \(x\)-coordinate \(x = \frac{3}{2}\) or \(1.5\).
A1: Correct exact \(y\)-coordinate \(y = -\frac{1}{2}e^3\) (do not accept decimal approximations).
M1: Finds the second derivative \(\frac{\text{d}^2y}{\text{d}x^2}\) and evaluates it at their \(x\)-value to determine the nature.
A1: Correctly concludes it is a minimum with a valid reason (e.g., \(2e^3 > 0\)).

(c)
B1: Identifies the intersection point with the \(y\)-axis as \((0, -2)\).
M1: Finds the gradient of the tangent at \(x = 0\) and takes the negative reciprocal to find the gradient of the normal.
A1: Correct normal gradient \(m = \frac{1}{3}\).
A1: Correct equation in the form \(ax + by + c = 0\) with integer coefficients (e.g., \(x - 3y - 6 = 0\) or \(3y - x + 6 = 0\)).

(d)
M1: Sets \(\frac{\text{d}^2y}{\text{d}x^2} = 0\) using their expression from part (b).
A1: Solves to show \(x = 1\).
B1: Explains or demonstrates a sign change of \(\frac{\text{d}^2y}{\text{d}x^2}\) around \(x = 1\) to confirm it is a point of inflection.

(a) The \(n\)-th term of an arithmetic series is \(T_{A,n} = a + (n-1)d\).
The \(n\)-th term of a geometric series is \(T_{G,n} = ar^{n-1}\).

From the given information:
\(T_{A,3} = T_{G,2} \Rightarrow a + 2d = ar\) [Equation 1]
\(T_{A,4} = T_{G,3} \Rightarrow a + 3d = ar^2\) [Equation 2]

From Equation 1:
\(2d = ar - a = a(r - 1) \Rightarrow d = \frac{a(r - 1)}{2}\)

Substitute this into Equation 2:
\(a + 3\left(\frac{a(r - 1)}{2}\right) = ar^2\)

Since \(a \neq 0\), we can divide the entire equation by \(a\):
\(1 + \frac{3(r - 1)}{2} = r^2\)

Multiply by 2 to clear the fraction:
\(2 + 3(r - 1) = 2r^2\)
\(2 + 3r - 3 = 2r^2\)
\(2r^2 - 3r + 1 = 0\) (as required)

(b) Solving \(2r^2 - 3r + 1 = 0\):
\((2r - 1)(r - 1) = 0\)
So \(r = 0.5\) or \(r = 1\).
Since we are given \(r \neq 1\), \(r = 0.5\).

(c) Substitute \(r = 0.5\) back into the expression for \(d\):
\(d = \frac{a(0.5 - 1)}{2} = \frac{-0.5a}{2} = -0.25a\)

(d) The sum to infinity of a geometric series is \(S_{\infty} = \frac{a}{1 - r}\).
\(S_{\infty} = \frac{a}{1 - 0.5} = 2a\)

Given \(S_{\infty} = -8\):
\(2a = -8 \Rightarrow a = -4\)

(e) Using \(d = -0.25a\):
\(d = -0.25(-4) = 1\)

(f) The sum of the first \(n\) terms of an arithmetic series is:
\(S_n = \frac{n}{2}[2a + (n-1)d]\)

Substitute \(a = -4\) and \(d = 1\):
\(S_n = \frac{n}{2}[2(-4) + (n-1)(1)] = \frac{n}{2}[-8 + n - 1] = \frac{n}{2}(n - 9)\)

We want to find the least \(n\) such that \(S_n > 50\):
\(\frac{n}{2}(n - 9) > 50\)
\(n(n - 9) > 100\)
\(n^2 - 9n - 100 > 0\)

Consider the corresponding quadratic equation \(n^2 - 9n - 100 = 0\):
\(n = \frac{9 \pm \sqrt{(-9)^2 - 4(1)(-100)}}{2} = \frac{9 \pm \sqrt{81 + 400}}{2} = \frac{9 \pm \sqrt{481}}{2}\)

Using \(\sqrt{481} \approx 21.932\):
\(n \approx \frac{9 + 21.932}{2} \approx 15.47\) (taking the positive root since \(n > 0\))

Since \(n\) must be an integer, the least integer value for \(n\) is \(16\).

Let us check:
For \(n = 15\): \(S_{15} = \frac{15}{2}(15 - 9) = 45 \le 50\).
For \(n = 16\): \(S_{16} = \frac{16}{2}(16 - 9) = 56 > 50\).

So the least value of \(n\) is \(16\).

(a)
M1: Writes down the correct terms for the AP and GP using standard formulas for \(T_3\) and \(T_4\).
M1: Eliminates \(d\) to form an equation in \(a\) and \(r\).
A1: Divides by \(a\) (recognizing \(a \neq 0\)) and forms a correct quadratic equation in \(r\).
A1: Achieves the given equation \(2r^2 - 3r + 1 = 0\) with no errors in working.

(b)
B1: Correctly identifies \(r = 0.5\) (accept \(\frac{1}{2}\)).

(c)
M1: Substitutes their \(r\) into an expression relating \(d\) and \(a\).
A1: Correctly expresses \(d = -0.25a\) (or \(d = -\frac{1}{4}a\)).

(d)
M1: Uses the sum to infinity formula \(S_{\infty} = \frac{a}{1 - r}\) with their value of \(r\).
A1: Correctly finds \(a = -4\).

(e)
B1: Correctly finds \(d = 1\).

(f)
M1: Writes down the correct formula for \(S_n\) and substitutes their \(a\) and \(d\).
M1: Sets up the inequality \(\frac{n}{2}(n-9) > 50\) and simplifies to a 3-term quadratic inequality.
M1: Solves the quadratic equation to find the critical value of \(n\) (approx. 15.5).
A1: Identifies \(n = 16\).
B1: Shows verification (e.g. evaluating \(S_{15}\) and \(S_{16}\)) to confirm 16 is the least value.

To find the stationary point, we need to find \(\frac{dy}{dx}\) and set it to 0.

Using the product rule on \( y = (2x - 3)(4x + 1)^{\frac{1}{2}} \):
\[ \frac{dy}{dx} = \frac{d}{dx}(2x - 3) \cdot (4x + 1)^{\frac{1}{2}} + (2x - 3) \cdot \frac{d}{dx}(4x + 1)^{\frac{1}{2}} \]
\[ \frac{dy}{dx} = 2(4x + 1)^{\frac{1}{2}} + (2x - 3) \cdot \frac{1}{2}(4x + 1)^{-\frac{1}{2}} \cdot 4 \]
\[ \frac{dy}{dx} = 2\sqrt{4x + 1} + \frac{2(2x - 3)}{\sqrt{4x + 1}} \]

Combine into a single fraction:
\[ \frac{dy}{dx} = \frac{2(4x + 1) + 2(2x - 3)}{\sqrt{4x + 1}} = \frac{8x + 2 + 4x - 6}{\sqrt{4x + 1}} = \frac{12x - 4}{\sqrt{4x + 1}} = \frac{4(3x - 1)}{\sqrt{4x + 1}} \]

At the stationary point, \(\frac{dy}{dx} = 0\):
\[ 4(3x - 1) = 0 \implies x = \frac{1}{3} \]

Substitute \( x = \frac{1}{3} \) back into the equation for \( y \):
\[ y = \left(2\left(\frac{1}{3}\right) - 3\right)\sqrt{4\left(\frac{1}{3}\right) + 1} = \left(\frac{2}{3} - 3\right)\sqrt{\frac{7}{3}} = -\frac{7}{3}\sqrt{\frac{7}{3}} \]

Rationalize the denominator:
\[ \sqrt{\frac{7}{3}} = \frac{\sqrt{21}}{3} \]
\[ y = -\frac{7}{3} \cdot \frac{\sqrt{21}}{3} = -\frac{7\sqrt{21}}{9} \]

Thus, the exact coordinates of the stationary point are \(\left(\frac{1}{3}, -\frac{7\sqrt{21}}{9}\right)\).

M1: Attempting to differentiate using the product rule or chain rule.
A1: Correct derivative, e.g., \(\frac{dy}{dx} = 2\sqrt{4x+1} + \frac{2(2x-3)}{\sqrt{4x+1}}\) (any equivalent form).
M1: Setting their \(\frac{dy}{dx} = 0\) and attempting to solve for \(x\).
A1: \(x = \frac{1}{3}\) (or exact equivalent).
M1: Substituting their \(x\) back into the equation for \(y\).
A1: Correct y-coordinate, \(y = -\frac{7\sqrt{21}}{9}\) (or exact equivalent, e.g., \(-\frac{7}{3}\sqrt{\frac{7}{3}}\)).

We use the double-angle identity for cosine:
\[ \cos 2\theta = 1 - 2\sin^2\theta \]

Substitute this into the equation:
\[ 3(1 - 2\sin^2\theta) - \sin\theta - 2 = 0 \]
\[ 3 - 6\sin^2\theta - \sin\theta - 2 = 0 \]
\[ -6\sin^2\theta - \sin\theta + 1 = 0 \]
\[ 6\sin^2\theta + \sin\theta - 1 = 0 \]

Factorize the quadratic in \(\sin\theta\):
\[ (3\sin\theta - 1)(2\sin\theta + 1) = 0 \]

This gives two possible values for \(\sin\theta\):
\[ \sin\theta = \frac{1}{3} \quad \text{or} \quad \sin\theta = -\frac{1}{2} \]

For \( 0 \le \theta < 360^\circ \):

If \(\sin\theta = \frac{1}{3}\):
\[ \theta = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.47^\circ \implies \theta \approx 19.5^\circ \]
The second solution in the second quadrant is:
\[ \theta = 180^\circ - 19.47^\circ \approx 160.5^\circ \]

If \(\sin\theta = -\frac{1}{2}\):
The principal angle is \(-30^\circ\). The solutions in the third and fourth quadrants are:
\[ \theta = 180^\circ - (-30^\circ) = 210^\circ \]
\[ \theta = 360^\circ - 30^\circ = 330^\circ \]

So the complete set of solutions is \(\theta = 19.5^\circ, 160.5^\circ, 210^\circ, 330^\circ\).

M1: Using the identity \(\cos 2\theta = 1 - 2\sin^2\theta\) to form a quadratic equation in \(\sin\theta\).
A1: Obtains the correct quadratic equation: \(6\sin^2\theta + \sin\theta - 1 = 0\).
M1: Solves their quadratic equation to find two values for \(\sin\theta\).
A1: \(\sin\theta = \frac{1}{3}\) and \(\sin\theta = -\frac{1}{2}\).
A1: \(\theta \approx 19.5^\circ\) and \(160.5^\circ\) (accept 19.47 and 160.53, or standard rounding).
A1: \(\theta = 210^\circ\) and \(330^\circ\) (must be exact or rounded appropriately).

(a) The sum of the first \( n \) terms of an arithmetic series is given by:
\[ S_n = \frac{n}{2}[2a + (n-1)d] \]
Given \( S_8 = 80 \):
\[ \frac{8}{2}[2a + 7d] = 80 \implies 4(2a + 7d) = 80 \implies 2a + 7d = 20 \quad \text{(Equation 1)} \]

The \( n \)-th term is given by:
\[ u_n = a + (n-1)d \]
Given \( u_{12} = 25 \):
\[ a + 11d = 25 \quad \text{(Equation 2)} \]

Multiply Equation 2 by 2:
\[ 2a + 22d = 50 \quad \text{(Equation 3)} \]

Subtract Equation 1 from Equation 3:
\[ (2a + 22d) - (2a + 7d) = 50 - 20 \]
\[ 15d = 30 \implies d = 2 \]

Substitute \( d = 2 \) into Equation 2:
\[ a + 11(2) = 25 \implies a + 22 = 25 \implies a = 3 \]

(b) The sum of the first 20 terms is:
\[ S_{20} = \frac{20}{2}[2a + 19d] \]
\[ S_{20} = 10[2(3) + 19(2)] = 10[6 + 38] = 10(44) = 440 \]

Part (a):
M1: Applies the sum formula to write a correct equation in terms of \( a \) and \( d \) (e.g., \(4(2a+7d)=80\)).
M1: Applies the term formula to write a correct equation for the 12th term (e.g., \(a+11d=25\)).
M1: Solves the two simultaneous equations to find either \( a \) or \( d \).
A1: Correct values: \( a = 3 \) and \( d = 2 \).

Part (b):
M1: Applies the sum formula for \( S_{20} \) with their values of \( a \) and \( d \).
A1: Correct sum of 440.

First, find the midpoint \( M \) of the line segment \( AB \):
\[ M = \left(\frac{-2 + 4}{2}, \frac{5 - 3}{2}\right) = (1, 1) \]

Next, find the gradient of \( AB \):
\[ m_{AB} = \frac{-3 - 5}{4 - (-2)} = \frac{-8}{6} = -\frac{4}{3} \]

The gradient of the perpendicular bisector is:
\[ m_{\perp} = -\frac{1}{m_{AB}} = \frac{3}{4} \]

The equation of the perpendicular bisector of \( AB \) (passing through \( M(1,1) \)) is:
\[ y - 1 = \frac{3}{4}(x - 1) \implies y = \frac{3}{4}x + \frac{1}{4} \]

Let the coordinates of \( C \) be \(\left(t, \frac{3}{4}t + \frac{1}{4}\right)\).

Find the length of the base \( AB \):
\[ AB = \sqrt{(4 - (-2))^2 + (-3 - 5)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{100} = 10 \]

The area of triangle \( ABC \) is 25. Since \( C \) lies on the perpendicular bisector, the height \( h \) of the triangle is the distance from \( C \) to the midpoint \( M \):
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
\[ 25 = \frac{1}{2} \times 10 \times MC \implies MC = 5 \]

Using the distance formula for \( MC = 5 \):
\[ MC^2 = (t - 1)^2 + \left(\left(\frac{3}{4}t + \frac{1}{4}\right) - 1\right)^2 = 25 \]
\[ (t - 1)^2 + \left(\frac{3}{4}t - \frac{3}{4}\right)^2 = 25 \]
\[ (t - 1)^2 + \frac{9}{16}(t - 1)^2 = 25 \]
\[ \frac{25}{16}(t - 1)^2 = 25 \]
\[ (t - 1)^2 = 16 \implies t - 1 = \pm 4 \]

This gives two values for \( t \):
1) \( t - 1 = 4 \implies t = 5 \). Then \( y = \frac{3}{4}(5) + \frac{1}{4} = 4 \).
2) \( t - 1 = -4 \implies t = -3 \). Then \( y = \frac{3}{4}(-3) + \frac{1}{4} = -2 \).

Thus, the possible coordinates of \( C \) are \((5, 4)\) and \((-3, -2)\).

M1: Finds the midpoint \(M(1,1)\) and the gradient of \(AB\) (\(-\frac{4}{3}\)).
A1: Correct equation of the perpendicular bisector: \(y = \frac{3}{4}x + \frac{1}{4}\) (or equivalent).
M1: Finds the length of \(AB = 10\) and uses the area formula to find the height \(MC = 5\).
M1: Forms an equation for the distance from \(C(t, y)\) to \(M(1,1)\) using their perpendicular bisector equation.
A1: Solves to find the two possible x-coordinates: \(t = 5\) and \(t = -3\).
A1: Gives both correct coordinate pairs: \((5, 4)\) and \((-3, -2)\).

For the given equation \( 2x^2 - 5x + 4 = 0 \), we have:
\[ \alpha + \beta = \frac{5}{2} \]
\[ \alpha\beta = \frac{4}{2} = 2 \]

First, find the sum of the new roots, \( S \):
\[ S = (\alpha^2 + \beta) + (\beta^2 + \alpha) = (\alpha^2 + \beta^2) + (\alpha + \beta) \]
We know that \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\). So:
\[ \alpha^2 + \beta^2 = \left(\frac{5}{2}\right)^2 - 2(2) = \frac{25}{4} - 4 = \frac{9}{4} \]
Thus:
\[ S = \frac{9}{4} + \frac{5}{2} = \frac{19}{4} \]

Next, find the product of the new roots, \( P \):
\[ P = (\alpha^2 + \beta)(\beta^2 + \alpha) = \alpha^2\beta^2 + \alpha^3 + \beta^3 + \alpha\beta = (\alpha\beta)^2 + (\alpha^3 + \beta^3) + \alpha\beta \]
We need to find \(\alpha^3 + \beta^3\):
\[ \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) \]
\[ \alpha^3 + \beta^3 = \left(\frac{5}{2}\right)^3 - 3(2)\left(\frac{5}{2}\right) = \frac{125}{8} - 15 = \frac{5}{8} \]
Now, substitute this back into the expression for \( P \):
\[ P = (2)^2 + \frac{5}{8} + 2 = 4 + \frac{5}{8} + 2 = \frac{53}{8} \]

The required quadratic equation is given by:
\[ x^2 - Sx + P = 0 \]
\[ x^2 - \frac{19}{4}x + \frac{53}{8} = 0 \]

Multiplying by 8 to get integer coefficients:
\[ 8x^2 - 38x + 53 = 0 \]

M1: Identifies \(\alpha + \beta = \frac{5}{2}\) and \(\alpha\beta = 2\).
M1: Expresses the sum of the new roots \(S\) in terms of \(\alpha+\beta\) and \(\alpha\beta\), and finds \(S = \frac{19}{4}\).
M1: Expands the product of the new roots \(P\) to get \(\alpha^2\beta^2 + \alpha^3 + \beta^3 + \alpha\beta\).
M1: Uses a correct identity for \(\alpha^3 + \beta^3\) (e.g. \((\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)\)) to find its value (\(\frac{5}{8}\)).
A1: Correctly calculates the product \(P = \frac{53}{8}\).
A1: Forms the final equation \( 8x^2 - 38x + 53 = 0 \) (or any integer multiple thereof, must have "= 0").

(a) Since \( OP : PA = 2 : 1 \), we have:
\[ \vec{OP} = \frac{2}{3}\mathbf{a} \]

Since \( AQ : QB = 3 : 2 \), the point \( Q \) divides \( AB \) in the ratio \( 3 : 2 \):
\[ \vec{OQ} = \vec{OA} + \vec{AQ} = \mathbf{a} + \frac{3}{5}\vec{AB} \]
Since \( \vec{AB} = \mathbf{b} - \mathbf{a} \):
\[ \vec{OQ} = \mathbf{a} + \frac{3}{5}(\mathbf{b} - \mathbf{a}) = \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b} \]

Now, find \( \vec{PQ} \):
\[ \vec{PQ} = \vec{OQ} - \vec{OP} = \left(\frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\right) - \frac{2}{3}\mathbf{a} \]
\[ \vec{PQ} = \left(\frac{2}{5} - \frac{2}{3}\right)\mathbf{a} + \frac{3}{5}\mathbf{b} = -\frac{4}{15}\mathbf{a} + \frac{3}{5}\mathbf{b} \]

(b) We are given \( \vec{PR} = k\vec{PQ} \). Therefore:
\[ \vec{OR} = \vec{OP} + \vec{PR} = \frac{2}{3}\mathbf{a} + k\left(-\frac{4}{15}\mathbf{a} + \frac{3}{5}\mathbf{b}\right) \]
\[ \vec{OR} = \left(\frac{2}{3} - \frac{4}{15}k\right)\mathbf{a} + \frac{3}{5}k\mathbf{b} \]

Since \( \vec{OR} = \lambda\mathbf{b} \), and \(\mathbf{a}\) and \(\mathbf{b}\) are non-parallel vectors, the coefficient of \(\mathbf{a}\) must be 0:
\[ \frac{2}{3} - \frac{4}{15}k = 0 \implies \frac{4}{15}k = \frac{2}{3} \implies k = \frac{2}{3} \times \frac{15}{4} = \frac{5}{2} \]

Now equate the coefficient of \(\mathbf{b}\):
\[ \lambda = \frac{3}{5}k = \frac{3}{5}\left(\frac{5}{2}\right) = \frac{3}{2} \]

So, \( k = \frac{5}{2} \) and \( \lambda = \frac{3}{2} \).

Part (a):
M1: Finds \( \vec{OP} = \frac{2}{3}\mathbf{a} \) or \( \vec{OQ} = \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b} \).
A1: Expresses \( \vec{OQ} \) correctly in terms of \( \mathbf{a} \) and \( \mathbf{b} \).
A1: Correct expression for \( \vec{PQ} = -\frac{4}{15}\mathbf{a} + \frac{3}{5}\mathbf{b} \) (or equivalent fraction form).

Part (b):
M1: Expresses \( \vec{OR} \) in terms of \( \mathbf{a} \), \( \mathbf{b} \) and \( k \).
M1: Sets the coefficient of \( \mathbf{a} \) to 0 and solves for \( k \).
A1: Correct values: \( k = \frac{5}{2} \) (or 2.5) and \( \lambda = \frac{3}{2} \) (or 1.5).

(a) Using the double angle identities \(\cos 2\theta = 1 - 2\sin^2\theta\), \(\cos 2\theta = 2\cos^2\theta - 1\), and \(\sin 2\theta = 2\sin\theta\cos\theta\):
\(\text{LHS} = \frac{1 - (1 - 2\sin^2\theta) + 2\sin\theta\cos\theta}{1 + (2\cos^2\theta - 1) + 2\sin\theta\cos\theta}\)
\(\text{LHS} = \frac{2\sin^2\theta + 2\sin\theta\cos\theta}{2\cos^2\theta + 2\sin\theta\cos\theta}\)
Factorising the numerator and the denominator:
\(\text{LHS} = \frac{2\sin\theta(\sin\theta + \cos\theta)}{2\cos\theta(\cos\theta + \sin\theta)}\)
Cancelling the common factor \(2(\sin\theta + \cos\theta)\):
\(\text{LHS} = \frac{\text{sin}\theta}{\cos\theta} = \tan\theta = \text{RHS}\) [Proven]

(b) Using the identity proven in part (a), the equation simplifies to:
\(\tan\theta = 3\cot\theta - 2\)
Since \ ext{cot}\theta = \frac{1}{\tan\theta}\):
\(\tan\theta = \frac{3}{\tan\theta} - 2\)
Multiplying throughout by \(\tan\theta\):
\(\tan^2\theta + 2\tan\theta - 3 = 0\)
Factorising the quadratic equation:
\((\tan\theta - 1)(\tan\theta + 3) = 0\)
This gives:
\(\tan\theta = 1\) or \(\tan\theta = -3\)
For \(0 < \theta < \pi\):
If \(\tan\theta = 1\), \(\theta = \frac{\pi}{4} \approx 0.785\)
If \(\tan\theta = -3\), \(\theta = \pi - \arctan(3) \approx 3.14159 - 1.24905 = 1.89\)

(a)
M1: Attempts to substitute double angle formulae for \(\cos 2\theta\) and \(\sin 2\theta\) into LHS.
A1: Correctly simplifies the numerator to \(2\sin^2\theta + 2\sin\theta\cos\theta\) and the denominator to \(2\cos^2\theta + 2\sin\theta\cos\theta\).
M1: Factorises both the numerator and denominator to isolate \((\sin\theta + \cos\theta)\).
A1: Fully correct proof showing cancellation and concluding with \(\tan\theta\).

(b)
M1: Rewrites the equation using the identity in (a) to obtain \(\tan\theta = 3\cot\theta - 2\).
M1: Substitutes \(\cot\theta = \frac{1}{\tan\theta}\) and forms a three-term quadratic in \(\tan\theta\).
A1: Obtains the correct quadratic equation: \(\tan^2\theta + 2\tan\theta - 3 = 0\).
M1: Solves the quadratic to find two values for \(\tan\theta\).
A1: Obtains \(\tan\theta = 1\) and \(\tan\theta = -3\).
A1: \(\theta \approx 0.785\) (or \(\frac{\pi}{4}\)) (accept 0.79 with working).
A1: \(\theta \approx 1.89\) (accept 1.9 with working). Deduct 1 mark overall if any extra solutions are within the range.

(a) Using the product rule to differentiate \(y = x^2 e^{-2x}\):
\(\frac{dy}{dx} = 2x e^{-2x} + x^2 (-2e^{-2x}) = 2x e^{-2x}(1 - x)\)
At stationary points, \(\frac{dy}{dx} = 0\):
\(2x e^{-2x}(1 - x) = 0\)
Since \(e^{-2x} \ne 0\), we have \(x = 0\) or \(x = 1\).
When \(x = 0\), \(y = 0^2 e^{0} = 0\). Stationary point is \((0, 0)\).
When \(x = 1\), \(y = 1^2 e^{-2} = e^{-2}\). Stationary point is \((1, e^{-2})\).

(b) Differentiating \(\frac{dy}{dx} = (2x - 2x^2)e^{-2x}\) to find the second derivative:
\(\frac{d^2y}{dx^2} = (2 - 4x)e^{-2x} - 2(2x - 2x^2)e^{-2x} = (4x^2 - 8x + 2)e^{-2x}\)
At \(x = 0\):
\(\frac{d^2y}{dx^2} = (0 - 0 + 2)e^{0} = 2 > 0 \implies\) Local Minimum at \((0, 0)\).
At \(x = 1\):
\(\frac{d^2y}{dx^2} = (4(1)^2 - 8(1) + 2)e^{-2} = -2e^{-2} < 0 \implies\) Local Maximum at \((1, e^{-2})\).

(c) At \(x = 2\), the y-coordinate is:
\(y = 2^2 e^{-2(2)} = 4e^{-4}\)
The gradient of the tangent at \(x = 2\) is:
\(m = \left.\frac{dy}{dx}\right|_{x=2} = 2(2)e^{-4}(1 - 2) = -4e^{-4}\)
The equation of the tangent is:
\(y - y_1 = m(x - x_1)\)
\(y - 4e^{-4} = -4e^{-4}(x - 2)\)
\(y = -4e^{-4}x + 8e^{-4} + 4e^{-4}\)
\(y = -4e^{-4}x + 12e^{-4}\)

(a)
M1: Correct use of product rule to differentiate \(y = x^2 e^{-2x}\).
A1: Correct derivative \(\frac{dy}{dx} = 2x e^{-2x}(1 - x)\) (any equivalent form).
M1: Equating \(\frac{dy}{dx}\) to 0 and solving for \(x\).
A1: Obtains \(x = 0\) and \(x = 1\).
A1: Coordinates \((0, 0)\) and \((1, e^{-2})\) (or \((1, \frac{1}{e^2})\)).

(b)
M1: Differentiates \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2}\). (Accept first derivative sign test with values shown).
A1: Correct second derivative expression \((4x^2 - 8x + 2)e^{-2x}\).
A1: Evaluates derivative sign correctly to conclude \((0, 0)\) is a minimum and \((1, e^{-2})\) is a maximum.

(c)
M1: Calculates both the y-coordinate \(4e^{-4}\) and gradient \(-4e^{-4}\) at \(x = 2\).
M1: Substitutes their coordinates and gradient into the equation of a straight line.
A1: Correctly simplifies to \(y = -4e^{-4}x + 12e^{-4}\) (or equivalent form matching \(y=mx+c\)).

(a) For the geometric series \(G\), the sum to infinity is given by:
\(S_{\infty} = \frac{a}{1 - r} = 3a\)
Since \(a \ne 0\), we can divide by \(a\):
\(\frac{1}{1 - r} = 3 \implies 1 - r = \frac{1}{3} \implies r = \frac{2}{3}\)
The 4th term of \(A\) is \(a + 3d\). The 2nd term of \(G\) is \(ar\).
Since these are equal:
\(a + 3d = ar\)
\(a + 3d = \frac{2}{3}a\)
\(3d = -\frac{1}{3}a \implies a = -9d\) [Proven]

(b) For the arithmetic series \(A\), the sum of the first 10 terms is given by:
\(S_{10} = \frac{10}{2}(2a + 9d) = 120\)
\(5(2a + 9d) = 120 \implies 2a + 9d = 24\)
Substitute \(a = -9d\) into the equation:
\(2(-9d) + 9d = 24\)
\(-18d + 9d = 24\)
\(-9d = 24 \implies d = -\frac{24}{9} = -\frac{8}{3}\)
Now find \(a\):
\(a = -9\left(-\frac{8}{3}\right) = 24\)

(c) For \(G\), \(a = 24\) and \(r = \frac{2}{3}\).
The sum of the first 8 terms is:
\(S_8 = \frac{a(1 - r^8)}{1 - r} = \frac{24(1 - (2/3)^8)}{1 - 2/3} = 72\left(1 - \frac{256}{6561}\right)\)
\(S_8 = 72 \times \frac{6305}{6561} = \frac{50440}{729} \approx 69.190672\)
To 3 decimal places, \(S_8 = 69.191\)

(d) The sum of the first \(n\) terms of \(A\) is:
\(S_n = \frac{n}{2}(2a + (n-1)d) < 0\)
Since \(n > 0\), this requires:
\(2a + (n-1)d < 0\)
Substitute \(a = 24\) and \(d = -\frac{8}{3}\):
\(2(24) + (n-1)\left(-\frac{8}{3}\right) < 0\)
\(48 - \frac{8}{3}(n-1) < 0\)
\(48 < \frac{8}{3}(n-1)\)
\(18 < n - 1 \implies n > 19\)
The smallest integer value of \(n\) is 20.

(a)
M1: Equates sum to infinity of \(G\) to \(3a\) and solves to find \(r\).
A1: Finds \(r = \frac{2}{3}\).
A1: Equates \(a+3d\) to \(ar\) and correctly rearranges to show \(a = -9d\).

(b)
M1: Uses the sum of arithmetic series formula to set up an equation: \(5(2a + 9d) = 120\).
A1: Solves simultaneous equations to find \(d = -\frac{8}{3}\) (or equivalent).
A1: Finds \(a = 24\).

(c)
M1: Substitutes their \(a\) and \(r\) into the geometric sum formula for \(n=8\).
A1: Correct numerical substitution: \(24\left(1 - (2/3)^8\right) / (1/3)\).
A1: Evaluates to \(69.191\) (correct to 3 d.p.).

(d)
M1: Sets up the inequality \(S_n < 0\) with their values of \(a\) and \(d\).
A1: Solves the inequality to find \(n > 19\) and states the smallest integer value is \(n = 20\).

(a) We find the first derivative of \(y\) with respect to \(x\): \(\frac{dy}{dx} = 3x^2 - 12x + 9\). To find the stationary points, set \(\frac{dy}{dx} = 0\): \(3(x^2 - 4x + 3) = 0 \implies 3(x-1)(x-3) = 0\), which gives \(x = 1\) or \(x = 3\). When \(x = 1\), \(y = 1^3 - 6(1)^2 + 9(1) - 4 = 0\). The first stationary point is \((1, 0)\). When \(x = 3\), \(y = 3^3 - 6(3)^2 + 9(3) - 4 = -4\). The second stationary point is \((3, -4)\). To determine the nature of these points, find the second derivative: \(\frac{d^2y}{dx^2} = 6x - 12\). At \(x = 1\), \(\frac{d^2y}{dx^2} = -6 < 0\), so \((1, 0)\) is a local maximum. At \(x = 3\), \(\frac{d^2y}{dx^2} = 6 > 0\), so \((3, -4)\) is a local minimum. (b) At the point \(P(0, -4)\), the gradient of the tangent to \(C\) is \(\frac{dy}{dx} = 3(0)^2 - 12(0) + 9 = 9\). The gradient of the normal to \(C\) at \(P\) is \(-\frac{1}{9}\). The equation of the normal is \(y - (-4) = -\frac{1}{9}(x - 0) \implies y + 4 = -\frac{1}{9}x \implies 9y + 36 = -x \implies x + 9y + 36 = 0\). (c) The normal intersects the \(x\)-axis when \(y = 0\): \(x + 36 = 0 \implies x = -36\). The curve \(C\) can be written as \(y = (x-1)^2(x-4)\), which intersects the \(x\)-axis at \(x = 1\) and \(x = 4\). For \(x < 4\), the curve lies on or below the \(x\)-axis. The finite region \(R\) lies entirely below the \(x\)-axis and is bounded by the normal from \(x = -36\) to \(x = 0\), the curve from \(x = 0\) to \(x = 4\), and the \(x\)-axis. The area of the triangular region under the normal is \(\frac{1}{2} \times 36 \times 4 = 72\). The area under the curve is \(\int_{0}^{4} -y \, dx = \int_{0}^{4} (-x^3 + 6x^2 - 9x + 4) \, dx = \left[ -\frac{1}{4}x^4 + 2x^3 - \frac{9}{2}x^2 + 4x \right]_{0}^{4} = \left( -64 + 128 - 72 + 16 \right) - 0 = 8\). The total area of \(R\) is \(72 + 8 = 80\).

(a) M1: Differentiates to find \(\frac{dy}{dx} = 3x^2 - 12x + 9\). M1: Sets \(\frac{dy}{dx} = 0\) and solves for \(x\). A1: Finds both \(x\)-coordinates \(x=1, 3\) and evaluates \(y\)-coordinates to get \((1, 0)\) and \((3, -4)\). M1: Finds the second derivative \(\frac{d^2y}{dx^2} = 6x - 12\) and substitutes at least one \(x\)-value. A1: Correctly identifies \((1, 0)\) as maximum and \((3, -4)\) as minimum with complete working. (b) M1: Substitutes \(x=0\) into \(\frac{dy}{dx}\) to find the tangent gradient (9). M1: Uses perpendicular gradient rule to get normal gradient of \(-\frac{1}{9}\). M1: Sets up the equation of the line using \((0, -4)\) and their normal gradient. A1: Correctly obtains \(x + 9y + 36 = 0\) (or any integer multiple). (c) M1: Finds the \(x\)-intercept of the normal as \(x = -36\). M1: Calculates the area of the triangle under the normal as 72 (or integrates the normal equation from \(-36\) to \(0\)). M1: Sets up integration of \(-y\) with correct limits from 0 to 4. A1: Correct integration to get \(-\frac{1}{4}x^4 + 2x^3 - \frac{9}{2}x^2 + 4x\) and evaluates to 8. A1: Adds both areas to get 80.

(a) Since \(OP : PA = 2 : 1\), we have \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\). Since \(OQ : QB = 1 : 3\), we have \(\overrightarrow{OQ} = \frac{1}{4}\mathbf{b}\). Since \(X\) lies on \(AQ\), we can write \(\overrightarrow{AX} = \lambda \overrightarrow{AQ} = \lambda (\overrightarrow{OQ} - \overrightarrow{OA}) = \lambda (\frac{1}{4}\mathbf{b} - \mathbf{a})\). Thus, \(\overrightarrow{OX} = \overrightarrow{OA} + \overrightarrow{AX} = (1 - \lambda)\mathbf{a} + \frac{1}{4}\lambda \mathbf{b}\). Since \(X\) lies on \(BP\), we can write \(\overrightarrow{PX} = \mu \overrightarrow{PB} = \mu (\overrightarrow{OB} - \overrightarrow{OP}) = \mu (\mathbf{b} - \frac{2}{3}\mathbf{a})\). Thus, \(\overrightarrow{OX} = \overrightarrow{OP} + \overrightarrow{PX} = \frac{2}{3}\mathbf{a} + \mu (\mathbf{b} - \frac{2}{3}\mathbf{a}) = \frac{2}{3}(1 - \mu)\mathbf{a} + \mu \mathbf{b}\). Equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\), we get: \(1 - \lambda = \frac{2}{3}(1 - \mu)\) and \(\frac{1}{4}\lambda = \mu\). Substituting \(\mu = \frac{1}{4}\lambda\) into the first equation: \(1 - \lambda = \frac{2}{3}(1 - \frac{1}{4}\lambda) \implies 1 - \lambda = \frac{2}{3} - \frac{1}{6}\lambda \implies \frac{5}{6}\lambda = \frac{1}{3} \implies \lambda = \frac{2}{5}\). This gives \(\mu = \frac{1}{10}\). Thus, \(\overrightarrow{OX} = \frac{3}{5}\mathbf{a} + \frac{1}{10}\mathbf{b}\). (b) Since \(Y\) lies on \(OX\) extended, \(\overrightarrow{OY} = k \overrightarrow{OX} = \frac{3}{5}k\mathbf{a} + \frac{1}{10}k\mathbf{b}\). Since \(Y\) lies on the line \(AB\), the sum of the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) must be 1: \(\frac{3}{5}k + \frac{1}{10}k = 1 \implies \frac{7}{10}k = 1 \implies k = \frac{10}{7}\). Thus, \(\overrightarrow{OY} = \frac{6}{7}\mathbf{a} + \frac{1}{7}\mathbf{b}\). (c) (i) Since \(\overrightarrow{OY} = \frac{10}{7}\overrightarrow{OX}\), we have \(OX : OY = 7 : 10\), which means \(OX : XY = 7 : 3\). (ii) From \(\overrightarrow{OY} = \frac{6}{7}\mathbf{a} + \frac{1}{7}\mathbf{b}\), the section formula implies that \(Y\) divides \(AB\) in the ratio \(1 : 6\), so \(AY : YB = 1 : 6\). (d) (i) Since \(\lambda = \frac{2}{5}\), \(AX : XQ = 2 : 3\), so the area of triangle \(OAX\) is \(\frac{2}{5}\) of the area of triangle \(OAQ\). Given \(\text{Area}(OAX) = 12\), we find \(\text{Area}(OAQ) = 12 \times \frac{5}{2} = 30\). Since \(OQ = \frac{1}{4}OB\), the area of triangle \(OAQ\) is \(\frac{1}{4}\) of the area of triangle \(OAB\). Thus, \(\text{Area}(OAB) = 30 \times 4 = 120 \text{ cm}^2\). (ii) Since \(OX : XY = 7 : 3\), we have \(XY = \frac{3}{10}OY\). Since triangles \(ABX\) and \(ABO\) share the base \(AB\), their areas are proportional to the distance of their third vertex from \(AB\). Thus, \(\text{Area}(ABX) = \frac{3}{10} \text{Area}(OAB) = \frac{3}{10} \times 120 = 36 \text{ cm}^2\).

(a) M1: Writes \(\overrightarrow{OX}\) in terms of \(\mathbf{a}\), \(\mathbf{b}\) and a parameter (e.g., \(\lambda\)) using the line \(AQ\). M1: Writes \(\overrightarrow{OX}\) in terms of \(\mathbf{a}\), \(\mathbf{b}\) and another parameter (e.g., \(\mu\)) using the line \(BP\). M1: Equates coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to set up two simultaneous equations. M1: Solves these equations to find the values of both parameters. A1: Correctly finds \(\lambda = \frac{2}{5}\) (and/or \(\mu = \frac{1}{10}\)). A1: Writes the final correct vector expression for \(\overrightarrow{OX}\). (b) M1: Sets \(\overrightarrow{OY} = k \overrightarrow{OX}\) and uses the collinearity condition that sum of coefficients equals 1. M1: Solves for \(k = \frac{10}{7}\). A1: Correctly obtains \(\overrightarrow{OY} = \frac{6}{7}\mathbf{a} + \frac{1}{7}\mathbf{b}\). (c) B1: Correctly states \(OX : XY = 7 : 3\). B1: Correctly states \(AY : YB = 1 : 6\). (d) M1: Connects the area of \(OAX\) to \(OAQ\) using the ratio \(AX : XQ = 2 : 3\) to get \(30\). M1: Connects the area of \(OAQ\) to \(OAB\) using the ratio \(OQ : QB = 1 : 3\). A1: Correctly finds the area of \(OAB\) is 120. M1: Uses the ratio of \(XY : OY = 3 : 10\) to find the area of \(ABX\) (or alternative complete subtraction method). A1: Correctly finds the area of \(ABX\) is 36.