2024 Edexcel IGCSE Further Pure Mathematics Practice Paper with Answers

(a) The volume of a cylinder is given by \( V = \pi r^2 h = 128\pi \), which gives \( h = \frac{128}{r^2} \). The total surface area is \( S = 2\pi r^2 + 2\pi r h \). Substituting \( h \) into the area formula gives \( S = 2\pi r^2 + 2\pi r \left(\frac{128}{r^2}\right) = 2\pi r^2 + \frac{256\pi}{r} \). (b) Differentiating \( S \) with respect to \( r \) gives \( \frac{dS}{dr} = 4\pi r - \frac{256\pi}{r^2} \). Setting \( \frac{dS}{dr} = 0 \) to find the stationary point: \( 4\pi r = \frac{256\pi}{r^2} \Rightarrow r^3 = 64 \Rightarrow r = 4 \text{ cm} \). The minimum surface area is \( S = 2\pi(4)^2 + \frac{256\pi}{4} = 32\pi + 64\pi = 96\pi \text{ cm}^2 \). (c) Differentiating a second time gives \( \frac{d^2S}{dr^2} = 4\pi + \frac{512\pi}{r^3} \). When \( r = 4 \), \( \frac{d^2S}{dr^2} = 4\pi + \frac{512\pi}{64} = 12\pi \). Since \( 12\pi > 0 \), the stationary value is a minimum.

(a) M1: Uses volume formula to express h in terms of r. M1: Substitutes h into the surface area formula. A1: Correctly simplifies to show the given expression. (b) M1: Correctly differentiates S with respect to r. M1: Sets the derivative to 0 and solves for r. A1: Obtains r = 4. M1: Substitutes r = 4 back into S. A1: Obtains 96\pi. (c) M1: Differentiates again to find the second derivative. A1: Shows that the second derivative is positive and concludes it is a minimum.

(a) Using the double angle identity \( \cos(2\theta) = 1 - 2\sin^2\theta \), we substitute this into the equation: \( 3(1 - 2\sin^2\theta) - 7\sin\theta = 0 \Rightarrow 3 - 6\sin^2\theta - 7\sin\theta = 0 \). Multiplying the entire equation by \( -1 \) gives \( 6\sin^2\theta + 7\sin\theta - 3 = 0 \). (b) Factorising the quadratic equation: \( (2\sin\theta + 3)(3\sin\theta - 1) = 0 \). This gives two possible cases: \( 2\sin\theta + 3 = 0 \Rightarrow \sin\theta = -1.5 \) which has no real solutions because \( -1 \le \sin\theta \le 1 \). Alternatively, \( 3\sin\theta - 1 = 0 \Rightarrow \sin\theta = \frac{1}{3} \). The basic angle is \( \sin^{-1}(\frac{1}{3}) \approx 19.47^\circ \). In the range \( 0^\circ \le \theta \le 360^\circ \), the solutions are in the first and second quadrants: \( \theta = 19.5^\circ \) and \( \theta = 180^\circ - 19.47^\circ = 160.5^\circ \).

(a) M1: Uses the identity \( \cos(2\theta) = 1 - 2\sin^2\theta \). M1: Expands and groups terms. A1: Obtains the correct quadratic form as shown. (b) M1: Factorises or uses formula on the quadratic in sin. A1: Identifies \( \sin\theta = -1.5 \) has no solution with brief reason. A1: Identifies \( \sin\theta = 1/3 \). M1: Calculates primary angle 19.5 (or 19.47). M1: Finds second quadrant angle. A1: Both answers correct to 1 decimal place (19.5 and 160.5).

(a) Let the first term of \( G \) be \( a \) and the common ratio be \( r \). We have \( ar = 6 \) and \( ar^4 = 48 \). Dividing the two equations gives \( r^3 = 8 \Rightarrow r = 2 \). Substituting back gives \( a(2) = 6 \Rightarrow a = 3 \). (b) The sum of the first 10 terms of \( G \) is \( S_{10} = \frac{a(r^{10} - 1)}{r - 1} = \frac{3(2^{10} - 1)}{2 - 1} = 3(1023) = 3069 \). (c) For the arithmetic series \( A \), the first term is \( a = 3 \). The sum of the first \( n \) terms is \( S_n = \frac{n}{2}[2a + (n-1)d] \). Therefore, \( S_{20} = 10(6 + 19d) = 60 + 190d \) and \( S_{10} = 5(6 + 9d) = 30 + 45d \). Given \( S_{20} = 4S_{10} \), we have \( 60 + 190d = 4(30 + 45d) \Rightarrow 60 + 190d = 120 + 180d \Rightarrow 10d = 60 \Rightarrow d = 6 \).

(a) M1: Set up two simultaneous equations for G. M1: Solve for r. A1: a = 3 and r = 2. (b) M1: Apply GP sum formula with n = 10. A1: 3069. (c) M1: Write expression for S_20 of AP. M1: Write expression for S_10 of AP. M1: Equate S_20 = 4 * S_10. A1: Solve for d to obtain d = 6.

(a) First find the midpoint of \( AB \): \( M = \left(\frac{0+4}{2}, \frac{1+3}{2}\right) = (2, 2) \). The gradient of \( AB \) is \( m = \frac{3-1}{4-0} = \frac{2}{4} = \frac{1}{2} \). The gradient of the perpendicular bisector is the negative reciprocal: \( m_{\perp} = -2 \). The equation of the perpendicular bisector is \( y - 2 = -2(x - 2) \Rightarrow y - 2 = -2x + 4 \Rightarrow y + 2x - 6 = 0 \). (b) The perpendicular bisector meets the \( x \)-axis when \( y = 0 \). Substituting \( y = 0 \) into the equation: \( 0 + 2x - 6 = 0 \Rightarrow 2x = 6 \Rightarrow x = 3 \). So the coordinates of \( C \) are \( (3, 0) \). (c) The coordinates of the vertices of the triangle are \( A(0, 1) \), \( B(4, 3) \), and \( C(3, 0) \). Using the shoelace formula: \( \text{Area} = \frac{1}{2} |(0\times3 + 4\times0 + 3\times1) - (1\times4 + 3\times3 + 0\times0)| = \frac{1}{2} |3 - 13| = \frac{1}{2} |-10| = 5 \).

(a) M1: Finds midpoint of AB. M1: Finds gradient of AB. M1: Finds perpendicular gradient and uses it in line equation. A1: Obtains y + 2x - 6 = 0. (b) M1: Sets y = 0. A1: C(3, 0). (c) M1: Recognises a valid method to find the area of a triangle given coordinates. A1: Computes terms correctly. A1: Area = 5.

From the given quadratic equation, we have: \( \alpha + \beta = \frac{5}{2} \) and \( \alpha\beta = 2 \). (a) \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{5}{2}\right)^2 - 2(2) = \frac{25}{4} - 4 = \frac{9}{4} \). (b) \( \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) = \left(\frac{5}{2}\right)\left(\frac{9}{4} - 2\right) = \frac{5}{2} \times \frac{1}{4} = \frac{5}{8} \). (c) Let the new roots be \( p = \frac{\alpha}{\beta^2} \) and \( q = \frac{\beta}{\alpha^2} \). Sum of new roots: \( S = p + q = \frac{\alpha}{\beta^2} + \frac{\beta}{\alpha^2} = \frac{\alpha^3 + \beta^3}{\alpha^2\beta^2} = \frac{\alpha^3 + \beta^3}{(\alpha\beta)^2} = \frac{5/8}{2^2} = \frac{5/8}{4} = \frac{5}{32} \). Product of new roots: \( P = pq = \frac{\alpha}{\beta^2} \times \frac{\beta}{\alpha^2} = \frac{\alpha\beta}{(\alpha\beta)^3} = \frac{1}{(\alpha\beta)^2} = \frac{1}{4} \) (Wait, let us check: \( pq = \frac{\alpha\beta}{\alpha^2\beta^2} = \frac{1}{\alpha\beta} = \frac{1}{2} \)). The new equation is \( x^2 - Sx + P = 0 \Rightarrow x^2 - \frac{5}{32}x + \frac{1}{2} = 0 \). Multiplying by 32 to get integer coefficients: \( 32x^2 - 5x + 16 = 0 \).

M1: States correct sum and product of original roots. (a) M1: Uses algebraic identity for sum of squares. A1: 9/4. (b) M1: Uses algebraic identity for sum of cubes. A1: 5/8. (c) M1: Formulates the sum of new roots in terms of original sum of cubes and product. M1: Formulates product of new roots as 1/(alpha * beta). A1: Evaluates S = 5/32 and P = 1/2. M1: Sets up quadratic equation. A1: 32x^2 - 5x + 16 = 0.

(a)(i) Since \( Q \) is the midpoint of \( AB \), \( \vec{OQ} = \frac{1}{2}(\mathbf{a} + \mathbf{b}) \). (ii) \( OP : PA = 2 : 1 \Rightarrow \vec{OP} = \frac{2}{3}\mathbf{a} \). Thus, \( \vec{BP} = \vec{OP} - \vec{OB} = \frac{2}{3}\mathbf{a} - \mathbf{b} \). (b) We write \( \vec{OX} \) in two ways: Method 1: \( \vec{OX} = \mu \vec{OQ} = \frac{1}{2}\mu\mathbf{a} + \frac{1}{2}\mu\mathbf{b} \). Method 2: \( \vec{OX} = \vec{OB} + \vec{BX} = \mathbf{b} + \lambda \vec{BP} = \mathbf{b} + \lambda\left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2}{3}\lambda\mathbf{a} + (1 - \lambda)\mathbf{b} \). Since \( \mathbf{a} \) and \( \mathbf{b} \) are non-parallel vectors, we equate coefficients: For \( \mathbf{a} \): \( \frac{1}{2}\mu = \frac{2}{3}\lambda \Rightarrow \mu = \frac{4}{3}\lambda \). For \( \mathbf{b} \): \( \frac{1}{2}\mu = 1 - \lambda \). Substituting \( \mu \) gives: \( \frac{2}{3}\lambda = 1 - \lambda \Rightarrow \frac{5}{3}\lambda = 1 \Rightarrow \lambda = \frac{3}{5} \). Then \( \mu = \frac{4}{3}\left(\frac{3}{5}\right) = \frac{4}{5} \). (c) Since \( \vec{BX} = \frac{3}{5}\vec{BP} \), the point \( X \) is \( \frac{3}{5} \) of the way along \( BP \) from \( B \). This means \( BX : XP = 3 : 2 \), so the ratio \( PX : XB = 2 : 3 \).

(a) B1: Correct expression for \( \vec{OQ} \). B2: Correct expression for \( \vec{BP} \) (B1 for \( \vec{OP} = 2/3\mathbf{a} \)). (b) M1: Expresses \( \vec{OX} \) using \( \mu \). M1: Expresses \( \vec{OX} \) using \( \lambda \). M1: Equates coefficients to form two simultaneous equations. A1: Finds \( \lambda = 3/5 \). A1: Finds \( \mu = 4/5 \). (c) M1: Interprets \( \lambda \) to find division of the line segment. A1: Ratio is 2:3.

(a) Using the laws of logarithms: \( \log_3((x-4)(x+4)) = 2 \Rightarrow \log_3(x^2 - 16) = 2 \). Converting to exponential form: \( x^2 - 16 = 3^2 \Rightarrow x^2 - 16 = 9 \Rightarrow x^2 = 25 \Rightarrow x = \pm 5 \). Since the arguments of log must be positive (i.e., \( x - 4 > 0 \Rightarrow x > 4 \)), we reject \( x = -5 \). Thus, the only solution is \( x = 5 \). (b) Equation 1: \( 2^{2x} \times (2^2)^y = 2^5 \Rightarrow 2^{2x} \times 2^{2y} = 2^5 \Rightarrow 2^{2x+2y} = 2^5 \Rightarrow 2x + 2y = 5 \). Equation 2: \( \log_2(x+2y) = 2 \Rightarrow x + 2y = 2^2 \Rightarrow x + 2y = 4 \). Subtracting Equation 2 from Equation 1: \( (2x + 2y) - (x + 2y) = 5 - 4 \Rightarrow x = 1 \). Substitute \( x = 1 \) into Equation 2: \( 1 + 2y = 4 \Rightarrow 2y = 3 \Rightarrow y = 1.5 \).

(a) M1: Combines logarithms using product rule. M1: Converts log equation to quadratic equation. A1: Solves quadratic to find x = 5 and x = -5. A1: Explains rejection of x = -5 and states x = 5. (b) M1: Expresses first equation in terms of base 2 indices. A1: Forms linear equation 2x + 2y = 5. M1: Converts second log equation to linear form. A1: Forms linear equation x + 2y = 4. M1: Solves simultaneous linear equations. A1: x = 1, y = 1.5.

(a) Using the binomial expansion formula for rational powers: \( (1 + y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \dots \) with \( n = -\frac{1}{2} \) and \( y = -4x \): \( (1 - 4x)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)(-4x) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}(-4x)^2 + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{6}(-4x)^3 + \dots \) \( = 1 + 2x + \frac{3/4}{2}(16x^2) + \frac{-15/8}{6}(-64x^3) = 1 + 2x + 6x^2 + 20x^3 \). The expansion is valid for \( |-4x| < 1 \Rightarrow |x| < \frac{1}{4} \). (b) Substitute \( x = 0.01 \) into the LHS: \( f(0.01) = (1 - 0.04)^{-\frac{1}{2}} = (0.96)^{-\frac{1}{2}} = \frac{1}{\sqrt{0.96}} = \frac{1}{\sqrt{24/25}} = \frac{5}{\sqrt{24}} = \frac{5}{2\sqrt{6}} = \frac{5\sqrt{6}}{12} \). Substitute \( x = 0.01 \) into the expansion: \( 1 + 2(0.01) + 6(0.01)^2 + 20(0.01)^3 = 1 + 0.02 + 0.0006 + 0.00002 = 1.02062 \). Thus, \( \frac{5\sqrt{6}}{12} \approx 1.02062 \Rightarrow \sqrt{6} \approx 1.02062 \times \frac{12}{5} = 2.449488 \). To 5 decimal places, this is 2.44949.

(a) M1: Attempts binomial expansion with n = -1/2. M1: Correctly processes the third term (coefficient of x^2). M1: Correctly processes the fourth term (coefficient of x^3). A1: Final simplified expression 1 + 2x + 6x^2 + 20x^3. B1: States validity range |x| < 1/4. (b) M1: Evaluates expansion at x = 0.01 to get 1.02062. M1: Shows LHS simplifies to a rational multiple of \( \sqrt{6} \). A1: Obtains \( 5\sqrt{6}/12 \). M1: Rearranges equation to find \( \sqrt{6} \). A1: Obtains 2.44949.

(a) The total volume \( V \) of the solid is the sum of the volume of the cylinder and the hemisphere:
\[ V = \pi r^2 h + \frac{2}{3}\pi r^3 = 360\pi \]
Dividing the entire equation by \( \pi \):
\[ r^2 h + \frac{2}{3}r^3 = 360 \implies h = \frac{360}{r^2} - \frac{2}{3}r \]

The total surface area \( S \) of the solid consists of the curved surface of the hemisphere, the curved surface of the cylinder, and the base of the cylinder:
\[ S = 2\pi r^2 + 2\pi r h + \pi r^2 = 3\pi r^2 + 2\pi r h \]

Substituting \( h \) into the expression for \( S \):
\[ S = 3\pi r^2 + 2\pi r \left( \frac{360}{r^2} - \frac{2}{3}r \right) \]
\[ S = 3\pi r^2 + \frac{720\pi}{r} - \frac{4}{3}\pi r^2 \]
\[ S = \frac{5}{3}\pi r^2 + \frac{720\pi}{r} \]

(b) Differentiating \( S \) with respect to \( r \):
\[ \frac{\text{d}S}{\text{d}r} = \frac{10}{3}\pi r - \frac{720\pi}{r^2} \]

For a stationary value, set \( \frac{\text{d}S}{\text{d}r} = 0 \):
\[ \frac{10}{3}\pi r - \frac{720\pi}{r^2} = 0 \implies \frac{10}{3}r^3 = 720 \implies r^3 = 216 \implies r = 6 \]

To justify that this is a minimum, find the second derivative:
\[ \frac{\text{d}^2S}{\text{d}r^2} = \frac{10}{3}\pi + \frac{1440\pi}{r^3} \]

Substitute \( r = 6 \):
\[ \frac{\text{d}^2S}{\text{d}r^2} = \frac{10}{3}\pi + \frac{1440\pi}{216} = 10\pi > 0 \]

Since the second derivative is positive, \( r = 6 \) gives a minimum value for \( S \).

(c) Substitute \( r = 6 \) back into the formula for \( S \):
\[ S = \frac{5}{3}\pi(6)^2 + \frac{720\pi}{6} = 60\pi + 120\pi = 180\pi \]

(a) [4 Marks]
- M1: Attempts to write an expression for the volume, \( V = \pi r^2 h + \frac{2}{3}\pi r^3 = 360\pi \).
- A1: Rearranges correctly to express \( h \) in terms of \( r \), e.g., \( h = \frac{360}{r^2} - \frac{2}{3}r \).
- M1: Writes the total surface area formula as \( S = 3\pi r^2 + 2\pi r h \).
- A1*: Completes the algebraic proof to show \( S = \frac{5}{3}\pi r^2 + \frac{720\pi}{r} \) clearly and convincingly.

(b) [4 Marks]
- M1: Attempts to differentiate \( S \) with respect to \( r \) (at least one term correct).
- A1: Correctly finds \( \frac{\text{d}S}{\text{d}r} = \frac{10}{3}\pi r - \frac{720\pi}{r^2} \).
- A1: Sets derivative to 0 and correctly finds \( r = 6 \).
- M1: Differentiates again to find \( \frac{\text{d}^2S}{\text{d}r^2} \) and evaluates it at \( r = 6 \) to show it is positive (minimum).

(c) [2 Marks]
- M1: Substitutes their value of \( r \) into the expression for \( S \).
- A1: Obtains \( 180\pi \).

(a) Express \( 3\theta \) as \( 2\theta + \theta \) and apply the addition formula:
\[ \cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta \cos \theta - \sin 2\theta \sin \theta \]

Using double-angle formulas \( \cos 2\theta = 2\cos^2\theta - 1 \) and \( \sin 2\theta = 2\sin\theta\cos\theta \):
\[ \cos 3\theta = (2\cos^2\theta - 1)\cos\theta - (2\sin\theta\cos\theta)\sin\theta \]
\[ \cos 3\theta = 2\cos^3\theta - \cos\theta - 2\sin^2\theta\cos\theta \]

Substitute \( \sin^2\theta = 1 - \cos^2\theta \):
\[ \cos 3\theta = 2\cos^3\theta - \cos\theta - 2(1 - \cos^2\theta)\cos\theta \]
\[ \cos 3\theta = 2\cos^3\theta - \cos\theta - 2\cos\theta + 2\cos^3\theta \]
\[ \cos 3\theta = 4\cos^3\theta - 3\cos\theta \]

(b) Substitute the identity from part (a) into the given equation:
\[ 2(4\cos^3\theta - 3\cos\theta) + \cos\theta = 0 \]
\[ 8\cos^3\theta - 6\cos\theta + \cos\theta = 0 \]
\[ 8\cos^3\theta - 5\cos\theta = 0 \]
\[ \cos\theta(8\cos^2\theta - 5) = 0 \]

This yields two cases:
1. \( \cos\theta = 0 \)
Since \( 0 \le \theta \le \pi \), we get:
\[ \theta = \frac{\pi}{2} \approx 1.57 \text{ radians} \]

2. \( 8\cos^2\theta - 5 = 0 \implies \cos^2\theta = \frac{5}{8} \implies \cos\theta = \pm\sqrt{\frac{5}{8}} \)

For \( \cos\theta = \sqrt{\frac{5}{8}} \):
\[ \theta = \arccos\left(\sqrt{\frac{5}{8}}\right) \approx 0.659 \text{ radians} \]

For \( \cos\theta = -\sqrt{\frac{5}{8}} \):
\[ \theta = \pi - 0.659058... \approx 2.48 \text{ radians} \]

Thus, the solutions in the interval \( 0 \le \theta \le \pi \) are \( \theta \approx 0.659, 1.57, 2.48 \).

(a) [4 Marks]
- M1: Writes \( \cos 3\theta = \cos(2\theta + \theta) \) and applies compound angle formula.
- M1: Uses double angle identities for \( \cos 2\theta \) and \( \sin 2\theta \) to express terms in \( \sin \theta \) and \( \cos \theta \).
- M1: Uses \( \sin^2 \theta = 1 - \cos^2 \theta \) to eliminate all sine terms.
- A1*: Achieves the final printed result with no errors in working.

(b) [6 Marks]
- M1: Substitutes the result of (a) into the given equation to form an expression in \( \cos \theta \).
- A1: Obtains simplified cubic equation \( 8\cos^3\theta - 5\cos\theta = 0 \).
- M1: Factorizes to find solutions for \( \cos\theta = 0 \) and \( \cos^2\theta = \frac{5}{8} \).
- B1: Identifies \( \theta = \frac{\pi}{2} \) (or \( 1.57 \)).
- A1: Finds \( \theta \approx 0.659 \) (allow 3 s.f.).
- A1: Finds \( \theta \approx 2.48 \) (allow 3 s.f.). Deduct 1 mark overall if extra solutions within the interval are present.

The volume \(V\) of the solid of revolution is given by:
\(V = \pi \int_{1}^{5} y^2 \, dx\)

First, we find \(y^2\):
\(y^2 = (3 + \sqrt{2x - 1})^2 = 9 + 6\sqrt{2x - 1} + (2x - 1) = 8 + 2x + 6(2x - 1)^{1/2}\)

Now integrate with respect to \(x\):
\(\int (8 + 2x + 6(2x - 1)^{1/2}) \, dx = 8x + x^2 + 6 \cdot \frac{(2x - 1)^{3/2}}{\frac{3}{2} \cdot 2} = 8x + x^2 + 2(2x - 1)^{3/2}\)

Evaluate the definite integral from \(1\) to \(5\):
At \(x = 5\):
\(8(5) + 5^2 + 2(2(5) - 1)^{3/2} = 40 + 25 + 2(9)^{3/2} = 65 + 2(27) = 119\)

At \(x = 1\):
\(8(1) + 1^2 + 2(2(1) - 1)^{3/2} = 8 + 1 + 2(1)^{3/2} = 11\)

Subtracting the lower limit from the upper limit:
\(V = \pi (119 - 11) = 108\pi\)

M1: Attempt to square \(y\) to find \(y^2\), getting at least two terms correct including the middle term.
A1: Correct expansion: \(y^2 = 8 + 2x + 6(2x - 1)^{1/2}\).
M1: Attempt to integrate \(y^2\), showing \(x \to x^2\) and \((2x - 1)^{1/2} \to k(2x - 1)^{3/2}\).
A1: Correct integration: \(8x + x^2 + 2(2x - 1)^{3/2}\).
M1: Substitute limits of \(5\) and \(1\) into their integrated expression and subtract.
A1: Exact volume is \(108\pi\).

Using the sum-to-product identity:
\(\cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)\)

We get:
\(\cos 3\theta + \cos \theta = 2 \cos 2\theta \cos \theta\)

The equation becomes:
\(2 \cos 2\theta \cos \theta = \cos 2\theta\)

Rearranging gives:
\(\cos 2\theta (2 \cos \theta - 1) = 0\)

This yields two sets of solutions:

1) \(\cos 2\theta = 0\)
Since \(0 \le \theta < \pi\), we have \(0 \le 2\theta < 2\pi\).
\(2\theta = \frac{\pi}{2}, \frac{3\pi}{2} \implies \theta = \frac{\pi}{4}, \frac{3\pi}{4}\)

2) \(2 \cos \theta - 1 = 0 \implies \cos \theta = \frac{1}{2}\)
For \(0 \le \theta < \pi\):
\(\theta = \frac{\pi}{3}\)

Therefore, the solutions in the interval are:
\(\theta = \frac{\pi}{4}, \frac{\pi}{3}, \frac{3\pi}{4}\)

M1: Use sum-to-product formula or expand \(\cos 3\theta\) to rewrite the equation in terms of \(\cos \theta\) and \(\cos 2\theta\).
A1: Obtain the factored form \(\cos 2\theta (2 \cos \theta - 1) = 0\) or equivalent.
M1: Set \(\cos 2\theta = 0\) and find at least one correct value of \(\theta\).
A1: Obtain \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{3\pi}{4}\).
M1: Set \(\cos \theta = \frac{1}{2}\) and solve for \(\theta\).
A1: Obtain \(\theta = \frac{\pi}{3}\) and no other values in the interval.

(a) The sum of the first three terms of the arithmetic series is:
\(S_3 = a + (a+d) + (a+2d) = 3a + 3d\)
Given \(a = 16\) and \(S_3 = 42\):
\(3(16) + 3d = 42 \implies 48 + 3d = 42 \implies 3d = -6 \implies d = -2\)

(b) The 7th term of the arithmetic series is:
\(u_7 = a + 6d = 16 + 6(-2) = 16 - 12 = 4\)

The 3rd term of the geometric series is:
\(v_3 = a r^2 = 16 r^2\)

Setting them equal:
\(16 r^2 = 4 \implies r^2 = \frac{1}{4} \implies r = \pm \frac{1}{2}\)

(c) For \(r = -\frac{1}{2}\):
\(S_\infty = \frac{a}{1 - r} = \frac{16}{1 - (-1/2)} = \frac{16}{1.5} = \frac{32}{3}\)

Part (a):
M1: Write an expression for the sum of the first 3 terms and set equal to 42.
A1: Correctly solve to find \(d = -2\).

Part (b):
M1: Find the 7th term of the arithmetic series.
M1: Write the 3rd term of the geometric series as \(16r^2\) and set equal to their 7th term.
A1: Correct values \(r = \pm \frac{1}{2}\) (must have both).

Part (c):
M1: Correct use of the sum to infinity formula with \(a = 16\) and their negative \(r\).
A1: Correct exact sum to infinity \(\frac{32}{3}\) or \(10\frac{2}{3}\).

(a) The vertical asymptote occurs where the denominator is zero:
\(x + 2 = 0 \implies x = -2\)

The horizontal asymptote is found by taking the limit as \(x \to \pm \infty\):
\(y = \lim_{x\to\infty} \frac{3 - \frac{2}{x}}{1 + \frac{2}{x}} = 3 \implies y = 3\)

(b) Intersect \(y\)-axis (set \(x=0\)):
\(y = \frac{3(0) - 2}{0 + 2} = -1 \implies (0, -1)\)

Intersect \(x\)-axis (set \(y=0\)):
\(\frac{3x - 2}{x + 2} = 0 \implies 3x - 2 = 0 \implies x = \frac{2}{3} \implies (\frac{2}{3}, 0)\)

(c) To find the gradient, differentiate \(y\) with respect to \(x\) using the quotient rule:
\(u = 3x - 2 \implies u' = 3\)
\(v = x + 2 \implies v' = 1\)
\(\frac{dy}{dx} = \frac{u'v - uv'}{v^2} = \frac{3(x + 2) - (3x - 2)(1)}{(x + 2)^2} = \frac{3x + 6 - 3x + 2}{(x + 2)^2} = \frac{8}{(x + 2)^2}\)

Since \(8 > 0\) and \((x + 2)^2 > 0\) for all \(x \ne -2\), the gradient \(\frac{dy}{dx}\) is always positive.

Part (a):
A1: Correct vertical asymptote \(x = -2\).
A1: Correct horizontal asymptote \(y = 3\).

Part (b):
A1: Correct \(y\)-intercept \((0, -1)\).
A1: Correct \(x\)-intercept \((\frac{2}{3}, 0)\).

Part (c):
M1: Attempt to differentiate \(y\) using quotient rule or product rule.
A1: Correct derivative \(\frac{dy}{dx} = \frac{8}{(x+2)^2}\).
A1: Clear conclusion stating why \(\frac{dy}{dx} > 0\) for all \(x \ne -2\).

(a) Using the ratio formula for \(P\) dividing \(AB\) in the ratio \(3 : 1\):
\(x_P = \frac{1(-2) + 3(6)}{3 + 1} = \frac{16}{4} = 4\)
\(y_P = \frac{1(5) + 3(1)}{3 + 1} = \frac{8}{4} = 2\)
So, \(P = (4, 2)\).

(b) The gradient of \(AB\) is:
\(m_{AB} = \frac{1 - 5}{6 - (-2)} = \frac{-4}{8} = -\frac{1}{2}\)

Since \(L\) is perpendicular to \(AB\), its gradient \(m_L\) is:
\(m_L = -\frac{1}{m_{AB}} = 2\)

The equation of \(L\) is:
\(y - 2 = 2(x - 4) \implies y - 2 = 2x - 8 \implies 2x - y - 6 = 0\)

(c) To find \(Q\), set \(x = 0\) in the equation of \(L\):
\(2(0) - y - 6 = 0 \implies y = -6 \implies Q = (0, -6)\)

The vertices of triangle \(OAQ\) are \(O(0, 0)\), \(A(-2, 5)\), and \(Q(0, -6)\).
Using the coordinates, the area is:
\(\text{Area} = \frac{1}{2} | 0(5 - (-6)) + (-2)(-6 - 0) + 0(0 - 5) | = \frac{1}{2} | 12 | = 6\)

Part (a):
M1: Use the division of a line segment formula with correct weights.
A1: Correct coordinates \(P(4, 2)\).

Part (b):
M1: Find gradient of \(AB\).
M1: Find negative reciprocal gradient and write down equation of line through their \(P\).
A1: Correct equation in required form, e.g., \(2x - y - 6 = 0\) (or any integer multiple).

Part (c):
M1: Find coordinates of \(Q\) by setting \(x = 0\) in their line equation.
M1: Use a correct method to find the area of triangle \(OAQ\).
A1: Correct area of \(6\).

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

(a) \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{5}{2}\right)^2 - 2(2) = \frac{25}{4} - 4 = \frac{9}{4}\)

(b) \(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = \left(\frac{5}{2}\right)^3 - 3(2)\left(\frac{5}{2}\right) = \frac{125}{8} - 15 = \frac{5}{8}\)

(c) Let the new roots be \(\gamma = \frac{\alpha}{\beta^2}\) and \(\delta = \frac{\beta}{\alpha^2}\).

Sum of new roots:
\(S = \gamma + \delta = \frac{\alpha}{\beta^2} + \frac{\beta}{\alpha^2} = \frac{\alpha^3 + \beta^3}{\alpha^2 \beta^2} = \frac{\alpha^3 + \beta^3}{(\alpha\beta)^2}\)
Using values from (b) and the product:
\(S = \frac{\frac{5}{8}}{2^2} = \frac{5}{32}\)

Product of new roots:
\(P = \gamma\delta = \frac{\alpha}{\beta^2} \cdot \frac{\beta}{\alpha^2} = \frac{\alpha\beta}{(\alpha\beta)^2} = \frac{1}{\alpha\beta} = \frac{1}{2}\)

The quadratic equation is given by:
\(x^2 - Sx + P = 0 \implies x^2 - \frac{5}{32}x + \frac{1}{2} = 0\)

Multiplying by 32 to get integer coefficients:
\(32x^2 - 5x + 16 = 0\)

Part (a):
M1: Correct algebraic expression for \(\alpha^2 + \beta^2\).
A1: Correct value \(\frac{9}{4}\).

Part (b):
M1: Correct algebraic expression for \(\alpha^3 + \beta^3\).
A1: Correct value \(\frac{5}{8}\).

Part (c):
M1: Express the sum of the new roots in terms of \(\alpha^3+\beta^3\) and \(\alpha\beta\) and substitute.
M1: Express the product of the new roots in terms of \(\alpha\beta\) and substitute.
A1: Correctly form the quadratic equation in integer form: \(32x^2 - 5x + 16 = 0\).

(a) Since \(AP : PB = 2 : 3\), \(P\) divides \(AB\) in the ratio \(2:3\).
\(\overrightarrow{OP} = \overrightarrow{OA} + \frac{2}{5}\overrightarrow{AB} = \mathbf{a} + \frac{2}{5}(\mathbf{b} - \mathbf{a}) = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\)

(b) Since \(X\) lies on \(OP\), we can write:
\(\overrightarrow{OX} = \mu \overrightarrow{OP} = \mu \left(\frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\right) = \frac{3}{5}\mu \mathbf{a} + \frac{2}{5}\mu \mathbf{b}\)

Since \(X\) lies on \(AQ\):
\(\overrightarrow{OQ} = \frac{3}{4}\mathbf{b}\)
\(\overrightarrow{AQ} = \overrightarrow{OQ} - \overrightarrow{OA} = \frac{3}{4}\mathbf{b} - \mathbf{a}\)
\(\overrightarrow{OX} = \overrightarrow{OA} + \lambda \overrightarrow{AQ} = \mathbf{a} + \lambda \left(\frac{3}{4}\mathbf{b} - \mathbf{a}\right) = (1 - \lambda)\mathbf{a} + \frac{3}{4}\lambda \mathbf{b}\)

Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\):
\(\frac{3}{5}\mu = 1 - \lambda\)
\(\frac{2}{5}\mu = \frac{3}{4}\lambda \implies \lambda = \frac{8}{15}\mu\)

Substitute \(\lambda\):
\(\frac{3}{5}\mu = 1 - \frac{8}{15}\mu \implies \frac{17}{15}\mu = 1 \implies \mu = \frac{15}{17}\)

Substitute \(\mu = \frac{15}{17}\) into the expression for \(\overrightarrow{OX}\):
\(\overrightarrow{OX} = \frac{15}{17}\left(\frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\right) = \frac{9}{17}\mathbf{a} + \frac{6}{17}\mathbf{b}\)

(c) Since \(\overrightarrow{OX} = \frac{15}{17}\overrightarrow{OP}\), the ratio of the lengths is:
\(OX : XP = 15 : 2\)

Part (a):
M1: Correct use of ratio theorem or vector addition to find \(\overrightarrow{OP}\).
A1: Correct vector \(\frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\).

Part (b):
M1: Write \(\overrightarrow{OX}\) as a multiple of \(\overrightarrow{OP}\).
M1: Find \(\overrightarrow{AQ}\) and express \(\overrightarrow{OX}\) using a second parameter along \(AQ\).
M1: Set up simultaneous equations by equating coefficients.
A1: Solve for one parameter correctly.
A1: Correct vector expression for \(\overrightarrow{OX}\).

Part (c):
A1: Correct deduction of the ratio \(15 : 2\).

From the first equation, use the laws of logarithms:
\(\log_2 \left(\frac{y}{x}\right) = 3 \implies \frac{y}{x} = 2^3 = 8 \implies y = 8x\)

From the second equation, express both sides with base 3:
\(3^y = (3^3)^{x + 1} = 3^{3x + 3} \implies y = 3x + 3\)

Equating the two expressions for \(y\):
\(8x = 3x + 3 \implies 5x = 3 \implies x = \frac{3}{5} = 0.6\)

Now solve for \(y\):
\(y = 8(0.6) = 4.8\)

M1: Apply log subtraction law to rewrite first equation.
A1: Obtain \(y = 8x\) or equivalent.
M1: Express both sides of second equation with base 3.
A1: Obtain \(y = 3x + 3\) or equivalent.
M1: Solve the simultaneous equations for \(x\) and \(y\).
A1: Correct solutions \(x = 0.6\) and \(y = 4.8\).

(a) Using the product rule for differentiation with \(u = x^2\) and \(v = \ln(2x)\):
\(\frac{\mathrm{d}u}{\mathrm{d}x} = 2x\)
\(\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{2}{2x} = \frac{1}{x}\)
Therefore,
\(\frac{\mathrm{d}y}{\mathrm{d}x} = u \frac{\mathrm{d}v}{\mathrm{d}x} + v \frac{\mathrm{d}u}{\mathrm{d}x} = x^2 \left(\frac{1}{x}\right) + \ln(2x) (2x) = x + 2x \ln(2x)\).

(b) From part (a), we have:
\(\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2 \ln(2x)\right) = 2x \ln(2x) + x\)
Integrating both sides with respect to \(x\):
\(x^2 \ln(2x) = \int (2x \ln(2x) + x) \mathrm{d}x = 2 \int x \ln(2x) \mathrm{d}x + \int x \mathrm{d}x\)
\(x^2 \ln(2x) = 2 \int x \ln(2x) \mathrm{d}x + \frac{1}{2}x^2 + c'\)
Rearranging to make \(\int x \ln(2x) \mathrm{d}x\) the subject:
\(2 \int x \ln(2x) \mathrm{d}x = x^2 \ln(2x) - \frac{1}{2}x^2 - c'\)
\(\int x \ln(2x) \mathrm{d}x = \frac{1}{2}x^2 \ln(2x) - \frac{1}{4}x^2 + C\) (where \(C = -\frac{1}{2}c'\)).

(c) Using the result from (b):
\(\int_{1}^{2} x \ln(2x) \mathrm{d}x = \left[ \frac{1}{2}x^2 \ln(2x) - \frac{1}{4}x^2 \right]_{1}^{2}\)
Substituting the upper limit \(x = 2\):
\(\left(\frac{1}{2}(2)^2 \ln(4) - \frac{1}{4}(2)^2\right) = 2 \ln(4) - 1 = 2 \ln(2^2) - 1 = 4 \ln 2 - 1\)
Substituting the lower limit \(x = 1\):
\(\left(\frac{1}{2}(1)^2 \ln(2) - \frac{1}{4}(1)^2\right) = \frac{1}{2} \ln 2 - \frac{1}{4}\)
Subtracting the lower limit value from the upper limit value:
\((4 \ln 2 - 1) - \left(\frac{1}{2} \ln 2 - \frac{1}{4}\right) = \left(4 - \frac{1}{2}\right) \ln 2 - 1 + \frac{1}{4} = \frac{7}{2} \ln 2 - \frac{3}{4}\)
Hence, \(a = \frac{7}{2}\) and \(b = -\frac{3}{4}\).

M1: Applies product rule correctly to differentiate \(x^2\ln(2x)\) with one term correct.
A1: Obtains the correct derivative \(x + 2x\ln(2x)\).
M1: Integrates both sides of the identity and rearranges to isolate \(\int x\ln(2x)\mathrm{d}x\).
A1: Correctly shows the given expression with constant of integration included.
M1: Substitutes limits 2 and 1 into their integrated expression.
A1: Obtains correct evaluations for limits, e.g., \(4\ln 2 - 1\) and \(\frac{1}{2}\ln 2 - \frac{1}{4}\).
A1: Simplifies to find \(a = \frac{7}{2}\) and \(b = -\frac{3}{4}\).

(a) Using the double angle formulae \(\sin 2\theta = 2\sin\theta\cos\theta\) and \(\cos 2\theta = 2\cos^2\theta - 1\):
\(\frac{\sin 2\theta}{1 + \cos 2\theta} = \frac{2\sin\theta\cos\theta}{1 + (2\cos^2\theta - 1)} = \frac{2\sin\theta\cos\theta}{2\cos^2\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta\).

(b) Using the compound angle formula for \(\cos(A + B)\) where \(A = 2\theta\) and \(B = \theta\):
\(\cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta \cos\theta - \sin 2\theta \sin\theta\)
Substitute double angle formulas \(\cos 2\theta = 2\cos^2\theta - 1\) and \(\sin 2\theta = 2\sin\theta\cos\theta\):
\(\cos 3\theta = (2\cos^2\theta - 1)\cos\theta - (2\sin\theta\cos\theta)\sin\theta\)
\(= 2\cos^3\theta - \cos\theta - 2\sin^2\theta\cos\theta\)
Using the identity \(\sin^2\theta = 1 - \cos^2\theta\):
\(\cos 3\theta = 2\cos^3\theta - \cos\theta - 2(1 - \cos^2\theta)\cos\theta\)
\(= 2\cos^3\theta - \cos\theta - 2\cos\theta + 2\cos^3\theta\)
\(= 4\cos^3\theta - 3\cos\theta\).

(c) Substituting the expression for \(\cos 3\theta\) from part (b) into \(\cos 3\theta + \cos \theta = 0\):
\((4\cos^3\theta - 3\cos\theta) + \cos\theta = 0\)
\(4\cos^3\theta - 2\cos\theta = 0\)
Factorising the equation:
\(2\cos\theta(2\cos^2\theta - 1) = 0\)
This yields \(\cos\theta = 0\) or \(\cos^2\theta = \frac{1}{2} \implies \cos\theta = \pm \frac{1}{\sqrt{2}}\).
For the interval \(0 \le \theta \le \pi\):
If \(\cos\theta = 0\), then \(\theta = \frac{\pi}{2}\).
If \(\cos\theta = \frac{1}{\sqrt{2}}\), then \(\theta = \frac{\pi}{4}\).
If \(\cos\theta = -\frac{1}{\sqrt{2}}\), then \(\theta = \frac{3\pi}{4}\).
Hence, the solutions are \(\theta = \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}\).

M1: Applies correct double angle identities to the numerator and denominator in (a).
A1: Correctly simplifies the algebraic expression to show \(\tan\theta\).
M1: Uses compound angle formula for \(\cos(2\theta + \theta)\) in (b).
M1: Expresses \(\sin^2\theta\) in terms of \(\cos^2\theta\) and expands correctly.
A1: Reaches the correct identity \(4\cos^3\theta - 3\cos\theta\).
M1: Obtains a factorised expression in terms of \(\cos\theta\) in (c).
A1: Identifies critical values \(\cos\theta = 0\) and \(\cos\theta = \pm \frac{1}{\sqrt{2}}\).
A1: Finds all three correct solutions \(\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}\) within the specified interval and no extras.

(a) The sum of the first \(n\) terms of an arithmetic series is \(S_n = \frac{n}{2}(2a + (n-1)d)\).
For \(n = 4\):
\(S_4 = \frac{4}{2}(2a + 3d) = 2(2a + 3d) = 34 \implies 2a + 3d = 17\) (Equation 1)
The \(n\)-th term of an arithmetic series is \(u_n = a + (n-1)d\).
For \(n = 7\):
\(u_7 = a + 6d = 22\) (Equation 2)
Multiplying Equation 2 by 2:
\(2a + 12d = 44\)
Subtracting Equation 1 from this product:
\(9d = 27 \implies d = 3\)
Substituting \(d = 3\) back into Equation 2:
\(a + 6(3) = 22 \implies a + 18 = 22 \implies a = 4\).

(b) The third term of the arithmetic series is:
\(u_3 = a + 2d = 4 + 2(3) = 10\)
The sum to infinity of the geometric series is \(S_{\infty} = \frac{a}{1-r}\).
Given that \(S_{\infty} = 3 \times u_3\):
\(\frac{4}{1-r} = 3 \times 10 = 30\)
\(30(1-r) = 4 \implies 1-r = \frac{4}{30} = \frac{2}{15} \implies r = 1 - \frac{2}{15} = \frac{13}{15}\).

(c) The sum of the first 5 terms of the geometric series is given by \(S_5 = \frac{a(1-r^5)}{1-r\)}:
\(S_5 = \frac{4\left(1 - \left(\frac{13}{15}\right)^5\right)}{1 - \frac{13}{15}}
= \frac{4}{\frac{2}{15}} \left(1 - \left(\frac{13}{15}\right)^5\right) = 30 \left(1 - \left(\frac{13}{15}\right)^5\right)\)
Calculating \(\left(\frac{13}{15}\right)^5 \approx 0.4889456\):
\(S_5 \approx 30 \times (1 - 0.4889456) = 30 \times 0.5110544 = 15.33163...\)
Rounding to 3 decimal places gives \(15.332\).

M1: Formulates two linear equations in terms of \(a\) and \(d\) using arithmetic series formulae.
A1: Correctly solves the linear system to find \(d = 3\).
A1: Obtains the correct first term \(a = 4\).
M1: Computes the 3rd term of the arithmetic series \(u_3 = 10\) and sets up the sum to infinity equation.
A1: Correctly evaluates \(r = \frac{13}{15}\).
M1: Applies the sum formula for the first 5 terms of a geometric series.
A1: Correctly substitutes \(a = 4\) and \(r = \frac{13}{15}\) into the sum formula.
A1: Provides the final calculation of \(15.332\) rounded correctly to 3 decimal places.