2025 Edexcel IAL Pure Mathematics (YPM01) Practice Paper with Answers

First, rearrange the equation of \(l_2\) to find its gradient: \(3y = -2x + 5\) which gives \(y = -\frac{2}{3}x + \frac{5}{3}\). So the gradient of \(l_2\) is \(-\frac{2}{3}\). Since \(l_1\) is perpendicular to \(l_2\), the gradient of \(l_1\), denoted \(m_1\), satisfies: \(m_1 \times \left(-\frac{2}{3}\right) = -1 \implies m_1 = \frac{3}{2}\). (a) Using the gradient formula for the points \(A(-2, k)\) and \(B(4, 8)\): \(\frac{8 - k}{4 - (-2)} = \frac{3}{2}\) which simplifies to \(\frac{8 - k}{6} = \frac{3}{2}\), so \(8 - k = 9\) and thus \(k = -1\). (b) Using the point \(B(4, 8)\) and the gradient \(m_1 = \frac{3}{2}\), the equation of \(l_1\) is: \(y - 8 = \frac{3}{2}(x - 4)\). Multiplying by 2 to clear the fraction gives \(2y - 16 = 3x - 12\). Rearranging into the required form: \(3x - 2y + 4 = 0\).

M1: Identifies the gradient of \(l_2\) as \(-\frac{2}{3}\) and uses the perpendicular rule to find the gradient of \(l_1\) is \(\frac{3}{2}\). M1: Sets up an equation using the gradient of \(l_1\) with coordinates \(A(-2, k)\) and \(B(4, 8)\). A1: Correct value of \(k = -1\). M1: Uses their gradient for \(l_1\) and either point \(A\) or \(B\) to form a linear equation. A1: Correct equation in the form \(ax + by + c = 0\) with integer coefficients, e.g., \(3x - 2y + 4 = 0\).

Express the terms of \(f'(x)\) in index form: \(f'(x) = 5x^{3/2} - 3x^{-1/2}\). Now integrate \(f'(x)\) with respect to \(x\): \(f(x) = \int (5x^{3/2} - 3x^{-1/2}) \, dx = \frac{5}{\frac{5}{2}}x^{5/2} - \frac{3}{\frac{1}{2}}x^{1/2} + C = 2x^{5/2} - 6x^{1/2} + C\). Since the curve passes through \((4, 50)\), substitute \(x = 4\) and \(f(x) = 50\) to find the constant of integration \(C\): \(50 = 2(4)^{5/2} - 6(4)^{1/2} + C\). Since \(4^{5/2} = 32\) and \(4^{1/2} = 2\), this becomes \(50 = 2(32) - 6(2) + C \implies 50 = 64 - 12 + C \implies 50 = 52 + C\), which gives \(C = -2\). Thus, the equation of the curve is \(f(x) = 2x^{5/2} - 6x^{1/2} - 2\).

M1: Rewrites \(5x\sqrt{x}\) as \(5x^{3/2}\) or \(\frac{3}{\sqrt{x}}\) as \(3x^{-1/2}\). M1: Integrates at least one term correctly by increasing the power by 1. A1: Correct integration with constant of integration: \(2x^{5/2} - 6x^{1/2} + C\). M1: Substitutes \(x = 4\) and \(y = 50\) into their integrated expression to find \(C\). A1: Fully correct simplified expression for \(f(x)\): \(2x^{5/2} - 6x^{1/2} - 2\).

Use the trigonometric identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to express the entire equation in terms of \(\sin \theta\): \(6(1 - \sin^2 \theta) - \sin \theta - 5 = 0 \implies 6 - 6\sin^2 \theta - \sin \theta - 5 = 0 \implies 6\sin^2 \theta + \sin \theta - 1 = 0\). Factorize the quadratic equation: \((3\sin \theta - 1)(2\sin \theta + 1) = 0\). This gives two cases: Case 1: \(\sin \theta = \frac{1}{3} \implies \theta = \sin^{-1}(1/3) \approx 19.5^\circ\). The second solution in the interval is \(\theta = 180^\circ - 19.47^\circ = 160.5^\circ\). Case 2: \(\sin \theta = -\frac{1}{2} \implies \theta = 180^\circ - (-30^\circ) = 210^\circ\) and \(\theta = 360^\circ - 30^\circ = 330^\circ\). Thus, the complete set of solutions in the given range is \(\theta = 19.5^\circ, 160.5^\circ, 210^\circ, 330^\circ\).

M1: Substitutes \(\cos^2 \theta = 1 - \sin^2 \theta\) to get a quadratic in \(\sin \theta\). A1: Correct quadratic equation: \(6\sin^2 \theta + \sin \theta - 1 = 0\). M1: Solves their quadratic to obtain \(\sin \theta = \frac{1}{3}\) and \(\sin \theta = -\frac{1}{2}\) (or equivalent). A1: Any two correct angles from the set \(\{19.5^\circ, 160.5^\circ, 210^\circ, 330^\circ\}\). A1: All four angles correct: \(19.5^\circ, 160.5^\circ, 210^\circ, 330^\circ\). Deduct 1 mark for any extra solutions within the range.

(a) Equating the expressions for \( y \): \( kx - 3 = x^2 - 5x + 1 \). Rearranging gives \( x^2 - 5x - kx + 1 + 3 = 0 \), which simplifies to \( x^2 - (k+5)x + 4 = 0 \). (b) For no intersection, the quadratic equation has no real roots, so the discriminant \( b^2 - 4ac < 0 \). Here, \( a = 1 \), \( b = -(k+5) \), and \( c = 4 \). Thus, \( (-(k+5))^2 - 4(1)(4) < 0 \), which simplifies to \( (k+5)^2 - 16 < 0 \). Expanding or factoring gives \( k^2 + 10k + 9 < 0 \), which factors as \( (k+9)(k+1) < 0 \). The critical values are \( k = -9 \) and \( k = -1 \). Since the inequality is less than zero, the solution is \( -9 < k < -1 \).

(a) M1: Equating the line and the curve expressions. A1: Correctly simplifying to the given quadratic equation. (b) M1: Realizing that no intersection implies \( b^2 - 4ac < 0 \). A1: Correct substitution of \( a, b, c \) into the discriminant. M1: Finding critical values of \( k = -9 \) and \( k = -1 \). A1: Correct final inequality range \( -9 < k < -1 \).

(a) Using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \), substitute into the equation: \( 3(1 - \cos^2 \theta) + 7\cos \theta - 5 = 0 \). Expanding gives \( 3 - 3\cos^2 \theta + 7\cos \theta - 5 = 0 \). Simplifying and multiplying by \( -1 \) gives \( 3\cos^2 \theta - 7\cos \theta + 2 = 0 \) as required. (b) Let \( y = \cos \theta \). The equation is \( 3y^2 - 7y + 2 = 0 \). Factoring gives \( (3y - 1)(y - 2) = 0 \), so \( y = \frac{1}{3} \) or \( y = 2 \). Since \( \cos \theta = 2 \) has no solutions, we solve \( \cos \theta = \frac{1}{3} \). The principal value is \( \theta = \arccos(1/3) \approx 70.5^\circ \). The other value in the interval is \( 360^\circ - 70.5^\circ = 289.5^\circ \).

(a) M1: Applying \( \sin^2 \theta = 1 - \cos^2 \theta \). A1: Expanding and simplifying to get the correct quadratic in \( \cos \theta \). (b) M1: Solving the quadratic equation to find values for \( \cos \theta \). A1: Identifying \( \cos \theta = 1/3 \) and rejecting \( \cos \theta = 2 \). M1: Finding one correct angle (approx. 70.5). A1: Finding both correct angles to 1 decimal place: \( 70.5^\circ \) and \( 289.5^\circ \).

(a) Differentiating \( y = 2x^3 - 5x^2 + 3x + 4 \) term by term with respect to \( x \) gives \( \frac{dy}{dx} = 6x^2 - 10x + 3 \). (b) At the point \( P \) where \( x = 2 \), the gradient of the tangent is \( \frac{dy}{dx} = 6(2)^2 - 10(2) + 3 = 24 - 20 + 3 = 7 \). The gradient of the normal is the negative reciprocal, \( m_n = -\frac{1}{7} \). The equation of the normal is \( y - y_1 = m_n(x - x_1) \), which gives \( y - 6 = -\frac{1}{7}(x - 2) \). Multiplying by 7: \( 7y - 42 = -x + 2 \). Rearranging gives \( x + 7y - 44 = 0 \).

(a) M1: Attempting to differentiate at least two terms of the cubic curve. A1: Correct derivative \( 6x^2 - 10x + 3 \). (b) M1: Substituting \( x = 2 \) into their derivative to find the tangent gradient. M1: Using \( m_1 m_2 = -1 \) to find the gradient of the normal. M1: Setting up the linear equation with their normal gradient and point \( (2,6) \). A1: Correct equation in the required integer form \( x + 7y - 44 = 0 \) (or any integer multiple).

(a) Integrating term by term: \( \int (9x^{1/2} - 3x - 6) \, dx = 9\frac{x^{3/2}}{3/2} - 3\frac{x^2}{2} - 6x + C = 6x^{3/2} - 1.5x^2 - 6x + C \). (b) The area is given by the definite integral from \( x = 1 \) to \( x = 4 \): \( \int_{1}^{4} (9x^{1/2} - 3x - 6) \, dx = [6x^{3/2} - 1.5x^2 - 6x]_{1}^{4} \). Substituting the upper limit \( x = 4 \): \( 6(4)^{3/2} - 1.5(4)^2 - 6(4) = 6(8) - 1.5(16) - 24 = 48 - 24 - 24 = 0 \). Substituting the lower limit \( x = 1 \): \( 6(1)^{3/2} - 1.5(1)^2 - 6(1) = 6 - 1.5 - 6 = -1.5 \). The area is \( 0 - (-1.5) = 1.5 \).

(a) M1: Attempting to integrate at least two terms (power increased by 1). A1: Any two terms correct. A1: Fully correct simplified expression including constant \( +C \). (b) M1: Formulating the definite integral with limits 1 and 4. M1: Substituting the limits 4 and 1 into their integrated expression. A1: Finding the correct value of \( 1.5 \) (accept positive area only).

(a) The gradient of \( AB \) is \( m = \frac{7-5}{4-(-2)} = \frac{2}{6} = \frac{1}{3} \). Using the point-gradient formula with \( A(-2,5) \): \( y - 5 = \frac{1}{3}(x + 2) \). Multiplying by 3: \( 3y - 15 = x + 2 \). Rearranging gives \( x - 3y + 17 = 0 \). (b) The midpoint \( M \) of \( AB \) is \( \left(\frac{-2+4}{2}, \frac{5+7}{2}\right) = (1, 6) \). The gradient of the perpendicular line is \( -\frac{1}{1/3} = -3 \). The equation of the perpendicular bisector is \( y - 6 = -3(x - 1) \), which simplifies to \( y = -3x + 9 \). (c) The line \( y = -3x + 9 \) intersects the \( y \)-axis at \( (0,9) \) and the \( x \)-axis at \( (3,0) \) (since \( 0 = -3x + 9 \Rightarrow x = 3 \)). The area of the right-angled triangle is \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 9 = 13.5 \).

(a) M1: Finding the gradient of \( AB \). M1: Attempting to find the equation of the line through \( A \) and \( B \). A1: Form \( x - 3y + 17 = 0 \) or any integer multiple. (b) M1: Finding the midpoint of \( AB \). M1: Finding the perpendicular gradient. A1: Correct equation \( y = -3x + 9 \). (c) M1: Finding both intercepts of their perpendicular bisector with the axes. A1: Correct area of \( 13.5 \) (or \( \frac{27}{2} \)).

(a) Using rules of indices, \( 2^{2x+1} = 2^1 \times 2^{2x} = 2(2^x)^2 \). Substituting \( y = 2^x \) yields \( 2^{2x+1} = 2y^2 \). Therefore, the equation \( 2^{2x+1} - 9(2^x) + 4 = 0 \) becomes \( 2y^2 - 9y + 4 = 0 \). (b) Factoring the quadratic equation: \( (2y-1)(y-4) = 0 \), which gives solutions \( y = \frac{1}{2} \) or \( y = 4 \). Returning to \( x \): if \( 2^x = \frac{1}{2} \), then \( x = -1 \). If \( 2^x = 4 \), then \( x = 2 \). The solutions are \( x = -1 \) and \( x = 2 \).

(a) M1: Expressing \( 2^{2x+1} \) as \( 2 \times (2^x)^2 \) or similar index law use. A1: Substituting \( y = 2^x \) to correctly obtain the quadratic expression. (b) M1: Attempting to solve the quadratic equation to find two values of \( y \). A1: Correct values \( y = 0.5 \) and \( y = 4 \). M1: Setting up \( 2^x = \text{their } y \) and attempting to solve for \( x \). A1: Both solutions \( x = -1 \) and \( x = 2 \) correct.

(a) To find the stationary points, we solve \( \frac{dy}{dx} = 0 \). First, find \( \frac{dy}{dx} = 6x^2 - 18x - 24 \). Setting this to 0: \( 6(x^2 - 3x - 4) = 0 \Rightarrow 6(x-4)(x+1) = 0 \). This gives stationary points at \( x = 4 \) and \( x = -1 \). When \( x = 4 \), \( y = 2(4)^3 - 9(4)^2 - 24(4) + 15 = 128 - 144 - 96 + 15 = -97 \). When \( x = -1 \), \( y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 15 = -2 - 9 + 24 + 15 = 28 \). The coordinates are \( (4, -97) \) and \( (-1, 28) \). (b) Find the second derivative: \( \frac{d^2y}{dx^2} = 12x - 18 \). At \( x = 4 \), \( \frac{d^2y}{dx^2} = 12(4) - 18 = 30 > 0 \), so \( (4, -97) \) is a local minimum. At \( x = -1 \), \( \frac{d^2y}{dx^2} = 12(-1) - 18 = -30 < 0 \), so \( (-1, 28) \) is a local maximum.

(a) M1: Attempting to differentiate and setting \( \frac{dy}{dx} = 0 \). A1: Finding critical values \( x = 4 \) and \( x = -1 \). M1: Substituting both \( x \)-values into the original curve equation to find \( y \)-coordinates. A1: Both points correctly identified: \( (4, -97) \) and \( (-1, 28) \). (b) M1: Finding the second derivative \( \frac{d^2y}{dx^2} = 12x - 18 \). A1: Showing \( \frac{d^2y}{dx^2} > 0 \) at \( x=4 \) and concluding it is a minimum point. A1: Showing \( \frac{d^2y}{dx^2} < 0 \) at \( x=-1 \) and concluding it is a maximum point.

Using the power law of logarithms on the first term:\
\[ 2\log_2(x - 1) = \log_2((x - 1)^2) \]\
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Using the subtraction law of logarithms:\
\[ \log_2((x - 1)^2) - \log_2(x - 2) = \log_2\left(\frac{(x - 1)^2}{x - 2}\right) \]\
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So the equation becomes:\
\[ \log_2\left(\frac{(x - 1)^2}{x - 2}\right) = 3 \]\
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Convert from logarithmic to exponential form:\
\[ \frac{(x - 1)^2}{x - 2} = 2^3 \]\
\[ \frac{x^2 - 2x + 1}{x - 2} = 8 \]\
\[ x^2 - 2x + 1 = 8(x - 2) \]\
\[ x^2 - 2x + 1 = 8x - 16 \]\
\[ x^2 - 10x + 17 = 0 \]\
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Solve the quadratic equation using the quadratic formula:\
\[ x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(17)}}{2(1)} \]\
\[ x = \frac{10 \pm \sqrt{100 - 68}}{2} \]\
\[ x = \frac{10 \pm \sqrt{32}}{2} \]\
\[ x = \frac{10 \pm 4\sqrt{2}}{2} \]\
\[ x = 5 \pm 2\sqrt{2} \]\
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We must check if both solutions are valid by ensuring the arguments of the logarithms are positive:\
For both solutions, we require \( x - 1 > 0 \) and \( x - 2 > 0 \), which means \( x > 2 \).\
Since \( 2\sqrt{2} \approx 2.83 \):\
- \( 5 + 2\sqrt{2} \approx 7.83 > 2 \) (valid)\
- \( 5 - 2\sqrt{2} \approx 2.17 > 2 \) (valid)\
\
Thus, the exact solutions are \( x = 5 + 2\sqrt{2} \) and \( x = 5 - 2\sqrt{2} \).

- **M1**: Uses the power law of logarithms to write \( 2\log_2(x - 1) \) as \( \log_2(x - 1)^2 \).\
- **M1**: Uses the subtraction law of logarithms to express the LHS as a single logarithm, resulting in \( \log_2\left(\frac{(x-1)^2}{x-2}\right) = 3 \).\
- **M1**: Eliminates the logarithm to form an equation of the form \( \frac{(x-1)^2}{x-2} = 2^3 \) (or 8).\
- **A1**: Simplifies to obtain the correct 3-term quadratic equation \( x^2 - 10x + 17 = 0 \).\
- **A1**: Solves the quadratic equation to obtain the exact answers \( x = 5 \pm 2\sqrt{2} \) (or equivalent exact forms) and shows or states that both solutions are valid since \( x > 2 \).

The sum of the first two terms of a geometric series is given by:\
\[ S_2 = a + ar = a(1 + r) = 15 \quad \text{--- (Equation 1)} \]\
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The sum to infinity of a geometric series is given by:\
\[ S_{\infty} = \frac{a}{1 - r} = 27 \quad \text{--- (Equation 2)} \]\
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From Equation 2, we can express \( a \) in terms of \( r \):\
\[ a = 27(1 - r) \]\
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Substitute this expression for \( a \) into Equation 1:\
\[ 27(1 - r)(1 + r) = 15 \]\
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Using the difference of two squares:\
\[ 27(1 - r^2) = 15 \]\
\[ 1 - r^2 = \frac{15}{27} \]\
\[ 1 - r^2 = \frac{5}{9} \]\
\[ r^2 = 1 - \frac{5}{9} \]\
\[ r^2 = \frac{4}{9} \]\
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Since \( r > 0 \), we take the positive square root:\
\[ r = \frac{2}{3} \]\
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Substitute \( r = \frac{2}{3} \) back into the expression for \( a \):\
\[ a = 27\left(1 - \frac{2}{3}\right) \]\
\[ a = 27 \times \frac{1}{3} = 9 \]\
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Thus, the first term is \( a = 9 \) and the common ratio is \( r = \frac{2}{3} \).

- **M1**: Sets up a correct equation for the sum of the first two terms, e.g., \( a(1 + r) = 15 \).\
- **M1**: Sets up a correct equation for the sum to infinity, e.g., \( \frac{a}{1-r} = 27 \).\
- **M1**: Eliminates one variable to find a quadratic equation in \( r \) or \( a \), e.g., \( 27(1 - r^2) = 15 \).\
- **A1**: Solves the equation to find \( r = \frac{2}{3} \) (must reject the negative root \( r = -\frac{2}{3} \) due to \( r > 0 \)).\
- **A1**: Finds the correct value of the first term \( a = 9 \).

(a) We substitute \(x = 3\) into \(f(x)\):
\[f(3) = 2(3)^3 - 3(3)^2 - 11(3) + 6\]
\[f(3) = 2(27) - 3(9) - 33 + 6 = 54 - 27 - 33 + 6 = 0\]
Since \(f(3) = 0\), by the factor theorem, \((x-3)\) is a factor of \(f(x)\).

(b) Dividing \(f(x)\) by \((x-3)\):
\[2x^3 - 3x^2 - 11x + 6 = (x-3)(2x^2 + kx - 2)\]
Comparing coefficients of \(x^2\):
\[-3 = k - 6 \implies k = 3\]
So,
\[2x^3 - 3x^2 - 11x + 6 = (x-3)(2x^2 + 3x - 2)\]
Factorising the quadratic \(2x^2 + 3x - 2 = (2x - 1)(x + 2)\):
\[f(x) = (x-3)(2x-1)(x+2)\]

(c) The denominator can be factorised as a difference of two squares:
\[x^2 - 9 = (x-3)(x+3)\]
The expression becomes:
\[\frac{(x-3)(2x-1)(x+2)}{(x-3)(x+3)}\]
Cancelling the common factor \((x-3)\) gives:
\[\frac{(2x-1)(x+2)}{x+3} = \frac{2x^2 + 3x - 2}{x+3}\]

**Part (a)**
M1: Attempting to evaluate \(f(3)\) by substitution of \(x = 3\) into the polynomial.
A1: Correct evaluation to show \(f(3) = 0\) with a brief concluding statement (e.g., 'hence \((x-3)\) is a factor').

**Part (b)**
M1: Attempt to divide or factorise out \((x-3)\) to obtain a quadratic factor of the form \(2x^2 + kx + c\).
A1: Correct quadratic factor \(2x^2 + 3x - 2\).
A1: Fully factorised expression \((x-3)(2x-1)(x+2)\).

**Part (c)**
M1: Recognises \(x^2 - 9\) as a difference of two squares and factorises to \((x-3)(x+3)\).
M1: Substitutes the factorised form of \(f(x)\) from part (b) into the fraction.
A1: Cancelling the common factor \((x-3)\) to get \(\frac{(2x-1)(x+2)}{x+3}\).
A1: Final simplified expression \(\frac{2x^2 + 3x - 2}{x+3}\) or \(\frac{(2x-1)(x+2)}{x+3}\).

(a) Given:
\[\log_3(x - 2) + \log_3(x + 6) = 2\]
Using the addition law of logarithms:
\[\log_3((x - 2)(x + 6)) = 2\]
Converting to exponential form:
\[(x - 2)(x + 6) = 3^2\]
\[x^2 + 4x - 12 = 9\]
\[x^2 + 4x - 21 = 0\]
Factorising the quadratic equation:
\[(x + 7)(x - 3) = 0\]
This gives \(x = -7\) or \(x = 3\).

Since \(\log_3(x-2)\) is only defined for \(x - 2 > 0 \implies x > 2\), we reject \(x = -7\).
Therefore, the only valid solution is \(x = 3\).

(b) Given:
\[2^{3y + 1} = 5^{y}\]
Taking natural logarithms of both sides:
\[\ln(2^{3y + 1}) = \ln(5^{y})\]
\[(3y + 1)\ln 2 = y \ln 5\]
\[3y\ln 2 + \ln 2 = y \ln 5\]
\[y(3\ln 2 - \ln 5) = -\ln 2\]
\[y(\ln 8 - \ln 5) = \ln(1/2)\]
\[y \ln\left(\frac{8}{5}\right) = \ln\left(\frac{1}{2}\right)\]
\[y = \frac{\ln(1/2)}{\ln(8/5)} = \frac{-\ln 2}{-\ln(5/8)} = \frac{\ln 2}{\ln(5/8)}\]
Thus, \(a = 2\) and \(b = \frac{5}{8}\) (or \(0.625\)).

**Part (a)**
M1: Applies the addition law of logarithms: \(\log_3((x-2)(x+6)) = 2\).
M1: Correctly removes logarithms: \((x-2)(x+6) = 3^2\).
A1: Obtains the correct quadratic equation \(x^2 + 4x - 21 = 0\).
M1: Solves the quadratic equation to find \(x = 3\) and \(x = -7\).
A1: Rejects \(x = -7\) with a valid reason and identifies \(x = 3\) as the only valid solution.

**Part (b)**
M1: Takes natural logarithms (or logs with any base) of both sides and applies power laws correctly: \((3y+1)\ln 2 = y\ln 5\).
M1: Expands and collects terms containing \(y\): \(y(3\ln 2 - \ln 5) = -\ln 2\).
M1: Applies logarithmic rules to simplify coefficients or constants: e.g., \(3\ln 2 = \ln 8\) or \(-\ln 2 = \ln(1/2)\).
A1: Reaches the correct exact form, e.g., \(y = \frac{\ln 2}{\ln(5/8)}\) or equivalent (such as \(y = \frac{\ln(0.5)}{\ln(1.6)}\)).

(a) The \(n\)-th term of an arithmetic progression is given by:
\[u_n = a + (n - 1)d\]
Since the 3rd term is 20:
\[a + 2d = 20 \quad \text{(as required)}\]
The sum of the first \(n\) terms of an arithmetic progression is given by:
\[S_n = \frac{n}{2}(2a + (n - 1)d)\]
Given the sum of the first 10 terms is 325:
\[S_{10} = \frac{10}{2}(2a + 9d) = 325\]
\[5(2a + 9d) = 325\]
\[2a + 9d = 65\]

(b) We have two simultaneous equations:
1) \(a + 2d = 20 \implies 2a + 4d = 40\)
2) \(2a + 9d = 65\)
Subtracting (1) from (2):
\[(2a + 9d) - (2a + 4d) = 65 - 40\]
\[5d = 25 \implies d = 5\]
Substituting \(d = 5\) back into (1):
\[a + 2(5) = 20 \implies a + 10 = 20 \implies a = 10\]

(c) The sum of the first 20 terms is \(S_{20}\):
\[S_{20} = \frac{20}{2}(2a + 19d)\]
\[S_{20} = 10(2(10) + 19(5))\]
\[S_{20} = 10(20 + 95) = 10(115) = 1150\]

**Part (a)**
M1: Uses the \(n\)-th term formula with \(n=3\) to write \(a + 2d = 20\).
M1: Uses the sum formula for \(S_{10} = 325\): \(\frac{10}{2}(2a + 9d) = 325\).
A1: Correctly simplifies the sum equation to obtain \(2a + 9d = 65\) (or equivalent).

**Part (b)**
M1: Attempts to solve the simultaneous equations to eliminate one variable.
A1: Correct value of \(d = 5\).
A1: Correct value of \(a = 10\).

**Part (c)**
M1: Uses the sum formula with \(n = 20\) and their values of \(a\) and \(d\).
A1ft: Correct substitution of their values: \(10(2(10) + 19(5))\).
A1: Correct final answer of \(1150\).

(a) The radius \(r\) of the circle is the distance between the center \(A(3, -2)\) and point \(B(7, 1)\):
\[r^2 = (7 - 3)^2 + (1 - (-2))^2 = 4^2 + 3^2 = 16 + 9 = 25\]
So, the radius is \(r = 5\).
The equation of the circle \(C\) with center \((3, -2)\) and radius \(5\) is:
\[(x - 3)^2 + (y + 2)^2 = 25\]

(b) First, find the gradient of the radius \(AB\):
\[m_{AB} =
\frac{1 - (-2)}{7 - 3} = \frac{3}{4}\]
Since the tangent line at \(B\) is perpendicular to the radius \(AB\):
\[m_{\text{tangent}} = -\frac{1}{m_{AB}} = -\frac{4}{3}\]
The equation of the tangent line passing through \(B(7, 1)\) is:
\[y - 1 = -\frac{4}{3}(x - 7)\]
Multiplying both sides by 3 to eliminate the fraction:
\[3(y - 1) = -4(x - 7)\]
\[3y - 3 = -4x + 28\]
Rearranging into the form \(py + qx + r = 0\):
\[3y + 4x - 31 = 0\]

**Part (a)**
M1: Uses the distance formula to find \(r^2\) or \(r\) using coordinates of \(A\) and \(B\).
A1: Correct value of \(r^2 = 25\).
A1: Correct circle equation: \((x - 3)^2 + (y + 2)^2 = 25\) (accept equivalent expanded forms).

**Part (b)**
M1: Calculates the gradient of the radius \(AB\) using \(\frac{y_2 - y_1}{x_2 - x_1}\).
A1: Correct gradient of \(AB\) is \(\frac{3}{4}\).
M1: Uses perpendicular gradient relationship \(m_1 m_2 = -1\) to find the gradient of the tangent.
A1ft: Correct tangent gradient \(m_{\text{tangent}} = -\frac{4}{3}\) (follow through on their \(m_{AB}\)).
M1: Uses their tangent gradient and coordinates of \(B\) to construct a linear equation.
A1: Correct final equation in the required form: \(3y + 4x - 31 = 0\) (or any integer multiple thereof).

(a) Using the identity \(\cos^2 x = 1 - \sin^2 x\), substitute into the equation:
\[3(1 - \sin^2 x) - 2 \sin x - 2 = 0\]
\[3 - 3\sin^2 x - 2\sin x - 2 = 0\]
\[-3\sin^2 x - 2\sin x + 1 = 0\]
Multiplying through by \(-1\):
\[3\sin^2 x + 2\sin x - 1 = 0 \quad \text{(as required)}\]

(b) Let \(y = \sin x\). The equation becomes:
\[3y^2 + 2y - 1 = 0\]
Factorising this quadratic equation:
\[(3y - 1)(y + 1) = 0\]
So,
\[\sin x = \frac{1}{3} \quad \text{or} \quad \sin x = -1\]

Case 1: \(\sin x = \frac{1}{3}\)
The principal value is:
\[x = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.471^\circ \approx 19.5^\circ\]
The second quadrant angle is:
\[x = 180^\circ - 19.471^\circ = 160.529^\circ \approx 160.5^\circ\]

Case 2: \(\sin x = -1\)
In the interval \(0 \le x < 360^\circ\):
\[x = 270^\circ\]

Thus, the solutions are \(x = 19.5^\circ\), \(160.5^\circ\), and \(270^\circ\).

**Part (a)**
M1: Substitutes \(\cos^2 x = 1 - \sin^2 x\) into the given equation.
A1: Correctly simplifies and rearranges terms to obtain the requested equation.

**Part (b)**
M1: Solves the quadratic equation to find values for \(\sin x\).
A1: Obtains \(\sin x = \frac{1}{3}\) and \(\sin x = -1\).
M1: Finds one angle for \(\sin x = \frac{1}{3}\) (approx \(19.5^\circ\)).
A1: Finds the second quadrant angle (approx \(160.5^\circ\)).
M1: Finds \(x = 270^\circ\) for \(\sin x = -1\).
A2: Correct final solutions: \(19.5^\circ, 160.5^\circ, 270^\circ\) (deduct 1 mark for any extra incorrect solutions in the range, or for incorrect rounding).

(a) The volume \(V\) of a cuboid with a square base of side length \(x\) and height \(h\) is given by:
\[V = x^2 h = 216 \implies h = \frac{216}{x^2}\]
The total surface area \(S\) is the sum of the areas of the 6 faces (2 square faces and 4 rectangular side faces):
\[S = 2x^2 + 4xh\]
Substitute \(h = \frac{216}{x^2}\) into the surface area equation:
\[S = 2x^2 + 4x\left(\frac{216}{x^2}\right)\]
\[S = 2x^2 + \frac{864}{x} \quad \text{(as required)}\]

(b) To find the stationary point, differentiate \(S\) with respect to \(x\):
\[\frac{\text{d}S}{\text{d}x} = \frac{\text{d}}{\text{d}x}\left(2x^2 + 864x^{-1}\right) = 4x - 864x^{-2} = 4x - \frac{864}{x^2}\]
Set \(\frac{\text{d}S}{\text{d}x} = 0\):
\[4x - \frac{864}{x^2} = 0\]
\[4x^3 = 864 \implies x^3 = 216\]
\[x = \sqrt[3]{216} = 6\]
Substitute \(x = 6\) back into the equation for \(S\) to get the minimum surface area:
\[S = 2(6)^2 + \frac{864}{6} = 72 + 144 = 216\text{ cm}^2\]
To justify that it is a minimum, find the second derivative:
\[\frac{\text{d}^2S}{\text{d}x^2} = 4 + 1728x^{-3} = 4 + \frac{1728}{x^3}\]
At \(x = 6\):
\[\frac{\text{d}^2S}{\text{d}x^2} = 4 + \frac{1728}{216} = 4 + 8 = 12 > 0\]
Since \(\frac{\text{d}^2S}{\text{d}x^2} > 0\), the value \(S = 216\) is a minimum.

**Part (a)**
M1: Writes down a correct expression for the volume: \(V = x^2 h = 216\).
A1: Expresses \(h\) in terms of \(x\): \(h = \frac{216}{x^2}\).
M1: Formulates the correct total surface area expression: \(S = 2x^2 + 4xh\).
A1: Substitutes \(h\) and simplifies to show \(S = 2x^2 + \frac{864}{x}\) clearly.

**Part (b)**
M1: Differentiates \(S\) with respect to \(x\) to obtain \(4x - \frac{864}{x^2}\).
M1: Sets \(\frac{\text{d}S}{\text{d}x} = 0\) and solves for \(x\).
A1: Obtains \(x = 6\).
M1: Finds the second derivative \(\frac{\text{d}^2S}{\text{d}x^2} = 4 + \frac{1728}{x^3}\) and evaluates its sign at \(x = 6\).
A1: Obtains the correct minimum surface area value of \(216\) and states that \(\frac{\text{d}^2S}{\text{d}x^2} > 0\) confirms it is a minimum.

(a) For the given table, the step width \(h\) is:
\[h = 1\]
Applying the trapezium rule:
\[I \approx \frac{h}{2} \left[ y_0 + y_4 + 2(y_1 + y_2 + y_3) \right]\]
\[I \approx \frac{1}{2} \left[ 6 + 2 + 2(4 + 3 + 2.4) \right]\]
\[I \approx \frac{1}{2} \left[ 8 + 2(9.4) \right]\]
\[I \approx \frac{1}{2} \left[ 8 + 18.8 \right] = \frac{26.8}{2} = 13.4\]

(b) Using definite integration:
\[\int_{0}^{4} \frac{12}{x + 2} \text{ d}x = \left[ 12 \ln |x + 2| \right]_0^4\]
Substituting limits:
\[= 12 \ln(4 + 2) - 12 \ln(0 + 2)\]
\[= 12 \ln(6) - 12 \ln(2)\]
Using the logarithmic division rule \(\ln M - \ln N = \ln\left(\frac{M}{N}\right)\):
\[= 12 \ln\left(\frac{6}{2}\right) = 12 \ln 3\]
Here \(a = 12\) and \(b = 3\), which are both integers.

**Part (a)**
M1: Identifies the width of each strip as \(h = 1\).
M1: Applies the correct formula structure of the trapezium rule.
A1: Correct substitution of values into the formula: \(\frac{1}{2} [6 + 2 + 2(4 + 3 + 2.4)]\).
A1: Obtains \(13.4\).

**Part (b)**
M1: Integrates to obtain the form \(k \ln(x+2)\).
A1: Correctly integrates to \(12 \ln(x+2)\).
M1: Substitutes the upper limit (4) and lower limit (0) into their integrated term.
A1: Obtains \(12 \ln 6 - 12 \ln 2\).
A1: Simplifies to \(12 \ln 3\) using log properties.

(a) By algebraic manipulation: \(f(x) = \frac{3(x+2) - 7}{x+2} = 3 - \frac{7}{x+2}\). (b) Set \(y = \frac{3x-1}{x+2}\). Then \(y(x+2) = 3x - 1 \Rightarrow xy + 2y = 3x - 1 \Rightarrow x(y-3) = -2y-1 \Rightarrow x = \frac{2y+1}{3-y}\). Thus \(f^{-1}(x) = \frac{2x+1}{3-x}\). (c) The domain of \(f^{-1}\) is the range of \(f\). Since \(x > 1\), \(f(1) = \frac{2}{3}\) and as \(x \to \infty\), \(f(x) \to 3\). Thus, the range of \(f\) is \(\frac{2}{3} < y < 3\), so the domain of \(f^{-1}\) is \(\frac{2}{3} < x < 3\). (d) \(ff(x) = \frac{11}{7} \Rightarrow f(x) = f^{-1}\left(\frac{11}{7}\right) = \frac{2(11/7)+1}{3-11/7} = 2.9\). Solving \(f(x) = 2.9 \Rightarrow \frac{3x-1}{x+2} = 2.9 \Rightarrow 3x - 1 = 2.9x + 5.8 \Rightarrow 0.1x = 6.8 \Rightarrow x = 68\).

M1: Attempts to rewrite the numerator of the fraction to match the denominator. A1: Fully correct verification. M1: Sets up an equation with \(y\) and attempts to make \(x\) the subject. A1: Correctly finds \(f^{-1}(x)\) with correct variable notation. B1: States the correct domain \(\frac{2}{3} < x < 3\). M1: Sets up the equation \(f(x) = f^{-1}(11/7)\). A1: Evaluates to find \(f(x) = 2.9\). A1: Obtains the correct final answer \(x = 68\).

(a) LHS = \(\frac{1 - (1 - 2\sin^2\theta)}{2\sin\theta\cos\theta} = \frac{2\sin^2\theta}{2\sin\theta\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta\). (b) Let \(\theta = 2x\), then \(\frac{3 - 3\cos 4x}{\sin 4x} = 3\tan 2x\). The equation becomes \(3\tan 2x = 2\cos 2x \Rightarrow \frac{3\sin 2x}{\cos 2x} = 2\cos 2x \Rightarrow 3\sin 2x = 2\cos^2 2x = 2(1 - \sin^2 2x)\). This gives the quadratic equation \(2\sin^2 2x + 3\sin 2x - 2 = 0\), which factors to \((2\sin 2x - 1)(\sin 2x + 2) = 0\). Since \(\sin 2x\) must lie in \([-1, 1]\), we have \(\sin 2x = 0.5\). For \(0 \le x \le \pi\), \(0 \le 2x \le 2\pi\), so \(2x = \frac{\pi}{6}\) or \(2x = \frac{5\pi}{6}\). This yields \(x = \frac{\pi}{12}\) and \(x = \frac{5\pi}{12}\).

M1: Applies double angle formulas to both numerator and denominator. A1: Correctly simplifies to show \(\tan\theta\). M1: Recognises LHS of the equation is \(3\tan 2x\) using part (a). M1: Uses the relation \(\cos^2 2x = 1 - \sin^2 2x\) to form a quadratic equation in \(\sin 2x\). A1: Correct quadratic equation. M1: Solves the quadratic to find \(\sin 2x = 0.5\). A1: Finds at least one correct value of \(x\). A1: Finds both correct values of \(x\) and no extras in the interval.

(a) Using the quotient rule with \(u = \ln(3x-2)\) and \(v = x^2\), we have \(\frac{\text{d}u}{\text{d}x} = \frac{3}{3x-2}\) and \(\frac{\text{d}v}{\text{d}x} = 2x\). Thus, \(\frac{\text{d}y}{\text{d}x} = \frac{x^2 \left(\frac{3}{3x-2}\right) - 2x\ln(3x-2)}{x^4} = \frac{3x - 2(3x-2)\ln(3x-2)}{x^3(3x-2)}\). (b) When \(x = 1\), \(y = \frac{\ln 1}{1} = 0\). The gradient of the tangent at \(x = 1\) is \(\frac{\text{d}y}{\text{d}x} = \frac{3(1) - 2(1)(0)}{1^3(1)} = 3\). The gradient of the normal is therefore \(-\frac{1}{3}\). The equation of the normal is \(y - 0 = -\frac{1}{3}(x - 1) \Rightarrow 3y = -x + 1 \Rightarrow x + 3y - 1 = 0\).

M1: Applies quotient rule or product rule correctly. A1: Differentiates \(\ln(3x-2)\) correctly to obtain \(\frac{3}{3x-2}\). A1: Obtains a correct unsimplified expression for the derivative. A1: Simplifies to the final derivative. B1: States or finds the correct y-coordinate of 0 when \(x = 1\). M1: Evaluates the derivative at \(x = 1\) and finds the negative reciprocal. M1: Writes down the equation of a line using their point and gradient. A1: Obtains the correct integer form \(x + 3y - 1 = 0\) or integer multiples.

(a) When \(t = 0\), \(M = 150 \Rightarrow A = 150\). (b) When \(t = 45\), \(M = 75 \Rightarrow 75 = 150e^{-45k} \Rightarrow 0.5 = e^{-45k} \Rightarrow -45k = \ln 0.5 \Rightarrow -45k = -\ln 2 \Rightarrow k = \frac{1}{45}\ln 2\). (c) When \(M = 15\): \(15 = 150e^{-kt} \Rightarrow 0.1 = e^{-kt} \Rightarrow -kt = \ln 0.1 \Rightarrow t = \frac{\ln 10}{k} = \frac{45\ln 10}{\ln 2} \approx 149.49 \approx 149\) years. (d) The rate of decrease of the mass is given by \(-\frac{\text{d}M}{\text{d}t} = k M\). When \(t = 30\), \(M = 150 e^{-30 \left(\frac{\ln 2}{45}\right)} = 150 e^{-\frac{2}{3}\ln 2} = 150 \times 2^{-2/3} \approx 94.494\) grams. The rate of decrease is \(k M = \frac{\ln 2}{45} \times 94.494 \approx 1.4555 \approx 1.46\) grams per year.

B1: Identifies \(A = 150\). M1: Substitutes \(M = 75\) and \(t = 45\) and uses logarithms to isolate \(k\). A1: Correctly shows the given exact value of \(k\). M1: Formulates the equation \(15 = 150 e^{-kt}\) and attempts to solve for \(t\). A1: Correctly rounds to \(149\) years. M1: Differentiates \(M\) with respect to \(t\) to find an expression for \(\frac{\text{d}M}{\text{d}t}\). M1: Evaluates the rate of decrease at \(t = 30\). A1: Obtains \(1.46\) (to 3 s.f.).

(a) \(f(1.7) = (1.7)^3 - 3(1.7)^2 + 5(1.7) - 5 = -0.257 < 0\). \(f(1.8) = (1.8)^3 - 3(1.8)^2 + 5(1.8) - 5 = 0.112 > 0\). Since there is a change of sign and \(f(x)\) is a continuous function, a root \(\alpha\) exists in the interval \([1.7, 1.8]\). (b) \(x^3 - 3x^2 + 5x - 5 = 0 \Rightarrow x^3 + 5x = 3x^2 + 5 \Rightarrow x(x^2 + 5) = 3x^2 + 5 \Rightarrow x = \frac{3x^2+5}{x^2+5}\). (c) With \(x_0 = 1.8\): \(x_1 = \frac{3(1.8)^2+5}{1.8^2+5} \approx 1.786408 \Rightarrow 1.7864\). \(x_2 = \frac{3(1.786408)^2+5}{1.786408^2+5} \approx 1.779180 \Rightarrow 1.7792\). \(x_3 = \frac{3(1.779180)^2+5}{1.779180^2+5} \approx 1.775330 \Rightarrow 1.7753\). (d) Consider the interval \([1.7705, 1.7715]\). \(f(1.7705) = (1.7705)^3 - 3(1.7705)^2 + 5(1.7705) - 5 \approx -0.00159 < 0\). \(f(1.7715) = (1.7715)^3 - 3(1.7715)^2 + 5(1.7715) - 5 \approx 0.00217 > 0\). Since there is a change of sign and the function is continuous, \(1.7705 < \alpha < 1.7715\), hence \(\alpha = 1.771\) to 3 decimal places.

B1: Evaluates both boundary values and concludes with reference to change of sign and continuity. M1: Rearranges the cubic equation to group terms. A1: Correctly shows the given formula. M1: Correctly evaluates \(x_1\). A1: Calculates \(x_1, x_2, x_3\) to 4 d.p. correctly. M1: Chooses the correct boundary interval \([1.7705, 1.7715]\) and evaluates \(f(x)\) at both points. A1: Correctly obtains opposite signs and concludes.

(a) Let \(u = 2x+1 \Rightarrow \text{d}u = 2\text{d}x \Rightarrow \text{d}x = \frac{1}{2}\text{d}u\). Also \(x = \frac{u-1}{2}\). For limits: when \(x = 0\), \(u = 1\); when \(x = 4\), \(u = 9\). Substituting these yields: \(\int_{0}^{4} \frac{x}{\sqrt{2x+1}} \, \text{d}x = \int_{1}^{9} \frac{\frac{u-1}{2}}{\sqrt{u}} \cdot \frac{1}{2} \, \text{d}u = \frac{1}{4} \int_{1}^{9} \frac{u-1}{u^{1/2}} \, \text{d}u = \frac{1}{4} \int_{1}^{9} \left(u^{1/2} - u^{-1/2}\right) \, \text{d}u\). (b) Integrating gives: \(\frac{1}{4} \left[ \frac{2}{3}u^{3/2} - 2u^{1/2} \right]_{1}^{9} = \frac{1}{4} \left( \left( \frac{2}{3}(27) - 2(3) \right) - \left( \frac{2}{3}(1) - 2(1) \right) \right) = \frac{1}{4} \left( (18 - 6) - \left( \frac{2}{3} - 2 \right) \right) = \frac{1}{4} \left( 12 - \left( -\frac{4}{3} \right) \right) = \frac{1}{4} \left( \frac{40}{3} \right) = \frac{10}{3}\).

M1: Finds the relationship between \(\text{d}x\) and \(\text{d}u\). M1: Substitutes \(x\) and \(\text{d}x\) in terms of \(u\) into the integral. B1: Correctly updates the integration limits to 1 and 9. A1: Achieves the simplified target integral expression in (a). M1: Integrates terms correctly to obtain the form \(A u^{3/2} + B u^{1/2}\). A1: Correctly integrated expression. M1: Evaluates the integral between 9 and 1. A1: Obtains \(\frac{10}{3}\) or equivalent exact rational value.

(a) The graph of \(y = |2x-5|\) is a V-shape with its vertex at \((2.5, 0)\) and y-intercept at \((0, 5)\). The graph of \(y = x+1\) is a straight line with gradient 1, passing through \((0, 1)\) and \((-1, 0)\). (b) Case 1: \(2x - 5 = x + 1 \Rightarrow x = 6\). Case 2: \(-(2x - 5) = x + 1 \Rightarrow -2x + 5 = x + 1 \Rightarrow 3x = 4 \Rightarrow x = \frac{4}{3}\). (c) The V-shape is strictly above the line \(y = x+1\) for values of \(x\) to the left of the first intersection point and to the right of the second. Thus, the solution is \(x < \frac{4}{3}\) or \(x > 6\).

B1: Sketches the correct V-shape for the modulus graph on the positive x-axis. B1: Correct straight line, and all relevant axis intercepts labeled on both. M1: Sets up and solves the linear equation \(2x - 5 = x + 1\). M1: Sets up and solves the linear equation \(-(2x - 5) = x + 1\). A1: Finds both solutions \(x = 6\) and \(x = \frac{4}{3}\). M1: Uses their critical values to construct the correct inequalities. A1: Expresses the final answer as the two disjoint regions \(x < \frac{4}{3}\) or \(x > 6\).

(a) By the product rule: \(\frac{\text{d}y}{\text{d}x} = e^{2x}(3\cos 3x) + 2e^{2x}\sin 3x = e^{2x} (2\sin 3x + 3\cos 3x)\). (b) Setting the derivative to 0: \(e^{2x} (2\sin 3x + 3\cos 3x) = 0\). Since \(e^{2x} > 0\) for all real \(x\), we have \(2\sin 3x + 3\cos 3x = 0 \Rightarrow \tan 3x = -1.5\). Let \(\theta = 3x\), where \(0 \le \theta \le 3\pi\). The principal value of \(\arctan(-1.5) \approx -0.98279\). The solutions in the interval are: \(\theta_1 = -0.98279 + \pi \approx 2.1588 \Rightarrow x_1 \approx 0.7196 \Rightarrow 0.720\). \(\theta_2 = -0.98279 + 2\pi \approx 5.3004 \Rightarrow x_2 \approx 1.7668 \Rightarrow 1.77\). \(\theta_3 = -0.98279 + 3\pi \approx 8.4420 \Rightarrow x_3 \approx 2.8140 \Rightarrow 2.81\). Thus, \(x = 0.720, \, 1.77, \, 2.81\) to 3 s.f.

M1: Applies product rule correctly to differentiate the expression. A1: Correctly differentiates both terms. A1: Simplifies to the final derivative. M1: Sets their derivative to 0 and simplifies to \(\tan 3x = C\). A1: Obtains \(\tan 3x = -1.5\). M1: Solves to find at least one value of \(x\) within the interval. A1: Finds one correct value of \(x\) to 3 s.f. A1: Obtains all three correct solutions \(x = 0.720, 1.77, 2.81\) and no extras in range.

(a) Given \[f(x) = \frac{x^2 + 5}{x - 2}\] Using the quotient rule with \(u = x^2 + 5 \implies u' = 2x\) and \(v = x - 2 \implies v' = 1\): \[f'(x) = \frac{u'v - uv'}{v^2} = \frac{2x(x - 2) - 1(x^2 + 5)}{(x - 2)^2}\] \[f'(x) = \frac{2x^2 - 4x - x^2 - 5}{(x - 2)^2} = \frac{x^2 - 4x - 5}{(x - 2)^2}\] Factorising the numerator: \[x^2 - 4x - 5 = (x - 5)(x + 1)\] Thus: \[f'(x) = \frac{(x - 5)(x + 1)}{(x - 2)^2}\] where \(a = 5\) and \(b = -1\) (or vice-versa). (b) At a stationary point, \(f'(x) = 0\): \[\frac{(x - 5)(x + 1)}{(x - 2)^2} = 0 \implies (x - 5)(x + 1) = 0\] This gives \(x = 5\) or \(x = -1\). Since the domain of \(f\) is \(x > 2\), we reject \(x = -1\). For \(x = 5\): \[f(5) = \frac{5^2 + 5}{5 - 2} = \frac{30}{3} = 10\] Therefore, the coordinates of the stationary point are \((5, 10)\). (c) Since \(x > 2\), the only stationary point in the domain of \(f\) is the minimum at \((5, 10)\). As \(x \to 2^+\), \(f(x) \to \infty\), and as \(x \to \infty\), \(f(x) \to \infty\). Therefore, the minimum value of \(f(x)\) is \(10\), and it is unbounded above. Hence, the range of \(f\) is \(f(x) \ge 10\).

(a) * **M1**: Attempts quotient rule with correct structure: \[\frac{(x-2)(Ax) - (x^2+5)(B)}{(x-2)^2}\] where \(A \neq 0, B \neq 0\). * **A1**: Correct unsimplified derivative: \[\frac{2x(x-2) - (x^2+5)}{(x-2)^2}\] * **A1**: Correctly simplified and factorised derivative \[\frac{(x-5)(x+1)}{(x-2)^2}\] and identifies \(a = 5, b = -1\) (or vice-versa). (b) * **M1**: Sets their numerator equal to 0 and finds a value for \(x > 2\) (i.e. \(x = 5\)). * **A1**: Correct coordinates \((5, 10)\). (c) * **M1**: Recognises that the stationary point is a minimum and uses the \(y\)-coordinate \(10\). * **M1**: Fully justifies that the function is unbounded above (e.g., states \(f(x) \to \infty\) as \(x \to 2^+\) or as \(x \to \infty\)). * **A1**: Correct range \(f(x) \ge 10\) (accept \(y \ge 10\) or \([10, \infty)\)). Reject strict inequality \(f(x) > 10\) or range in terms of \(x\).

(a) Rewrite \(f(x)\) as \((1+ax)(4-x)^{-1/2}\).

Expand \((4-x)^{-1/2}\):

\[ (4-x)^{-1/2} = 4^{-1/2}\left(1 - \frac{x}{4}\right)^{-1/2} = \frac{1}{2}\left(1 - \frac{x}{4}\right)^{-1/2} \]

Using the binomial expansion formula:

\[ \left(1 - \frac{x}{4}\right)^{-1/2} = 1 + \left(-\frac{1}{2}\right)\left(-\frac{x}{4}\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(-\frac{x}{4}\right)^2 + \dots \]

\[ = 1 + \frac{x}{8} + \frac{3}{128}x^2 + \dots \]

So,

\[ (4-x)^{-1/2} = \frac{1}{2}\left(1 + \frac{x}{8} + \frac{3}{128}x^2 + \dots\right) = \frac{1}{2} + \frac{1}{16}x + \frac{3}{256}x^2 + \dots \]

Now multiply by \((1+ax)\):

\[ f(x) = (1+ax)\left(\frac{1}{2} + \frac{1}{16}x + \frac{3}{256}x^2 + \dots\right) \]

\[ = \frac{1}{2} + \frac{1}{16}x + \frac{3}{256}x^2 + \frac{a}{2}x + \frac{a}{16}x^2 + \dots \]

\[ = \frac{1}{2} + \left(\frac{1}{16} + \frac{a}{2}\right)x + \left(\frac{3}{256} + \frac{a}{16}\right)x^2 + \dots \]

Comparing coefficients with \(b + \frac{5}{16}x + cx^2\):

For the constant term:
\[ b = \frac{1}{2} \]

For the \(x\) term:
\[ \frac{1}{16} + \frac{a}{2} = \frac{5}{16} \implies \frac{a}{2} = \frac{4}{16} \implies a = \frac{1}{2} \]

For the \(x^2\) term:
\[ c = \frac{3}{256} + \frac{a}{16} = \frac{3}{256} + \frac{1/2}{16} = \frac{3}{256} + \frac{8}{256} = \frac{11}{256} \]

(b) The expansion is valid for \(|-\frac{x}{4}| < 1 \implies |x| < 4\).

(a)
- M1: Attempts to write \((4-x)^{-1/2} = \frac{1}{2}(1 - \frac{x}{4})^{-1/2}\) or equivalent.
- M1: Correct binomial expansion of \((1 - \frac{x}{4})^{-1/2}\) up to the term in \(x^2\).
- A1: Correctly obtains \(\frac{1}{2} + \frac{1}{16}x + \frac{3}{256}x^2\).
- dM1: Multiplies by \((1+ax)\) and sets the coefficient of \(x\) equal to \(\frac{5}{16}\) to find \(a\). Dependent on the first M mark.
- A1: Correct values for \(a\), \(b\) and \(c\): \(a = \frac{1}{2}\), \(b = \frac{1}{2}\), \(c = \frac{11}{256}\).

(b)
- B1: For \(|x| < 4\) (or \(-4 < x < 4\)).

(a) Since \(P\) lies on the line \(l\), the position vector of \(P\) is:

\[ \mathbf{r}_P = \begin{pmatrix} 2+\lambda \\ 1-2\lambda \\ -3+2\lambda \end{pmatrix} \]

The vector \(\mathbf{AP}\) is given by:

\[ \mathbf{AP} = \mathbf{r}_P - \mathbf{a} = \begin{pmatrix} 2+\lambda - 2 \\ 1-2\lambda - (-1) \\ -3+2\lambda - 4 \end{pmatrix} =
\begin{pmatrix} \lambda \\ 2-2\lambda \\ 2\lambda - 7 \end{pmatrix} \]

Since \(AP\) is perpendicular to \(l\), the scalar product of \(\mathbf{AP}\) and the direction vector of \(l\) is zero:

\[ \begin{pmatrix} \lambda \\ 2-2\lambda \\ 2\lambda - 7 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} = 0 \]

\[ \lambda(1) + (2-2\lambda)(-2) + (2\lambda - 7)(2) = 0 \]

\[ \lambda - 4 + 4\lambda + 4\lambda - 14 = 0 \]

\[ 9\lambda - 18 = 0 \implies \lambda = 2 \]

Substituting \(\lambda = 2\) into \(\mathbf{r}_P\):

\[ \mathbf{r}_P = \begin{pmatrix} 2+2 \\ 1-2(2) \\ -3+2(2) \end{pmatrix} = \begin{pmatrix} 4 \\ -3 \\ 1
\end{pmatrix} \]

So, the coordinates of \(P\) are \((4, -3, 1)\).

(b) The shortest distance from \(A\) to \(l\) is the length of the vector \(\mathbf{AP}\) when \(\lambda = 2\).

When \(\lambda = 2\):

\[ \mathbf{AP} = \begin{pmatrix} 2 \\ 2-2(2) \\ 2(2)-7 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ -3 \end{pmatrix} \]

The shortest distance is:

\[ |\mathbf{AP}| = \sqrt{2^2 + (-2)^2 + (-3)^2} = \sqrt{4 + 4 + 9} = \sqrt{17} \]

(a)
- M1: Writes down an expression for \(\mathbf{AP}\) in terms of \(\lambda\).
- M1: Uses the condition for perpendicularity by taking the dot product of \(\mathbf{AP}\) with the direction vector of \(l\) and setting it to 0.
- A1: Solves the equation to find \(\lambda = 2\).
- A1: Finds the correct coordinates of \(P\) as \((4, -3, 1)\).

(b)
- M1: Attempts to find the magnitude of \(\mathbf{AP}\) using their value of \(\lambda\) or coordinates of \(P\).
- A1: Obtains the exact distance \(\sqrt{17}\).

The volume of a solid of revolution about the \(x\)-axis is given by:

\[ V = \pi \int_{a}^{b} y^2 \, \mathrm{d}x \]

Here, the limits of integration are from \(x = 0\) to \(x = 4\).

We have:

\[ y^2 = \left( \frac{6}{\sqrt{3x+4}} \right)^2 = \frac{36}{3x+4} \]

Thus,

\[ V = \pi \int_{0}^{4} \frac{36}{3x+4} \, \mathrm{d}x \]

Integrate \(\frac{36}{3x+4}\):

\[ \int \frac{36}{3x+4} \, \mathrm{d}x = 12 \ln|3x+4| \]

Evaluating between the limits \(0\) and \(4\):

\[ V = \pi \left[ 12 \ln|3x+4| \right]_{0}^{4} \]

\[ V = 12\pi \left( \ln|3(4)+4| - \ln|3(0)+4| \right) \]

\[ V = 12\pi \left( \ln 16 - \ln 4 \right) \]

Using laws of logarithms:

\[ V = 12\pi \ln\left( \frac{16}{4} \right) = 12\pi \ln 4 \]

Since \(b\) must be a prime number:

\[ V = 12\pi \ln(2^2) = 24\pi \ln 2 \]

Thus, \(a = 24\) and \(b = 2\).

- M1: Uses the volume of revolution formula \(V = \pi \int y^2 \, \mathrm{d}x\) with correct limits \(0\) and \(4\) (limits can be implied by later work).
- A1: Correct expression for \(y^2 = \frac{36}{3x+4}\).
- M1: Integrates to get \(k \ln(3x+4)\), where \(k\) is a constant.
- A1: Correct integration resulting in \(12 \ln|3x+4|\) (constant \(\pi\) can be omitted here).
- M1: Substitutes limits \(4\) and \(0\) into their integrated expression and subtracts.
- A1: Simplifies to the exact form \(24\pi \ln 2\).

Assume for contradiction that \(\log_5 2\) is a rational number.

Then we can write \(\log_5 2 = \frac{p}{q}\), where \(p\) and \(q\) are positive integers.

By definition of logarithms:
\[ 5^{p/q} = 2 \]

Raising both sides to the power of \(q\):
\[ 5^p = 2^q \]

Since \(p\) is a positive integer, \(5^p\) is an odd integer (as any integer power of 5 is odd).

Since \(q\) is a positive integer, \(2^q\) is an even integer (as any positive integer power of 2 is even).

This implies that an odd integer is equal to an even integer, which is a contradiction.

Therefore, the assumption that \(\log_5 2\) is rational must be false, so \(\log_5 2\) is irrational.

M1: Sets up proof by contradiction by assuming \(\log_5 2 = \frac{p}{q}\) where \(p, q\) are positive integers.
M1: Correctly applies laws of logarithms and indices to write \(5^p = 2^q\).
A1: Explains clearly that \(5^p\) is always odd and \(2^q\) is always even for positive integers \(p, q\).
A1: Identifies the contradiction and concludes that therefore \(\log_5 2\) must be irrational.

(a) We use the trigonometric identity:
\[ \sec^2 \theta - \tan^2 \theta = 1 \]
From the parametric equations:
\[ \sec \theta = \frac{x}{3} \quad \text{and} \quad \tan \theta = \frac{y}{2} \]
Substituting these into the identity gives:
\[ \left(\frac{x}{3}\right)^2 - \left(\frac{y}{2}\right)^2 = 1 \implies \frac{x^2}{9} - \frac{y^2}{4} = 1 \]
Thus \(a = 3\) and \(b = 2\).

(b) To find the gradient of the tangent, we find \(\frac{dy}{dx}\):
\[ \frac{dx}{d\theta} = 3 \sec \theta \tan \theta \]
\[ \frac{dy}{d\theta} = 2 \sec^2 \theta \]
Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2 \sec^2 \theta}{3 \sec \theta \tan \theta} = \frac{2 \sec \theta}{3 \tan \theta} = \frac{2}{3 \sin \theta} \]
At \(\theta = \frac{\pi}{6}\):
\[ x = 3 \sec\left(\frac{\pi}{6}\right) = 3 \times \frac{2}{\sqrt{3}} = 2\sqrt{3} \]
\[ y = 2 \tan\left(\frac{\pi}{6}\right) = 2 \times \frac{1}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \]
The gradient of the tangent is:
\[ m = \frac{2}{3 \sin(\pi/6)} = \frac{2}{3 \times 0.5} = \frac{4}{3} \]
The equation of the tangent is:
\[ y - \frac{2\sqrt{3}}{3} = \frac{4}{3}(x - 2\sqrt{3}) \]
\[ y = \frac{4}{3}x - \frac{8\sqrt{3}}{3} + \frac{2\sqrt{3}}{3} \]
\[ y = \frac{4}{3}x - 2\sqrt{3} \]

Part (a):
M1: Uses the trigonometric identity \(\sec^2 \theta - \tan^2 \theta = 1\) with parametric definitions.
A1: Obtains the correct Cartesian equation \(\frac{x^2}{9} - \frac{y^2}{4} = 1\) (accepting explicit identification of \(a = 3, b = 2\)).

Part (b):
M1: Differentiates \(x\) and \(y\) with respect to \(\theta\) to find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\).
M1: Obtains a correct expression for \(\frac{dy}{dx}\) in terms of \(\theta\).
A1: Calculates the correct gradient \(m = \frac{4}{3}\) at \(\theta = \frac{\pi}{6}\).
M1: Finds the coordinates \(P\left(2\sqrt{3}, \frac{2\sqrt{3}}{3}\right)\) and applies the equation of a line formula.
A1: Arrives at the correct simplified exact equation \(y = \frac{4}{3}x - 2\sqrt{3}\).

(a) Rewrite \(f(x)\) as:
\[ f(x) = (4 - 9x)^{-1/2} \]
Factor out 4 to express it in the form \(K(1 + y)^n\):
\[ f(x) = 4^{-1/2} \left(1 - \frac{9}{4}x\right)^{-1/2} = \frac{1}{2} \left(1 - \frac{9}{4}x\right)^{-1/2} \]
Using the binomial expansion:
\[ (1 + y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \dots \]
where \(n = -\frac{1}{2}\) and \(y = -\frac{9}{4}x\):
\[ \left(1 - \frac{9}{4}x\right)^{-1/2} = 1 + \left(-\frac{1}{2}\right)\left(-\frac{9}{4}x\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(-\frac{9}{4}x\right)^2 + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{6}\left(-\frac{9}{4}x\right)^3 + \dots \]
\[ = 1 + \frac{9}{8}x + \frac{3}{8}\left(\frac{81}{16}x^2\right) - \frac{5}{16}\left(-\frac{729}{64}x^3\right) \]
\[ = 1 + \frac{9}{8}x + \frac{243}{128}x^2 + \frac{3645}{1024}x^3 \]
Multiply this expansion by the factor \(\frac{1}{2}\):
\[ f(x) \approx \frac{1}{2} + \frac{9}{16}x + \frac{243}{256}x^2 + \frac{3645}{2048}x^3 \]

(b) The expansion is valid when \(|y| < 1\):
\[ \left|-\frac{9}{4}x\right| < 1 \implies |x| < \frac{4}{9} \]

Part (a):
B1: Extracts \(4^{-1/2}\) outside to get the coefficient \(\frac{1}{2}\).
M1: Applies the binomial expansion with \(n = -\frac{1}{2}\) and a term of \(-\frac{9}{4}x\).
A1: Correct term in \(x\) which is \(\frac{9}{16}x\).
A1: Correct term in \(x^2\) which is \(\frac{243}{256}x^2\).
A1: Correct term in \(x^3\) which is \(\frac{3645}{2048}x^3\).

Part (b):
B1: Correctly states that the validity range is \(|x| < \frac{4}{9}\) or \(-\frac{4}{9} < x < \frac{4}{9}\).

(a) Differentiating both sides of the equation implicitly with respect to \(x\):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(3xy) - \frac{d}{dx}(2y^2) + \frac{d}{dx}(2x) = \frac{d}{dx}(-6) \]
Using the product rule on \(3xy\):
\[ 2x + \left(3y + 3x\frac{dy}{dx}\right) - 4y\frac{dy}{dx} + 2 = 0 \]
Group terms with \(\frac{dy}{dx}\):
\[ (3x - 4y)\frac{dy}{dx} = -2x - 3y - 2 \]
\[ (4y - 3x)\frac{dy}{dx} = 2x + 3y + 2 \]
\[ \frac{dy}{dx} = \frac{2x + 3y + 2}{4y - 3x} \]

(b) Substitute the coordinates of \(P(1, 3)\) into the gradient expression:
\[ m_{\text{tangent}} = \frac{2(1) + 3(3) + 2}{4(3) - 3(1)} = \frac{2 + 9 + 2}{12 - 3} = \frac{13}{9} \]
The gradient of the normal is:
\[ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{9}{13} \]
The equation of the normal is:
\[ y - 3 = -\frac{9}{13}(x - 1) \]
Multiply by 13 to clear the fraction:
\[ 13(y - 3) = -9(x - 1) \]
\[ 13y - 39 = -9x + 9 \]
Rearranging to the required form:
\[ 9x + 13y - 48 = 0 \]

Part (a):
M1: Differentiates \(x^2 + 2x\) to obtain \(2x + 2\), and \(-2y^2\) to obtain \(-4y\frac{dy}{dx}\).
M1: Uses the product rule correctly on the term \(3xy\) to obtain \(3y + 3x\frac{dy}{dx}\).
A1: Writes down the fully differentiated equation correctly: \(2x + 3y + 3x\frac{dy}{dx} - 4y\frac{dy}{dx} + 2 = 0\).
A1: Rearranges correctly to make \(\frac{dy}{dx}\) the subject: \(\frac{dy}{dx} = \frac{2x + 3y + 2}{4y - 3x}\).

Part (b):
M1: Substitutes \(x = 1\) and \(y = 3\) to find the numerical gradient of the tangent.
M1: Finds the negative reciprocal of their tangent gradient to get the normal gradient.
A1: Obtains the correct integer-form equation of the normal: \(9x + 13y - 48 = 0\) (or any integer multiple thereof).

Set up the algebraic identity:
\[ \frac{7x^2 - 13x + 4}{(x-1)^2(2x-1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{2x-1} \]
Multiply both sides of the identity by \((x-1)^2(2x-1)\):
\[ 7x^2 - 13x + 4 = A(x-1)(2x-1) + B(2x-1) + C(x-1)^2 \]
To find the constants, substitute strategic values of \(x\):

Let \(x = 1\):
\[ 7(1)^2 - 13(1) + 4 = B(2(1) - 1) \]
\[ -2 = B \implies B = -2 \]

Let \(x = \frac{1}{2}\):
\[ 7\left(\frac{1}{2}\right)^2 - 13\left(\frac{1}{2}\right) + 4 = C\left(\frac{1}{2} - 1\right)^2 \]
\[ \frac{7}{4} - \frac{13}{2} + 4 = C\left(-\frac{1}{2}\right)^2 \]
\[ -\frac{3}{4} = \frac{1}{4}C \implies C = -3 \]

To find \(A\), compare the coefficients of \(x^2\) on both sides:
\[ 7 = 2A + C \]
Since \(C = -3\):
\[ 7 = 2A - 3 \implies 2A = 10 \implies A = 5 \]

Therefore, the partial fraction decomposition is:
\[ f(x) = \frac{5}{x-1} - \frac{2}{(x-1)^2} - \frac{3}{2x-1} \]

M1: Sets up the correct form with the three partial fraction terms.
M1: Clears the fractions correctly to get the identity: \(7x^2 - 13x + 4 = A(x-1)(2x-1) + B(2x-1) + C(x-1)^2\).
A1: Finds \(B = -2\) (or equivalent value in their setup).
A1: Finds \(C = -3\) (or equivalent value in their setup).
M1: Uses a valid method (comparing coefficients or substituting another value like \(x=0\)) to find \(A\).
A1: Finds \(A = 5\).
A1: Writes the final correct expression: \(\frac{5}{x-1} - \frac{2}{(x-1)^2} - \frac{3}{2x-1}\).

Let \(u = \sqrt{2x + 1}\).
Squaring both sides:
\[ u^2 = 2x + 1 \implies x = \frac{u^2 - 1}{2} \]
Differentiating \(u^2 = 2x + 1\) with respect to \(u\) and \(x\):
\[ 2u \, du = 2 \, dx \implies dx = u \, du \]
Now, find the new integration limits for \(u\):
When \(x = 0\), \(u = \sqrt{2(0) + 1} = 1\).
When \(x = 4\), \(u = \sqrt{2(4) + 1} = 3\).

Substitute all parts into the integral:
\[ \int_{0}^{4} x \sqrt{2x + 1} \, dx = \int_{1}^{3} \left( \frac{u^2 - 1}{2} \right) (u) (u \, du) \]
\[ = \frac{1}{2} \int_{1}^{3} u^2(u^2 - 1) \, du \]
\[ = \frac{1}{2} \int_{1}^{3} (u^4 - u^2) \, du \]

Now integrate term by term:
\[ = \frac{1}{2} \left[ \frac{u^5}{5} - \frac{u^3}{3} \right]_{1}^{3} \]

Evaluate the limits:
At \(u = 3\):
\[ \frac{3^5}{5} - \frac{3^3}{3} = \frac{243}{5} - 9 = \frac{198}{5} \]

At \(u = 1\):
\[ \frac{1^5}{5} - \frac{1^3}{3} = \frac{1}{5} - \frac{1}{3} = -\frac{2}{15} \]

Subtract the lower limit evaluation from the upper limit evaluation:
\[ \text{Value} = \frac{1}{2} \left( \frac{198}{5} - \left(-\frac{2}{15}\right) \right) \]
\[ = \frac{1}{2} \left( \frac{594}{15} + \frac{2}{15} \right) = \frac{1}{2} \left( \frac{596}{15} \right) = \frac{298}{15} \]

B1: Obtains \(dx = u \, du\) or equivalent correctly.
M1: Expresses \(x\) in terms of \(u\) as \(\frac{u^2 - 1}{2}\) and substitutes successfully into the integral.
B1: Converts limits from \(x\) to \(u\) correctly, finding lower limit 1 and upper limit 3.
M1: Integrates \(u^4 - u^2\) correctly to get \(\frac{u^5}{5} - \frac{u^3}{3}\).
A1: Achieves the correct overall integrated expression with the coefficient \(\frac{1}{2}\).
M1: Integrates and substitutes the limits 3 and 1, showing subtraction.
A1: Reaches the correct simplified final exact value of \(\frac{298}{15}\).

(a) Equating the components of \(l_1\) and \(l_2\):
\[ 2 + \lambda = 5 - 2\mu \quad (1) \]
\[ 5 - 2\lambda = -1 + \mu \quad (2) \]
\[ -1 + 2\lambda = 5 + 4\mu \quad (3) \]
From (1), we have \(\lambda = 3 - 2\mu\).
Substitute this into (2):
\[ 5 - 2(3 - 2\mu) = -1 + \mu \]
\[ 5 - 6 + 4\mu = -1 + \mu \]
\[ -1 + 4\mu = -1 + \mu \implies 3\mu = 0 \implies \mu = 0 \]
If \(\mu = 0\), then \(\lambda = 3 - 2(0) = 3\).
Now check if these parameters satisfy the third equation (3):
LHS of (3): \(-1 + 2(3) = 5\)
RHS of (3): \(5 + 4(0) = 5\)
Since LHS = RHS, the lines intersect.
To find the coordinates of the point of intersection \(P\), substitute \(\mu = 0\) into the equation of \(l_2\):
\[ \mathbf{r}_P = \begin{pmatrix} 5 \\ -1 \\ 5 \end{pmatrix} \]
So the coordinates of \(P\) are \((5, -1, 5)\).

(b) The direction vectors of \(l_1\) and \(l_2\) are:
\[ \mathbf{d}_1 = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} \quad \text{and} \quad \mathbf{d}_2 = \begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix} \]
Calculate the scalar product \(\mathbf{d}_1 \cdot \mathbf{d}_2\):
\[ \mathbf{d}_1 \cdot \mathbf{d}_2 = (1)(-2) + (-2)(1) + (2)(4) = -2 - 2 + 8 = 4 \]
Calculate the magnitudes of \(\mathbf{d}_1\) and \(\mathbf{d}_2\):
\[ |\mathbf{d}_1| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3 \]
\[ |\mathbf{d}_2| = \sqrt{(-2)^2 + 1^2 + 4^2} = \sqrt{21} \]
The angle \(\theta\) between the lines is given by:
\[ \cos \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1| |\mathbf{d}_2|} = \frac{4}{3\sqrt{21}} \]
\[ \cos \theta \approx 0.290957 \]
\[ \theta = \arccos(0.290957) \approx 73.085^\circ \]
Thus, the acute angle to 1 decimal place is \(73.1^\circ\).

Part (a):
M1: Sets up simultaneous equations by equating the \(x\), \(y\), and \(z\) components.
M1: Solves two of the equations to find values for \(\lambda\) and \(\mu\).
A1: Shows that these values are consistent by testing them in the remaining third equation.
A1: Finds the correct coordinates of \(P\) which are \((5, -1, 5)\).

Part (b):
M1: Identifies correct direction vectors and calculates their dot product.
M1: Calculates the magnitudes of both direction vectors.
A1: Sets up the cosine relation: \(\cos \theta = \frac{4}{3\sqrt{21}}\).
A1: Obtains the correct acute angle \(73.1^\circ\).