2025 Edexcel IAL Mathematics (YMA01) Practice Paper with Answers

(a) For the quadratic equation to have no real roots, its discriminant must be negative: \(b^2 - 4ac < 0\). Here, \(a = k+1\), \(b = -4k\), and \(c = 2k+3\). Substituting these values: \((-4k)^2 - 4(k+1)(2k+3) < 0\) which expands to \(16k^2 - 4(2k^2 + 5k + 3) < 0\), simplifying to \(16k^2 - 8k^2 - 20k - 12 < 0\), and thus \(8k^2 - 20k - 12 < 0\). Dividing through by 4 gives the required inequality: \(2k^2 - 5k - 3 < 0\). (b) To solve \(2k^2 - 5k - 3 < 0\), find the critical values by solving \(2k^2 - 5k - 3 = 0\) which factorises to \((2k+1)(k-3) = 0\), giving critical values \(k = -1/2\) and \(k = 3\). Since the inequality is less than 0, the solution lies between these critical values: \(-1/2 < k < 3\).

(a) M1: Attempts to use the discriminant formula with correct a, b, and c values. A1: Correctly expands to obtain the simplified quadratic form 8k^2 - 20k - 12 < 0. A1*: Divides by 4 to obtain the target inequality with no errors. (b) M1: Factorises the quadratic expression to find the critical values. A1: Identifies critical values as k = -1/2 and k = 3. M1: Selects the region between the critical values for a less than inequality. A1.33: Correctly states the final range of values as -1/2 < k < 3.

(a) The gradient of \(l_1\) is \(m_1 = \frac{2 - 5}{4 - (-2)} = -\frac{3}{6} = -\frac{1}{2}\). Using the point-slope formula with \(B(4, 2)\): \(y - 2 = -\frac{1}{2}(x - 4)\), which simplifies to \(2y - 4 = -x + 4\), giving the equation \(x + 2y - 8 = 0\). (b) The midpoint \(M\) of \(AB\) is \(M = (\frac{-2+4}{2}, \frac{5+2}{2}) = (1, 3.5)\). The gradient of \(l_2\) is perpendicular to \(l_1\), so \(m_2 = -\frac{1}{m_1} = 2\). Using the point-slope formula with \(M(1, 3.5)\): \(y - 3.5 = 2(x - 1)\), which simplifies to \(y = 2x + 1.5\).

(a) M1: Computes the correct gradient of the line l1. A1: Gradient is -1/2. M1: Forms the equation of the line using one of the given points and their gradient. A1: Correctly expresses the line in the form ax + by + c = 0 with integer coefficients. (b) M1: Finds the correct coordinates of the midpoint. M1: Computes the perpendicular gradient using the negative reciprocal of their gradient. A1: Substitutes their midpoint and perpendicular gradient into the line formula. A1.33: Formulates the correct final equation in the form y = mx + c.

(a) Substitute \(\cos^2 x = 1 - \sin^2 x\) into the given equation: \(4(1 - \sin^2 x) + 9\sin x - 6 = 0\) which expands to \(4 - 4\sin^2 x + 9\sin x - 6 = 0\). This simplifies to \(-4\sin^2 x + 9\sin x - 2 = 0\). Multiplying the entire equation by -1 yields: \(4\sin^2 x - 9\sin x + 2 = 0\). (b) Let \(x = 2\theta - 30^\circ\). The equation becomes \(4\sin^2 x - 9\sin x + 2 = 0\), which factorises to \((4\sin x - 1)(\
sin x - 2) = 0\). Since \(\sin x \le 1\), \(\sin x = 2\) has no real solutions. We solve \(\sin x = 0.25\). Since \(0 \le \theta < 360^\circ\), the range for \(x\) is \(-30^\circ \le x < 690^\circ\). The principal value is \(x = 14.48^\circ\). The other solutions are \(x = 180^\circ - 14.48^\circ = 165.52^\circ\), and by adding \(360^\circ\) to these, we get \(x = 374.48^\circ\) and \(x = 525.52^\circ\). Substituting back to solve for \(\theta = \frac{x + 30^
\circ}{2}\) gives: \(\theta \approx 22.2^\circ, 97.8^\circ, 202.2^\circ, 277.8^\circ\).

(a) M1: Uses the identity cos^2(x) + sin^2(x) = 1 to express the equation entirely in terms of sin(x). A1: Correct expansion. A1*: Achieves the target equation with no algebraic errors. (b) M1: Factorises the quadratic equation to get sin(x) = 0.25. M1: Finds at least one correct principal value for the angle. M1: Determines further correct angles within the expanded range. A1: Finds at least two correct theta values. A1.33: Finds all four correct values to 1 decimal place.

(a) Rewrite \(y\) in index form: \(y = 2x^3 - 5x^{1/2} + 4x^{-1}\). Differentiating term by term gives: \(\frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - \frac{5}{2}x^{-1/2} - 4x^{-2}\). (b) At point \(P(1, 1)\), the gradient of the tangent is \(m = 6(1)^2 - \frac{5}{2}(1)^{-1/2} - 4(1)^{-2} = 6 - 2.5 - 4 = -0.5\). Using \(y - y_1 = m(x - x_1)\): \(y - 1 = -0.5(x - 1)\) which yields \(2y - 2 = -x + 1\), rearranging to \(x + 2y - 3 = 0\).

(a) M1: Attempts to differentiate at least one term. A1: Differentiates two terms correctly. A1.33: Fully correct derivative. (b) M1: Substitutes x = 1 into their derivative to find the tangent gradient. A1: Gradient is -0.5. M1: Sets up the equation of the line using P(1,1). A1: Correctly expresses the final equation in the requested integer-coefficient format.

(a) Set \(y = 0\) to find the intersections: \(9x^2 - 4x^3 = 0\) factorises to \(x^2(9 - 4x) = 0\). Thus \(x = 0\) or \(x = 2.25\). The coordinates of the intersection points are \((0, 0)\) and \((2.25, 0)\). (b) The area is given by the definite integral: \(A = \int_{0}^{2.25} (9x^2 - 4x^3) \mathrm{d}x = [3x^3 - x^4]_{0}^{2.25}\). Substituting the upper limit: \(A = 3(\frac{9}{4})^3 - (\frac{9}{4})^4 = 3(\frac{729}{64}) - \frac{6561}{256} = \frac{2187}{64} - \frac{6561}{256} = \frac{8748 - 6561}{256} = \frac{2187}{256}\).

(a) M1: Sets y = 0 and factorises the equation. A1: Finds correct x coordinates. A1: States correct coordinate pairs. (b) M1: Formulates the integral with limits from 0 to 2.25. M1: Integrates term-by-term correctly. A1: Obtains the correct integrated function [3x^3 - x^4]. M1: Substitutes the limits into the integrated function. A1.33: Finds the correct exact value as a fraction or decimal equivalent.

(a) The graph of \(y = f(x+1)\) is a translation of \(y = f(x)\) by 1 unit to the left. The vertical asymptote is \(x = 1\) and the horizontal asymptote remains \(y = 3\). The equation is \(y = \frac{4}{x-1} + 3\). Setting \(x = 0\) gives \(y = -4 + 3 = -1\), so the \(y\)-intercept is \((0, -1)\). Setting \(y = 0\) gives \(0 = \frac{4}{x-1} + 3 \Rightarrow x - 1 = -\frac{4}{3} \Rightarrow x = -\frac{1}{3}\), so the \(x\)-intercept is \((-\frac{1}{3}, 0)\). (b) We solve \(\frac{4}{x-2} + 3 > 5 \Rightarrow \frac{4}{x-2} > 2\). Since \(x - 2\) must be positive for the fraction to be greater than 2, we have \(x > 2\). Multiplying through gives \(4 > 2x - 4 \Rightarrow 2x < 8 \Rightarrow x < 4\). Combining these results yields \(2 < x < 4\).

(a) M1: Correct shape of translated hyperbola in correct quadrants. A1: Correct asymptotes drawn and clearly labelled. A1: Correct y-intercept. A1.33: Correct x-intercept. (b) M1: Sets up inequality to solve for critical value. A1: Identifies critical boundary x = 4. M1: Recognises vertical asymptote at x = 2 as lower limit. A1: Correct inequality 2 < x < 4.

(a) Simplify \(g'(x) = 6x - 5 + 3x^{-3/2}\). Integrating with respect to \(x\) gives: \(g(x) = \int (6x - 5 + 3x^{-3/2}) \mathrm{d}x = 3x^2 - 5x - 6x^{-1/2} + C\). Using point \(P(1, 8)\): \(8 = 3(1)^2 - 5(1) - 6(1) + C \Rightarrow 8 = -8 + C \Rightarrow C = 16\). Thus \(g(x) = 3x^2 - 5x - \frac{6}{\sqrt{x}} + 16\). (b) The gradient of the tangent at \(x = 1\) is \(g'(1) = 6(1) - 5 + 3(1) = 4\). The gradient of the normal is therefore \(m_n = -\frac{1}{4}\). Using \(P(1, 8)\): \(y - 8 = -\frac{1}{4}(x - 1)\), which simplifies to \(4y - 32 = -x + 1\), and rearranging gives \(x + 4y - 33 = 0\).

(a) M1: Writes g'(x) in standard index form. A1: Correct simplified derivative. M1: Integrates term-by-term. A1: Obtains correct integration with an arbitrary constant. M1: Substitutes (1, 8) to find C. A1.33: Gives fully correct equation for g(x). (b) M1: Evaluates derivative at x = 1 and takes perpendicular gradient. A1: Correct normal gradient of -1/4. A1: Formulates equation of the normal in requested form.

From the first equation, rewrite \(y\) as: \(y = 2x + 3\). Substitute this into the second equation: \(x^2 + x(2x + 3) - (2x + 3)^2 = -19\). Expanding: \(x^2 + 2x^2 + 3x - (4x^2 + 12x + 9) = -19\), which simplifies to \(-x^2 - 9x - 9 = -19\). Rearranging gives the quadratic: \(x^2 + 9x - 10 = 0\). Factorising: \((x - 1)(x + 10) = 0\), yielding solutions \(x = 1\) and \(x = -10\). Corresponding \(y\) values are found using \(y = 2x + 3\): For \(x = 1\), \(y = 2(1) + 3 = 5\). For \(x = -10\), \(y = 2(-10) + 3 = -17\).

M1: Rearranges the linear equation to express y in terms of x. M1: Substitutes this expression into the quadratic equation. A1: Expands the substitution terms correctly. A1: Forms a correct quadratic equation. M1: Solves the resulting quadratic equation. A1: Correctly identifies x = 1 and x = -10. M1: Substitutes x back to find corresponding values of y. A2.33: Correctly pairs both sets of coordinates.

**(a)**
First, express the derivative in index form:
\[\frac{\text{d}y}{\text{d}x} = 3x^{-\frac{1}{2}} - 4x\]

Integrate with respect to \(x\) to find \(y = \text{f}(x)\):
\[y = \int \left(3x^{-\frac{1}{2}} - 4x\right) \text{d}x\]
\[y = \frac{3x^{\frac{1}{2}}}{\frac{1}{2}} - \frac{4x^2}{2} + C\]
\[y = 6x^{\frac{1}{2}} - 2x^2 + C\]

Since the curve passes through \(P(4, -13)\), substitute \(x = 4\) and \(y = -13\\right):\n\[-13 = 6\sqrt{4} - 2(4)^2 + C\]\n\[-13 = 6(2) - 2(16) + C\]\n\[-13 = 12 - 32 + C\]\n\[-13 = -20 + C\]\n\[C = 7\]\n\nTherefore, the equation of the curve is:\n\[\text{f}(x) = 6x^{\frac{1}{2}} - 2x^2 + 7\quad \text{or}\quad \text{f}(x) = 6\sqrt{x} - 2x^2 + 7\]\n\n**(b)**\nFind the gradient of the curve at \)P(4, -13)\) by substituting \(x = 4\) into \(\frac{\text{d}y}{\text{d}x}\):
\[m = \frac{3}{\sqrt{4}} - 4(4) = \frac{3}{2} - 16 = -\frac{29}{2}\]

Using the straight line formula \(y - y_1 = m(x - x_1)\) with the point \(P(4, -13)\) and gradient \(m = -\frac{29}{2}\):
\[y - (-13) = -\frac{29}{2}(x - 4)\]
\[y + 13 = -\frac{29}{2}(x - 4)\]

Multiply by 2 to clear the fraction:
\[2(y + 13) = -29(x - 4)\]
\[2y + 26 = -29x + 116\]

Rearranging into the form \(ax + by + c = 0\):
\[29x + 2y - 90 = 0\]

**(c)**
To find where the tangent crosses the \(y\)-axis, substitute \(x = 0\) into the equation of the tangent:
\[29(0) + 2y - 90 = 0\]
\[2y = 90 \implies y = 45\]

Thus, the coordinates of the intersection point are \((0, 45)\).

**(a)**
* **M1**: Attempts to integrate \(3x^{-1/2} - 4x\) with at least one term integrated correctly (power increased by 1).
* **A1**: Correct integrated expression \(6x^{1/2} - 2x^2\) (constant of integration not required for this mark).
* **M1**: Substitutes \(x = 4\) and \(y = -13\) into their integrated expression containing a constant of integration \(C\) to find \(C\\right).\n* **A1**: Fully correct simplified equation, \)\\text{f}(x) = 6\\sqrt{x} - 2x^2 + 7\) or equivalent.

**(b)**
* **M1**: Substitutes \(x = 4\) into \(\frac{\text{d}y}{\text{d}x}\) to obtain a value for the gradient.
* **M1**: Writes down the equation of the line using their gradient and the point \(P(4, -13)\).
* **A1**: \(29x + 2y - 90 = 0\) or any non-zero integer multiple of this equation (e.g. \(-29x - 2y + 90 = 0\).

**(c)**
* **B1**: Coordinates \((0, 45)\) or clearly states \(x=0, y=45\). Accept \(y=45\) with a correct method shown, but coordinates are preferred.

(a) Since \( (x-2) \) is a factor of \( \mathrm{f}(x) \), by the factor theorem:
\[ \mathrm{f}(2) = 0 \implies 2(2)^3 + a(2)^2 + b(2) - 10 = 0 \]
\[ 16 + 4a + 2b - 10 = 0 \implies 4a + 2b = -6 \implies 2a + b = -3 \quad \text{(Equation 1)} \]

Since the remainder when divided by \( (x+1) \) is \( -18 \), by the remainder theorem:
\[ \mathrm{f}(-1) = -18 \implies 2(-1)^3 + a(-1)^2 + b(-1) - 10 = -18 \]
\[ -2 + a - b - 10 = -18 \implies a - b = -6 \quad \text{(Equation 2)} \]

Adding Equation 1 and Equation 2:
\[ (2a + b) + (a - b) = -3 + (-6) \implies 3a = -9 \implies a = -3 \]

Substituting \( a = -3 \) into Equation 2:
\[ -3 - b = -6 \implies b = 3 \]

(b) Substituting \( a = -3 \) and \( b = 3 \) into \( \mathrm{f}(x) \):
\[ \mathrm{f}(x) = 2x^3 - 3x^2 + 3x - 10 \]
Dividing \( \mathrm{f}(x) \) by \( (x-2) \) using algebraic division or equating coefficients:
\[ 2x^3 - 3x^2 + 3x - 10 = (x-2)(2x^2 + kx + 5) \]
Comparing coefficients of the \( x^2 \) term:
\[ -4 + k = -3 \implies k = 1 \]
Thus, \( \mathrm{f}(x) = (x-2)(2x^2 + x + 5) \), which gives \( p = 2 \), \( q = 1 \), and \( r = 5 \).

(a)
M1: Attempts to use the factor theorem with \( \mathrm{f}(2) = 0 \) to obtain a linear equation in \( a \) and \( b \).
A1: Correct equation, e.g., \( 2a + b = -3 \) or equivalent.
M1: Attempts to use the remainder theorem with \( \mathrm{f}(-1) = -18 \) to obtain a second linear equation in \( a \) and \( b \).
A1: Correct equation, e.g., \( a - b = -6 \) or equivalent.
A1: Solves simultaneously to find \( a = -3 \) and \( b = 3 \). Both values must be correct.

(b)
M1: Attempts algebraic long division or equates coefficients to find a quadratic expression.
A1: Correct factorization \( (x-2)(2x^2 + x + 5) \) (or states correct values of \( p = 2 \), \( q = 1 \), \( r = 5 \)).

(a) For any real numbers \( a \) and \( b \), the square of their difference is always non-negative:
\[ (a - b)^2 \ge 0 \]
Expanding the left-hand side:
\[ a^2 - 2ab + b^2 \ge 0 \]
Adding \( 2ab \) to both sides:
\[ a^2 + b^2 \ge 2ab \]
This completes the proof.

(b) To prove by exhaustion, we must test all integers in the given domain, \( n \in \{1, 2, 3, 4, 5\} \).
- Case \( n = 1 \): \( 1^2 + 3 = 4 \), which is not a multiple of 8.
- Case \( n = 2 \): \( 2^2 + 3 = 7 \), which is not a multiple of 8.
- Case \( n = 3 \): \( 3^2 + 3 = 12 \), which is not a multiple of 8.
- Case \( n = 4 \): \( 4^2 + 3 = 19 \), which is not a multiple of 8.
- Case \( n = 5 \): \( 5^2 + 3 = 28 \), which is not a multiple of 8.
Since all possible cases have been tested and none of the resulting values are multiples of 8, the statement is proved by exhaustion.

(a)
M1: Starts with a valid inequality such as \( (a-b)^2 \ge 0 \) or equivalent complete logical step.
A1: Correct expansion and rearrangement to show \( a^2 + b^2 \ge 2ab \) with a concluding statement.

(b)
M1: Recognises that proof by exhaustion requires testing all five integer values \( n = 1, 2, 3, 4, 5 \).
M1: Evaluates \( n^2 + 3 \) for at least three of the values.
A2: Evaluates all five cases correctly and shows none is a multiple of 8 (1 mark for four correct, 2 marks for all five correct).
A1: Concludes clearly that the statement has been proved by exhaustion.

(a) Complete the square for the equation of the circle \( C \):
\[ x^2 - 6x + y^2 + 8y = 0 \]
\[ (x - 3)^2 - 9 + (y + 4)^2 - 16 = 0 \]
\[ (x - 3)^2 + (y + 4)^2 = 25 \]
(i) The coordinates of the centre are \( (3, -4) \).
(ii) The radius is \( \sqrt{25} = 5 \).

(b) Substitute \( y = 2x - 5 \) into the equation of the circle:
\[ x^2 + (2x - 5)^2 - 6x + 8(2x - 5) = 0 \]
\[ x^2 + 4x^2 - 20x + 25 - 6x + 16x - 40 = 0 \]
\[ 5x^2 - 10x - 15 = 0 \]
Divide by 5:
\[ x^2 - 2x - 3 = 0 \]
The discriminant of this quadratic is:
\[ \Delta = b^2 - 4ac = (-2)^2 - 4(1)(-3) = 4 + 12 = 16 \]
Since \( \Delta = 16 > 0 \), the quadratic has two distinct real roots. Thus, the line \( L \) intersects the circle \( C \) at two distinct points.

(c) Solve the quadratic to find the \( x \)-coordinates of the intersection points:
\[ x^2 - 2x - 3 = 0 \implies (x - 3)(x + 1) = 0 \implies x = 3 \text{ or } x = -1 \]
Find the corresponding \( y \)-coordinates:
- When \( x = 3 \), \( y = 2(3) - 5 = 1 \). Point is \( A(3, 1) \).
- When \( x = -1 \), \( y = 2(-1) - 5 = -7 \). Point is \( B(-1, -7) \).

Use the distance formula to find \( AB \):
\[ AB = \sqrt{(3 - (-1))^2 + (1 - (-7))^2} = \sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80} \]
\[ AB = \sqrt{16 \times 5} = 4\sqrt{5} \]
Thus, \( k = 4 \).

(a)
M1: Attempts to complete the square for both \( x \) and \( y \) terms.
A1: Centre is \( (3, -4) \).
A1: Radius is \( 5 \) (accept \( \sqrt{25} \)).

(b)
M1: Substitutes \( y = 2x - 5 \) into the circle equation to obtain a quadratic in terms of \( x \).
A1: Correct simplified quadratic equation, e.g., \( 5x^2 - 10x - 15 = 0 \) or \( x^2 - 2x - 3 = 0 \).
A1: Calculates discriminant \( \Delta = 16 \) (or solves to find two distinct roots) and concludes that \( \Delta > 0 \) implies two distinct intersection points.

(c)
M1: Finds both points of intersection: \( (3, 1) \) and \( (-1, -7) \).
A1: Correct distance of \( 4\sqrt{5} \) (accept \( k = 4 \)).

(a) The second term of the geometric series is given by:
\[ u_2 = ar = 12 \implies a = \frac{12}{r} \]

The sum to infinity is given by:
\[ S_\infty = \frac{a}{1-r} = 50 \implies a = 50(1-r) \]

Equating the two expressions for \( a \):
\[ \frac{12}{r} = 50(1-r) \]
\[ 12 = 50r(1-r) \]
\[ 12 = 50r - 50r^2 \]
\[ 50r^2 - 50r + 12 = 0 \]
Dividing both sides of the equation by 2 gives:
\[ 25r^2 - 25r + 6 = 0 \quad \text{(as required)} \]

(b) Solve the quadratic equation by factoring:
\[ 25r^2 - 25r + 6 = 0 \]
\[ (5r - 2)(5r - 3) = 0 \]
This gives:
\[ r = \frac{2}{5} \quad \text{or} \quad r = \frac{3}{5} \]

(c) Substitute each value of \( r \) back into the formula \( a = \frac{12}{r} \):
- If \( r = \frac{2}{5} \), then \( a = \frac{12}{2/5} = 30 \).
- If \( r = \frac{3}{5} \), then \( a = \frac{12}{3/5} = 20 \).

(a)
M1: Expresses second term as \( ar = 12 \) and sum to infinity as \( \frac{a}{1-r} = 50 \).
M1: Eliminates \( a \) to obtain an equation in \( r \).
M1: Multiplies out brackets and rearranges to form a quadratic equation.
A1*: Reaches \( 25r^2 - 25r + 6 = 0 \) with no errors in working.

(b)
M1: Attempts to solve the quadratic equation by factoring, formula, or completing the square.
A1: Correct values: \( r = \frac{2}{5} \) and \( r = \frac{3}{5} \).

(c)
M1: Attempts to find \( a \) using at least one of their values of \( r \).
A1: Both \( a = 30 \) and \( a = 20 \) correctly paired with their respective \( r \) values.

(a) Take the natural logarithm of both sides of the equation:
\[ \ln(3^{2x+1}) = \ln(5^{x-1}) \]
Use the power law of logarithms:
\[ (2x+1)\ln 3 = (x-1)\ln 5 \]
Expand the brackets:
\[ 2x\ln 3 + \ln 3 = x\ln 5 - \ln 5 \]
Rearrange to group terms with \( x \) on one side:
\[ x\ln 5 - 2x\ln 3 = \ln 3 + \ln 5 \]
\[ x(\ln 5 - \ln 3^2) = \ln(3 \times 5) \]
\[ x(\ln 5 - \ln 9) = \ln 15 \]
Using the subtraction law for logarithms:
\[ x\ln\left(\frac{5}{9}\right) = \ln 15 \]
\[ x = \frac{\ln 15}{\ln(5/9)} \]
Thus, \( a = 15 \) and \( b = \frac{5}{9} \).

(b) Combine the logarithms using the division rule:
\[ \log_2\left(\frac{y + 3}{y - 1}\right) = 3 \]
Convert the logarithmic equation into exponential form:
\[ \frac{y + 3}{y - 1} = 2^3 \]
\[ \frac{y + 3}{y - 1} = 8 \]
Multiply both sides by \( (y - 1) \):
\[ y + 3 = 8(y - 1) \]
\[ y + 3 = 8y - 8 \]
Rearrange to solve for \( y \):
\[ 7y = 11 \implies y = \frac{11}{7} \]

(a)
M1: Takes natural logs of both sides and applies power law to bring down powers.
M1: Expands brackets and groups all \( x \) terms on one side.
M1: Applies log laws to simplify \( 2\ln 3 \) to \( \ln 9 \) and combines terms using subtraction or addition rules.
A1: Obtains \( x = \frac{\ln 15}{\ln(5/9)} \) or equivalent form matching the required template (e.g. \( a=15 \), \( b=5/9 \)).

(b)
M1: Uses subtraction law of logs to write as a single log.
M1: Removes log correctly by writing \( \frac{y+3}{y-1} = 2^3 = 8 \).
A1: Solves to find \( y = \frac{11}{7} \).

(a) Use the trigonometric identity \( \cos^2 \theta = 1 - \sin^2 \theta \):
\[ 3(1 - \sin^2 \theta) + 11\sin \theta = 9 \]
\[ 3 - 3\sin^2 \theta + 11\sin \theta = 9 \]
Rearrange the terms to one side:
\[ 3\sin^2 \theta - 11\sin \theta + 6 = 0 \quad \text{(as required)} \]

(b) Let \( \theta = 2x - 30^\circ \).
The equation becomes:
\[ 3\sin^2 \theta - 11\sin \theta + 6 = 0 \]
Factor the quadratic:
\[ (3\sin \theta - 2)(\sin \theta - 3) = 0 \]
Since \( \sin \theta \le 1 \), the equation \( \sin \theta - 3 = 0 \implies \sin \theta = 3 \) has no real solutions.
Thus:
\[ \sin \theta = \frac{2}{3} \]

We need to solve \( \sin(2x - 30^\circ) = \frac{2}{3} \) for \( 0 \le x < 360^\circ \).
The range of values for \( \theta = 2x - 30^\circ \) is:
\[ 0 \le x < 360^\circ \implies 0 \le 2x < 720^\circ \implies -30^\circ \le 2x - 30^\circ < 690^\circ \]

Find the principal value of \( \theta \):
\[ \theta = \arcsin\left(\frac{2}{3}\right) \approx 41.8103^\circ \]
Other values for \( \theta \) in the interval \( [-30^\circ, 690^\circ) \):
- Quadrant 1: \( \theta = 41.8103^\circ \)
- Quadrant 2: \( \theta = 180^\circ - 41.8103^\circ = 138.1897^\circ \)
- Quadrant 1 (+360): \( \theta = 41.8103^\circ + 360^\circ = 401.8103^\circ \)
- Quadrant 2 (+360): \( \theta = 138.1897^\circ + 360^\circ = 538.1897^\circ \)

Now calculate the corresponding values of \( x \) using \( x = \frac{\theta + 30^\circ}{2} \):
1) \( x = \frac{41.8103^\circ + 30^\circ}{2} \approx 35.9^\circ \)
2) \( x = \frac{138.1897^\circ + 30^\circ}{2} \approx 84.1^\circ \)
3) \( x = \frac{401.8103^\circ + 30^\circ}{2} \approx 215.9^\circ \)
4) \( x = \frac{538.1897^\circ + 30^\circ}{2} \approx 284.1^\circ \)

(a)
M1: Substitutes identity \( \cos^2 \theta = 1 - \sin^2 \theta \) into the equation.
A1*: Expands and rearranges correctly to show the given quadratic in \( \sin \theta \).

(b)
M1: Solves the quadratic to find \( \sin(2x - 30^\circ) = \frac{2}{3} \) or equivalent, and rejects \( \sin \theta = 3 \).
M1: Finds the principal value \( \approx 41.8^\circ \) (or \( 138.2^\circ \)) of the angle.
M1: Correctly identifies the interval for \( 2x - 30^\circ \) as \( [-30^\circ, 690^\circ) \) and finds at least two correct values of \( \theta \).
A1: Finds any two correct values of \( x \) from \( 35.9^\circ, 84.1^\circ, 215.9^\circ, 284.1^\circ \).
A2: Obtains all four correct solutions: \( x = 35.9^\circ, 84.1^\circ, 215.9^\circ, 284.1^\circ \).

(a) The volume of the rectangular box is:
\[ V = \text{length} \times \text{width} \times \text{height} = (2x)(x)(h) = 2x^2 h \]
Since the volume is \( 36\text{ cm}^3 \):
\[ 2x^2 h = 36 \implies h = \frac{18}{x^2} \]

The box has an open top. Therefore, the surface area \( A \) consists of the area of the base and the four vertical sides:
\[ A = \text{Base Area} + 2(\text{width} \times h) + 2(\text{length} \times h) \]
\[ A = 2x^2 + 2(xh) + 2(2xh) \]
\[ A = 2x^2 + 6xh \]

Substitute \( h = \frac{18}{x^2} \) into this expression:
\[ A = 2x^2 + 6x\left(\frac{18}{x^2}\right) \]
\[ A = 2x^2 + \frac{108}{x} \quad \text{(as required)} \]

(b) Differentiate \( A \) with respect to \( x \):
\[ \frac{\mathrm{d}A}{\mathrm{d}x} = 4x - \frac{108}{x^2} \]
At the stationary point, \( \frac{\mathrm{d}A}{\mathrm{d}x} = 0 \):
\[ 4x - \frac{108}{x^2} = 0 \]
\[ 4x^3 = 108 \implies x^3 = 27 \implies x = 3 \]

Calculate the minimum value of \( A \) at \( x = 3 \):
\[ A = 2(3)^2 + \frac{108}{3} = 18 + 36 = 54 \]

To prove that this stationary value is a minimum, find the second derivative:
\[ \frac{\mathrm{d}^2 A}{\mathrm{d}x^2} = 4 + \frac{216}{x^3} \]
At \( x = 3 \):
\[ \frac{\mathrm{d}^2 A}{\mathrm{d}x^2} = 4 + \frac{216}{27} = 4 + 8 = 12 \]
Since \( \frac{\mathrm{d}^2 A}{\mathrm{d}x^2} = 12 > 0 \), the value of \( A \) is indeed a minimum.

(a)
M1: Formulates equation for the volume: \( 2x^2 h = 36 \) and expresses \( h \) in terms of \( x \).
M1: Writes down a correct expression for the surface area of the open box, e.g., \( A = 2x^2 + 6xh \).
M1: Substitutes their expression for \( h \) into their surface area expression.
A1*: Reaches \( A = 2x^2 + \frac{108}{x} \) with no algebraic errors.

(b)
M1: Differentiates \( A \) to find \( \frac{\mathrm{d}A}{\mathrm{d}x} \) (at least one term correct).
A1: Sets derivative to 0 and finds the correct critical value \( x = 3 \).
M1: Finds the value of \( A \) at this critical point and differentiates again to find \( \frac{\mathrm{d}^2 A}{\mathrm{d}x^2} \).
A1: Obtains \( A = 54 \), evaluates second derivative as \( 12 > 0 \) and states that this indicates a minimum.

(a)
- For \( x = 1.5 \): \( y = 4^{1.5} - 1.5 = 8 - 1.5 = 6.5 \).
- For \( x = 2 \): \( y = 4^2 - 2 = 16 - 2 = 14 \).

The completed table is:
| \( x \) | 0 | 0.5 | 1 | 1.5 | 2 |
|---|---|---|---|---|---|
| \( y \) | 1 | 1.5 | 3 | 6.5 | 14 |

(b) The step size is \( h = 0.5 \).
Using the trapezium rule:
\[ I \approx \frac{h}{2} [ y_0 + y_4 + 2(y_1 + y_2 + y_3) ] \]
\[ I \approx \frac{0.5}{2} [ 1 + 14 + 2(1.5 + 3 + 6.5) ] \]
\[ I \approx 0.25 [ 15 + 2(11) ] = 0.25 [ 15 + 22 ] = 0.25 [ 37 ] = 9.25 \]

(c) The exact area under the curve is given by:
\[ \int_{0}^{2} (4^x - x) \, \mathrm{d}x = \left[ \frac{4^x}{\ln 4} - \frac{x^2}{2} \right]_{0}^{2} \]
\[ = \left( \frac{4^2}{\ln 4} - \frac{2^2}{2} \right) - \left( \frac{4^0}{\ln 4} - \frac{0^2}{2} \right) \]
\[ = \left( \frac{16}{\ln 4} - 2 \right) - \left( \frac{1}{\ln 4} - 0 \right) \]
\[ = \frac{15}{\ln 4} - 2 \]
Since \( \ln 4 = \ln(2^2) = 2\ln 2 \), this can be written as:
\[ \frac{15}{2\ln 2} - 2 = \frac{7.5}{\ln 2} - 2 \]
Thus, \( a = 7.5 \) and \( b = 2 \).

(a)
B1: Both \( y \) values correct (\( 6.5 \) and \( 14 \)).

(b)
M1: Correct use of the trapezium rule structure with \( h = 0.5 \) and five y-values.
A1: Correct substitution of values inside brackets.
A1: Correct answer of \( 9.25 \).

(c)
M1: Integrates \( 4^x - x \) to obtain \( \frac{4^x}{\ln 4} - \frac{x^2}{2} \).
M1: Substitutes limits \( 2 \) and \( 0 \) and subtracts.
A1: Obtains exact area in the form \( \frac{7.5}{\ln 2} - 2 \) (or equivalent matching the requested form).

(a) Let \( u = 3^y \). Then \( 3^{2y+1} = 3(3^{2y}) = 3u^2 \). Substituting this into the equation gives:
\( 3u^2 - 10u + 3 = 0 \)
Factoring the quadratic equation:
\( (3u - 1)(u - 3) = 0 \)
So \( u = \frac{1}{3} \) or \( u = 3 \).
Since \( u = 3^y \):
For \( u = \frac{1}{3} \):
\( 3^y = 3^{-1} \implies y = -1 \)
For \( u = 3 \):
\( 3^y = 3^1 \implies y = 1 \)

(b) Using the laws of logarithms:
\( 2 \log_2 (x + 1) - \log_2 (2x - 1) = 2 \)
\( \log_2 (x + 1)^2 - \log_2 (2x - 1) = 2 \)
\( \log_2 \left( \frac{(x+1)^2}{2x-1} \right) = 2 \)
Now, rewrite in exponential form:
\( \frac{(x+1)^2}{2x-1} = 2^2 = 4 \)
\( (x+1)^2 = 4(2x-1) \)
\( x^2 + 2x + 1 = 8x - 4 \)
\( x^2 - 6x + 5 = 0 \)
Factoring the quadratic:
\( (x - 1)(x - 5) = 0 \)
So \( x = 1 \) or \( x = 5 \).
Both values make the arguments of the logarithms positive (\( x+1 > 0 \) and \( 2x-1 > 0 \)), so both are valid solutions.

(a)
- M1: Let \( u = 3^y \) or equivalent and write the equation as \( 3u^2 - 10u + 3 = 0 \).
- A1: Solve the quadratic to find \( u = \frac{1}{3} \) and \( u = 3 \) (or \( 3^y = \frac{1}{3} \) and \( 3^y = 3 \)).
- M1: Express \( 3^y = \frac{1}{3} \) and \( 3^y = 3 \) in terms of \( y \) and attempt to solve.
- A1: Correct answers \( y = -1 \) and \( y = 1 \) only.

(b)
- M1: Apply the power law to obtain \( \log_2 (x+1)^2 \).
- M1: Apply the division/subtraction law to combine logs and remove the logarithm to get \( \frac{(x+1)^2}{2x-1} = 4 \).
- A1: Correctly expand and simplify to the quadratic equation \( x^2 - 6x + 5 = 0 \) (or equivalent).
- A1: Solve to find \( x = 1 \) and \( x = 5 \). Must show that both are valid by reference to the domain or by explicitly presenting both as solutions.

(a) Using the trigonometric identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\( \cos^2 \theta = 1 - \sin^2 \theta \)
Substitute this into the given equation:
\( 4(1 - \sin^2 \theta) + 15 \sin \theta = 13 \)
\( 4 - 4 \sin^2 \theta + 15 \sin \theta = 13 \)
Rearranging all terms to the right side of the equation:
\( 0 = 4 \sin^2 \theta - 15 \sin \theta + 13 - 4 \)
\( 4 \sin^2 \theta - 15 \sin \theta + 9 = 0 \) (as required).

(b) Let \( \theta = 2x + 10^\circ \).
Since \( 0 \le x < 180^\circ \), the range of \( \theta \) is:
\( 0 \le 2x < 360^\circ \implies 10^\circ \le 2x + 10^\circ < 370^\circ \)
So \( 10^\circ \le \theta < 370^\circ \).
From part (a), we have:
\( 4 \sin^2 \theta - 15 \sin \theta + 9 = 0 \)
Factoring the quadratic equation:
\( (4 \sin \theta - 3)(\sin \theta - 3) = 0 \)
This gives:
\( \sin \theta = \frac{3}{4} \) or \( \sin \theta = 3 \)
Since \( -1 \le \sin \theta \le 1 \), the equation \( \sin \theta = 3 \) has no real solutions.
Thus, we solve:
\( \sin \theta = 0.75 \)
Finding the principal value:
\( \theta = \arcsin(0.75) \approx 48.59^\circ \)
Finding the second value in the interval \( [10^\circ, 370^\circ) \):
\( \theta = 180^\circ - 48.59^\circ \approx 131.41^\circ \)
Any further values like \( 360^\circ + 48.59^\circ = 408.59^\circ \) lie outside the interval.
Now find the values of \( x \):
For \( \theta = 48.59^\circ \):
\( 2x + 10^\circ = 48.59^\circ \implies 2x = 38.59^\circ \implies x \approx 19.3^\circ \)
For \( \theta = 131.41^\circ \):
\( 2x + 10^\circ = 131.41^\circ \implies 2x = 121.41^\circ \implies x \approx 60.7^\circ \)
So the solutions are \( x = 19.3^\circ \) and \( x = 60.7^\circ \) (to 1 d.p.).

(a)
- M1: Uses the identity \( \cos^2 \theta = 1 - \sin^2 \theta \) to substitute into the equation.
- A1*: Completes the proof to arrive at \( 4 \sin^2 \theta - 15 \sin \theta + 9 = 0 \) with no errors in working.

(b)
- M1: Attempts to factor or solve the quadratic in terms of \( \sin(2x + 10^\circ) \).
- A1: Obtains \( \sin(2x+10^\circ) = 0.75 \) and rejects \( \sin(2x+10^\circ) = 3 \).
- M1: Finds at least one correct value for \( 2x+10^\circ \) (e.g., \( 48.6^\circ \) or \( 131.4^\circ \)).
- M1: Sets up two equations \( 2x + 10 = \text{angle} \) and attempts to solve for \( x \).
- A1: Both answers \( x = 19.3^\circ \) and \( x = 60.7^\circ \) (accept awrt 19.3 and awrt 60.7). Deduct 1 mark for any extra solutions in range or failure to round correctly.

(a) Since \( \mathrm{f}(x) = \frac{2x+3}{x-1} = 2 + \frac{5}{x-1} \) and \( x > 1 \), we have \( x - 1 > 0 \), which means \( \frac{5}{x-1} > 0 \). Therefore, \( \mathrm{f}(x) > 2 \). (b) Let \( y = \frac{2x+3}{x-1} \). Rearranging gives: \( y(x-1) = 2x+3 \implies yx - y = 2x+3 \implies x(y-2) = y+3 \implies x = \frac{y+3}{y-2} \). Thus, \( \mathrm{f}^{-1}(x) = \frac{x+3}{x-2} \). The domain of \( \mathrm{f}^{-1} \) is the range of \( \mathrm{f} \), which is \( x > 2 \). (c) \( \mathrm{g}(\mathrm{f}(x)) = 8 \implies (\mathrm{f}(x))^2 - 1 = 8 \implies (\mathrm{f}(x))^2 = 9 \). Since the range of \( \mathrm{f} \) is \( \mathrm{f}(x) > 2 \), we discard the negative root to get \( \mathrm{f}(x) = 3 \). Solving \( \frac{2x+3}{x-1} = 3 \implies 2x+3 = 3x-3 \implies x = 6 \). Since \( 6 > 1 \), this is a valid solution.

(a) M1: Attempts to find the horizontal asymptote or reasons about the behavior of \( \mathrm{f}(x) \) as \( x \to \infty \) or \( x \to 1 \). A1: Correct range written as \( \mathrm{f}(x) > 2 \) or \( y > 2 \). (b) M1: Sets \( y = \mathrm{f}(x) \) and attempts to make \( x \) the subject. A1: Correct expression for \( \mathrm{f}^{-1}(x) = \frac{x+3}{x-2} \). B1ft: Correct domain \( x > 2 \) (must follow from their range in part a). (c) M1: Sets up the composite equation \( (\mathrm{f}(x))^2 - 1 = 8 \) or substitutes \( \mathrm{f}(x) \) into \( \mathrm{g} \). A1: Obtains \( \mathrm{f}(x) = 3 \) (explaining why \( -3 \) is rejected is not required but must only work with positive root). A1: Correct final answer \( x = 6 \).

(a) The graph of \( y = |3x-5| \) is a V-shape with its vertex on the x-axis at \( (\frac{5}{3}, 0) \) and y-intercept at \( (0, 5) \). The graph of \( y = x+3 \) is a straight line with a positive gradient, crossing the y-axis at \( (0, 3) \) and the x-axis at \( (-3, 0) \). (b) We find the critical values by solving \( |3x - 5| = x + 3 \). Case 1: \( 3x - 5 = x + 3 \implies 2x = 8 \implies x = 4 \). Case 2: \( -(3x - 5) = x + 3 \implies -3x + 5 = x + 3 \implies 4x = 2 \implies x = 0.5 \). Looking at the sketch, the V-shaped graph is strictly above the straight line when \( x < 0.5 \) or \( x > 4 \).

(a) B1: Correct V-shape graph with vertex on the positive x-axis and y-intercept marked. B1: Correct straight line with positive gradient, crossing both negative x-axis and positive y-axis. B1: All intercepts correctly labeled: \( (\frac{5}{3}, 0) \), \( (0, 5) \), \( (0, 3) \), and \( (-3, 0) \). (b) M1: Formulates two algebraic equations to find the intersection points. A1: Correct intersection points at \( x = 0.5 \) and \( x = 4 \). M1: Uses their graph to determine the correct regions (outside the critical values). A1: Correct final inequality: \( x < 0.5 \) or \( x > 4 \) (accept equivalent set notation; do not accept \( 4 < x < 0.5 \)).

(a) When \( t = 0 \), \( M = 150 \), so \( 150 = A \mathrm{e}^{0} \implies A = 150 \). (b) When \( t = 5 \), \( M = 120 \), so \( 120 = 150 \mathrm{e}^{-5k} \implies \mathrm{e}^{-5k} = \frac{120}{150} = \frac{4}{5} \). Taking the natural logarithm of both sides: \( -5k = \ln \left(\frac{4}{5}\right) \implies 5k = -\ln \left(\frac{4}{5}\right) = \ln \left(\frac{5}{4}\right) \). Thus, \( k = \frac{1}{5} \ln \left(\frac{5}{4}\right) \). (c) The rate of change of mass is \( \frac{\mathrm{d}M}{\mathrm{d}t} = -kA \mathrm{e}^{-kt} \). The rate of decrease is \( -\frac{\mathrm{d}M}{\mathrm{d}t} = kA \mathrm{e}^{-kt} \). When \( t = 10 \): \( \mathrm{e}^{-10k} = (\mathrm{e}^{-5k})^2 = \left(\frac{4}{5}\right)^2 = \frac{16}{25} \). Since \( k = \frac{1}{5} \ln(1.25) \approx 0.044629 \), the rate of decrease is \( 0.044629 \times 150 \times \frac{16}{25} = 96 \times 0.044629 \approx 4.2843 \) grams per year. To 3 s.f., the rate of decrease is 4.28.

(a) B1: States \( A = 150 \). (b) M1: Substitutes \( M = 120 \) and \( t = 5 \) with their \( A \) to get \( 120 = 150 \mathrm{e}^{-5k} \). M1: Isolates \( \mathrm{e}^{-5k} \) and applies natural logarithm to both sides. A1*: Shows convincing steps using log laws (e.g., \( -\ln(4/5) = \ln(5/4) \)) to reach the printed answer. (c) M1: Differentiates \( M \) with respect to \( t \) to get \( \pm kA \mathrm{e}^{-kt} \). M1: Substitutes \( t = 10 \), \( A = 150 \) and their value of \( k \) into their expression. A1: Obtains the exact multiplier value \( 96k \) or equivalent. A1: Evaluates to 4.28 (accept 4.28 or 4.28 grams per year; do not accept negative values since the rate of decrease is requested).

(a) LHS \( = \sec^2 \theta + \csc^2 \theta = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta} \). Combining the fractions: \( LHS = \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta \sin^2 \theta} \). Using the identity \( \sin^2 \theta + \cos^2 \theta \equiv 1 \): \( LHS = \frac{1}{\cos^2 \theta \sin^2 \theta} = \frac{1}{\cos^2 \theta} \cdot \frac{1}{\sin^2 \theta} = \sec^2 \theta \csc^2 \theta = RHS \). (b) Using the identity from part (a): \( \sec^2 \theta \csc^2 \theta = 4 \implies \frac{1}{\cos^2 \theta \sin^2 \theta} = 4 \implies \sin^2 \theta \cos^2 \theta = \frac{1}{4} \). Since \( \sin 2\theta = 2\sin \theta \cos \theta \), we can rewrite this as \( (\sin\theta\cos\theta)^2 = \frac{1}{4} \implies \left(\frac{1}{2}\sin 2\theta\right)^2 = \frac{1}{4} \implies \frac{1}{4}\sin^2 2\theta = \frac{1}{4} \implies \sin^2 2\theta = 1 \implies \sin 2\theta = \pm 1 \). Since \( 0 < \theta < \pi \), we have \( 0 < 2\theta < 2\pi \). For \( \sin 2\theta = 1 \), \( 2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4} \). For \( \sin 2\theta = -1 \), \( 2\theta = \frac{3\pi}{2} \implies \theta = \frac{3\pi}{4} \).

(a) M1: Converts \( \sec^2\theta \) and \( \csc^2\theta \) into terms of sine and cosine. M1: Puts fractions over a common denominator. A1*: Employs \( \sin^2\theta + \cos^2\theta = 1 \) correctly and reaches the RHS with no algebraic errors. (b) M1: Uses part (a) to substitute \( \sec^2\theta \csc^2\theta = 4 \). M1: Rearranges to obtain \( \sin\theta\cos\theta = \pm \frac{1}{2} \) or \( \sin^2 2\theta = 1 \). A1: Correctly identifies \( \theta = \frac{\pi}{4} \). A1: Correctly identifies \( \theta = \frac{3\pi}{4} \) and no other solutions within the range.

(a) We write \( 5\cos x - 12\sin x = R\cos x \cos \alpha - R\sin x \sin \alpha \). Comparing coefficients: \( R\cos\alpha = 5 \) and \( R\sin\alpha = 12 \). Thus, \( R^2 = 5^2 + 12^2 = 169 \implies R = 13 \). Also, \( \tan\alpha = \frac{12}{5} \implies \alpha = \arctan(2.4) \approx 1.176005 \) radians. So \( 5\cos x - 12\sin x \approx 13\cos(x + 1.176) \). (b) We can write \( \mathrm{h}(x) \) as: \( \mathrm{h}(x) = \frac{24}{2(5\cos x - 12\sin x) + 30} = \frac{24}{2(13\cos(x + 1.176)) + 30} = \frac{24}{26\cos(x + 1.176) + 30} \). The maximum value of \( \mathrm{h}(x) \) occurs when the denominator is minimized. The minimum value of the denominator is achieved when \( \cos(x + 1.176) = -1 \). In this case, the denominator is \( 26(-1) + 30 = 4 \). Thus, the maximum value of \( \mathrm{h}(x) \) is \( \frac{24}{4} = 6 \). (c) The maximum occurs when \( \cos(x + 1.176005) = -1 \). For the smallest positive value of \( x \): \( x + 1.176005 = \pi \implies x = \pi - 1.176005 \approx 3.141593 - 1.176005 = 1.965588 \) radians. To 2 d.p., \( x \approx 1.97 \).

(a) B1: Correctly identifies \( R = 13 \). M1: Uses \( \tan \alpha = \pm \frac{12}{5} \) or equivalent trig expression. A1: Correctly identifies \( \alpha \approx 1.176 \) (accept 1.18). (b) M1: Identifies that the minimum of the denominator occurs when \( \cos(x + \alpha) = -1 \) and substitutes this value. A1: Correctly obtains the maximum value of 6. (c) M1: Sets \( x + \text{their } \alpha = \pi \) (or \( (2n+1)\pi \)). A1: Obtains \( x = \pi - \text{their } \alpha \). A1: Evaluates correctly to 1.97 (or 1.96 depending on early rounding; accept interval [1.96, 1.97]).

(a) We use the quotient rule with \( u = \ln x \) and \( v = x^2+3 \). Differentiating, we get: \( u' = \frac{1}{x} \) and \( v' = 2x \). Applying the rule: \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{u'v - uv'}{v^2} = \frac{\frac{1}{x}(x^2 + 3) - 2x\ln x}{(x^2 + 3)^2} \). Multiplying the numerator and denominator by \( x \): \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^2 + 3 - 2x^2\ln x}{x(x^2 + 3)^2} \). (b) At the stationary point, \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \). Since \( x > 0 \), the denominator cannot be zero, so: \( x^2 + 3 - 2x^2\ln x = 0 \implies 3 = 2x^2\ln x - x^2 \implies 3 = x^2(2\ln x - 1) \implies x^2 = \frac{3}{2\ln x - 1} \). Taking the positive square root since \( x > 0 \): \( x = \sqrt{\frac{3}{2\ln x - 1}} \).

(a) M1: Applies the quotient rule formula \( \frac{u'v - uv'}{v^2} \) with correct identification of terms. A1: Correctly differentiates \( u \to \frac{1}{x} \) and \( v \to 2x \). M1: Simplifies the fraction by multiplying numerator and denominator by \( x \) (or equivalent algebraic step). A1: Correct final expression for the derivative. (b) M1: Sets their derivative numerator to 0. M1: Rearranges the equation to collect \( x^2 \) terms on one side. M1: Factorizes \( x^2 \) and isolates it. A1*: Concludes with the requested formula, showing all necessary steps clearly.

(a) Integrating: \( \int \frac{6}{2x-1} \\, \mathrm{d}x = 6 \times \frac{1}{2} \ln(2x-1) = 3 \ln(2x-1) \) (no modulus is needed since \( x > \frac{1}{2} \)). Applying limits from \( 1 \) to \( k \): \( \left[ 3 \ln(2x-1) \right]_1^k = 3 \ln(2k-1) - 3 \ln(2(1)-1) = 3 \ln(2k-1) - 3 \ln(1) \). Since \( \ln(1) = 0 \), the integral simplifies to \( 3 \ln(2k-1) \). (b) Since the area is \( \ln 64 \), we have: \( 3 \ln(2k-1) = \ln 64 \implies \ln(2k-1)^3 = \ln 64 \implies (2k-1)^3 = 64 \). Taking the cube root of both sides: \( 2k-1 = 4 \implies 2k = 5 \implies k = 2.5 \) (or \( \frac{5}{2} \)).

(a) M1: Integrates to obtain an expression of the form \( A \ln(2x-1) \). A1: Correct integration yielding \( 3 \ln(2x-1) \). A1*: Correctly substitutes the limits \( 1 \) and \( k \) and shows the step leading to \( 3 \ln(2k-1) \). (b) M1: Equates the given integral to \( \ln 64 \). M1: Uses logarithm laws to rewrite \( 3 \ln(2k-1) \) as \( \ln((2k-1)^3) \) or \( \ln 64 \) as \( 6 \ln 2 \). A1: Obtains the linear equation \( 2k - 1 = 4 \). A1: Correctly states \( k = 2.5 \) (or \( \frac{5}{2} \)).

(a) Evaluating at the endpoints of the interval: \( \mathrm{f}(2) = 2^3 - 5(2) + 1 = 8 - 10 + 1 = -1 \). \( \mathrm{f}(3) = 3^3 - 5(3) + 1 = 27 - 15 + 1 = 13 \). Since \( \mathrm{f}(x) \) is a continuous function and there is a change of sign between \( x = 2 \) and \( x = 3 \), there is a root \( \alpha \in [2, 3] \). (b) From \( x^3 - 5x + 1 = 0 \), we get \( x^3 = 5x - 1 \). Dividing both sides by \( x \) (which is valid as \( x \neq 0 \) in the interval): \( x^2 = 5 - \frac{1}{x} \). Since \( x > 0 \) in the interval, we take the positive square root: \( x = \sqrt{5 - \frac{1}{x}} \). (c) Using \( x_1 = 2.2 \): \( x_2 = \sqrt{5 - \frac{1}{2.2}} \approx 2.132007 \approx 2.1320 \). \( x_3 = \sqrt{5 - \frac{1}{2.132007}} \approx 2.128605 \approx 2.1286 \). \( x_4 = \sqrt{5 - \frac{1}{2.128605}} \approx 2.128429 \approx 2.1284 \).

(a) M1: Evaluates both \( \mathrm{f}(2) \) and \( \mathrm{f}(3) \) (at least one evaluation must be correct). A1: Obtains \( -1 \) and \( 13 \), and notes that the function is continuous with a change of sign, concluding there is a root in the interval. (b) M1: Attempts to isolate \( x^2 \) or \( x^3 \) from the equation. A1*: Shows clear algebraic progression to reach \( x = \sqrt{5 - \frac{1}{x}} \) with no errors. (c) M1: Demonstrates a correct substitution of \( x_1 = 2.2 \) into the iterative formula to find \( x_2 \). A1: Any two of \( x_2, x_3, x_4 \) correct to 4 decimal places. A1: All three of \( x_2 = 2.1320 \), \( x_3 = 2.1286 \), and \( x_4 = 2.1284 \) correct to 4 decimal places.

(a) We use the product rule to differentiate \(y = (x^2 - 3)\text{e}^{-2x}\). Let \(u = x^2 - 3\) which gives \(\dfrac{\text{d}u}{\text{d}x} = 2x\). Let \(v = \text{e}^{-2x}\) which gives \(\dfrac{\text{d}v}{\text{d}x} = -2\text{e}^{-2x}\). Using the product rule, \(\dfrac{\text{d}y}{\text{d}x} = u \dfrac{\text{d}v}{\text{d}x} + v \dfrac{\text{d}u}{\text{d}x} = (x^2 - 3)(-2\text{e}^{-2x}) + (2x)\text{e}^{-2x} = (-2x^2 + 2x + 6)\text{e}^{-2x}\). Thus, \(g(x) = -2x^2 + 2x + 6\). (b) Stationary points occur when \(\dfrac{\text{d}y}{\text{d}x} = 0\). Since \(\text{e}^{-2x} \neq 0\), we have \(-2x^2 + 2x + 6 = 0\), which simplifies to \(x^2 - x - 3 = 0\). Using the quadratic formula, \(x = \dfrac{1 \pm \sqrt{13}}{2}\). To find the corresponding y-coordinates, we substitute \(x\) back into \(y = (x^2 - 3)\text{e}^{-2x}\). Since \(x^2 - x - 3 = 0\) implies \(x^2 - 3 = x\), we can simplify the expression to \(y = x\text{e}^{-2x}\). Substituting our values of \(x\), we get \(y = \left(\dfrac{1+\sqrt{13}}{2}\right)\text{e}^{-(1+\sqrt{13})}\) and \(y = \left(\dfrac{1-\sqrt{13}}{2}\right)\text{e}^{-(1-\sqrt{13})}\). Therefore, the exact coordinates of the stationary points are \(\left(\dfrac{1+\sqrt{13}}{2}, \, \left(\dfrac{1+\sqrt{13}}{2}\right)\text{e}^{-1-\sqrt{13}}\right)\) and \(\left(\dfrac{1-\sqrt{13}}{2}, \, \left(\dfrac{1-\sqrt{13}}{2}\right)\text{e}^{-1+\sqrt{13}}\right)\).

Part (a): M1: Attempts the product rule to differentiate \(y = (x^2 - 3)\text{e}^{-2x}\) obtaining at least one correct term of the form \(A x \text{e}^{-2x}\) or \(B (x^2 - 3)\text{e}^{-2x}\). A1: Fully correct differentiation: \((2x)\text{e}^{-2x} - 2(x^2 - 3)\text{e}^{-2x}\). A1: Correctly factors out \(\text{e}^{-2x}\) to obtain \(\dfrac{\text{d}y}{\text{d}x} = (-2x^2 + 2x + 6)\text{e}^{-2x}\) or equivalent. Part (b): M1: Sets their \(\dfrac{\text{d}y}{\text{d}x} = 0\) and attempts to solve the quadratic equation \(g(x) = 0\). A1: Obtains the correct exact x-values: \(x = \dfrac{1 \pm \sqrt{13}}{2}\). M1: Substitutes at least one of their x-values into the equation of the curve to find the corresponding y-coordinate. A1: Obtains both pairs of coordinates fully correct in exact simplified form.

(a) Let \(f(x) = \ln(3x - 1) - x^2 + 4\). Evaluating at the endpoints: \(f(2) = \ln(5) - 4 + 4 = \ln(5) \approx 1.6094 > 0\). \(f(2.5) = \ln(6.5) - 6.25 + 4 \approx -0.3782 < 0\). Since there is a change of sign and \(f(x)\) is continuous on the interval \([2, \, 2.5]\), there is at least one root \(\alpha \in [2, \, 2.5]\). (b) Rearranging \(\ln(3x - 1) - x^2 + 4 = 0\) gives \(x^2 = \ln(3x - 1) + 4\). Since \(x > \dfrac{1}{3}\) and we are looking for a root in the range \([2, \, 2.5]\), \(x\) is positive. Taking the positive square root gives \(x = \sqrt{\ln(3x - 1) + 4}\). (c) Using \(x_1 = 2\): \(x_2 = \sqrt{\ln(5) + 4} \approx 2.368425 \approx 2.3684\) (to 4 d.p.). Using the unrounded value of \(x_2\), \(x_3 = \sqrt{\ln(3(2.368425) - 1) + 4} \approx 2.410218 \approx 2.4102\) (to 4 d.p.). (d) To show that \(\alpha = 2.415\) correct to 3 decimal places, we choose the interval \([2.4145, \, 2.4155]\). Evaluating \(f(x)\) at these bounds: \(f(2.4145) = \ln(3(2.4145) - 1) - (2.4145)^2 + 4 \approx 0.0017 > 0\). \(f(2.4155) = \ln(3(2.4155) - 1) - (2.4155)^2 + 4 \approx -0.0026 < 0\). Since there is a change of sign in the interval \([2.4145, \, 2.4155]\) and \(f(x)\) is continuous, the root \(\alpha\) lies in \((2.4145, \, 2.4155)\). Since any value in this interval rounds to \(2.415\) to 3 decimal places, we conclude \(\alpha = 2.415\) to 3 d.p.

Part (a): M1: Attempts to evaluate \(f(2)\) and \(f(2.5)\) for a continuous function \(f(x)\). A1: Correct evaluations \(f(2) \approx 1.61\) and \(f(2.5) \approx -0.38\) with a valid conclusion noting the sign change. Part (b): M1: Isolates \(x^2 = \ln(3x - 1) + 4\). A1: Explains taking the positive square root because \(x > \dfrac{1}{3}\) to obtain the given form. Part (c): M1: Attempts to calculate \(x_2\) using the iterative formula. A1: Both \(x_2 \approx 2.3684\) and \(x_3 \approx 2.4102\) correct to 4 decimal places. Part (d): M1: Chooses the correct interval \([2.4145, \, 2.4155]\) and attempts to evaluate \(f(x)\) at both bounds. A1: Obtains \(f(2.4145) \approx 0.0017 > 0\) and \(f(2.4155) \approx -0.0026 < 0\), notes the change of sign and continuity, and correctly concludes \(\alpha = 2.415\) to 3 d.p.

((a)) Applying the binomial expansion formula:
\((1 + y)^{n} = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \dots\n\nHere, \)n = \frac{1}{2}\) and \(y = -8x\):

\((1 - 8x)^{1/2} = 1 + \left(\frac{1}{2}\right)(-8x) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2}(-8x)^2 + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{6}(-8x)^3 + \dots\n\n\)(1 - 8x)^{1/2} = 1 - 4x - 8x^2 - 32x^3 - \dots\n\n((b)) Substituting \(x = \frac{1}{100}\) into \((1 - 8x)^{1/2}\):

\((1 - 0.08)^{1/2} = \sqrt{0.92} = \sqrt{\frac{92}{100}} = \frac{\sqrt{23}}{5}\n\nUsing the expansion from part (a) with \)x = 0.01\):

\(\frac{\sqrt{23}}{5} \approx 1 - 4(0.01) - 8(0.01)^2 - 32(0.01)^3\n\n\)\frac{\sqrt{23}}{5} \approx 1 - 0.04 - 0.0008 - 0.000032 = 0.959168\n\nMultiplying by 5 gives:\n\n\ \sqrt{23} \approx 5 \times 0.959168 = 4.79584 = \frac{479584}{100000} = \frac{14987}{3125}\)

((a))
M1: Starts binomial expansion with correct structure and substitution of \(n = \frac{1}{2}\) and \(y = -8x\).
A1: Correct second term \(-4x\).
A1: Correct third term \(-8x^2\).
A1: Correct fourth term \(-32x^3\) (allow unsimplified coefficients for M1 but A marks require simplified terms).

((b))
M1: Recognises that substituting \(x = 0.01\) results in \(\frac{\sqrt{23}}{5}\) or equivalent.
M1: Evaluates the expansion at \(x = 0.01\).
A1: Obtains the correct approximation of \(\frac{14987}{3125}\) or \(4.79584\).

((a)) Write the algebraic identity:
\(7x^2 + 16x + 11 = A(x+1)(2x+3) + B(2x+3) + C(x+1)^2\n\nSubstitute \)x = -1\):
\(7(-1)^2 + 16(-1) + 11 = B(2(-1) + 3)\n\n2 = B \implies B = 2\n\nSubstitute \)x = -1.5\):
\(7(-1.5)^2 + 16(-1.5) + 11 = C(-1.5 + 1)^2\n\n15.75 - 24 + 11 = 0.25 C\n\n2.75 = 0.25 C \implies C = 11\n\nCompare the coefficients of \)x^2\):

7 = 2A + C

7 = 2A + 11 \implies 2A = -4 \implies A = -2

So, \(f(x) = \frac{-2}{x + 1} + \frac{2}{(x + 1)^2} + \frac{11}{2x + 3}\).

((b)) Rewrite \(f(x)\) in index form:
\(f(x) = -2(x + 1)^{-1} + 2(x + 1)^{-2} + 11(2x + 3)^{-1}\n\nDifferentiate with respect to \)x\):
\(f'(x) = 2(x + 1)^{-2} - 4(x + 1)^{-3} - 22(2x + 3)^{-2}\n\nSubstitute \)x = 0\):
\(f'(0) = 2(1)^{-2} - 4(1)^{-3} - 22(3)^{-2}\n\n\)f'(0) = 2 - 4 - \frac{22}{9} = -2 - \frac{22}{9} = -\frac{40}{9}\)

((a))
M1: Forms correct identity by multiplying through by the denominator.
M1: Uses a valid method (substitution or comparing coefficients) to find at least one constant.
A1: Correct value of \(B = 2\).
A1: Correct value of \(C = 11\).
A1: Correct value of \(A = -2\).

((b))
M1: Correctly differentiates at least two terms of their partial fraction representation.
A1: Correct derivative expression \(f'(x) = 2(x + 1)^{-2} - 4(x + 1)^{-3} - 22(2x + 3)^{-2}\) or equivalent.
A1: Correct evaluation of \(f'(0) = -\frac{40}{9}\) (accept equivalent fractions).

((a)) First, find the coordinates of the point when \(t = 2\):
\(x = 2^2 - 3(2) = -2\)
\(y = \frac{4}{2-1} = 4\)

Now differentiate both parametric equations with respect to \(t\):
\(\frac{dx}{dt} = 2t - 3\)
\(\frac{dy}{dt} = -\frac{4}{(t-1)^2}\)

At \(t = 2\):
\(\frac{dx}{dt} = 2(2) - 3 = 1\)
\(\frac{dy}{dt} = -\frac{4}{(2-1)^2} = -4\)

Calculate the gradient of the tangent \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-4}{1} = -4\)

Using the equation of a line with gradient \(-4\) passing through \((-2, 4)\):
\(y - 4 = -4(x - (-2))\)
\(y - 4 = -4x - 8 \implies 4x + y + 4 = 0\)

((b)) Rearrange the parametric equation for \(y\) to express \(t\) in terms of \(y\):
\(y = \frac{4}{t-1} \implies t - 1 = \frac{4}{y} \implies t = 1 + \frac{4}{y}\)

Substitute \(t = 1 + \frac{4}{y}\) into the parametric equation for \(x\):
\(x = \left(1 + \frac{4}{y}\right)^2 - 3\left(1 + \frac{4}{y}\right)\)

\(x = 1 + \frac{8}{y} + \frac{16}{y^2} - 3 - \frac{12}{y} = -2 - \frac{4}{y} + \frac{16}{y^2}\)

Multiply both sides by \(y^2\):
\(y^2 x = -2y^2 - 4y + 16\)

Rearrange into the form \(y^2(x + k) = f(y)\):
\(y^2 x + 2y^2 = 16 - 4y \implies y^2(x + 2) = 16 - 4y\)

Here, \(k = 2\) and \(f(y) = 16 - 4y\).

((a))
M1: Evaluates both coordinates \(x\) and \(y\) at \(t = 2\).
M1: Differentiates \(x\) and \(y\) with respect to \(t\).
A1: Correct derivatives \(\frac{dx}{dt} = 2t - 3\) and \(\frac{dy}{dt} = -\frac{4}{(t-1)^2}\).
M1: Employs the chain rule to find \(\frac{dy}{dx} = -4\) at \(t=2\).
A1: Correct final equation of the tangent in the specified integer-coefficient format: \(4x + y + 4 = 0\).

((b))
M1: Correctly rearranges the formula for \(y\) to make \(t\) the subject.
M1: Substitutes their expression for \(t\) into the equation for \(x\).
A1: Correct final Cartesian equation \(y^2(x + 2) = 16 - 4y\).

((a)) Differentiate implicitly with respect to \(x\):
\(\frac{d}{dx}(3^x) + \frac{d}{dx}(2xy) - \frac{d}{dx}(y^2) = \frac{d}{dx}(4)\)

Recall that \(\frac{d}{dx}(3^x) = 3^x \ln 3\).
For the term \(2xy\), apply the product rule:
\(\frac{d}{dx}(2xy) = 2y + 2x \frac{dy}{dx}\)

For \(y^2\), apply the chain rule:
\(\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}\)

Combining these terms, we obtain:
\(3^x \ln 3 + 2y + 2x \frac{dy}{dx} - 2y \frac{dy}{dx} = 0\)

Factor out \(\frac{dy}{dx}\):
\(\frac{dy}{dx}(2x - 2y) = -(3^x \ln 3 + 2y)\)

\(\frac{dy}{dx} = \frac{3^x \ln 3 + 2y}{2y - 2x}\)

((b)) At the point \((2, 5)\), substitute \(x = 2\) and \(y = 5\) into the derivative:
\(\frac{dy}{dx} = \frac{3^2 \ln 3 + 2(5)}{2(5) - 2(2)} = \frac{9\ln 3 + 10}{6}\)

The equation of the tangent line is:
\(y - 5 = \left(\frac{9\ln 3 + 10}{6}\right)(x - 2)\)

Multiply through by 6:
\(6y - 30 = (9\ln 3 + 10)(x - 2)\)

\(6y - 30 = (9\ln 3 + 10)x - 18\ln 3 - 20\)

Rearranging to group terms:
\((9\ln 3 + 10)x - 6y - 18\ln 3 + 10 = 0\)

((a))
M1: Attempts implicit differentiation, with at least one term correctly differentiated containing \(\frac{dy}{dx}\).
A1: Correct derivative for \(3^x\) as \(3^x \ln 3\).
M1: Correctly applies the product rule to differentiate \(2xy\).
A1: Fully correct differentiated equation.
M1: Rearranges to isolate \(\frac{dy}{dx}\) as the subject.
A1: Correct final expression for \(\frac{dy}{dx}\).

((b))
M1: Substitutes \(x = 2\) and \(y = 5\) to find the numerical gradient of the curve.
A1: Correct gradient value \(\frac{9\ln 3 + 10}{6}\).
M1: Forms the equation of the line passing through \((2, 5)\) with their gradient.
A1: Correct equation in any equivalent form, e.g., \((9\ln 3 + 10)x - 6y - 18\ln 3 + 10 = 0\).

((a)) Let \(u = 1 + \sqrt{x} \implies u - 1 = \sqrt{x}\), so \(x = (u - 1)^2\).
Differentiating with respect to \(u\):
\(\frac{dx}{du} = 2(u - 1) \implies dx = 2(u - 1) \, du\).

Substitute these into the integral:
\(\int \frac{1}{1 + \sqrt{x}} \, dx = \int \frac{1}{u} \cdot 2(u - 1) \, du = \int \left(2 - \frac{2}{u}\right) \, du\)

Integrating this expression yields:
\(2u - 2\ln|u| + C\)

Substitute back \(u = 1 + \sqrt{x}\):
\(2(1 + \sqrt{x}) - 2\ln(1 + \sqrt{x}) + C = 2\sqrt{x} - 2\ln(1 + \sqrt{x}) + C'\) (where \(C' = C + 2\) is an arbitrary constant).

((b)) Let \(u = 1 + \sqrt{x}\).
Then \(\frac{du}{dx} = \frac{1}{2\sqrt{x}} \implies 2 \, du = \frac{1}{\sqrt{x}} \, dx\).

Change the limits of integration:
When \(x = 1\), \(u = 1 + \sqrt{1} = 2\).
When \(x = 9\), \(u = 1 + \sqrt{9} = 4\).

Rewrite the integral using substitution:
\(\int_{1}^{9} \frac{\ln(1 + \sqrt{x})}{\sqrt{x}} \, dx = \int_{2}^{4} \ln(u) \cdot 2 \, du = 2 \int_{2}^{4} \ln(u) \, du\)

Using integration by parts (or standard integral result) for \(\int \ln(u) \, du\):
\(\int \ln(u) \, du = u\ln(u) - u\)

Evaluating between the limits \(2\) and \(4\):
\(2 [u\ln(u) - u]_{2}^{4} = 2 \left( (4\ln 4 - 4) - (2\ln 2 - 2) \right)\)

Since \(\ln 4 = 2\ln 2\):
\(2 \left( 4(2\ln 2) - 4 - 2\ln 2 + 2 \right) = 2 \left( 8\ln 2 - 2\ln 2 - 2 \right) = 2 (6\ln 2 - 2) = 12\ln 2 - 4\)

Thus, \(A = 12\) and \(B = -4\).

((a))
M1: Correctly relates \(dx\) and \(du\) using the substitution.
M1: Substitutes expressions for \(x\) and \(dx\) to obtain an integral in terms of \(u\) only.
A1: Correctly simplified integral \(\int (2 - \frac{2}{u}) \, du\).
M1: Performs integration on both terms correctly to obtain \(2u - 2\ln|u|\).
A1: Correct back-substitution to obtain \(2\sqrt{x} - 2\ln(1 + \sqrt{x}) + C\) (accept equivalent forms, e.g. including the \(+2\)).

((b))
M1: Employs the substitution \(u = 1 + \sqrt{x}\) to rewrite the integrand, transforming limits of integration correctly.
A1: Obtains the simplified integral \(2 \int_{2}^{4} \ln(u) \, du\) with correct limits.
M1: Uses integration by parts correctly to integrate \(\ln(u)\).
A1: Correctly evaluates using limits and properties of logarithms to obtain \(12\ln 2 - 4\).

((a)) Separate variables in the differential equation:
\(\int \frac{10}{h(5 - h)} \, dh = \int 1 \, dt\)

Express the LHS integrand in partial fractions:
\(\frac{10}{h(5 - h)} = \frac{P}{h} + \frac{Q}{5 - h}\)

\(10 = P(5 - h) + Qh\)

Let \(h = 0\):
\(10 = 5P \implies P = 2\)

Let \(h = 5\):
\(10 = 5Q \implies Q = 2\)

So the integral becomes:
\(\int \left( \frac{2}{h} + \frac{2}{5 - h} \right) dh = \int 1 \, dt\)

Integrate both sides:
\(2\ln|h| - 2\ln|5 - h| = t + C\)

Combine the logarithmic terms:
\(2\ln\left|\frac{h}{5-h}\right| = t + C\)

Apply the initial condition: \(h = 1\) when \(t = 0\):
\(2\ln\left|\frac{1}{4}\right| = 0 + C \implies C = 2\ln(0.25) = -2\ln(4)\)

Substitute \(C\) back into the equation:
\(2\ln\left|\frac{h}{5-h}\right| = t - 2\ln(4)\)

Rearrange for \(t\):
\(t = 2\ln\left(\frac{h}{5-h}\right) + 2\ln(4) = 2 \left[ \ln\left(\frac{h}{5-h}\right) + \ln(4) \right]\)

\(t = 2 \ln \left( \frac{4h}{5-h} \right)\) (since \(1 \le h < 5\), the absolute values are positive).

((b)) When \(h = 4\):
\(t = 2 \ln \left( \frac{4(4)}{5-4} \right) = 2 \ln(16)\)

\(t = 8\ln(2) \approx 5.54517... \text{ seconds}\)

To 3 significant figures, \(t = 5.55\) seconds.

((a))
M1: Separates variables to form two integrals.
M1: Correctly sets up a partial fraction decomposition for the LHS integrand.
A1: Obtains correct partial fractions: \(\frac{2}{h} + \frac{2}{5-h}\).
M1: Integrates both sides with correct logarithmic forms (note the negative sign for \(-\ln|5-h|\)).
A1: Fully correct integrated expression including a constant of integration.
M1: Applies the initial condition \(h=1\), \(t=0\) to evaluate their constant.
A1*: Reaches the target equation through rigorous algebraic manipulation.

((b))
M1: Substitutes \(h = 4\) into the given equation.
A1: Evaluates correctly to \(5.55\) seconds (to 3 s.f.).

((a)) Equate the components of \(l_1\) and \(l_2\):
1) \(1 + 2\lambda = 6 - \mu \implies 2\lambda + \mu = 5\)
2) \(-2 + \lambda = -3 + 3\mu \implies \lambda - 3\mu = -1\)
3) \(3 - \lambda = -1 + 2\mu \implies \lambda + 2\mu = 4\)

Solve the system of equations 1) and 2):
From 1), \u0009\mu = 5 - 2\lambda\.
Substitute into 2):
\(\lambda - 3(5 - 2\lambda) = -1\)
\(\lambda - 15 + 6\lambda = -1\)
\(7\lambda = 14 \implies \lambda = 2\)

Substitute \(\lambda = 2\) back into the expression for \(\mu\):
\(\mu = 5 - 2(2) = 1\)

Now check if \(\lambda = 2\) and \(\mu = 1\) satisfy equation 3):
LHS of 3): \(\lambda + 2\mu = 2 + 2(1) = 4\).
RHS of 3): \(4\).
Since LHS = RHS, the equations are consistent, which confirms that the lines intersect.

To find the coordinates of \(P\), substitute \(\lambda = 2\) into the equation for \(l_1\):
\(\mathbf{r} = \begin{pmatrix} 1 + 2(2) \\ -2 + 2 \\ 3 - 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \\ 1 \end{pmatrix}\)

So the point of intersection \(P\) is \((5, 0, 1)\).

((b)) The direction vectors of the lines are:
\(\mathbf{d}_1 = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix}\)

Calculate their dot product:
\(\mathbf{d}_1 \cdot \mathbf{d}_2 = 2(-1) + 1(3) + (-1)(2) = -2 + 3 - 2 = -1\)

Calculate the magnitudes of \(\mathbf{d}_1\) and \(\mathbf{d}_2\):
\(|\mathbf{d}_1| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{6}\)
\(|\mathbf{d}_2| = \sqrt{(-1)^2 + 3^2 + 2^2} = \sqrt{14}\)

Let \(\theta\) be the acute angle between the lines:
\(\cos\theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|} = \frac{|-1|}{\sqrt{6}\sqrt{14}} = \frac{1}{\sqrt{84}}\)

\(\theta = \arccos\left(\frac{1}{\sqrt{84}}\right) \approx 83.737^\circ\)

Thus, the acute angle is \(83.7^\circ\) to 1 decimal place.

((a))
M1: Sets up simultaneous equations by equating the components of the two lines.
M1: Solves two equations to find values for \(\lambda\) and \(\mu\).
A1: Correct values: \(\lambda = 2\) and \(\mu = 1\).
M1: Substitutes \(\lambda\) and \(\mu\) into the third equation to verify consistency.
A1: Obtains the correct coordinates \(P(5, 0, 1)\).

((b))
M1: Identifies and utilizes correct direction vectors for both lines to calculate the dot product.
M1: Calculates the product of the magnitudes of the direction vectors.
A1: Obtains the correct acute angle of \(83.7^\circ\).

((a)) The initial assumption is:
Assume that \(\log_5 2\) is a rational number.

((b)) Since we assumed \(\log_5 2\) is rational, we can write:
\(\log_5 2 = \frac{a}{b}\) where \(a\) and \(b\) are positive integers (since \(\log_5 2 > 0\)), and \(b \neq 0\).

From the definition of logarithms, we can write this in exponential form:
\(5^{a/b} = 2\)

Raise both sides of the equation to the power of \(b\):
\(5^a = 2^b\)

Now examine the two sides of the equation:
- The left-hand side \(5^a\) is a positive power of 5. Any positive integer power of an odd number is always odd. Hence, \(5^a\) is an odd integer.
- The right-hand side \(2^b\) is a positive power of 2. Any positive integer power of 2 is always even. Hence, \(2^b\) is an even integer.

This means:
An odd integer = An even integer, which is a contradiction.

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

((a))
B1: States clearly the contradictory assumption that \(\log_5 2\) is a rational number.

((b))
M1: Translates their rational expression \(\frac{a}{b}\) into exponential form \(5^{a/b} = 2\) (where \(a, b\) are integers, \(b \neq 0\)).
M1: Correctly manipulates the indices to obtain \(5^a = 2^b\).
A1: Explains why \(5^a\) is odd for positive integers \(a\).
A1: Explains why \(2^b\) is even for positive integers \(b\).
A1*: Concludes the argument rigorously by identifying the contradiction (odd equals even) and stating that this proves \(\log_5 2\) is irrational.

(a) The area \(A\) of the region under the parametric curve is given by:
\[ A = \int y \, \text{d}x = \int y \frac{\text{d}x}{\text{d}t} \, \text{d}t \]
We are given \(x = \ln(t+1)\), so differentiating with respect to \(t\) gives:
\[ \frac{\text{d}x}{\text{d}t} = \frac{1}{t+1} \]
Now we find the limits for \(t\):
- When \(x = 0\):
\[ \ln(t+1) = 0 \implies t+1 = 1 \implies t = 0 \]
- When \(x = \ln 3\):
\[ \ln(t+1) = \ln 3 \implies t+1 = 3 \implies t = 2 \]
Substituting these into the area integral:
\[ A = \int_{0}^{2} \left(\frac{t}{t+2}\right) \left(\frac{1}{t+1}\right) \, \text{d}t = \int_{0}^{2} \frac{t}{(t+1)(t+2)} \, \text{d}t \quad \text{(as required)} \]

(b) Let:
\[ \frac{t}{(t+1)(t+2)} = \frac{A}{t+1} + \frac{B}{t+2} \]
Multiplying by \((t+1)(t+2)\):
\[ t = A(t+2) + B(t+1) \]
- Setting \(t = -1\):
\[ -1 = A(-1+2) \implies A = -1 \]
- Setting \(t = -2\):
\[ -2 = B(-2+1) \implies B = 2 \]
Thus, the partial fraction decomposition is:
\[ \frac{t}{(t+1)(t+2)} = \frac{2}{t+2} - \frac{1}{t+1} \]

(c) Using the result from part (b), we integrate:
\[ \int_{0}^{2} \left( \frac{2}{t+2} - \frac{1}{t+1} \right) \, \text{d}t = \left[ 2\ln(t+2) - \ln(t+1) \right]_{0}^{2} \]
Substituting the upper limit \(t = 2\):
\[ 2\ln(4) - \ln(3) = \ln(16) - \ln(3) \]
Substituting the lower limit \(t = 0\):
\[ 2\ln(2) - \ln(1) = \ln(4) - 0 = \ln(4) \]
Subtracting the lower limit from the upper limit:
\[ A = (\ln 16 - \ln 3) - \ln 4 = \ln\left(\frac{16}{3 \times 4}\right) = \ln\left(\frac{4}{3}\right) \]
Therefore, \(k = \frac{4}{3}\) and the exact area is \(\ln\left(\frac{4}{3}\right)\).

(a)
- M1: Applies the parametric area formula by finding \(\frac{\text{d}x}{\text{d}t} = \frac{1}{t+1}\) and substituting \(y\) and \(\frac{\text{d}x}{\text{d}t}\).
- M1: Converts the limits of integration from \(x = 0\) and \(x = \ln 3\) to \(t = 0\) and \(t = 2\) respectively.
- A1*: Completes the proof to arrive at the given integral, including \(\text{d}t\), with no errors in working.

(b)
- M1: Expresses the fraction in the form \(\frac{A}{t+1} + \frac{B}{t+2}\) and attempts to find at least one constant.
- A1: Correctly obtains \(\frac{2}{t+2} - \frac{1}{t+1}\) (or equivalent form).

(c)
- M1: Integrates their partial fractions to obtain \(P \ln(t+2) + Q \ln(t+1)\).
- M1: Substitutes their limits \(t=2\) and \(t=0\) and attempts to combine the logarithms using log rules.
- A1: Correct final exact answer in the form \(\ln\left(\frac{4}{3}\right)\) (or clearly states \(k = \frac{4}{3}\)).

**(a)**
First, we find \(\sin \alpha\) and \(\cos \alpha\) from \(\tan \alpha = \frac{3}{4}\):
\[\sin \alpha = 0.6, \quad \cos \alpha = 0.8\]

Since \(A\) has mass \(4\text{ kg}\), the normal reaction force \(R\) on \(A\) is:
\[R = 4g \cos \alpha = 4g(0.8) = 3.2g\]

The maximum frictional force \(F_{\max}\) is:
\[F_{\max} = \mu R = 0.25(3.2g) = 0.8g\]

When \(B\) is on the point of moving downwards, particle \(A\) is on the point of moving up the plane. The frictional force \(F_{\max}\) opposes this motion and acts down the plane.
For particle \(B\):
\[T = mg\]

For particle \(A\) (resolving parallel to the plane):
\[T - 4g \sin \alpha - F_{\max} = 0\]
\[T = 4g(0.6) + 0.8g = 2.4g + 0.8g = 3.2g\]

Substituting \(T = mg\):
\[mg = 3.2g \implies m = 3.2\]

**(b)**
If \(m = 6\text{ kg}\), then \(m > 3.2\), so the system accelerates. Particle \(B\) moves downwards with acceleration \(a\) and particle \(A\) moves up the plane with acceleration \(a\). Frictional force is \(F = 0.8g\) acting down the plane.

Equation of motion for \(B\):
\[6g - T = 6a \quad \text{--- (1)}\]

Equation of motion for \(A\):
\[T - 4g \sin \alpha - F = 4a\]
\[T - 2.4g - 0.8g = 4a \implies T - 3.2g = 4a \quad \text{--- (2)}\]

Adding (1) and (2):
\[6g - 3.2g = 10a\]
\[2.8g = 10a \implies a = 0.28g\]

**(a)**
- M1: For calculating \(\sin \alpha\) and \(\cos \alpha\) correctly.
- M1: For resolving forces perpendicular to the plane to find \(R = 3.2g\) or equivalent.
- M1: For applying \(F = \mu R\) with their \(R\).
- M1: For setting up the equations of equilibrium for both \(A\) and \(B\), resolving parallel to the plane.
- A1: Correct expression for \(T = 3.2g\).
- A1: Correct value of \(m = 3.2\).

**(b)**
- M1: For setting up the equation of motion for particle \(B\).
- M1: For setting up the equation of motion for particle \(A\) with friction acting down the plane.
- M1: For solving the simultaneous equations to find \(a\) in terms of \(g\).
- A1: Correct equation \(2.8g = 10a\) or equivalent.
- A1: Correct final value \(a = 0.28g\) (accept \(2.7\text{ m s}^{-2}\) or \(2.74\text{ m s}^{-2}\)).

**(a)**
When the rod is on the point of tilting about \(C\), the reaction at the support \(D\) is zero (\(R_D = 0\)).
Taking moments about \(C\) for the rod and the particle at \(A\):
\[\text{Counterclockwise Moments} = \text{Clockwise Moments}\]
\[4g \times AC = 12g \times CG\]
We are given \(AC = 1.5\text{ m}\). Since \(AG = x\), the distance \(CG = AG - AC = x - 1.5\).
Substituting these values:
\[4g(1.5) = 12g(x - 1.5)\]
\[6g = 12g(x - 1.5)\]
Divide both sides by \(6g\):
\[1 = 2(x - 1.5)\]
\[1 = 2x - 3 \implies 2x = 4 \implies x = 2 \quad \text{(shown)}\]

**(b)**
With the \(4\text{ kg}\) particle removed and a particle of mass \(M\text{ kg}\) placed at \(B\), the rod is on the point of tilting about \(D\), so the reaction at \(C\) is zero (\(R_C = 0\)).
Taking moments about \(D\):
\[\text{Clockwise Moments} = \text{Counterclockwise Moments}\]
\[Mg \times BD = 12g \times GD\]
We have \(BD = 1\text{ m}\).
The distance \(AD = 6 - BD = 5\text{ m}\).
Since \(AG = 2\text{ m}\), the distance \(GD = AD - AG = 5 - 2 = 3\text{ m}\).
Substituting these into the moments equation:
\[Mg(1) = 12g(3)\]
\[Mg = 36g \implies M = 36\]

**(a)**
- M1: For identifying that \(R_D = 0\) when tilting about \(C\).
- M1: For attempting to take moments about \(C\) (or another valid point with a consistent equation).
- A1: Correct terms for moments: \(4g \times 1.5 = 12g \times CG\).
- M1: Writing \(CG\) as \(x - 1.5\).
- A1: Clearly showing that \(x = 2\) with no errors in algebra.

**(b)**
- M1: For identifying that \(R_C = 0\) when tilting about \(D\).
- M1: For attempting to take moments about \(D\).
- A1: Correct moment due to the mass of the rod: \(12g \times GD\).
- M1: Finding the correct distance \(GD = 3\text{ m}\).
- A1: Correct moment equation: \(Mg(1) = 12g(3)\).
- A1: Correct value of \(M = 36\).

**(a)**
The speed-time graph consists of three straight-line segments:
1. A line starting at \((0, 12)\) and rising to \((8, V)\).
2. A horizontal line from \((8, V)\) to \((8+T, V)\).
3. A line falling from \((8+T, V)\) to \((28, 0)\).

**(b)**
Let \(t_3\) be the time taken for the third stage (deceleration).
Since deceleration is \(2\text{ m s}^{-2}\):
\[t_3 = \frac{V}{2}\]
The total time is \(28\text{ seconds}\):
\[8 + T + t_3 = 28 \implies T = 20 - \frac{V}{2}\]

Now we write expressions for the distance in each stage:
- First stage (trapezium):
\[s_1 = \frac{12 + V}{2} \times 8 = 4(12 + V) = 48 + 4V\]
- Second stage (rectangle):
\[s_2 = V T = V\left(20 - \frac{V}{2}\right) = 20V - \frac{V^2}{2}\]
- Third stage (triangle):
\[s_3 = \frac{1}{2} \times V \times t_3 = \frac{1}{2} V \left(\frac{V}{2}\right) = \frac{V^2}{4}\]

The total distance is \(428\text{ m}\):
\[s_1 + s_2 + s_3 = 428\]
\[(48 + 4V) + \left(20V - \frac{V^2}{2}\right) + \frac{V^2}{4} = 428\]
\[48 + 24V - \frac{V^2}{4} = 428\]
Multiply the entire equation by \(-4\) to clear fractions and rearrange:
\[-192 - 96V + V^2 = -1712\]
\[V^2 - 96V + 1520 = 0\]
Factorising this quadratic equation:
\[(V - 20)(V - 76) = 0\]
Since \(V = 76\) would give \(T = 20 - 38 = -18\text{ s}\) (which is impossible as \(T > 0\)), we must have:
\[V = 20 \quad \text{(shown)}\]

**(c)**
The acceleration in the first stage is given by:
\[a = \frac{V - 12}{8} = \frac{20 - 12}{8} = 1\text{ m s}^{-2}\]

**(d)**
Using the expression for \(T\):
\[T = 20 - \frac{20}{2} = 10\text{ seconds}\]

**(a)**
- G1: Correct shape for the 3 stages: trapezoidal profile starting at a non-zero velocity.
- G1: Correctly labeled coordinates/points on axes (\(12\), \(V\), \(8\), \(28\) or \(8+T\)).

**(b)**
- M1: Expressing the deceleration time \(t_3\) in terms of \(V\) (i.e., \(t_3 = V/2\)).
- M1: Attempting to write a distance equation using the area under the graph.
- A1: Correct distance expressions for all three stages in terms of \(V\).
- M1: Substituting \(T = 20 - V/2\) into the total distance equation and simplifying to a quadratic.
- A1: Solving the quadratic to show that \(V = 20\) is the only physically valid root.

**(c)**
- M1: Using \(a = (V-u)/t\).
- A1: Correct value of \(a = 1\).

**(d)**
- M1: Substituting \(V = 20\) into \(T = 20 - V/2\).
- A1: Correct value of \(T = 10\).

**(a)**
Since the force is constant, the acceleration \(\mathbf{a}\) is also constant:
\[\mathbf{a} = \frac{\mathbf{v}(4) - \mathbf{v}(0)}{4}\]
\[\mathbf{a} = \frac{(5\mathbf{i} - 4\mathbf{j}) - (-3\mathbf{i} + 2\mathbf{j})}{4} = \frac{8\mathbf{i} - 6\mathbf{j}}{4} = (2\mathbf{i} - 1.5\mathbf{j})\text{ m s}^{-2}\]

**(b)**
Using Newton's Second Law:
\[\mathbf{F} = m\mathbf{a} = 0.5(2\mathbf{i} - 1.5\mathbf{j}) = (1\mathbf{i} - 0.75\mathbf{j})\text{ N}\]

The magnitude of \(\mathbf{F}\) is:
\[|\mathbf{F}| = \sqrt{1^2 + (-0.75)^2} = \sqrt{1 + 0.5625} = \sqrt{1.5625} = 1.25\text{ N}\]

**(c)**
The displacement \(\mathbf{s}\) at \(t = 6\text{ s}\) relative to \(t = 0\) is:
\[\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\]
\[\mathbf{s} = 6(-3\mathbf{i} + 2\mathbf{j}) + \frac{1}{2}(2\mathbf{i} - 1.5\mathbf{j})(6^2)\]
\[\mathbf{s} = (-18\mathbf{i} + 12\mathbf{j}) + 18(2\mathbf{i} - 1.5\mathbf{j})\]
\[\mathbf{s} = (-18\mathbf{i} + 12\mathbf{j}) + (36\mathbf{i} - 27\mathbf{j}) = (18\mathbf{i} - 15\mathbf{j})\text{ m}\]

The distance from the starting position is the magnitude of the displacement:
\[|\mathbf{s}| = \sqrt{18^2 + (-15)^2} = \sqrt{324 + 225} = \sqrt{549} \approx 23.4\text{ m}\quad \text{(to 3 s.f.)}\]

**(d)**
The velocity at \(t = 6\text{ s}\) is:
\[\mathbf{v} = \mathbf{u} + \mathbf{a}t = (-3\mathbf{i} + 2\mathbf{j}) + 6(2\mathbf{i} - 1.5\mathbf{j}) = (9\mathbf{i} - 7\mathbf{j})\text{ m s}^{-1}\]

To find the bearing of the direction of motion, we look at the vector \(9\mathbf{i} - 7\mathbf{j}\), which points East and South:
Let \(\theta\) be the angle the vector makes with the south direction (measured clockwise from South):
\[\tan \theta = \frac{9}{7} \implies \theta \approx 52.13^{\circ}\]

Therefore, the bearing is:
\[180^{\circ} - 52.13^{\circ} = 127.87^{\circ} \approx 128^{\circ}\quad \text{(to the nearest degree)}\]

**(a)**
- M1: For attempting to find \(\mathbf{a}\) using \(\frac{\mathbf{v} - \mathbf{u}}{t}\).
- A1: Correct acceleration vector \((2\mathbf{i} - 1.5\mathbf{j})\text{ m s}^{-2}\).

**(b)**
- M1: For applying \(\mathbf{F} = m\mathbf{a}\).
- M1: For attempting to find the magnitude of their force vector.
- A1: Correct magnitude of \(1.25\text{ N}\).

**(c)**
- M1: For utilizing \(\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) with \(t = 6\).
- A1: Correct displacement vector \((18\mathbf{i} - 15\mathbf{j})\text{ m}\right).
- M1: Finding the magnitude of their displacement vector.
- A1: Correct distance of \)23.4\text{ m}\) (or \(3\sqrt{61}\)).

**(d)**
- M1: Finding the velocity vector at \(t = 6\) and attempting a trigonometric ratio for the direction.
- A1: Correct bearing of \(128^{\circ}\) (accept \(128\) or \(127.9\)).

**(a)**
Let \(R\) be the normal reaction force. Resolving perpendicular to the inclined plane:
\[R = mg \cos 30^{\circ} + H \sin 30^{\circ}\]
Since \(m = 5\text{ kg}\) and \(g = 9.8\text{ m s}^{-2}\flip:
\[R = 5(9.8)\cos 30^{\circ} + H \sin 30^{\circ}\]
\[R = 49 \left(\frac{\sqrt{3}}{2}\right) + 0.5H\]
\[R = 24.5\sqrt{3} + 0.5H \approx 42.4 + 0.5H\]

**(b)**
When \)H\) is at its maximum value, the particle \(P\) is on the point of slipping UP the inclined plane.
Therefore, the frictional force \(F\) acts DOWN the plane.
Since the motion is limiting:
\[F = F_{\max} = \mu R = 0.4(24.5\sqrt{3} + 0.5H)\]
\[F = 9.8\sqrt{3} + 0.2H\]

Resolving parallel to the inclined plane (upwards direction):
\[H \cos 30^{\circ} - mg \sin 30^{\circ} - F = 0\]
\[H \left(\frac{\sqrt{3}}{2}\right) - 5(9.8) \sin 30^{\circ} - (9.8\sqrt{3} + 0.2H) = 0\]
\[H \frac{\sqrt{3}}{2} - 24.5 - 9.8\sqrt{3} - 0.2H = 0\]
\[H \left(\frac{\sqrt{3}}{2} - 0.2\right) = 24.5 + 9.8\sqrt{3}\]

Using \(\frac{\sqrt{3}}{2} \approx 0.86603\) and \(9.8\sqrt{3} \approx 16.974\) :
\[H(0.86603 - 0.2) = 24.5 + 16.974\]
\[0.66603H = 41.474\]
\[H = \frac{41.474}{0.66603} \approx 62.3\text{ N} \quad \text{(to 3 s.f.)}\]

**(a)**
- M1: For resolving forces perpendicular to the plane.
- A1: Correct terms: \(mg \cos 30^{\circ}\) and \(H \sin 30^{\circ}\) acting in the same direction perpendicular to the plane.
- A1: Correct expression \(R = 24.5\sqrt{3} + 0.5H\) (or \(42.4 + 0.5H\)).

**(b)**
- M1: For stating or implying that \(F = \mu R\).
- M1: For setting up the equation of equilibrium parallel to the inclined plane.
- A1: Correct parallel terms: \(H \cos 30^{\circ}\), \(mg \sin 30^{\circ}\), and \(F\).
- M1: For realizing that \(F\) acts down the plane (opposing up-plane slip).
- A1: Correct substituted equation, e.g., \(H \cos 30^{\circ} - 5g \sin 30^{\circ} - 0.4 R = 0\).
- M1: For collecting terms of \(H\) and solving the linear equation.
- A1: Obtaining \(H = 62.3\text{ N}\) or \(62\text{ N}\).
- A1: Correct units and rounding (accept \(62\) or \(62.3\)).

**(a)**
Let the direction of motion of \(A\) before the collision be positive.
- Initial velocity of \(A\): \(u_A = 2u\)
- Initial velocity of \(B\): \(u_B = -u\) (since moving in the opposite direction)
- Final velocity of \(A\): \(v_A = -\frac{1}{3}u\) (since reversed)
- Final velocity of \(B\): \(v_B = \frac{4}{3}u\) (since reversed)

By Conservation of Linear Momentum:
\[m_A u_A + m_B u_B = m_A v_A + m_B v_B\]
\[(3m)(2u) + (km)(-u) = (3m)\left(-\frac{1}{3}u\right) + (km)\left(\frac{4}{3}u\right)\]
Divide through by \(mu\):
\[6 - k = -1 + \frac{4}{3}k\]
\[7 = \frac{7}{3}k \implies k = 3 \quad \text{(shown)}\]

**(b)**
The impulse \(I\) exerted by \(B\) on \(A\) is equal to the change in momentum of \(A\):
\[I = m_A v_A - m_A u_A\]
\[I = 3m\left(-\frac{1}{3}u\right) - 3m(2u) = -mu - 6mu = -7mu\]

The magnitude of the impulse is \(7mu\).

**(c)**
By modeling the bodies as particles:
1. We assume their mass acts at a single point.
2. We can neglect air resistance.
3. We assume they do not rotate or spin during the collision.

**(a)**
- M1: Attempting to use the Conservation of Linear Momentum.
- A1: Correct initial momentum expression: \(6mu - kmu\).
- A1: Correct final momentum expression: \(-mu + \frac{4}{3}kmu\).
- M1: Setting up the linear equation in \(k\).
- A1: Correctly showing \(k = 3\) with clear algebraic steps.

**(b)**
- M1: Attempting to use \(\text{Impulse} = \text{change in momentum}\) for particle \(A\).
- A1: Correct calculation showing \(-7mu\) or \(7mu\).
- A1: Stating the magnitude clearly as \(7mu\).

**(c)**
- B1: Mass acts at a single point.
- B1: Neglect rotation / spin.
- B1: Collisions occur along a single straight line / no dimensions to consider.

**(a)**
For runner \(P\), starting from rest (\(u = 0\)) and accelerating at \(a = 0.5\text{ m s}^{-2}\) to \(v = 4\text{ m s}^{-1}\):
\[v = u + at \implies 4 = 0 + 0.5t \implies t = 8\text{ seconds}\]

**(b)**
The distance traveled by \(P\) during acceleration is:
\[s_P = \frac{u+v}{2} t = \frac{0+4}{2} \times 8 = 16\text{ m}\]

**(c)**
Let \(s_P(t)\) and \(s_Q(t)\) be the distances traveled by \(P\) and \(Q\) from their respective starting positions at time \(t\).
They meet when:
\[s_P(t) + s_Q(t) = 150\]

For runner \(P\):
- For \(0 \le t \le 8\): \(s_P(t) = \frac{1}{2}(0.5)t^2 = 0.25t^2\)
- For \(t > 8\): \(s_P(t) = 16 + 4(t - 8) = 4t - 16\)

For runner \(Q\) (who starts at \(t = 2\)):
- Acceleration phase: \(Q\) accelerates from \(t=2\) at \(0.8\text{ m s}^{-2}\) until it reaches \(6\text{ m s}^{-1}\).
Time taken to reach \(6\text{ m s}^{-1}\) is:
\[t_{\text{acc}} = \frac{6}{0.8} = 7.5\text{ seconds}\]
So \(Q\) reaches constant speed at \(t = 2 + 7.5 = 9.5\text{ s}\).
The distance \(Q\) travels during acceleration is:
\[s_{Q\text{,acc}} = \frac{0+6}{2} \times 7.5 = 22.5\text{ m}\]
- For \(t > 9.5\):
\[s_Q(t) = 22.5 + 6(t - 9.5) = 6t - 57 + 22.5 = 6t - 34.5\]

Let's test if they meet before \(t = 9.5\):
At \(t = 9.5\):
\[s_P(9.5) = 4(9.5) - 16 = 22\text{ m}\]
\[s_Q(9.5) = 22.5\text{ m}\]
\[s_P(9.5) + s_Q(9.5) = 22 + 22.5 = 44.5\text{ m} < 150\text{ m}\]
Since \(44.5 < 150\), they must meet at a time \(t > 9.5\).

For \(t > 9.5\):
\[s_P(t) + s_Q(t) = 150\]
\[(4t - 16) + (6t - 34.5) = 150\]
\[10t - 50.5 = 150\]
\[10t = 200.5 \implies t = 20.05\text{ s}\]

**(a)**
- M1: For utilizing \(v = u + at\).
- A1: Correct time \(t = 8\text{ s}\).

**(b)**
- M1: For using a correct SUVAT equation to find distance (e.g., \(s = vt - 0.5at^2\) or \(s = 0.5(u+v)t\)).
- A1: Correct distance of \(16\text{ m}\).

**(c)**
- M1: Stating the condition for meeting: \(s_P + s_Q = 150\).
- M1: Finding the time \(Q\) reaches constant speed (\(t = 9.5\text{ s}\)) and the distance \(Q\) travels in this phase (\(22.5\text{ m}\)).
- A1: Establishing the distance equation for \(P\) when \(t > 8\) as \(s_P = 4t - 16\).
- A1: Establishing the distance equation for \(Q\) when \(t > 9.5\) as \(s_Q = 6t - 34.5\).
- M1: Setting up and solving the equation \((4t - 16) + (6t - 34.5) = 150\).
- A1: Finding \(10t = 200.5\).
- A1: Correct final answer \(t = 20.05\text{ s}\) (accept \(20\text{ s}\) or \(20.1\text{ s}\)).

(a) Let \(P\), \(V\), and \(F\) represent the sets of students playing Piano, Violin, and Flute respectively. The number in the intersection of all three is 8: \(n(P \cap V \cap F) = 8\). The number playing Piano and Violin only is \(20 - 8 = 12\). The number playing Violin and Flute only is \(15 - 8 = 7\). The number playing Piano and Flute only is \(12 - 8 = 4\). The number playing only Piano is \(65 - 12 - 8 - 4 = 41\). The number playing only Violin is \(45 - 12 - 8 - 7 = 18\). The number playing only Flute is \(40 - 4 - 8 - 7 = 21\). Total students inside the three circles = \(41 + 18 + 21 + 12 + 7 + 4 + 8 = 111\). The number playing none of these is \(120 - 111 = 9\). (b)(i) Piano but not Flute: These are students in the 'only Piano' region (41) plus the 'Piano and Violin only' region (12). Number of students = \(41 + 12 = 53\). Probability = \(\frac{53}{120}\). (b)(ii) At least two instruments: These are in the regions of intersections: \(12 + 4 + 7 + 8 = 31\). Probability = \(\frac{31}{120}\). (c) Given Violin, also plays Piano: \(\mathrm{P}(P|V) = \frac{n(P \cap V)}{n(V)} = \frac{12 + 8}{45} = \frac{20}{45} = \frac{4}{9}\).

Part (a): M1: For 3 overlapping circles with '8' in the center intersection. A1: Correct values for double intersections (12, 7, 4). A1: Correct values for single regions (41, 18, 21). A1: Correct value for the region outside the circles (9) with the box labelled. Part (b)(i): M1: For (41 + 12)/120. A1: Correct probability 53/120 (or 0.442). Part (b)(ii): M1: For sum of intersections divided by 120. A1: Correct probability 31/120 (or 0.258). Part (c): M1: Use of conditional probability formula with a denominator of 45. A1: Numerator is 20. A1: Correct probability 4/9 (or 0.444).

(a) Using the standard normal distribution: \(\mathrm{P}(X < 800) = 0.05 \implies \mathrm{P}(Z < \frac{800 - \mu}{\sigma}) = 0.05\). Since 0.05 is in the lower tail, the \(z\)-value is negative: \(\frac{800 - \mu}{\sigma} = -1.6449\). \(\mathrm{P}(X > 1200) = 0.15 \implies \mathrm{P}(Z > \frac{1200 - \mu}{\sigma}) = 0.15\). Since 0.15 is in the upper tail, the \(z\)-value is positive: \(\frac{1200 - \mu}{\sigma} = 1.0364\). (b) Rearranging the equations: \(\mu = 800 + 1.6449\sigma\) and \(\mu = 1200 - 1.0364\sigma\). Equating the two expressions: \(800 + 1.6449\sigma = 1200 - 1.0364\sigma \implies 2.6813\sigma = 400 \implies \sigma \approx 149.18 \approx 149\). Substituting back: \(\mu = 800 + 1.6449(149.18) = 1045.39 \approx 1050\). (c) \(\mathrm{P}(X > 1000) = \mathrm{P}(Z > \frac{1000 - 1045.39}{149.18}) = \mathrm{P}(Z > -0.3043) = \mathrm{P}(Z < 0.3043) \approx 0.620\).

Part (a): M1: Equating standardisation formula to a z-value for 0.05 (accept \(\pm 1.64\) or \(\pm 1.6449\)). A1: \(\frac{800 - \mu}{\sigma} = -1.6449\). M1: Equating standardisation formula to a z-value for 0.15 (accept \(\pm 1.04\) or \(\pm 1.0364\)). A1: \(\frac{1200 - \mu}{\sigma} = 1.0364\). Part (b): M1: Eliminating one variable from their two equations. M1: Solving to find a value for \(\sigma\) or \(\mu\). A1: \(\sigma \approx 149\). A1: \(\mu \approx 1050\). B1: Both answers given to 3 s.f. Part (c): M1: Standardising 1000 using their \(\mu\) and \(\sigma\). A1: Correct probability, awrt 0.620 (or 0.618 if using 1.64 and 1.04).

(a) \(S_{tt} = \sum t^2 - \frac{(\sum t)^2}{n} = 2720 - \frac{144^2}{8} = 128\). \(S_{ts} = \sum ts - \frac{\sum t \sum s}{n} = 2280 - \frac{144 \times 120}{8} = 120\). (b) First find \(S_{ss} = \sum s^2 - \frac{(\sum s)^2}{n} = 1950 - \frac{120^2}{8} = 150\). Now PMCC \(r = \frac{S_{ts}}{\sqrt{S_{tt} S_{ss}}} = \frac{120}{\sqrt{128 \times 150}} \approx 0.866\). (c) Yes, because \(r = 0.866\) is close to 1, which indicates a strong positive linear correlation. (d) \(b = \frac{S_{ts}}{S_{tt}} = \frac{120}{128} = 0.9375\). \(\bar{t} = 18\), \(\bar{s} = 15\). \(a = \bar{s} - b\bar{t} = 15 - 0.9375(18) = -1.875\). Thus, \(s = -1.875 + 0.9375t\). (e) When \(t = 20\), \(s = -1.875 + 0.9375(20) = 16.875\). Total sales = \(16.875 \times 100 = £1687.50\).

Part (a): M1: For correct formula used for \(S_{tt}\) or \(S_{ts}\). A1: \(S_{tt} = 128\). A1: \(S_{ts} = 120\). Part (b): M1: Correct calculation of \(S_{ss} = 150\). M1: Correct formula for \(r\) with their values substituted. A1: \(r \approx 0.866\). Part (c): B1: Yes/supports, with mention of strong positive correlation or value close to 1. Part (d): M1: Calculating \(b = S_{ts} / S_{tt}\). M1: Calculating \(a = \bar{s} - b\bar{t}\). A1: \(s = -1.875 + 0.9375t\). Part (e): B1: Correct estimate of sales, £1687.50 (accept £1690).

(a) Since the sum of all probabilities is 1: \(a + b + c + 0.1 = 1 \implies a + b + c = 0.9\). We are given \(\mathrm{P}(X \le 2) = a + b = 0.6\). Substituting this: \(0.6 + c = 0.9 \implies c = 0.3\). (b) Using \(\mathrm{E}(X) = 2.2\): \(\mathrm{E}(X) = 1(a) + 2(b) + 3(c) + 4(0.1) = 2.2\). Substituting \(c = 0.3\): \(a + 2b + 0.9 + 0.4 = 2.2 \implies a + 2b = 0.9\). We also have \(a + b = 0.6 \implies a = 0.6 - b\). Substituting this: \((0.6 - b) + 2b = 0.9 \implies b = 0.3\). Then \(a = 0.6 - 0.3 = 0.3\). (c) \(\mathrm{E}(X^2) = 1^2(0.3) + 2^2(0.3) + 3^2(0.3) + 4^2(0.1) = 0.3 + 1.2 + 2.7 + 1.6 = 5.8\). \(\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2 = 5.8 - 2.2^2 = 5.8 - 4.84 = 0.96\). (d) \(\mathrm{Var}(3X - 2) = 9 \mathrm{Var}(X) = 9 \times 0.96 = 8.64\).

Part (a): M1: Using sum of probabilities equals 1. M1: Using \(a + b = 0.6\). A1: Showing \(c = 0.3\) clearly. Part (b): M1: Setting up expectation equation. A1: Obtaining \(a + 2b = 0.9\). M1: Solving simultaneous equations. A1: Both \(a = 0.3\) and \(b = 0.3\). Part (c): M1: Formula for \(\mathrm{E}(X^2)\). A1: \(\mathrm{E}(X^2) = 5.8\). A1: \(\mathrm{Var}(X) = 0.96\). Part (d): B1: 8.64 (or \(9 \times\) their \(\mathrm{Var}(X)\)).

(a) Total frequency = 80, so median is at the 40th value. Cumulative frequencies are: 8, 23, 48. The median lies in class \(45 \le t < 50\). By linear interpolation: \(Q_2 = 45 + \frac{40 - 23}{25} \times 5 = 45 + 3.4 = 48.4\) minutes. (b) Midpoints (\(x\)): 35, 42.5, 47.5, 55, 70. \(\sum f x = 8(35) + 15(42.5) + 25(47.5) + 20(55) + 12(70) = 4045\). Mean = \(4045 / 80 = 50.5625 \approx 50.6\) minutes. (c) \(\mathrm{IQR} = Q_3 - Q_1 = 54.5 - 43.0 = 11.5\). Upper boundary = \(Q_3 + 1.5 \times \mathrm{IQR} = 54.5 + 1.5(11.5) = 71.75\). Since 75 > 71.75, the runner's time of 75 minutes is considered an outlier. (d) For class \(40 \le t < 45\): width 5 units is 1 cm (so 1 unit = 0.2 cm), frequency density is \(15 / 5 = 3\) (represented by 6 cm height, so height scale is 2). For class \(60 \le t < 80\): width 20 units is represented by \(20 \times 0.2 = 4\) cm; frequency density is \(12 / 20 = 0.6\), represented by height of \(0.6 \times 2 = 1.2\) cm.

Part (a): M1: For identifying median class and correct fractional part (40-23)/25. M1: Fully correct linear interpolation equation. A1: 48.4. Part (b): M1: Correct midpoints. M1: Correct calculation of sum of fx divided by 80. A1: 50.6. Part (c): M1: Calculating IQR = 11.5. A1: Upper boundary of 71.75. A1: Correct comparison of 75 to 71.75. Part (d): B1: Width = 4 cm. B1: Height = 1.2 cm.

(a) The tree diagram should show: Branches for Bag \(A\) with \(\mathrm{P}(A) = \frac{1}{3}\) and Bag \(B\) with \(\mathrm{P}(B) = \frac{2}{3}\). From \(A\), branches for Gold with \(\mathrm{P}(G|A) = \frac{3}{8}\) and Silver with \(\mathrm{P}(S|A) = \frac{5}{8}\). From \(B\), branches for Gold with \(\mathrm{P}(G|B) = \frac{4}{6} = \frac{2}{3}\) and Silver with \(\mathrm{P}(S|B) = \frac{1}{3}\). (b) \(\mathrm{P}(G) = \mathrm{P}(A) \mathrm{P}(G|A) + \mathrm{P}(B) \mathrm{P}(G|B) = (\frac{1}{3} \times \frac{3}{8}) + (\frac{2}{3} \times \frac{2}{3}) = \frac{1}{8} + \frac{4}{9} = \frac{41}{72}\). (c) \(\mathrm{P}(A|G) = \frac{\mathrm{P}(A \cap G)}{\mathrm{P}(G)} = \frac{1/8}{41/72} = \frac{9}{41}\). (d) To get two gold coins from the same bag: From Bag \(A\): \(\mathrm{P}(A \cap G_1 \cap G_2) = \frac{1}{3} \times \frac{3}{8} \times \frac{2}{7} = \frac{1}{28}\). From Bag \(B\): \(\mathrm{P}(B \cap G_1 \cap G_2) = \frac{2}{3} \times \frac{4}{6} \times \frac{3}{5} = \frac{4}{15}\). Total probability = \(\frac{1}{28} + \frac{4}{15} = \frac{127}{420}\).

Part (a): M1: First branches with correct probabilities 1/3 and 2/3. A1: Correct branches/probabilities for Bag A. A1: Correct branches/probabilities for Bag B. Part (b): M1: Identifying both paths. A1: Correct substitution. A1: 41/72. Part (c): M1: Correct conditional probability formula. A1: Correct substitution. A1: 9/41. Part (d): M1: Correct calculation of at least one path with conditional probability. A1: Correct total probability 127/420 (or awrt 0.302).

(a)(i) \(\mathrm{P}(W > 30) = \mathrm{P}(Z > \frac{30 - 24}{6}) = \mathrm{P}(Z > 1) = 1 - 0.8413 = 0.1587\). (a)(ii) \(\mathrm{P}(15 < W < 27) = \mathrm{P}(\frac{15 - 24}{6} < Z < \frac{27 - 24}{6}) = \mathrm{P}(-1.5 < Z < 0.5) = \Phi(0.5) - [1 - \Phi(1.5)] = 0.6915 - (1 - 0.9332) = 0.6247\). (b) \(\mathrm{P}(W < k) = 0.75 \implies \mathrm{P}(Z < \frac{k - 24}{6}) = 0.75 \implies \frac{k - 24}{6} = 0.6745 \implies k = 28.0\). (c) \(\mathrm{P}(W > 30 | W > 24) = \frac{\mathrm{P}(W > 30)}{\mathrm{P}(W > 24)} = \frac{0.1587}{0.5} = 0.3174\).

Part (a)(i): M1: For standardising 30. A1: 0.1587 (or awrt 0.159). Part (a)(ii): M1: For standardising 15 and 27. A1: Both z-values correct. M1: Correct method to find area between two z-values. A1: 0.6247 (or awrt 0.625). Part (b): M1: Standardising k and equating to a positive z-value. A1: Equation with 0.6745. A1: k = 28.0. Part (c): M1: Correct conditional probability statement. A1: 0.3174.