2025 Edexcel IAS-Level Pure Mathematics (XPM01) Practice Paper with Answers

From the first equation, we can express \( y \) in terms of \( x \): \( y = 2x + 3 \). Substituting this into the second equation: \( x^2 + 2x(2x + 3) - (2x + 3)^2 = -7 \). Expanding the terms: \( x^2 + 4x^2 + 6x - (4x^2 + 12x + 9) = -7 \). Simplifying: \( x^2 - 6x - 9 = -7 \implies x^2 - 6x - 2 = 0 \). Solving the quadratic using the quadratic formula: \( x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-2)}}{2} = \frac{6 \pm \sqrt{44}}{2} = 3 \pm \sqrt{11} \). Substitute these back to find \( y \): If \( x = 3 + \sqrt{11} \), then \( y = 2(3 + \sqrt{11}) + 3 = 9 + 2\sqrt{11} \). If \( x = 3 - \sqrt{11} \), then \( y = 2(3 - \sqrt{11}) + 3 = 9 - 2\sqrt{11} \).

M1: Expresses \( y \) in terms of \( x \) (or vice versa) correctly.
M1: Substitutes their expression into the quadratic equation.
A1: Obtains a correct simplified quadratic in one variable, e.g., \( x^2 - 6x - 2 = 0 \).
M1: Attempts to solve their quadratic equation by completing the square or quadratic formula.
A1: Correct values of \( x \) (or \( y \)): \( x = 3 \pm \sqrt{11} \).
M1: Substitutes at least one of their values back to find the other coordinate.
A1: First pair: \( x = 3 + \sqrt{11}, y = 9 + 2\sqrt{11} \).
A1: Second pair: \( x = 3 - \sqrt{11}, y = 9 - 2\sqrt{11} \).

(a) Write \( y \) as: \( y = 4x^{3/2} - 8x^{-1/2} + 5 \). Differentiate with respect to \( x \): \( \frac{\mathrm{d}y}{\mathrm{d}x} = 4 \left(\frac{3}{2}\right)x^{1/2} - 8 \left(-\frac{1}{2}\right)x^{-3/2} = 6x^{1/2} + 4x^{-3/2} \). (b) At the point \( P(4, 33) \), \( x = 4 \). Substitute \( x = 4 \) into \( \frac{\mathrm{d}y}{\mathrm{d}x} \): \( m = 6(4^{1/2}) + 4(4^{-3/2}) = 6(2) + 4\left(\frac{1}{8}\right) = 12 + 0.5 = 12.5 \) or \( \frac{25}{2} \). Use the equation of a straight line with gradient \( m = \frac{25}{2} \) and point \( (4, 33) \): \( y - 33 = \frac{25}{2}(x - 4) \implies 2y - 66 = 25x - 100 \implies 25x - 2y - 34 = 0 \).

(a)
M1: Attempts to differentiate at least one non-constant term using \( x^n \to n x^{n-1} \).
A1: One term of the derivative correct.
A1: Fully correct derivative: \( 6x^{1/2} + 4x^{-3/2} \).
(b)
M1: Substitutes \( x = 4 \) into their derivative to find the gradient of the tangent.
A1: Gradient is \( 12.5 \) or \( \frac{25}{2} \).
M1: Uses their gradient and the point \( (4, 33) \) to find the equation of the line.
A1: Correct equation in the required form: \( 25x - 2y - 34 = 0 \) (or equivalent integer multiples).

(a) For \( x = 1 \): \( y = 3\sqrt{1} - 1 - 2 = 3 - 3 = 0 \). For \( x = 4 \): \( y = 3\sqrt{4} - 4 - 2 = 3(2) - 6 = 0 \). Thus the curve crosses the \( x \)-axis at \( x = 1 \) and \( x = 4 \). (b) The area is given by the definite integral: \( \int_{1}^{4} (3x^{1/2} - x - 2) \, \mathrm{d}x = \left[ \frac{3x^{3/2}}{3/2} - \frac{x^2}{2} - 2x \right]_{1}^{4} = \left[ 2x^{3/2} - \frac{1}{2}x^2 - 2x \right]_{1}^{4} \). Substituting the upper limit \( x = 4 \): \( 2(4^{3/2}) - \frac{1}{2}(4^2) - 2(4) = 2(8) - 8 - 8 = 0 \). Substituting the lower limit \( x = 1 \): \( 2(1^{3/2}) - \frac{1}{2}(1^2) - 2(1) = 2 - 0.5 - 2 = -0.5 \). The area is \( 0 - (-0.5) = 0.5 \) or \( \frac{1}{2} \).

(a)
M1: Substitutes both \( x = 1 \) and \( x = 4 \) into the equation of the curve to evaluate \( y \).
A1: Confirms both values yield \( y = 0 \).
(b)
M1: Integrates with at least one power raised by 1.
A1: Integrates at least two terms correctly.
A1: Fully correct integration: \( 2x^{3/2} - \frac{1}{2}x^2 - 2x \).
M1: Substitutes limits 4 and 1 into their integrated expression.
M1: Evaluates both limits correctly: \( 0 \) and \( -0.5 \).
A1: Subtracts and obtains \( \frac{1}{2} \) (or \( 0.5 \)).

(a) For the equation to have real roots, the discriminant \( \Delta = b^2 - 4ac \ge 0 \). Here, \( a = k+1 \), \( b = 3k+4 \), and \( c = k+2 \). So: \( \Delta = (3k+4)^2 - 4(k+1)(k+2) \ge 0 \). Expanding: \( 9k^2 + 24k + 16 - 4(k^2 + 3k + 2) \ge 0 \implies 9k^2 + 24k + 16 - 4k^2 - 12k - 8 \ge 0 \implies 5k^2 + 12k + 8 \ge 0 \). (b) Complete the square on \( 5k^2 + 12k + 8 \): \( 5\left(k^2 + \frac{12}{5}k\right) + 8 = 5\left( k + \frac{6}{5} \right)^2 - 5\left(\frac{36}{25}\right) + 8 = 5\left( k + \frac{6}{5} \right)^2 - \frac{36}{5} + \frac{40}{5} = 5\left( k + \frac{6}{5} \right)^2 + \frac{4}{5} \). Since \( \left( k + \frac{6}{5} \right)^2 \ge 0 \) for all real \( k \), \( 5\left( k + \frac{6}{5} \right)^2 + \frac{4}{5} \ge \frac{4}{5} > 0 \). Thus, the inequality is satisfied for all real values of \( k \).

(a)
M1: Attempts \( b^2 - 4ac \) with correct expressions for \( a \), \( b \), and \( c \).
M1: Correctly expands both \( (3k+4)^2 \) and \( (k+1)(k+2) \).
A1: Simplifies to find the correct discriminant expression \( 5k^2 + 12k + 8 \).
A1*: Completes the proof with a clear statement linking the discriminant \( \ge 0 \) to the presence of real roots.
(b)
M1: Attempts to complete the square on \( 5k^2 + 12k + 8 \).
A1: Obtains \( 5\left( k + \frac{6}{5} \right)^2 + \frac{4}{5} \) (or equivalent).
A1: Provides a clear conclusion stating that since a squared term is always non-negative, the expression is always positive (minimum value is \( 0.8 \)).

(a) First, find the midpoint of \( AB \): \( M = \left( \frac{-3 + 5}{2}, \frac{5 - 1}{2} \right) = (1, 2) \). Next, find the gradient of \( AB \): \( m_{AB} = \frac{-1 - 5}{5 - (-3)} = \frac{-6}{8} = -\frac{3}{4} \). The perpendicular gradient is \( m_L = -\frac{1}{m_{AB}} = \frac{4}{3} \). The equation of \( L \) is: \( y - 2 = \frac{4}{3}(x - 1) \implies 3y - 6 = 4x - 4 \implies 3y - 4x - 2 = 0 \). (b) Intersection with the \( x \)-axis (set \( y = 0 \)): \( -4x - 2 = 0 \implies x = -\frac{1}{2} \), so \( P = \left(-\frac{1}{2}, 0\right) \). Intersection with the \( y \)-axis (set \( x = 0 \)): \( 3y - 2 = 0 \implies y = \frac{2}{3} \), so \( Q = \left(0, \frac{2}{3}\right) \). The area of the right-angled triangle \( OPQ \) is: \( \text{Area} = \frac{1}{2} \times \left|-\frac{1}{2}\right| \times \frac{2}{3} = \frac{1}{6} \).

(a)
M1: Attempts to find the midpoint of \( AB \).
A1: Midpoint is \( (1, 2) \).
M1: Calculates the gradient of \( AB \) to be \( -\frac{3}{4} \).
M1: Uses the perpendicular gradient rule to find the gradient of \( L \) is \( \frac{4}{3} \).
A1: Obtains a correct equation in the required form: \( 3y - 4x - 2 = 0 \) (or equivalent).
(b)
M1: Sets \( x = 0 \) and \( y = 0 \) to find the coordinates of \( P \) and \( Q \).
M1: Applies the area formula \( \frac{1}{2} \times b \times h \) using their intercepts.
A1: Obtains area \( \frac{1}{6} \) (or equivalent).

(a) We integrate \( f'(x) \): \( f(x) = \int \left(6x^2 - 4x - 3x^{-2}\right) \, \mathrm{d}x = \frac{6x^3}{3} - \frac{4x^2}{2} - \frac{3x^{-1}}{-1} + C = 2x^3 - 2x^2 + \frac{3}{x} + C \). Since \( f(1) = 5 \), substitute \( x = 1 \) and \( y = 5 \): \( 2(1)^3 - 2(1)^2 + \frac{3}{1} + C = 5 \implies 2 - 2 + 3 + C = 5 \implies C = 2 \). So, \( f(x) = 2x^3 - 2x^2 + \frac{3}{x} + 2 \). (b) Substitute \( x = 3 \) into \( f(x) \): \( f(3) = 2(3^3) - 2(3^2) + \frac{3}{3} + 2 = 54 - 18 + 1 + 2 = 39 \).

(a)
M1: Integrates the expression, raising the power by 1 in at least two terms.
A1: At least two terms integrated correctly.
A1: All three terms integrated correctly, including the term with \( x^{-1} \).
M1: Uses the condition \( f(1) = 5 \) to find the constant of integration \( C \).
A1: Fully correct expression for \( f(x) = 2x^3 - 2x^2 + \frac{3}{x} + 2 \).
(b)
M1: Substitutes \( x = 3 \) into their expression.
A1: Obtains \( 39 \).

(a) Let \( f(x) = 3x^2 \). Then \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{3(x+h)^2 - 3x^2}{h} \). Expanding: \( f'(x) = \lim_{h \to 0} \frac{3(x^2 + 2xh + h^2) - 3x^2}{h} = \lim_{h \to 0} \frac{3x^2 + 6xh + 3h^2 - 3x^2}{h} = \lim_{h \to 0} \frac{6xh + 3h^2}{h} \). Dividing by \( h \): \( f'(x) = \lim_{h \to 0} (6x + 3h) \). As \( h \to 0 \), \( 3h \to 0 \), which leaves \( f'(x) = 6x \). (b) Let \( y = 3x^2 - 18x + 7 \). The gradient function is \( \frac{\mathrm{d}y}{\mathrm{d}x} = 6x - 18 \). We want the gradient to be positive: \( 6x - 18 > 0 \implies 6x > 18 \implies x > 3 \).

(a)
M1: Sets up the difference quotient \( \frac{3(x+h)^2 - 3x^2}{h} \).
M1: Expands \( (x+h)^2 \) correctly.
A1: Simplifies the numerator to get \( 6xh + 3h^2 \).
M1: Divides by \( h \) to obtain \( 6x + 3h \).
A1*: Applies the limit as \( h \to 0 \) and completes the proof, showing clear and consistent limit notation.
(b)
M1: Differentiates the quadratic to obtain \( 6x - 18 \).
M1: Sets their gradient function \( > 0 \).
A1: Obtains \( x > 3 \) (or equivalent set notation).

(a) The curve is a translation of \( y = \frac{1}{x} \) upwards by 2 units. The vertical asymptote is \( x = 0 \) and the horizontal asymptote is \( y = 2 \). There is no \( y \)-axis intercept. For the \( x \)-axis intercept, set \( y = 0 \): \( \frac{1}{x} + 2 = 0 \implies \frac{1}{x} = -2 \implies x = -\frac{1}{2} \). Thus, the intercept is at \( \left(-\frac{1}{2}, 0\right) \). (b) A translation by \( \begin{pmatrix} -3 \\ 1 \end{pmatrix} \) means \( h(x) = g(x+3) + 1 \). Substituting \( x+3 \) into \( g(x) \): \( h(x) = \frac{1}{x+3} + 2 + 1 = \frac{1}{x+3} + 3 \). Writing under a common denominator: \( h(x) = \frac{1 + 3(x+3)}{x+3} = \frac{3x+10}{x+3} \).

(a)
M1: Sketches a reciprocal curve with two branches in quadrants 1 and 2 (shifted upwards).
A1: Correct vertical asymptote \( x = 0 \) and horizontal asymptote \( y = 2 \) correctly indicated with equations.
A1: Correct \( x \)-intercept at \( \left(-\frac{1}{2}, 0\right) \) or \( -0.5 \) indicated on the axis.
A1: Curve does not cross the \( y \)-axis and the shape is correct.
(b)
M1: Correctly applies the translation: \( h(x) = \frac{1}{x+3} + 3 \).
M1: Puts the expression over a common denominator.
A1: Obtains \( h(x) = \frac{3x+10}{x+3} \).

(a) First rewrite the second term of the curve's equation in index form:
\(y = 2x^{\frac{3}{2}} - 8x^{-\frac{1}{2}} + 3\)

Differentiate with respect to \(x\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 2 \times \frac{3}{2}x^{\frac{1}{2}} - 8 \times \left(-\frac{1}{2}\right)x^{-\frac{3}{2}}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^{\frac{1}{2}} + 4x^{-\frac{3}{2}}\)

(b) To find the gradient of the tangent at \(P(4, 15)\), substitute \(x = 4\) into \(\frac{\mathrm{d}y}{\mathrm{d}x}\):
\(m = 3(4)^{\frac{1}{2}} + 4(4)^{-\frac{3}{2}}\)
\(m = 3(2) + 4 \times \frac{1}{8}\)
\(m = 6 + \frac{1}{2} = \frac{13}{2}\)

Using the equation of a straight line with gradient \(m = \frac{13}{2}\) through the point \((4, 15)\):
\(y - 15 = \frac{13}{2}(x - 4)\)
\(2(y - 15) = 13(x - 4)\)
\(2y - 30 = 13x - 52\)
\(13x - 2y - 22 = 0\)

(a)
- M1: Attempts to differentiate at least one non-fractional index term correctly (e.g. \(x^{\frac{3}{2}} \to k x^{\frac{1}{2}}\) or \(x^{-\frac{1}{2}} \to k x^{-\frac{3}{2}}\)).
- A1: One term differentiated correctly, either \(3x^{\frac{1}{2}}\) or \(4x^{-\frac{3}{2}}\).
- A1: Fully correct derivative: \(3x^{\frac{1}{2}} + 4x^{-\frac{3}{2}}\) (or equivalent simplified form).

(b)
- M1: Substitutes \(x = 4\) into their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find a numerical gradient.
- A1: Obtains gradient \(m = \frac{13}{2}\) (or 6.5).
- M1: Uses their gradient and the point \((4, 15)\) to form the equation of a straight line.
- A1: Correct equation in the required form: \(13x - 2y - 22 = 0\) (or any integer multiple, e.g. \(26x - 4y - 44 = 0\)).

(a) Express each term in index form:
\(\frac{3x^3 - 5}{x^2} = \frac{3x^3}{x^2} - \frac{5}{x^2} = 3x - 5x^{-2}\)

Also, \(4\sqrt{x} = 4x^{\frac{1}{2}}\).

Thus,
\(\mathrm{f}'(x) = 3x - 5x^{-2} + 4x^{\frac{1}{2}}\)
(So \(A = 3\), \(B = -5\), \(C = 4\)).

(b) Integrate \(\mathrm{f}'(x)\) with respect to \(x\):
\(\mathrm{f}(x) = \int \left( 3x - 5x^{-2} + 4x^{\frac{1}{2}} \right) \mathrm{d}x\)
\(\mathrm{f}(x) = \frac{3}{2}x^2 - \frac{5x^{-1}}{-1} + \frac{4x^{\frac{3}{2}}}{\frac{3}{2}} + K\)
\(\mathrm{f}(x) = \frac{3}{2}x^2 + \frac{5}{x} + \frac{8}{3}x^{\frac{3}{2}} + K\)

Since the curve passes through \((1, 8)\):
\(8 = \frac{3}{2}(1)^2 + \frac{5}{1} + \frac{8}{3}(1)^{\frac{3}{2}} + K\)
\(8 = \frac{3}{2} + 5 + \frac{8}{3} + K\)

Find a common denominator to sum the constants:
\(8 = \frac{9}{6} + \frac{30}{6} + \frac{16}{6} + K\)
\(8 = \frac{55}{6} + K\)
\(K = 8 - \frac{55}{6} = \frac{48}{6} - \frac{55}{6} = -\frac{7}{6}\)

So the equation of the curve is:
\(\mathrm{f}(x) = \frac{3}{2}x^2 + \frac{5}{x} + \frac{8}{3}x^{\frac{3}{2}} - \frac{7}{6}\)

(a)
- M1: Attempts to split the fraction term-by-term, showing at least one term with a correct index.
- A1: Correctly expresses the entire derivative as \(3x - 5x^{-2} + 4x^{\frac{1}{2}}\).

(b)
- M1: Integrates their expression from part (a) term-by-term (at least two terms with integrated power increased by 1).
- A1: Any two terms integrated correctly.
- A1: All three terms integrated correctly, with or without the constant of integration \(K\): \(\frac{3}{2}x^2 + 5x^{-1} + \frac{8}{3}x^{\frac{3}{2}}\).
- M1: Uses the boundary condition \(x=1\), \(y=8\) to form an equation in \(K\) (constant of integration).
- A1: Finds \(K = -\frac{7}{6}\).
- A1: Fully correct simplified final expression for \(\mathrm{f}(x)\).

(a) The first, third, and eleventh terms of the arithmetic sequence are:
\(T_1 = a\)
\(T_3 = a + 2d\)
\(T_{11} = a + 10d\)

Since these terms form a geometric sequence, the ratio between consecutive terms is equal:
\[\frac{a+2d}{a} = \frac{a+10d}{a+2d}\]

Cross-multiplying gives:
\[(a+2d)^2 = a(a+10d)\]
\[a^2 + 4ad + 4d^2 = a^2 + 10ad\]
\[4d^2 = 6ad\]

Since \(d \neq 0\), we can divide both sides by \(2d\) to obtain:
\[2d = 3a\] (as required).

(b) The sum of the first 15 terms of the arithmetic sequence is:
\[S_{15} = \frac{15}{2}(2a + 14d) = 345\]
\[15(a + 7d) = 345\]
\[a + 7d = 23\]

Using \(2d = 3a \implies d = 1.5a\), we substitute into the equation:
\[a + 7(1.5a) = 23\]
\[11.5a = 23 \implies a = 2\]

(c) Since \(a = 2\), the common difference is \(d = 1.5(2) = 3\).
The first three terms of the geometric sequence are:
\(G_1 = a = 2\)
\(G_2 = a + 2d = 2 + 6 = 8\)
\(G_3 = a + 10d = 2 + 30 = 32\)

The common ratio \(r\) is:
\[r = \frac{8}{2} = 4\]

(a)
M1: Sets up the geometric progression condition \((a+2d)^2 = a(a+10d)\) or equivalent ratio equation.
A1: Correctly expands to get \(a^2 + 4ad + 4d^2 = a^2 + 10ad\).
A1*: Completes the proof to show \(2d = 3a\) clearly with no errors.

(b)
M1: Uses \(S_{15} = 345\) with correct arithmetic sum formula to form an equation in \(a\) and \(d\).
M1: Substitutes \(d = 1.5a\) (or equivalent relation) into their equation.
A1: Obtains \(a = 2\).

(c)
M1: Identifies the first and second terms of the geometric sequence (or uses \(r = \frac{a+2d}{a}\) directly with their values).
A1: Obtains common ratio \(r = 4\).

We apply logarithm laws to simplify the equation:
\[ 2\log_2 \sqrt{x} = \log_2 (\sqrt{x})^2 = \log_2 x \]

Substitute this back into the original equation:
\[ \log_2 (3x + 8) - \log_2 x = 3 \]

Apply the subtraction law of logarithms:
\[ \log_2 \left( \frac{3x+8}{x} \right) = 3 \]

Rewrite the logarithmic equation in exponential form:
\[ \frac{3x+8}{x} = 2^3 \]
\[ \frac{3x+8}{x} = 8 \]

Multiply both sides by \(x\):
\[ 3x + 8 = 8x \]
\[ 5x = 8 \implies x = \frac{8}{5} \]

Since \(x = 1.6 > 0\), the solution is valid.

M1: Uses the power law of logarithms to rewrite \(2\log_2 \sqrt{x}\) as \(\log_2 x\).
M1: Uses the subtraction law to write the left-hand side as a single logarithm \(\log_2 \left(\frac{3x+8}{x}\right)\).
M1: Converts the logarithmic equation to an exponential one, writing \(\frac{3x+8}{x} = 2^3\).
A1: Obtains the correct simplified linear equation \(3x + 8 = 8x\).
M1: Solves the linear equation for \(x\).
A1: Obtains \(x = \frac{8}{5}\\ or equivalent decimal \)1.6\).
B1: Confirms the solution is valid by noting \(x > 0\).

(a) To find the center and radius, we complete the square for both \(x\) and \(y\):
\[(x-5)^2 - 25 + (y+3)^2 - 9 + 9 = 0\]
\[(x-5)^2 + (y+3)^2 = 25\]

Therefore, the center of the circle \(C\) has coordinates \((5, -3)\) and the radius is \(\sqrt{25} = 5\).

(b) Let the center of the circle be \(M(5, -3)\).
The gradient of the radius \(MP\) is:
\[m_{\text{radius}} = \frac{1 - (-3)}{8 - 5} = \frac{4}{3}\]

Since the tangent line is perpendicular to the radius at the point of contact:
\[m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}} = -\frac{3}{4}\]

Using the point-slope form with point \(P(8, 1)\):
\[y - 1 = -\frac{3}{4}(x - 8)\]
\[4(y - 1) = -3(x - 8)\]
\[4y - 4 = -3x + 24\]
\[3x + 4y - 28 = 0\]

(a)
M1: Attempts to complete the square for both \(x\) and \(y\).
A1: Correct coordinates of the center: \((5, -3)\).
A1: Correct radius: \(5\).

(b)
M1: Attempts to find the gradient of the radius \(MP\).
A1: Correct gradient of \(MP\) is \(\frac{4}{3}\).
M1: Uses the perpendicular gradient rule \(m_1 m_2 = -1\) to find the gradient of the tangent.
M1: Applies the straight line formula using point \(P(8, 1)\) and their perpendicular gradient.
A1: Obtains \(3x + 4y - 28 = 0\) (or any integer multiple thereof).

(a) Substitute \(\text{tan}\theta = \frac{\text{sin}\theta}{\text{cos}\theta}\):
\[ 2\text{sin}\theta \left( \frac{\text{sin}\theta}{\text{cos}\theta} \right) = 5 - 5\text{cos}\theta \]
\[ \frac{2\text{sin}^2\theta}{\text{cos}\theta} = 5 - 5\text{cos}\theta \]
\[ 2\text{sin}^2\theta = 5\text{cos}\theta - 5\text{cos}^2\theta \]

Using the identity \(\text{sin}^2\theta = 1 - \text{cos}^2\theta\):
\[ 2(1 - \text{cos}^2\theta) = 5\text{cos}\theta - 5\text{cos}^2\theta \]
\[ 2 - 2\text{cos}^2\theta = 5\text{cos}\theta - 5\text{cos}^2\theta \]
\[ 3\text{cos}^2\theta - 5\text{cos}\theta + 2 = 0 \] (as required).

(b) Let \(\theta = 2x - 30^\circ\). Since \(0 \le x < 180^\circ\), the interval for \(\theta\) is:
\[ -30^\circ \le \theta < 330^\circ \]

The quadratic in \(\text{cos}\theta\) factorizes to:
\[ (3\text{cos}\theta - 2)(\text{cos}\theta - 1) = 0 \]

This gives \(\text{cos}\theta = \frac{2}{3}\) or \(\text{cos}\theta = 1\).

If \(\text{cos}\theta = 1\):
Within \([-30^\circ, 330^\circ)\), the only solution is \(\theta = 0^\circ\).
\[ 2x - 30^\circ = 0^\circ \implies x = 15^\circ \]

If \(\text{cos}\theta = \frac{2}{3}\):
\[ \theta = \text{cos}^{-1}\left(\frac{2}{3}\right) \approx 48.19^\circ \]
Or \(\theta = 360^\circ - 48.19^\circ = 311.81^\circ\).

For \(\theta = 48.19^\circ\):
\[ 2x - 30^\circ = 48.19^\circ \implies 2x = 78.19^\circ \implies x \approx 39.1^\circ \]

For \(\theta = 311.81^\circ\):
\[ 2x - 30^\circ = 311.81^\circ \implies 2x = 341.81^\circ \implies x \approx 170.9^\circ \]

Thus, the solutions are \(x = 15.0^\circ, 39.1^\circ, 170.9^\circ\).

(a)
M1: Substitute \(\text{tan}\theta = \frac{\text{sin}\theta}{\text{cos}\theta}\) and clears the fraction.
M1: Uses identity \(\text{sin}^2\theta = 1 - \text{cos}^2\theta\) to obtain an equation containing only cosine terms.
A1*: Completes steps cleanly to show the required quadratic equation.

(b)
M1: Factorizes or uses formula to solve the quadratic equation, obtaining \(\text{cos}\theta = 1\) and \(\text{cos}\theta = 2/3\).
A1: Obtains at least one correct principal value for \(\theta\) (e.g., \(0^\circ\) or \(48.2^\circ\)).
M1: Identifies all valid values of \(\theta\) within the interval \([-30^\circ, 330^\circ)\) and sets up the linear equations to find \(x\).
A1: Obtains all three correct values: \(x = 15^\circ\), \(x \approx 39.1^\circ\), and \(x \approx 170.9^\circ\). (Deduct 1 mark for any additional solutions in the range).

(a) The area of the sector is \(A_{\text{cross}} = \frac{1}{2} r^2 \theta\).
The volume \(V\) of the prism is:
\[ V = A_{\text{cross}} \times h = \frac{1}{2} r^2 \theta h = 4 \implies h = \frac{8}{r^2 \theta} \]

The total surface area \(A\) of the closed prism consists of two sector faces, the curved surface, and two flat rectangular sides:
\[ A = 2\left( \frac{1}{2} r^2 \theta \right) + (r \theta)h + 2rh \]
\[ A = r^2 \theta + r h (\theta + 2) \]

Substitute \(h = \frac{8}{r^2 \theta}\) into the equation:
\[ A = r^2 \theta + r \left( \frac{8}{r^2 \theta} \right) (\theta + 2) \]
\[ A = r^2 \theta + \frac{8(\theta + 2)}{r \theta} \] (as required).

(b) Substitute \(\theta = 1.5\):
\[ A = 1.5 r^2 + \frac{8(1.5 + 2)}{1.5 r} = 1.5 r^2 + \frac{28}{1.5 r} = 1.5 r^2 + \frac{56}{3} r^{-1} \]

Differentiate \(A\) with respect to \(r\):
\[ \frac{\mathrm{d}A}{\mathrm{d}r} = 3r - \frac{56}{3} r^{-2} \]

At the minimum, \(\frac{\mathrm{d}A}{\mathrm{d}r} = 0\):
\[ 3r - \frac{56}{3r^2} = 0 \implies 3r = \frac{56}{3r^2} \]
\[ 9r^3 = 56 \implies r^3 = \frac{56}{9} \]
\[ r = \sqrt[3]{\frac{56}{9}} \approx 1.8391 \]

To 3 significant figures, \(r \approx 1.84\text{ m}\).

(a)
M1: Writes down the volume of the prism \(V = \frac{1}{2} r^2 \theta h = 4\) and expresses \(h\) in terms of \(r\) and \(\theta\).
M1: Formulates the total surface area equation \(A = r^2 \theta + r \theta h + 2 r h\).
M1: Substitutes \(h = \frac{8}{r^2 \theta}\) into their surface area expression.
A1*: Completes algebraic simplification to obtain the given formula.

(b)
M1: Substitute \(\theta = 1.5\) into the formula for \(A\).
M1: Differentiates \(A\) with respect to \(r\) (at least one term with index decreased by 1).
A1: Correctly obtains \(\frac{\mathrm{d}A}{\mathrm{d}r} = 3r - \frac{56}{3r^2}\) (or equivalent).
A1: Sets \(\frac{\mathrm{d}A}{\mathrm{d}r} = 0\) and solves for \(r\) to obtain \(r \approx 1.84\).

(a) Using the recurrence relation:
\[ u_2 = \frac{4}{u_1} + 2 = \frac{4}{k} + 2 = \frac{4 + 2k}{k} \]

Now, find \(u_3\):
\[ u_3 = \frac{4}{u_2} + 2 = \frac{4}{\frac{4+2k}{k}} + 2 \]
\[ u_3 = \frac{4k}{2k + 4} + 2 = \frac{2k}{k + 2} + 2 \]
\[ u_3 = \frac{2k + 2(k + 2)}{k + 2} = \frac{4k + 4}{k + 2} \]

(b) We are given \(u_3 = \frac{8}{3}\), so:
\[ \frac{4k + 4}{k + 2} = \frac{8}{3} \]

Cross-multiplying gives:
\[ 3(4k + 4) = 8(k + 2) \]
\[ 12k + 12 = 8k + 16 \]
\[ 4k = 4 \implies k = 1 \]

Since \(k = 1\) is a positive constant, this is the correct value.

(a)
M1: Attempts to find \(u_2\) in terms of \(k\).
M1: Substitutes their expression for \(u_2\) into the recurrence relation to find \(u_3\).
A1: Obtains the fully simplified fraction \(u_3 = \frac{4k + 4}{k + 2}\).

(b)
M1: Equates their expression for \(u_3\) to \(\frac{8}{3}\).
M1: Eliminates fractions to obtain a linear equation in \(k\).
A1: Simplifies to \(12k + 12 = 8k + 16\) or equivalent.
A1: Obtains \(k = 1\).

(a) Taking the common logarithm (base 10) of both sides:
\[ \log_{10} P = \log_{10}(A k^t) \]
\[ \log_{10} P = \log_{10} A + t \log_{10} k \]

This is in the linear form \(Y = m t + C\), where the vertical intercept is \(C = \log_{10} A\) and the gradient is \(m = \log_{10} k\).

From the point \((0, 2.5)\), the intercept is \(2.5\):
\[ \log_{10} A = 2.5 \implies A = 10^{2.5} = 10^2 \times 10^{0.5} = 100\sqrt{10} \]

From the points \((0, 2.5)\) and \((10, 3.1)\), the gradient \(m\) is:
\[ m = \frac{3.1 - 2.5}{10 - 0} = 0.06 \]

Thus:
\[ \log_{10} k = 0.06 \implies k = 10^{0.06} \approx 1.14815 \]
To 3 significant figures, \(k \approx 1.15\).

(b) The constant \(A\) represents the initial population of birds on the island when monitoring began (at \(t = 0\)).

(c) We want to find the value of \(t\) when \(P = 1000\).
Using the logarithmic equation:
\[ \log_{10} 1000 = 2.5 + 0.06 t \]
\[ 3 = 2.5 + 0.06 t \]
\[ 0.5 = 0.06 t \]
\[ t = \frac{0.5}{0.06} = \frac{50}{6} \approx 8.33\text{ years} \]

To the nearest year, it takes 8 years.

(a)
M1: Applies \(\log_{10}\) to both sides and uses laws of logs to obtain the linear equation \(\log_{10} P = \log_{10} A + t \log_{10} k\).
M1: Identifies \(\log_{10} A = 2.5\) from the vertical intercept.
A1: Obtains the exact value \(A = 100\sqrt{10}\) (or \(10^{2.5}\)).
M1: Computes the gradient of the line as \(0.06\) and equates it to \(\log_{10} k\).
A1: Obtains \(k \approx 1.15\).

(b)
B1: States that \(A\) represents the initial population of birds (at \(t = 0\)).

(c)
M1: Sets \(P = 1000\) or \(\log_{10} P = 3\) in their linear formula to solve for \(t\).
A1: Obtains \(t = 8\) (years) rounded to the nearest integer.

(a) Express the fractional term as a negative power:
\[ y = 10 - x^2 - 9x^{-2} \]

Integrate term-by-term:
\[ \int y \, \mathrm{d}x = \int \left( 10 - x^2 - 9x^{-2} \right) \mathrm{d}x \]
\[ = 10x - \frac{x^3}{3} - \frac{9x^{-1}}{-1} + c \]
\[ = 10x - \frac{x^3}{3} + \frac{9}{x} + c \]

(b) The area of region \(R\) is given by the definite integral from \(x = 1\) to \(x = 3\):
\[ \text{Area} = \int_{1}^{3} \left( 10 - x^2 - \frac{9}{x^2} \right) \mathrm{d}x \]
\[ = \left[ 10x - \frac{x^3}{3} + \frac{9}{x} \right]_{1}^{3} \]

Evaluate at the upper limit \(x = 3\):
\[ \left( 10(3) - \frac{3^3}{3} + \frac{9}{3} \right) = 30 - 9 + 3 = 24 \]

Evaluate at the lower limit \(x = 1\):
\[ \left( 10(1) - \frac{1^3}{3} + \frac{9}{1} \right) = 10 - \frac{1}{3} + 9 = \frac{56}{3} \]

Subtract the lower limit value from the upper limit value:
\[ \text{Area} = 24 - \frac{56}{3} = \frac{72 - 56}{3} = \frac{16}{3} \]

(a)
M1: Integrates at least one term correctly (increasing the index by 1 and dividing by the new index).
A1: Integrates at least two terms correctly.
A1: Correct indefinite integral with constant \(+c\).

(b)
M1: Sets up the definite integral with limits \(1\) and \(3\).
M1: Substitute the limits \(3\) and \(1\) into their integrated function and subtracts.
A1: Obtains the correct evaluated limits \(24\) and \(\frac{56}{3}\).
A1: Obtains the exact area of \(\frac{16}{3}\) (or equivalent mixed number \(5\frac{1}{3}\)).

(a) Since the terms are in arithmetic progression, the common difference \(d\) is constant. Therefore, \(d = \log_3(2x) - \log_3 x = \log_3\left(\frac{2x}{x}\right) = \log_3 2\). Also, \(d = \log_3(x^2 - 5) - \log_3(2x) = \log_3\left(\frac{x^2 - 5}{2x}\right)\). Equating the two expressions for \(d\) gives: \(\log_3 2 = \log_3\left(\frac{x^2 - 5}{2x}\right)\). Taking 3 to the power of both sides yields: \(2 = \frac{x^2 - 5}{2x}\), which simplifies to \(4x = x^2 - 5\), leading to \(x^2 - 4x - 5 = 0\).

(b) Solving the quadratic equation: \((x - 5)(x + 1) = 0\), which gives \(x = 5\) or \(x = -1\). Since the argument of a logarithm must be strictly positive, we require \(x > 0\) (for \(\log_3 x\) to be defined) and \(x^2 - 5 > 0\). The solution \(x = -1\) is rejected because it does not satisfy \(x > 0\). Hence, the only valid solution is \(x = 5\).

(c) For \(x = 5\), the first term is \(a = \log_3 5\) and the common difference is \(d = \log_3 2\). Using the sum formula for an arithmetic series, \(S_n = \frac{n}{2}[2a + (n - 1)d]\), for \(n = 10\): \(S_{10} = \frac{10}{2}[2\log_3 5 + 9\log_3 2] = 5[2\log_3 5 + 9\log_3 2] = 10\log_3 5 + 45\log_3 2\). Thus, \(p = 10\) and \(q = 45\).

(a) M1: Attempts to find the common difference by subtracting consecutive terms and equates them. M1: Uses laws of logarithms correctly to simplify \(\log_3(2x) - \log_3 x = \log_3 2\) or \(\log_3(x^2-5) - \log_3(2x) = \log_3((x^2-5)/(2x))\). A1*: Completes the proof to show \(x^2 - 4x - 5 = 0\) with no errors.
(b) M1: Solves the quadratic to find \(x = 5\) and \(x = -1\). A1: Correctly identifies \(x = 5\) and gives a valid reason for rejecting \(x = -1\) (e.g., \(x\) must be positive for \(\log_3 x\) to be defined).
(c) M1: Identifies the first term \(a = \log_3 5\) and common difference \(d = \log_3 2\), and uses the arithmetic series sum formula with \(n=10\). M1: Expands the expression to the form \(p\log_3 5 + q\log_3 2\). A1: Correctly identifies \(p = 10\) and \(q = 45\) (or writes the final sum as \(10\log_3 5 + 45\log_3 2\)).

(a) The cross-section is a right-angled triangle with base \(3x\) and height \(4x\). The area of the cross-section is \(A = \frac{1}{2} \times 3x \times 4x = 6x^2\). The volume \(V\) of the prism is given by \(V = A \times L = 6x^2 L\). Since \(V = 324\), we have: \(6x^2 L = 324 \implies L = \frac{54}{x^2}\). The total surface area \(S\) is the sum of the areas of the two triangular ends and the three rectangular faces: \(S = 2A + (3x + 4x + 5x)L = 12x^2 + 12x L\). Substituting \(L = \frac{54}{x^2}\) into the surface area formula gives: \(S = 12x^2 + 12x\left(\frac{54}{x^2}\right) = 12x^2 + \frac{648}{x}\) (as required).

(b) To find the stationary value, we differentiate \(S\) with respect to \(x\): \(\frac{dS}{dx} = 24x - \frac{648}{x^2}\). Setting \(\frac{dS}{dx} = 0\) gives: \(24x - \frac{648}{x^2} = 0 \implies 24x^3 = 648 \implies x^3 = 27\). Taking the cube root gives \(x = 3\).

(c) Substituting \(x = 3\) into the formula for \(S\): \(S_{\text{min}} = 12(3)^2 + \frac{648}{3} = 108 + 216 = 324\). To justify that this is a minimum, we find the second derivative: \(\frac{d^2S}{dx^2} = 24 + \frac{1296}{x^3}\). For \(x = 3\), \(\frac{d^2S}{dx^2} = 24 + \frac{1296}{27} = 24 + 48 = 72\). Since \(\frac{d^2S}{dx^2} > 0\), the stationary value is a minimum.

(a) M1: Finds the correct area of the triangular cross-section \(6x^2\) and expresses the volume \(V = 6x^2 L = 324\). M1: Expresses the total surface area \(S = 12x^2 + 12x L\). A1*: Eliminates \(L\) to obtain the given formula \(S = 12x^2 + \frac{648}{x}\) with no errors shown.
(b) M1: Differentiates the expression for \(S\) to find \(\frac{dS}{dx} = 24x - \frac{648}{x^2}\) (at least one term differentiated correctly). M1: Sets \(\frac{dS}{dx} = 0\) and solves for \(x^3\). A1: Obtains \(x = 3\).
(c) M1: Substitutes \(x = 3\) into \(S\) to find the minimum surface area of \(324\). A1: Finds the second derivative \(\frac{d^2S}{dx^2} = 24 + \frac{1296}{x^3}\), evaluates it at \(x = 3\) to get \(72 > 0\), and concludes that it is a minimum.