2025 Edexcel IAL Pure Mathematics (YPM01) 模擬試題連答案詳解

Substitute \( y = 2x + k \) into the equation for \( C \): \( x^2 + 2x(2x + k) + 3 = 0 \) which simplifies to \( 5x^2 + 2kx + 3 = 0 \). Since the line and curve do not intersect, this quadratic equation has no real roots. Therefore, the discriminant must be negative: \( b^2 - 4ac < 0 \) which gives \( (2k)^2 - 4(5)(3) < 0 \), leading to \( 4k^2 - 60 < 0 \), or \( k^2 < 15 \). Solving \( k^2 < 15 \) gives the final exact range of \( -\sqrt{15} < k < \sqrt{15} \).

M1: Attempts to substitute the linear equation into the quadratic curve equation to get an equation in one variable. A1: Obtains a correct simplified quadratic equation, e.g., \( 5x^2 + 2kx + 3 = 0 \). M1: Uses the discriminant condition \( b^2 - 4ac < 0 \) for no intersection. A1: Formulates the correct inequality in \( k \), e.g., \( 4k^2 - 60 < 0 \) or \( k^2 < 15 \). M1: Finds the critical values \( k = \pm\sqrt{15} \). A2.5: Correct final set of values in exact form, written as \( -\sqrt{15} < k < \sqrt{15} \) (or equivalent interval notation).

Find the gradient of \( AB \): \( m_{AB} = \frac{5 - (-3)}{8 - 2} = \frac{8}{6} = \frac{4}{3} \). Find the midpoint \( M \) of \( AB \): \( M = \left(\frac{2+8}{2}, \frac{-3+5}{2}\right) = (5, 1) \). The gradient of the perpendicular bisector is \( m_{\perp} = -\frac{1}{m_{AB}} = -\frac{3}{4} \). Using the point-gradient form of a straight line with point \( (5, 1) \) and gradient \( -\frac{3}{4} \): \( y - 1 = -\frac{3}{4}(x - 5) \) which simplifies to \( 4(y - 1) = -3(x - 5) \), giving \( 4y - 4 = -3x + 15 \), and thus \( 3x + 4y - 19 = 0 \).

M1: Attempts to find the gradient of \( AB \). A1: Correct gradient \( \frac{4}{3} \). M1: Attempts to find the midpoint of \( AB \). A1: Correct midpoint \( (5, 1) \). M1: Uses \( m_1 m_2 = -1 \) to find the perpendicular gradient. M1: Uses their midpoint and perpendicular gradient to write down an equation of the line. A1.5: Correct equation in the form \( ax + by + c = 0 \) with integer coefficients, e.g., \( 3x + 4y - 19 = 0 \).

Using the identity \( \cos^2\theta = 1 - \sin^2\theta \), we rewrite the equation as: \( 3(1 - \sin^2\theta) - 5\sin\theta - 1 = 0 \) which simplifies to \( 3\sin^2\theta + 5\sin\theta - 2 = 0 \). Factorising the quadratic in \( \sin\theta \): \( (3\sin\theta - 1)(\sin\theta + 2) = 0 \), which gives \( \sin\theta = \frac{1}{3} \) or \( \sin\theta = -2 \). Since \( -1 \le \sin\theta \le 1 \), the equation \( \sin\theta = -2 \) has no real solutions. For \( \sin\theta = \frac{1}{3} \), the solutions in the range are \( \theta = \arcsin(1/3) \approx 19.5^\circ \) and \( \theta = 180^\circ - 19.5^\circ = 160.5^\circ \).

M1: Substitutes \( \cos^2\theta = 1 - \sin^2\theta \) to obtain an equation in \( \sin\theta \). A1: Formulates the correct quadratic equation \( 3\sin^2\theta + 5\sin\theta - 2 = 0 \). M1: Attempts to solve the quadratic to find values of \( \sin\theta \). B1: Rejects the solution \( \sin\theta = -2 \) with brief justification. M1: Finds one correct angle (approx. \( 19.5^\circ \)). A2.5: Both \( 19.5^\circ \) and \( 160.5^\circ \) and no other solutions in the range.

First, rewrite the equation in index form: \( y = 2x^2 - 8x^{-1/2} + 3 \). Differentiate with respect to \( x \): \( \frac{dy}{dx} = 4x + 4x^{-3/2} \). At the point \( P(4, 31) \), where \( x = 4 \), the gradient of the tangent is \( m = 4(4) + 4(4)^{-3/2} = 16 + 4(\frac{1}{8}) = 16.5 = \frac{33}{2} \). The equation of the tangent is: \( y - 31 = \frac{33}{2}(x - 4) \) which simplifies to \( y - 31 = \frac{33}{2}x - 66 \), leading to \( y = \frac{33}{2}x - 35 \).

M1: Expresses \( \frac{8}{\sqrt{x}} \) as \( 8x^{-1/2} \). M1: Differentiates to find at least one term correct of their derivative. A1: Fully correct derivative \( \frac{dy}{dx} = 4x + 4x^{-3/2} \). M1: Substitutes \( x = 4 \) into their derivative to find the gradient of the tangent. A1: Obtains gradient \( \frac{33}{2} \). M1: Applies \( y - y_1 = m(x - x_1) \) using their gradient and point \( P(4, 31) \). A1.5: Correct tangent equation in the form \( y = mx + c \), i.e., \( y = \frac{33}{2}x - 35 \).

(a) Splitting the fraction: \( f'(x) = \frac{x^2}{x^2} - \frac{5x^{1/2}}{x^2} = 1 - 5x^{-3/2} \). (b) Integrate \( f'(x) \): \( f(x) = \int (1 - 5x^{-3/2}) dx = x - \frac{5x^{-1/2}}{-1/2} + C = x + 10x^{-1/2} + C \). Since the curve passes through \( (9, 2) \), substitute \( x = 9 \) and \( f(x) = 2 \): \( 2 = 9 + 10(9)^{-1/2} + C \) which gives \( 2 = 9 + \frac{10}{3} + C \), leading to \( C = 2 - \frac{37}{3} = -\frac{31}{3} \). Therefore, \( f(x) = x + \frac{10}{\sqrt{x}} - \frac{31}{3} \).

Part (a): M1: Attempts to split numerator term-by-term. A0.5: Obtains the given expression correctly. Part (b): M1: Attempts to integrate \( f'(x) \) by raising the power of \( x \) by 1. A1: Integrates \( 1 \to x \). A1: Integrates \( -5x^{-3/2} \to 10x^{-1/2} \). M1: Includes the constant of integration and substitutes \( x = 9 \), \( f(x) = 2 \) to find \( C \). A1: Obtains \( C = -\frac{31}{3} \). A1: Correct final expression for \( f(x) \).

(a) The transformation \( y = f(x + 2) \) is a translation 2 units to the left. The touch point \( (2, 0) \) translates to \( (0, 0) \) and the crossing point \( (-3, 0) \) translates to \( (-5, 0) \). Hence, the sketch touches the \( x \)-axis at the origin \( (0,0) \) and crosses the \( x \)-axis at \( (-5, 0) \). (b) Expand \( f(x) = x^3 - x^2 - 8x + 12 \). Differentiating gives \( f'(x) = 3x^2 - 2x - 8 \). Setting \( f'(x) = 0 \) yields \( (3x + 4)(x - 2) = 0 \). Since \( x = 2 \) is the minimum, the maximum is at \( x = -\frac{4}{3} \). Evaluating \( f\left(-\frac{4}{3}\right) = \left(-\frac{10}{3}\right)^2 \left(\frac{5}{3}\right) = \frac{500}{27} \). For this local maximum to lie on the \( x \)-axis, we must translate the graph vertically downwards by \( \frac{500}{27} \), so \( k = -\frac{500}{27} \).

Part (a): M1: Identifies the transformation as a translation of 2 units to the left. A1: Correct shape of a positive cubic graph with a local maximum to the left of the y-axis and a local minimum touching the origin. A1.5: Correctly labels/marks the coordinates of the intercepts on the axes: \( (0, 0) \) and \( (-5, 0) \). Part (b): M1: Attempts to differentiate \( f(x) \). A1: Correct derivative \( f'(x) = 3x^2 - 2x - 8 \). M1: Sets \( f'(x) = 0 \) and identifies \( x = -\frac{4}{3} \) as the local maximum. A1: Finds \( f\left(-\frac{4}{3}\right) = \frac{500}{27} \) and states \( k = -\frac{500}{27} \).

(a) The perimeter is \( P = 2r + r\theta = 30 \), so \( r\theta = 30 - 2r \). The area is \( A = \frac{1}{2}r^2\theta = \frac{1}{2}r(r\theta) = \frac{1}{2}r(30 - 2r) = 15r - r^2 \). (b) Differentiate \( A \) with respect to \( r \): \( \frac{dA}{dr} = 15 - 2r \). At a stationary point, \( \frac{dA}{dr} = 0 \implies 15 - 2r = 0 \implies r = 7.5 \). The maximum area is \( A = 15(7.5) - (7.5)^2 = 112.5 - 56.25 = 56.25 \) \( \text{cm}^2 \). To justify that it is a maximum, find \( \frac{d^2A}{dr^2} = -2 \). Since \( \frac{d^2A}{dr^2} < 0 \), the value is a maximum.

Part (a): M1: Formulates the perimeter equation \( 2r + r\theta = 30 \). M1: Writes down the area formula \( A = \frac{1}{2}r^2\theta \). A1.5: Eliminates \( \theta \) correctly and completes the proof to show \( A = 15r - r^2 \). Part (b): M1: Differentiates \( A \) to obtain \( \frac{dA}{dr} = 15 - 2r \). A1: Solves \( \frac{dA}{dr} = 0 \) to get \( r = 7.5 \). A1: Calculates the maximum area as \( 56.25 \). B1: Finds \( \frac{d^2A}{dr^2} = -2 \) and explains that since \( \frac{d^2A}{dr^2} < 0 \), the value is indeed a maximum.

(a) Equate equations: \( 9 - x^2 = x + 3 \implies x^2 + x - 6 = 0 \implies (x + 3)(x - 2) = 0 \). Thus, \( x = -3 \) or \( x = 2 \). The corresponding points are \( (-3, 0) \) and \( (2, 5) \). (b) The area of the region is \( \int_{-3}^{2} ((9 - x^2) - (x + 3)) dx = \int_{-3}^{2} (6 - x - x^2) dx = \left[ 6x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-3}^{2} \). At the upper limit \( x = 2 \), we have \( 12 - 2 - \frac{8}{3} = \frac{22}{3} \). At the lower limit \( x = -3 \), we have \( -18 - \frac{9}{2} + 9 = -\frac{27}{2} \). The area is \( \frac{22}{3} - (-\frac{27}{2}) = \frac{125}{6} \).

Part (a): M1: Equates equations of curve and line to form a quadratic. A1: Solves to find \( x = -3, 2 \). A1: Obtains the correct coordinates \( (-3, 0) \) and \( (2, 5) \). Part (b): M1: Formulates a correct integral representing the area. A1: Integrand simplified to \( 6 - x - x^2 \). M1: Integrates their expression, raising powers of \( x \) by 1. A1: Correct integration: \( 6x - \frac{x^2}{2} - \frac{x^3}{3} \). A0.5: Substitutes limits \( 2 \) and \( -3 \) correctly and simplifies to find the exact area as \( \frac{125}{6} \).

(a) The gradient of \(l_2\) is given by:
\[m_2 = \frac{-3 - (-1)}{1 - 0} = -2\]
Since the \(y\)-intercept is \((0, -1)\), the equation of \(l_2\) is:
\[y = -2x - 1\]

(b) Since \(l_1\) is perpendicular to \(l_2\), the gradient of \(l_1\) is:
\[m_1 = -\frac{1}{m_2} = -\frac{1}{-2} = \frac{1}{2}\]
The gradient of \(l_1\) can also be expressed using points \(A(-2, 5)\) and \(B(4, r)\):
\[m_1 = \frac{r - 5}{4 - (-2)} = \frac{r - 5}{6}\]
Equating these two expressions:
\[\frac{r - 5}{6} = \frac{1}{2} \implies r - 5 = 3 \implies r = 8\]

(c) Using \(r = 8\) and the gradient \(m_1 = \frac{1}{2}\), the equation of \(l_1\) is:
\[y - 5 = \frac{1}{2}(x - (-2)) \implies y - 5 = \frac{1}{2}x + 1 \implies y = \frac{1}{2}x + 6\]
To find the point of intersection, solve the equations of \(l_1\) and \(l_2\) simultaneously:
\[\frac{1}{2}x + 6 = -2x - 1\]
\[\frac{5}{2}x = -7 \implies x = -\frac{14}{5} = -2.8\]
Substitute \(x = -2.8\) back into the equation for \(l_2\):
\[y = -2(-2.8) - 1 = 5.6 - 1 = 4.6\]
Thus, the coordinates of the point of intersection are \((-2.8, 4.6)\) or \(\left(-\frac{14}{5}, \frac{23}{5}\right)\).

(a)
- M1: For an attempt to find the gradient of \(l_2\) using the given points \((1, -3)\) and \((0, -1)\).
- A1: \(y = -2x - 1\) (or equivalent, must be in the form \(y = mx + c\)).

(b)
- M1: Identifies the gradient of \(l_1\) as \(\frac{1}{2}\) by applying the perpendicular rule \(m_1 \cdot m_2 = -1\) to their gradient from (a).
- M1: Formulates an equation for the gradient of \(l_1\) using points \(A\) and \(B\) and equates it to their gradient.
- A1*: Completes the proof to show \(r = 8\) with no errors in the working. This is a given answer.

(c)
- M1: Finds a correct equation for \(l_1\), e.g., \(y - 5 = \frac{1}{2}(x + 2)\) or \(y = \frac{1}{2}x + 6\).
- M1: Equates their equations for \(l_1\) and \(l_2\) to find a value for \(x\) (or \(y\)).
- A1: Correct coordinates: \((-2.8, 4.6)\) or \(\left(-\frac{14}{5}, \frac{23}{5}\right)\).

(a) Express the derivative with fractional indices:
\[f'(x) = 3x^{-1/2} - 4x\]
Integrate to find \(f(x)\):
\[f(x) = \int \left(3x^{-1/2} - 4x\right) \text{d}x = \frac{3x^{1/2}}{1/2} - \frac{4x^2}{2} + C = 6x^{1/2} - 2x^2 + C\]
Since the point \(P(4, -15)\) lies on \(C\), substitute \(x = 4\) and \(y = -15\):
\[-15 = 6(4)^{1/2} - 2(4)^2 + C\]
\[-15 = 6(2) - 2(16) + C\]
\[-15 = 12 - 32 + C \implies -15 = -20 + C \implies C = 5\]
Thus, the equation of the curve is:
\[f(x) = 6\sqrt{x} - 2x^2 + 5\]

(b) The gradient of the tangent at \(P\) is \(f'(4)\):
\[f'(4) = \frac{3}{\sqrt{4}} - 4(4) = \frac{3}{2} - 16 = -\frac{29}{2}\]
The equation of the tangent at \((4, -15)\) is:
\[y - (-15) = -\frac{29}{2}(x - 4)\]
Multiply by 2 to clear the fraction:
\[2(y + 15) = -29(x - 4)\]
\[2y + 30 = -29x + 116\]
Rearrange into the required form:
\[29x + 2y - 86 = 0\]

(a)
- M1: Attempts to integrate the given expression. Correctly integrates at least one term: \(3x^{-1/2} \to kx^{1/2}\) or \(-4x \to -2x^2\).
- A1: Correct integration including the constant of integration \(C\), i.e., \(6x^{1/2} - 2x^2 + C\) (or equivalent).
- M1: Substitutes the coordinates of \(P(4, -15)\) into their integrated expression to find \(C\).
- A1: \(f(x) = 6\sqrt{x} - 2x^2 + 5\) (or equivalent simplified form, e.g. with \(x^{1/2}\)).

(b)
- M1: Substitutes \(x = 4\) into the given expression for \(f'(x)\) to find the gradient of the tangent.
- M1: Attempts to find the equation of the straight line using their gradient and the point \(P(4, -15)\).
- A1: Correct equation in the required form: \(29x + 2y - 86 = 0\) (or any integer multiple, e.g., \(-29x - 2y + 86 = 0\)).

(a) The sum of the first two terms is \(a + ar = 15\), which gives \(a(1+r) = 15\). The sum to infinity is \(\frac{a}{1-r} = 27\), which gives \(a = 27(1-r)\). Substituting \(a\) into the first equation: \(27(1-r)(1+r) = 15\), which simplifies to \(27(1-r^2) = 15\). Therefore, \(27 - 27r^2 = 15\), which gives \(27r^2 = 12\) as required. (b) From \(27r^2 = 12\), we get \(r^2 = \frac{12}{27} = \frac{4}{9}\). Since \(r > 0\), we have \(r = \frac{2}{3}\). Substituting this back into \(a = 27(1-r)\) gives \(a = 27\left(1 - \frac{2}{3}\right) = 9\).

(a) M1: Writes down the two equations \(a + ar = 15\) and \(\frac{a}{1-r} = 27\). M1: Eliminates \(a\) to obtain an equation in terms of \(r\). A1: Completes the proof to show \(27r^2 = 12\) with no errors seen. (b) M1: Solves \(27r^2 = 12\) to find \(r^2\) and takes the positive square root. A1: Obtains \(r = \frac{2}{3}\). M1: Substitutes their value of \(r\) into one of the equations to find \(a\). A1: Obtains \(a = 9\).

(a) Since \((x-2)\) is a factor, \(f(2) = 0\). This gives \(2(2)^3 + a(2)^2 + b(2) - 10 = 0 \implies 16 + 4a + 2b - 10 = 0 \implies 4a + 2b = -6 \implies 2a + b = -3\). Since the remainder when divided by \((x+1)\) is \(-12\), \(f(-1) = -12\). This gives \(2(-1)^3 + a(-1)^2 + b(-1) - 10 = -12 \implies -2 + a - b - 10 = -12 \implies a - b = 0 \implies a = b\). Substituting \(a = b\) into \(2a + b = -3\) gives \(3a = -3\), so \(a = -1\) and \(b = -1\). (b) Substituting the values of \(a\) and \(b\), \(f(x) = 2x^3 - x^2 - x - 10\). Since \((x-2)\) is a factor, we can write \(2x^3 - x^2 - x - 10 = (x-2)(2x^2 + cx + 5)\). Comparing the \(x^2\) term: \(-4 + c = -1 \implies c = 3\). Thus, \(f(x) = (x-2)(2x^2 + 3x + 5)\). Since the discriminant of \(2x^2 + 3x + 5\) is \(3^2 - 4(2)(5) = 9 - 40 = -31 < 0\), it cannot be factorised further over the real numbers.

(a) M1: Uses the factor theorem to set \(f(2) = 0\) and obtain a linear equation in \(a\) and \(b\). M1: Uses the remainder theorem to set \(f(-1) = -12\) and obtain a second linear equation in \(a\) and \(b\). M1: Solves the simultaneous equations to find \(a\) and \(b\). A1: Obtains \(a = -1\). A1: Obtains \(b = -1\). (b) M1: Attempts to divide or write \(f(x)\) as \((x-2)(2x^2 + cx + d)\) where \(d = 5\) or \(2d = -10\). A1: Finds the correct quadratic factor \(2x^2 + 3x + 5\). A1: States the final fully factorised form \((x-2)(2x^2 + 3x + 5)\).

(a) To find where the curve intersects the \(x\)-axis, set \(y = 0\): \(6x^{\frac{1}{2}} - 2x^{\frac{3}{2}} = 0 \implies 2x^{\frac{1}{2}}(3 - x) = 0\). This gives \(x = 0\) (the origin) or \(x = 3\). Since \(3 > 0\), the curve intersects the positive \(x\)-axis at the point \(A(3, 0)\). (b) The area is given by \(\int_{0}^{3} \left(6x^{\frac{1}{2}} - 2x^{\frac{3}{2}}\right) dx\). Integrating term by term: \(\int \left(6x^{\frac{1}{2}} - 2x^{\frac{3}{2}}\right) dx = \left[ 6 \left(\frac{2}{3}\right) x^{\frac{3}{2}} - 2 \left(\frac{2}{5}\right) x^{\frac{5}{2}} \right]_{0}^{3} = \left[ 4x^{\frac{3}{2}} - \frac{4}{5}x^{\frac{5}{2}} \right]_{0}^{3}\). Substituting the upper limit \(x = 3\): \(4(3)^{\frac{3}{2}} - \frac{4}{5}(3)^{\frac{5}{2}} = 4(3\sqrt{3}) - \frac{4}{5}(9\sqrt{3}) = 12\sqrt{3} - \frac{36}{5}\sqrt{3} = \frac{60 - 36}{5}\sqrt{3} = \frac{24\sqrt{3}}{5}\). Substituting the lower limit \(x = 0\) gives 0. Thus, the exact area is \(\frac{24\sqrt{3}}{5}\).

(a) M1: Sets \(y = 0\) and attempts to factorise the expression. A1: Obtains \(x = 3\) and states coordinates of \(A\). (b) M1: Attempts to integrate \(y\) with respect to \(x\), raising at least one power by 1. A1: Obtains the correct integrated term \(4x^{\frac{3}{2}}\). A1: Obtains the correct integrated term \(-\frac{4}{5}x^{\frac{5}{2}}\). M1: Substitutes the limits 3 and 0 into their integrated expression. A1: Obtains the exact area of \(\frac{24\sqrt{3}}{5}\) or an equivalent exact simplified form.

(a) The volume of the box is given by \(V = (2x)(x)(y) = 2x^2 y = 36\), which gives \(y = \frac{18}{x^2}\). The surface area of the open-topped box consists of one base of area \(2x \times x = 2x^2\), two sides of area \(2x \times y = 2xy\), and two sides of area \(x \times y = xy\). Thus, \(S = 2x^2 + 2(2xy) + 2(xy) = 2x^2 + 6xy\). Substituting \(y = \frac{18}{x^2}\) into the surface area equation: \(S = 2x^2 + 6x\left(\frac{18}{x^2}\right) = 2x^2 + \frac{108}{x}\) as required. (b) Differentiating \(S\) with respect to \(x\) gives \(\frac{dS}{dx} = 4x - \frac{108}{x^2}\). Setting \(\frac{dS}{dx} = 0\) to find the stationary point: \(4x - \frac{108}{x^2} = 0 \implies 4x^3 = 108 \implies x^3 = 27 \implies x = 3\). Substituting \(x = 3\) back into the equation for \(S\): \(S = 2(3)^2 + \frac{108}{3} = 18 + 36 = 54\). To justify that it is a minimum, find the second derivative: \(\frac{d^2S}{dx^2} = 4 + \frac{216}{x^3}\). When \(x = 3\), \(\frac{d^2S}{dx^2} = 4 + \frac{216}{27} = 4 + 8 = 12\). Since \(\frac{d^2S}{dx^2} > 0\), the surface area \(S = 54\text{ cm}^2\) is indeed a minimum.

(a) M1: Writes down a correct formula for volume \(V = 2x^2 y = 36\) and expresses \(y\) in terms of \(x\). M1: Writes down a correct expression for the surface area \(S = 2x^2 + 6xy\). A1: Eliminates \(y\) to obtain the given expression for \(S\) showing all steps clearly. (b) M1: Differentiates \(S\) with respect to \(x\) to obtain \(\frac{dS}{dx} = Ax - Bx^{-2}\). A1: Obtains the correct derivative \(4x - \frac{108}{x^2}\). M1: Sets their derivative to 0 and solves for \(x\) to find \(x = 3\). A1: Finds the minimum surface area \(S = 54\). M1: Finds the second derivative and evaluates it at \(x = 3\) to show it is positive, confirming a minimum.

Using the power law of logarithms, \(2\log_3(x-2) = \log_3(x-2)^2\). The equation becomes \(\log_3(x-2)^2 - \log_3(x+4) = 1\). Using the subtraction law of logarithms, \(\log_3\left(\frac{(x-2)^2}{x+4}\right) = 1\). Converting this from logarithmic to exponential form, \(\frac{(x-2)^2}{x+4} = 3^1 = 3\). Expanding and rearranging: \((x-2)^2 = 3(x+4) \implies x^2 - 4x + 4 = 3x + 12 \implies x^2 - 7x - 8 = 0\). Factorising the quadratic equation: \((x-8)(x+1) = 0\), which gives solutions \(x = 8\) or \(x = -1\). Since the original equation requires \(x > 2\) for the term \(\log_3(x-2)\) to be defined, we discard \(x = -1\). Thus, the only valid solution is \(x = 8\).

M1: Applies the power law of logarithms to write \(2\log_3(x-2) = \log_3(x-2)^2\). M1: Applies the subtraction law of logarithms to combine the terms into a single logarithm. M1: Correctly removes the logarithm by writing \(\frac{(x-2)^2}{x+4} = 3\). A1: Expands and simplifies to obtain the correct quadratic equation \(x^2 - 7x - 8 = 0\). M1: Solves the quadratic equation to find two roots. A1: Identifies \(x = 8\) as a solution. A1: Discards \(x = -1\) with reference to the domain \(x > 2\) to leave \(x = 8\) as the unique solution.

(a) The radius \(r\) of the circle is the distance between \(P(4, -3)\) and \(Q(8, 0)\): \(r = \sqrt{(8-4)^2 + (0 - (-3))^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5\). Therefore, the equation of the circle \(C\) is \((x-4)^2 + (y+3)^2 = 5^2\), which simplifies to \((x-4)^2 + (y+3)^2 = 25\). (b) The gradient of the radius \(PQ\) is \(m_{PQ} = \frac{0 - (-3)}{8 - 4} = \frac{3}{4}\). Since the tangent is perpendicular to the radius, the gradient of the tangent at \(Q\) is \(m = -\frac{1}{m_{PQ}} = -\frac{4}{3}\). Using the point-gradient formula for the tangent line at \(Q(8, 0)\): \(y - 0 = -\frac{4}{3}(x - 8) \implies 3y = -4(x - 8) \implies 3y = -4x + 32\). Rearranging this into the required form: \(4x + 3y - 32 = 0\).

(a) M1: Uses the distance formula to find the radius or radius squared of the circle. A1: Obtains \(r^2 = 25\) or \(r = 5\). A1: Writes down the correct equation of the circle \((x-4)^2 + (y+3)^2 = 25\). (b) M1: Calculates the gradient of the radius \(PQ\). A1: Obtains the correct gradient \(3/4\). M1: Finds the negative reciprocal gradient to get the gradient of the tangent. M1: Uses their tangent gradient and point \(Q(8,0)\) to form the equation of the line. A1: Obtains the correct equation in the form \(4x + 3y - 32 = 0\).

Using the identity \(\sin^2\theta = 1 - \cos^2\theta\), substitute into the equation: \(6(1 - \cos^2\theta) - \cos\theta - 4 = 0 \implies 6 - 6\cos^2\theta - \cos\theta - 4 = 0 \implies 6\cos^2\theta + \cos\theta - 2 = 0\). Let \(y = \cos\theta\), then the quadratic equation is \(6y^2 + y - 2 = 0\). Factorising this gives \((2y - 1)(3y + 2) = 0\), which has solutions \(y = \frac{1}{2}\) or \(y = -\frac{2}{3}\). Case 1: \(\cos\theta = \frac{1}{2}\). The solutions in the interval \(0 \le \theta < 360^{\circ}\) are \(\theta = 60^{\circ}\) and \(\theta = 360^{\circ} - 60^{\circ} = 300^{\circ}\). Case 2: \(\cos\theta = -\frac{2}{3}\). The basic angle is \(\arccos\left(\frac{2}{3}\right) \approx 48.19^{\circ}\). Since the cosine is negative, the solutions are in the second and third quadrants: \(\theta = 180^{\circ} - 48.19^{\circ} = 131.8^{\circ}\) (to 1 d.p.) and \(\theta = 180^{\circ} + 48.19^{\circ} = 228.2^{\circ}\) (to 1 d.p.). Thus, the solutions are \(\theta = 60.0^{\circ}, 131.8^{\circ}, 228.2^{\circ}, 300.0^{\circ}\).

M1: Uses the identity \(\sin^2\theta = 1 - \cos^2\theta\) to rewrite the equation in terms of \(\cos\theta\) only. A1: Obtains the correct quadratic equation \(6\cos^2\theta + \cos\theta - 2 = 0\). M1: Solves the quadratic equation to find values for \(\cos\theta\). A1: Obtains \(\cos\theta = \frac{1}{2}\) and \(\cos\theta = -\frac{2}{3}\). M1: Finds both solutions for \(\cos\theta = \frac{1}{2}\) (60 and 300). A1: Finds one correct solution for \(\cos\theta = -\frac{2}{3}\) (either 131.8 or 228.2). A1: Finds the other correct solution for \(\cos\theta = -\frac{2}{3}\) (both to 1 d.p.).

(a) For any real numbers \(x\) and \(y\), the square of their difference must be non-negative: \((x-y)^2 \ge 0\). Expanding the left-hand side gives \(x^2 - 2xy + y^2 \ge 0\). Adding \(2xy\) to both sides of the inequality gives \(x^2 + y^2 \ge 2xy\) as required. (b) To disprove the statement, we need a counterexample where \(x > y\) but \(x^2 \le y^2\). Let \(x = 1\) and \(y = -2\). Since \(1 > -2\), the condition \(x > y\) is met. However, \(x^2 = 1^2 = 1\) and \(y^2 = (-2)^2 = 4\). Since \(1\) is not greater than \(4\), \(x^2 > y^2\) is false. Thus, the statement is false.

(a) M1: Starts with the true statement \((x-y)^2 \ge 0\) for all real \(x, y\). A1: Correctly expands to obtain \(x^2 - 2xy + y^2 \ge 0\). A1: Completes the proof by adding \(2xy\) to both sides with no errors. (b) M1: Chooses a value for \(x\) and a value for \(y\) where \(x > y\) and at least one value is negative. A1: Evaluates \(x^2\) and \(y^2\) for their chosen values. A1: Clearly explains why their chosen values form a counterexample, showing that \(x > y\) but \(x^2 \le y^2\).

(a) The sum of the first two terms is \( S_2 = a + ar = a(1 + r) = 15 \). The sum to infinity is \( S_{\infty} = \frac{a}{1 - r} = 27 \), which gives \( a = 27(1 - r) \). Substituting this into the first equation yields \( 27(1 - r)(1 + r) = 15 \), so \( 27(1 - r^2) = 15 \). Thus, \( 1 - r^2 = \frac{15}{27} = \frac{5}{9} \), which simplifies to \( r^2 = \frac{4}{9} \). Since \( r > 0 \Offset, we have \) r = \frac{2}{3} \). (b) Substituting \( r = \frac{2}{3} \) into \( a = 27(1 - r) \) gives \( a = 27\left(1 - \frac{2}{3}\right) = 9 \). (c) The sum of the first 4 terms is \( S_4 = \frac{a(1 - r^4)}{1 - r} = \frac{9\left(1 - \left(\frac{2}{3}\right)^4\right)}{1 - \frac{2}{3}} = 27\left(1 - \frac{16}{81}\right) = \frac{65}{3} \). The difference between the sum to infinity and the sum of the first 4 terms is \( S_{\infty} - S_4 = 27 - \frac{65}{3} = \frac{16}{3} \).

(a) M1: Attempts to write down two correct equations for the sum of the first two terms and the sum to infinity, e.g., \( a + ar = 15 \) and \( \frac{a}{1-r} = 27 \). M1: Substitutes one equation into the other to eliminate \( a \). A1: Obtains a correct simplified equation in \( r \), such as \( r^2 = \frac{4}{9} \). A1*: Reaches \( r = \frac{2}{3} \) with no errors or omissions, and explicitly references the condition \( r > 0 \) to justify rejecting \( r = -\frac{2}{3} \). (b) B1: \( a = 9 \) (cao). (c) M1: Attempts to find the sum of the first 4 terms \( S_4 \) using the correct formula with their value of \( a \) and \( r = \frac{2}{3} \), or attempts to calculate \( S_{\infty} \times r^4 \). A1: Obtains \( \frac{16}{3} \) or equivalent exact fraction.

(a) To find the intersection points, equate the two equations: \( 6x^{\frac{1}{2}} - x^{\frac{3}{2}} = 2\sqrt{2}x \). For \( x = 0 \), both equations yield \( y = 0 \), so the origin is an intersection point. For \( x > 0 \), divide both sides by \( x^{\frac{1}{2}} \) to get \( 6 - x = 2\sqrt{2}x^{\frac{1}{2}} \). Letting \( u = x^{\frac{1}{2}} \) gives the quadratic equation \( u^2 + 2\sqrt{2}u - 6 = 0 \). Solving via the quadratic formula: \( u = \frac{-2\sqrt{2} \pm \sqrt{8 - 4(1)(-6)}}{2} = \frac{-2\sqrt{2} \pm \sqrt{32}}{2} = \frac{-2\sqrt{2} \pm 4\sqrt{2}}{2} \). Since \( u > 0 \), we have \( u = \sqrt{2} \), which implies \( x = u^2 = 2 \). (b) The area is given by the integral \( \int_{0}^{2} \left(6x^{\frac{1}{2}} - x^{\frac{3}{2}} - 2\sqrt{2}x\right) dx = \left[ 4x^{\frac{3}{2}} - \frac{2}{5}x^{\frac{5}{2}} - \sqrt{2}x^2 \right]_{0}^{2} \). Evaluating this at the limits: at \( x = 2 \), \( 4(2^{\frac{3}{2}}) - \frac{2}{5}(2^{\frac{5}{2}}) - \sqrt{2}(2^2) = 4(2\sqrt{2}) - \frac{2}{5}(4\sqrt{2}) - 4\sqrt{2} = 8\sqrt{2} - \frac{8}{5}\sqrt{2} - 4\sqrt{2} = \frac{12}{5}\sqrt{2} \). At \( x = 0 \), the value is 0. Thus, the area is \( \frac{12}{5}\sqrt{2} \), so \( k = \frac{12}{5} \) (or 2.4).

(a) M1: Equates the curve and line equations: \( 6x^{\frac{1}{2}} - x^{\frac{3}{2}} = 2\sqrt{2}x \). M1: Attempts to solve the equation, e.g., by substituting \( u = x^{\frac{1}{2}} \) to obtain a quadratic equation in \( u \). (Alternatively: M1 for verifying both points by substituting \( x = 0 \) and \( x = 2 \) into both equations). A1: Completes the algebraic proof to show that \( x = 0 \) and \( x = 2 \) are the intersection points. (b) M1: Attempts to set up the integration of the difference of the functions, i.e., \( \int (6x^{\frac{1}{2}} - x^{\frac{3}{2}} - 2\sqrt{2}x) dx \) with limits \( 0 \) and \( 2 \). M1: Integrates at least two terms correctly (increasing power by 1). A1: Fully correct integration: \( 4x^{\frac{3}{2}} - \frac{2}{5}x^{\frac{5}{2}} - \sqrt{2}x^2 \). M1: Substitutes the limits \( 2 \) and \( 0 \) into their integrated expression. A1: Obtains \( \frac{12}{5}\sqrt{2} \) or equivalent exact form.

(a) Let \(y = \frac{2x+3}{x-1}\).
\(y(x-1) = 2x+3 \implies xy - y = 2x+3\)
\(xy - 2x = y + 3 \implies x(y-2) = y+3\)
\(x = \t\t\frac{y+3}{y-2}\).
Therefore, \(\mathrm{f}^{-1}(x) = \frac{x+3}{x-2}\).
To find the domain of \(\mathrm{f}^{-1}\), we find the range of \(\mathrm{f}\) for \(x > 1\).
As \(x \to \infty\), \(\mathrm{f}(x) \to 2\).
As \(x \to 1^+\), \(\mathrm{f}(x) \to \infty\).
Since \(\mathrm{f}(x)\) is decreasing and continuous, the range of \(\mathrm{f}\) is \(y > 2\).
So the domain of \(\mathrm{f}^{-1}\) is \(x > 2\).

(b) \(\mathrm{fg}(x) = \mathrm{f}(x^2 - 2) = \frac{2(x^2-2)+3}{(x^2-2)-1} = \frac{2x^2-4+3}{x^2-3} = \frac{2x^2-1}{x^2-3}\).

(c) Set \(\mathrm{fg}(x) = 3\):
\(\frac{2x^2-1}{x^2-3} = 3 \implies 2x^2-1 = 3x^2-9\)
\(x^2 = 8 \implies x = \pm \sqrt{8} = \pm 2\sqrt{2}\).
Since the domain of \(\mathrm{g}\) is \(x > 2\), we must have \(x = 2\sqrt{2}\) (as \(-2\sqrt{2} \le 2\)).

(a)
- M1: Attempts to make \(x\) the subject of \(y = \frac{2x+3}{x-1}\).
- A1: Obtains \(x = \frac{y+3}{y-2}\) or equivalent.
- A1: Correct expression \(\mathrm{f}^{-1}(x) = \frac{x+3}{x-2}\).
- B1: Correct domain \(x > 2\) (accept \(\mathrm{f}^{-1}(x) > 2\) if clearly identified as domain).

(b)
- M1: Substitutes \(\mathrm{g}(x)\) into \(\mathrm{f}(x)\).
- A1: Correct substitution \(\frac{2(x^2-2)+3}{(x^2-2)-1}\).
- A1: Fully simplified expression \(\frac{2x^2-1}{x^2-3}\).

(c)
- M1: Sets their \(\mathrm{fg}(x) = 3\) and attempts to solve for \(x^2\).
- A1.375: Obtains \(x = 2\sqrt{2}\) (or \(\sqrt{8}\)) only, explicitly rejecting the negative root due to the domain \(x > 2\).

(a) The graph is a V-shape.
- Vertex: occurs when \(2x - 5 = 0 \implies x = 2.5\). At this point, \(y = -3\). So, the vertex is at \((2.5, -3)\).
- \(y\)-intercept: occurs when \(x = 0 \implies y = |-5| - 3 = 2\). So, \((0, 2)\).
- \(x\)-intercepts: occur when \(|2x - 5| - 3 = 0 \implies |2x - 5| = 3\).
Either \(2x - 5 = 3 \implies x = 4\) or \(2x - 5 = -3 \implies x = 1\). So, \((1, 0)\) and \((4, 0)\).

(b) To solve \(|2x - 5| - 3 < \frac{1}{2}x + 1\), we first find the intersection points:
\(|2x - 5| = \frac{1}{2}x + 4\)

Case 1 (positive branch):
\(2x - 5 = \frac{1}{2}x + 4 \implies \frac{3}{2}x = 9 \implies x = 6\)

Case 2 (negative branch):
\(-(2x - 5) = \frac{1}{2}x + 4 \implies -2x + 5 = \frac{1}{2}x + 4 \implies \frac{5}{2}x = 1 \implies x = 0.4\)

From the graph, the line \(y = \frac{1}{2}x + 1\) lies above the V-shaped curve between these two intersection points.
Thus, the solution set is \(0.4 < x < 6\).

(a)
- B1: Correct V-shape graph in quadrants 1, 3, and 4.
- B1: Vertex coordinates \((2.5, -3)\) correctly labelled.
- B1: \(y\)-intercept at \((0, 2)\) correctly labelled.
- B1: \(x\)-intercepts at \((1, 0)\) and \((4, 0)\) correctly labelled.

(b)
- M1: Attempts to solve the positive branch equation \(2x - 5 = \frac{1}{2}x + 4\).
- A1: Obtains \(x = 6\).
- M1: Attempts to solve the negative branch equation \(-(2x - 5) = \frac{1}{2}x + 4\).
- A1: Obtains \(x = 0.4\) (or \(\frac{2}{5}\)).
- A1.375: Combines intervals correctly to give \(0.4 < x < 6\) (or equivalent interval notation).

(a) Using double angle formula: \
\(\sin 4\theta = 2\sin 2\theta \cos 2\theta\)

LHS = \(\frac{2\sin 2\theta + 2\sin 2\theta \cos 2\theta}{2\sin 2\theta - 2\sin 2\theta \cos 2\theta}\)

Factorising \(2\sin 2\theta\) from the numerator and denominator:
LHS = \(\frac{2\sin 2\theta (1 + \cos 2\theta)}{2\sin 2\theta (1 - \cos 2\theta)} = \frac{1 + \cos 2\theta}{1 - \cos 2\theta}\)

Using double angle identities for cosine:
\(\cos 2\theta = 2\cos^2\theta - 1 \implies 1 + \cos 2\theta = 2\cos^2\theta\)
\(\cos 2\theta = 1 - 2\sin^2\theta \implies 1 - \cos 2\theta = 2\sin^2\theta\)

Substitute these back:
LHS = \(\frac{2\cos^2\theta}{2\sin^2\theta} = \frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta\) = RHS.

(b) Using the identity proved in part (a), the equation simplifies to:
\(\cot^2 x = 5 - \csc x\)

We know that \(\cot^2 x = \csc^2 x - 1\).
\(\csc^2 x - 1 = 5 - \csc x \implies \csc^2 x + \csc x - 6 = 0\)

Factorising the quadratic in terms of \(\csc x\):
\((\csc x + 3)(\csc x - 2) = 0\)

This gives:
\(\csc x = -3 \implies \sin x = -\frac{1}{3}\)
\(\csc x = 2 \implies \sin x = \frac{1}{2}\)

For \(0 < x < \pi\), \(\sin x\) must be positive. Therefore, \(\sin x = -\frac{1}{3}\) has no solutions in this range.
For \(\sin x = \frac{1}{2}\):
\(x = \frac{\pi}{6}\) or \(x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\).

(a)
- M1: Uses double angle formula \(\text{sin } 4\theta = 2\sin 2\theta \cos 2\theta\).
- M1: Factorises the numerator and denominator to get \(\frac{1 + \cos 2\theta}{1 - \cos 2\theta}\).
- M1: Uses double angle formulas for \(\cos 2\theta\) to express \(1 + \cos 2\theta\) as \(2\cos^2\theta\) and \(1 - \cos 2\theta\) as \(2\sin^2\theta\).
- A1*: Completely correct proof with no errors showing identity holds.

(b)
- M1: Uses the identity from (a) to rewrite the equation as \(\cot^2 x = 5 - \csc x\).
- M1: Applies identity \(\cot^2 x = \csc^2 x - 1\) to form a quadratic in \(\csc x\).
- A1: Obtains the correct quadratic equation \(\csc^2 x + \csc x - 6 = 0\) and solves to find \(\csc x = 2\) or \(\csc x = -3\).
- B1: Explains why \(\sin x = -\frac{1}{3}\) has no solutions in the interval \(0 < x < \pi\).
- A1.375: Correctly obtains both \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\).

(a) \(R\cos(\theta + \alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha\).
Comparing coefficients with \(7\cos\theta - 24\sin\theta\):
\(R\cos\alpha = 7\)
\(R\sin\alpha = 24\)

\(R = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = 25\).
\(\tan\alpha = \frac{24}{7} \implies \alpha = \arctan\left(\frac{24}{7}\right) \approx 1.287002 \approx 1.2870\) rad.
So, \(\mathrm{f}(\theta) = 25\cos(\theta + 1.2870)\).

(b) (i) \(H(\theta) = 30 + 25\cos(\theta + 1.2870)\).
The maximum value of \(\cos(\theta + 1.2870)\) is 1.
Maximum height is \(30 + 25 = 55\) metres.

(ii) The maximum height occurs when \(\cos(\theta + 1.2870) = 1\).
\(\theta + 1.2870 = 2k\pi, \quad k \in \mathbb{Z}\).
For the smallest \(\theta \ge 0\):
\(\theta + 1.2870 = 2\pi \implies \theta = 2\pi - 1.2870 \approx 6.2832 - 1.2870 = 4.996 \approx 5.00\) rad (to 3 s.f.).

(c) Set \(H(\theta) = 15\):
\(30 + 25\cos(\theta + 1.2870) = 15\)
\(25\cos(\theta + 1.2870) = -15\)
\(\cos(\theta + 1.2870) = -0.6\)

Let \(\theta + 1.2870 = \arccos(-0.6) \approx 2.2143\).
\(\theta = 2.2143 - 1.2870 = 0.9273 \approx 0.927\) rad (to 3 s.f.).

(a)
- B1: \(R = 25\).
- M1: Attempts \(\tan\alpha = \pm \frac{24}{7}\).
- A1: \(̅\alpha = 1.2870\).

(b)
- B1: Correct maximum height of 55 metres.
- M1: Set \(\theta + 1.2870 = 2\pi\) (or equivalent rotation value).
- A1.375: Correct value of \(\theta \approx 5.00\) (accept 4.996 or 5.00).

(c)
- M1: Equates expression to 15 and makes \(\cos(\theta + \alpha)\) the subject.
- M1: Finds the inverse cosine to get \(\approx 2.21\) and subtracts their \(\alpha\).
- A1: Correct answer \(\theta \approx 0.927\) only.

(a) \(\mathrm{f}(1.5) = \mathrm{e}^{0} + 4.5 - 8 = -2.5\)
\(\mathrm{f}(2.0) = \mathrm{e}^{1} + 6 - 8 = \mathrm{e} - 2 \approx 0.7183\)
Since there is a sign change and \(\mathrm{f}(x)\) is continuous on \([1.5, 2.0]\), there is at least one root \(\alpha\) in this interval.

(b) \(\mathrm{e}^{2x-3} + 3x - 8 = 0 \implies \mathrm{e}^{2x-3} = 8 - 3x\)
Taking natural logarithms on both sides:
\(2x - 3 = \ln(8 - 3x)\)
\(2x = 3 + \ln(8 - 3x)\)
\(x = \frac{1}{2}\left(3 + \ln(8 - 3x)\right)\).

(c) Using \(x_0 = 1.7\):
\(x_1 = \frac{1}{2}\left(3 + \ln(8 - 3(1.7))\right) = \frac{1}{2}\left(3 + \ln(2.9)\right) \approx 2.0324\)
\(x_2 = \frac{1}{2}\left(3 + \ln(8 - 3(2.03235))\right) \approx 1.8217\)
\(x_3 = \frac{1}{2}\left(3 + \ln(8 - 3(1.82172))\right) \approx 1.9651\).

(d) We choose the interval \([1.9095, 1.9105]\) which represents the boundary bounds for rounding to 1.910.
\(\mathrm{f}(1.9095) = \mathrm{e}^{2(1.9095)-3} + 3(1.9095) - 8 = \mathrm{e}^{0.819} + 5.7285 - 8 \approx -0.0033\)
\(\mathrm{f}(1.9105) = \mathrm{e}^{2(1.9105)-3} + 3(1.9105) - 8 = \mathrm{e}^{0.821} + 5.7315 - 8 \approx +0.0043\)
Since there is a change of sign and the function is continuous, the root \(\alpha\) lies in the interval \([1.9095, 1.9105]\). Hence, \(\alpha = 1.910\) to 3 decimal places.

(a)
- M1: Evaluates both boundary values \(\mathrm{f}(1.5)\) and \(\mathrm{f}(2.0)\).
- A1: Obtains \(-2.5\) and \(\approx 0.718\) (or \(\mathrm{e}-2\)) and concludes with continuous function + sign change.

(b)
- M1: Makes \(\mathrm{e}^{2x-3}\) the subject.
- A1*: Takes natural logarithms and shows rearrangement to the given form with no errors.

(c)
- B1: \(x_1 \approx 2.0324\).
- B1: \(x_2 \approx 1.8217\).
- B1: \(x_3 \approx 1.9651\).

(d)
- M1: Selects the bounds \([1.9095, 1.9105]\) and evaluates \(\mathrm{f}(1.9095)\) and \(\mathrm{f}(1.9105)\).
- A1.375: Correctly calculates values (approx \(-0.003\) and \(0.004\)) and gives a conclusion indicating the sign change.

(a) Using the given conditions:
\(1200 = A \mathrm{e}^{2k}\) (1)
\(3600 = A \mathrm{e}^{6k}\) (2)

Dividing equation (2) by equation (1):
\(\frac{3600}{1200} = \frac{A \mathrm{e}^{6k}}{A \mathrm{e}^{2k}} \implies 3 = \mathrm{e}^{4k}\)
Taking natural logarithms:
\(4k = \ln 3 \implies k = \frac{1}{4}\ln 3\).

(b) Substituting \(k\) back into equation (1):
\(1200 = A \mathrm{e}^{2\left(\frac{1}{4}\ln 3\right)} = A \mathrm{e}^{\frac{1}{2}\ln 3} = A \sqrt{3}\)
\(A = \frac{1200}{\sqrt{3}} = 400\sqrt{3} \approx 692.82 \approx 693\) (nearest integer).

(c) The rate of increase is given by \(\frac{\mathrm{d}N}{\mathrm{d}t} = k A \mathrm{e}^{kt} = k N\).
First, find the number of bacteria \(N\) when \(t = 10\):
\(N(10) = 400\sqrt{3} \mathrm{e}^{10\left(\frac{1}{4}\ln 3\right)} = 400\sqrt{3} \mathrm{e}^{2.5\ln 3} = 400\sqrt{3} \times 3^{2.5} = 400\sqrt{3} \times 9\sqrt{3} = 400 \times 27 = 10800\).

Now calculate the rate of increase:
\(\frac{\mathrm{d}N}{\mathrm{d}t} = k N(10) = \left(\frac{1}{4}\ln 3\right) \times 10800 = 2700\ln 3 \approx 2966.25 \approx 2966\) bacteria per hour.

(a)
- M1: Formulates both equations and attempts division to eliminate \(A\).
- M1: Solves \(\mathrm{e}^{4k} = 3\) by taking logarithms.
- A1: Correct exact value \(k = \frac{1}{4}\ln 3\) (or equivalent).

(b)
- M1: Substitutes their \(k\) into \(1200 = A\mathrm{e}^{2k}\) to solve for \(A\).
- A1: \(A = 400\sqrt{3} \approx 693\).

(c)
- M1: Differentiates \(N\) with respect to \(t\) to find \(\frac{\mathrm{d}N}{\mathrm{d}t} = k N\).
- M1: Substitutes \(t = 10\) to evaluate \(N(10)\).
- A1: Obtains \(N(10) = 10800\) (or equivalent).
- A1.375: Correct rate of increase \(2966\) (accept range 2960 to 2970).

(a) Using the quotient rule with \(u = \t\t\ln(2x - 3)\) and \(v = x^2\):
\(\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{2}{2x - 3}\)
\(\frac{\mathrm{d}v}{\mathrm{d}x} = 2x\)

\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v\frac{\mathrm{d}u}{\mathrm{d}x} - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} = \frac{x^2 \left(\frac{2}{2x-3}\right) - 2x\ln(2x-3)}{x^4}\)
Multiplying numerator and denominator by \(2x-3\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x^2 - 2x(2x-3)\ln(2x-3)}{x^4(2x-3)} = \frac{2x - 2(2x-3)\ln(2x-3)}{x^3(2x-3)}\).

(b) At the stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(\frac{2x - 2(2x-3)\ln(2x-3)}{x^3(2x-3)} = 0\)
\(2x - 2(2x-3)\ln(2x-3) = 0 \implies 2x = 2(2x-3)\ln(2x-3)\)
Dividing by 2:
\(x = (2x-3)\ln(2x-3)\).

(c) At \(x = 2\):
\(y = \frac{\ln(4-3)}{4} = \frac{\ln(1)}{4} = 0\). So the point is \((2, 0)\).

The gradient of the tangent at \(x = 2\):
\(m_t = \frac{2(2) - 2(4-3)\ln(4-3)}{2^3(4-3)} = \frac{4 - 0}{8(1)} = \frac{1}{2}\).

The gradient of the normal is:
\(m_n = -\frac{1}{m_t} = -2\).

Using the equation of the line:
\(y - 0 = -2(x - 2) \implies y = -2x + 4\)
Re-arranging into the required form:
\(2x + y - 4 = 0\).

(a)
- M1: Correct use of the quotient rule (or product rule) with correct form.
- M1: Correctly differentiates \(\ln(2x-3)\) to get \(\frac{2}{2x-3}\).
- A1: Correct unsimplified derivative.
- A1: Fully simplified derivative to the given answer (or equivalent single fraction).

(b)
- M1: Sets their numerator equal to 0.
- A1*: Shows rearrangement to \(x = (2x - 3)\ln(2x - 3)\) clearly with no errors.

(c)
- B1: Identifies the coordinate point \((2, 0)\).
- M1: Substitutes \(x=2\) into their derivative to find tangent gradient, and takes negative reciprocal for normal gradient.
- A1.375: Obtains \(2x + y - 4 = 0\) (or any integer multiple thereof).

(a) Let \(u = \mathrm{e}^x + 3 \implies \mathrm{e}^x = u - 3\).
Differentiating with respect to \(x\):
\(\frac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^x \implies \mathrm{d}x = \frac{\mathrm{d}u}{\mathrm{e}^x} = \frac{\mathrm{d}u}{u - 3}\).

We rewrite the integral:
\(\int \frac{\mathrm{e}^{2x}}{\mathrm{e}^x + 3} \mathrm{d}x = \int \frac{(\mathrm{e}^x)^2}{u} \frac{\mathrm{d}u}{\mathrm{e}^x} = \int \frac{\mathrm{e}^x}{u} \mathrm{d}u\)

Substitute \(\mathrm{e}^x = u - 3\):
\(\int \frac{u - 3}{u} \mathrm{d}u = \int \left(1 - \frac{3}{u}\right) \mathrm{d}u\).

(b) Change the limits of integration:
When \(x = 0 \implies u = \mathrm{e}^0 + 3 = 4\).
When \(x = \ln 2 \implies u = \mathrm{e}^{\ln 2} + 3 = 5\).

Now, perform the integration:
\(\int_4^5 \left(1 - \frac{3}{u}\right) \mathrm{d}u = [u - 3\ln u]_4^5\)
\(= (5 - 3\t\t\ln 5) - (4 - 3\ln 4)\)
\(= 1 - 3\ln 5 + 3\ln 4\)
\(= 1 - \ln(5^3) + \ln(4^3)\)
\(= 1 - \ln 125 + \ln 64\)
\(= 1 + \ln\left(\frac{64}{125}\right)\).

(a)
- M1: Finds \(\frac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^x\).
- M1: Replaces \(\mathrm{e}^{2x}\) with \((u-3)^2\) or \(\mathrm{e}^x(u-3)\).
- M1: Substitutes all terms to get an integrand purely in terms of \(u\).
- A1*: Correct simplification to \(\int (1 - \frac{3}{u}) \mathrm{d}u\).

(b)
- B1: Changes limits of integration correctly to \(4\) and \(5\).
- M1: Integrates correctly to get \(u - 3\ln |u|\).
- M1: Correctly substitutes limits \(4\) and \(5\) into their integrated term.
- A1: Obtains \(1 - 3\ln 5 + 3\ln 4\) or equivalent.
- A1.375: Applies log laws to arrive at the given form \(1 + \ln\left(\frac{64}{125}\right)\) with no errors.

Rewrite \( f(x) \) as:
\[ f(x) = 4(4 - 3x)^{-1/2} = 4 \times 4^{-1/2} \left(1 - \frac{3}{4}x\right)^{-1/2} = 2 \left(1 - \frac{3}{4}x\right)^{-1/2} \]
Now apply the general binomial expansion formula:
\[ (1 + y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \dots \]
with \( y = -\frac{3}{4}x \) and \( n = -\frac{1}{2} \):
\[ \left(1 - \frac{3}{4}x\right)^{-1/2} = 1 + \left(-\frac{1}{2}\right)\left(-\frac{3}{4}x\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(-\frac{3}{4}x\right)^2 + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{6}\left(-\frac{3}{4}x\right)^3 + \dots \]
\[ = 1 + \frac{3}{8}x + \frac{3}{8}\left(\frac{9}{16}x^2\right) - \frac{5}{16}\left(-\frac{27}{64}x^3\right) + \dots \]
\[ = 1 + \frac{3}{8}x + \frac{27}{128}x^2 + \frac{135}{1024}x^3 + \dots \]
Multiply this expansion by 2:
\[ f(x) = 2 \left(1 + \frac{3}{8}x + \frac{27}{128}x^2 + \frac{135}{1024}x^3 + \dots \right) \]
\[ f(x) = 2 + \frac{3}{4}x + \frac{27}{64}x^2 + \frac{135}{512}x^3 \]

M1: Attempts to extract 4 to write in the form \(A(1 - kx)^{-1/2}\).
A1: Correctly identifies \(A = 2\) and \(k = \frac{3}{4}\).
M1: Applies binomial expansion formula for \((1 - kx)^{-1/2}\) with at least two terms correct (excluding the 1).
A1: Correct term in \(x\) (which is \(\frac{3}{8}x\) before multiplying by 2, or \(\frac{3}{4}x\) after multiplying).
A1: Correct term in \(x^2\) (which is \(\frac{27}{128}x^2\) before multiplying by 2, or \(\frac{27}{64}x^2\) after multiplying).
A1: Correct term in \(x^3\) (which is \(\frac{135}{1024}x^3\) before multiplying by 2, or \(\frac{135}{512}x^3\) after multiplying).
A1.333: Fully correct simplified final expansion.

Let \(u = \sqrt{2x+1}\). Then \(u^2 = 2x+1\), which gives \(x = \frac{1}{2}(u^2-1)\).
Differentiating \(u^2 = 2x+1\) with respect to \(u\) gives:
\[ 2u \, du = 2 \, dx \implies dx = u \, du \]
Change the limits of integration:
- When \(x = 0\), \(u = \sqrt{1} = 1\).
- When \(x = 1\), \(u = \sqrt{3}\).
Substitute these into the integral:
\[ \int_{0}^{1} x\sqrt{2x+1} \, dx = \int_{1}^{\sqrt{3}} \frac{1}{2}(u^2-1) \cdot u \cdot (u \, du) \]
\[ = \frac{1}{2} \int_{1}^{\sqrt{3}} (u^4 - u^2) \, du \]
\[ = \frac{1}{2} \left[ \frac{u^5}{5} - \frac{u^3}{3} \right]_{1}^{\sqrt{3}} \]
Substitute the upper limit \(u = \sqrt{3}\):
\[ \frac{(\sqrt{3})^5}{5} - \frac{(\sqrt{3})^3}{3} = \frac{9\sqrt{3}}{5} - \frac{3\sqrt{3}}{3} = \frac{9\sqrt{3}}{5} - \sqrt{3} = \frac{4\sqrt{3}}{5} \]
Substitute the lower limit \(u = 1\):
\[ \frac{1}{5} - \frac{1}{3} = -\frac{2}{15} \]
Now evaluate the definite integral:
\[ \frac{1}{2} \left[ \frac{4\sqrt{3}}{5} - \left(-\frac{2}{15}\right) \right] = \frac{1}{2} \left( \frac{4\sqrt{3}}{5} + \frac{2}{15} \right) = \frac{2\sqrt{3}}{5} + \frac{1}{15} \]
Thus, \(a = \frac{2}{5}\) and \(b = \frac{1}{15}\).

M1: Attempts substitution \(u = \sqrt{2x+1}\) to express \(x\) and \(dx\) in terms of \(u\).
A1: Correct expression for the integrand in terms of \(u\): \(\frac{1}{2}(u^4 - u^2)\).
B1: Correctly changes the limits of integration to \(1\) and \(\sqrt{3}\).
M1: Integrates the expression to obtain \(A u^5 + B u^3\) (any constants \(A, B \neq 0\)).
A1: Correct integration: \(\frac{1}{2} \left( \frac{u^5}{5} - \frac{u^3}{3} \right)\).
M1: Substitutes their limits \(1\) and \(\sqrt{3}\) into their integrated expression and subtracts.
A1.333: Correct final answer in the form \(\frac{2}{5}\sqrt{3} + \frac{1}{15}\).

(a) To show \(A(7, 1, 0)\) lies on \(l_1\), we check if there exists a value of \(\lambda\) that satisfies the coordinates:
\[ 1 + 2\lambda = 7 \implies 2\lambda = 6 \implies \lambda = 3 \]
Check the \(y\)-coordinate with \(\lambda = 3\):
\[ -2 + 3 = 1 \quad \text{(Correct)} \]
Check the \(z\)-coordinate with \(\lambda = 3\):
\[ 3 - 3 = 0 \quad \text{(Correct)} \]
Since all coordinates are satisfied for \(\lambda = 3\), \(A\) lies on \(l_1\).

(b) A general point \(B\) on \(l_1\) has coordinates:
\[ \overrightarrow{OB} = \begin{pmatrix} 1+2\lambda \\ -2+\lambda \\ 3-\lambda \end{pmatrix} \]
The direction vector of \(l_1\) is \(\mathbf{d} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\).
For \(\overrightarrow{OB}\) to be perpendicular to \(\mathbf{d}\), their scalar product must be zero:
\[ \overrightarrow{OB} \cdot \mathbf{d} = 0 \]
\[ (1+2\lambda)(2) + (-2+\lambda)(1) + (3-\lambda)(-1) = 0 \]
\[ 2 + 4\lambda - 2 + \lambda - 3 + \lambda = 0 \]
\[ 6\lambda - 3 = 0 \implies \lambda = \frac{1}{2} \]
Substitute \(\lambda = \frac{1}{2}\) back into \(\overrightarrow{OB}\) to find the coordinates of \(B\):
\[ x = 1 + 2\left(\frac{1}{2}\right) = 2 \]
\[ y = -2 + \frac{1}{2} = -\frac{3}{2} \]
\[ z = 3 - \frac{1}{2} = \frac{5}{2} \]
Thus, \(B\) has coordinates \(\left(2, -\frac{3}{2}, \frac{5}{2}\right)\).

M1 (part a): Sets up equations for coordinates to find \(\lambda\).
A1 (part a): Correctly shows that \(\lambda = 3\) satisfies all three component equations.
M1 (part b): Expresses \(\overrightarrow{OB}\) in terms of \(\lambda\).
M1 (part b): Uses \(\overrightarrow{OB} \cdot \mathbf{d} = 0\) to set up an equation in \(\lambda\).
A1 (part b): Correctly solves for \(\lambda = \frac{1}{2}\).
A2.333 (part b): Correctly finds the coordinates of \(B\) as \(\left(2, -\frac{3}{2}, \frac{5}{2}\right)\).

(a) First, find the derivatives of \(x\) and \(y\) with respect to \(t\):
\[ \frac{dx}{dt} = 2t \]
\[ \frac{dy}{dt} = 3t^2 - 3 \]
By the chain rule:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2 - 3}{2t} \]

(b) When \(t = 2\), find the coordinates of the point on \(C\):
\[ x = 2^2 + 2 = 6 \]
\[ y = 2^3 - 3(2) = 8 - 6 = 2 \]
Find the gradient \(m\) of the tangent at \(t = 2\):
\[ m = \frac{3(2)^2 - 3}{2(2)} = \frac{12 - 3}{4} = \frac{9}{4} \]
The equation of the tangent is:
\[ y - 2 = \frac{9}{4}(x - 6) \]
Multiply both sides by 4 to clear the fraction:
\[ 4(y - 2) = 9(x - 6) \]
\[ 4y - 8 = 9x - 54 \]
Rearrange into the form \(ax + by + c = 0\):
\[ 9x - 4y - 46 = 0 \]

M1 (part a): Differentiates both \(x\) and \(y\) with respect to \(t\).
A1 (part a): Correct \(\frac{dx}{dt} = 2t\) and \(\frac{dy}{dt} = 3t^2 - 3\).
M1 (part a): Applies chain rule \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).
A1 (part a): Correct expression \(\frac{3t^2 - 3}{2t}\).
M1 (part b): Substitutes \(t = 2\) to find coordinates \((6, 2)\) and gradient \(m = \frac{9}{4}\).
M1 (part b): Uses \(y - y_1 = m(x - x_1)\) with their coordinates and gradient.
A2.333 (part b): Correct final equation in the form \(ax + by + c = 0\) (e.g., \(9x - 4y - 46 = 0\)).

Differentiate implicitly with respect to \(x\):
\[ \frac{d}{dx}(x^2 y + 2y^2) = \frac{d}{dx}(12) \]
Using the product rule on \(x^2 y\):
\[ 2xy + x^2 \frac{dy}{dx} + 4y \frac{dy}{dx} = 0 \]
Factor out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} (x^2 + 4y) = -2xy \]
\[ \frac{dy}{dx} = \frac{-2xy}{x^2 + 4y} \]
For the tangent to be horizontal, the gradient must be zero (\(\frac{dy}{dx} = 0\)):
\[ -2xy = 0 \implies x = 0 \quad \text{or} \quad y = 0 \]

Case 1: \(x = 0\)
Substitute \(x = 0\) into the equation of \(C\):
\[ 0^2(y) + 2y^2 = 12 \implies 2y^2 = 12 \implies y^2 = 6 \implies y = \pm\sqrt{6} \]
So the coordinates are \((0, \sqrt{6})\) and \((0, -\sqrt{6})\). At both points, the denominator \(x^2 + 4y = 4y \neq 0\), so these are valid horizontal tangents.

Case 2: \(y = 0\)
Substitute \(y = 0\) into the equation of \(C\):
\[ x^2(0) + 2(0)^2 = 12 \implies 0 = 12 \]
This equation has no solutions.

Thus, the points are \((0, \sqrt{6})\) and \((0, -\sqrt{6})\).

M1: Attempts implicit differentiation of the equation.
A1: Correctly applies product rule to \(x^2 y\) to get \(2xy + x^2 \frac{dy}{dx}\).
A1: Correctly differentiates \(2y^2\) to get \(4y \frac{dy}{dx}\).
M1: Sets \(\frac{dy}{dx} = 0\) and concludes that \(x = 0\) or \(y = 0\).
A1: Substitutes \(x = 0\) into the original equation to obtain \(2y^2 = 12\).
M1: Solves \(2y^2 = 12\) to find two values of \(y\).
A2.333: Correctly identifies the two points \((0, \sqrt{6})\) and \((0, -\sqrt{6})\) and shows/states that \(y = 0\) yields no solutions.

Assume, for contradiction, that there exist positive real numbers \(a\) and \(b\) such that:
\[ \frac{a}{b} + \frac{b}{a} < 2 \]
Since \(a\) and \(b\) are positive real numbers, their product \(ab\) is also positive (\(ab > 0\)). We can multiply both sides of the inequality by \(ab\) without changing the direction of the inequality sign:
\[ \left(\frac{a}{b} + \frac{b}{a}\right)ab < 2ab \]
\[ a^2 + b^2 < 2ab \]
Rearrange the inequality by subtracting \(2ab\) from both sides:
\[ a^2 - 2ab + b^2 < 0 \]
Factor the left-hand side of the inequality:
\[ (a - b)^2 < 0 \]
However, \(a\) and \(b\) are real numbers, and the square of any real number is always non-negative (\((a-b)^2 \ge 0\)). This contradicts \((a-b)^2 < 0\).
Therefore, our initial assumption must be false. It follows that for all positive real numbers \(a\) and \(b\):
\[ \frac{a}{b} + \frac{b}{a} \ge 2 \]

B1: States the assumption for contradiction clearly: "Assume there exist positive real numbers \(a, b\) such that \(\frac{a}{b} + \frac{b}{a} < 2\)".
M1: Multiplies by \(ab\) (justifying that \(ab > 0\)).
A1: Obtains \(a^2 + b^2 < 2ab\).
M1: Rearranges and attempts to factorise to obtain a perfect square.
A1: Correctly obtains \((a - b)^2 < 0\).
M1: Explains that this is a contradiction because the square of any real number is non-negative.
A2.333: Concludes the proof clearly, stating that the original assertion must therefore be true.

Separate the variables of the differential equation:
\[ \frac{1}{y^2} \, dy = \sin(3x) \, dx \]
Integrate both sides:
\[ \int y^{-2} \, dy = \int \sin(3x) \, dx \]
\[ -y^{-1} = -\frac{1}{3} \cos(3x) + C \]
where \(C\) is an arbitrary constant of integration.
Multiply both sides by \(-1\):
\[ \frac{1}{y} = \frac{1}{3} \cos(3x) - C \]
Let \(K = -3C\) be another arbitrary constant. Then:
\[ \frac{1}{y} = \frac{\cos(3x) + K}{3} \]
Take the reciprocal of both sides to express \(y\) in terms of \(x\):
\[ y = \frac{3}{\cos(3x) + K} \]

M1: Separates variables to get \(\int \frac{1}{y^2} \, dy = \int \sin(3x) \, dx\).
M1: Integrates LHS to obtain \(-y^{-1}\) or equivalent.
M1: Integrates RHS to obtain \(-k \cos(3x)\) (where \(k\) is any constant).
A1: Correct integration of RHS: \(-\frac{1}{3} \cos(3x)\).
B1: Includes an arbitrary constant of integration on one side.
M1: Attempts to make \(y\) the subject of the formula.
A2.333: Obtains a correct expression for \(y\) in terms of \(x\), such as \(y = \frac{3}{\cos(3x) + K}\) (where \(K\) is an arbitrary constant).

Let:
\[ \frac{3x^2 + 6x - 6}{(x+1)^2(2x-1)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{2x-1} \]
Multiply both sides by the denominator \((x+1)^2(2x-1)\):
\[ 3x^2 + 6x - 6 = A(x+1)(2x-1) + B(2x-1) + C(x+1)^2 \]
Substitute \(x = -1\) to eliminate \(A\) and \(C\):
\[ 3(-1)^2 + 6(-1) - 6 = B(2(-1) - 1) \]
\[ 3 - 6 - 6 = -3B \implies -9 = -3B \implies B = 3 \]
Substitute \(x = \frac{1}{2}\) to eliminate \(A\) and \(B\):
\[ 3\left(\frac{1}{2}\right)^2 + 6\left(\frac{1}{2}\right) - 6 = C\left(\frac{1}{2} + 1\right)^2 \]
\[ \frac{3}{4} + 3 - 6 = C\left(\frac{3}{2}\right)^2 \]
\[ -\frac{9}{4} = \frac{9}{4}C \implies C = -1 \]
To find \(A\), compare the coefficients of \(x^2\) on both sides of the identity:
\[ 3 = 2A + C \]
Substitute \(C = -1\):
\[ 3 = 2A - 1 \implies 2A = 4 \implies A = 2 \]
Thus, the partial fraction decomposition is:
\[ \frac{3x^2 + 6x - 6}{(x+1)^2(2x-1)} = \frac{2}{x+1} + \frac{3}{(x+1)^2} - \frac{1}{2x-1} \]

M1: Sets up partial fractions form with correct denominators: \(\frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{2x-1}\).
M1: Multiplies by denominator to form the identity \(3x^2 + 6x - 6 = A(x+1)(2x-1) + B(2x-1) + C(x+1)^2\).
A1: Substitutes \(x = -1\) to find \(B = 3\).
A1: Substitutes \(x = \frac{1}{2}\) to find \(C = -1\).
M1: Equates coefficients of \(x^2\) or substitutes another value of \(x\) to find \(A\).
A1: Correctly finds \(A = 2\).
A2.333: Expresses the final answer correctly as \(\frac{2}{x+1} + \frac{3}{(x+1)^2} - \frac{1}{2x-1}\).

(a) Differentiating \(x\) and \(y\) with respect to \(\theta\):
\[ \frac{dx}{d\theta} = 2 - 2\cos(2\theta) \]
\[ \frac{dy}{d\theta} = -4\sin(\theta) \]
Using the double-angle identity \(\cos(2\theta) = 1 - 2\sin^2(\theta)\), we can rewrite \(\frac{dx}{d\theta}\):
\[ \frac{dx}{d\theta} = 2 - 2(1 - 2\sin^2(\theta)) = 4\sin^2(\theta) \]
Using the chain rule for parametric differentiation:
\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-4\sin(\theta)}{4\sin^2(\theta)} = -\frac{1}{\sin(\theta)} = -\csc(\theta) \]

(b) At \(\theta = \frac{\pi}{3}\):
\[ x = 2\left(\frac{\pi}{3}\right) - \sin\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \]
\[ y = 4\cos\left(\frac{\pi}{3}\right) = 4\left(\frac{1}{2}\right) = 2 \]

The gradient of the tangent at this point is:
\[ m = -\csc\left(\frac{\pi}{3}\right) = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3} \]

Using the equation of a straight line, \(y - y_1 = m(x - x_1)\):
\[ y - 2 = -\frac{2\sqrt{3}}{3}\left(x - \left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right)\right) \]
\[ y - 2 = -\frac{2\sqrt{3}}{3}x + \frac{4\sqrt{3}\pi}{9} - \frac{2\sqrt{3}}{3}\left(-\frac{\sqrt{3}}{2}\right) \]
\[ y - 2 = -\frac{2\sqrt{3}}{3}x + \frac{4\sqrt{3}\pi}{9} + 1 \]
\[ y = -\frac{2\sqrt{3}}{3}x + \frac{4\sqrt{3}\pi}{9} + 1 \]

Part (a):
- M1: Attempts to differentiate both \(x\) and \(y\) with respect to \(\theta\). Award for achieving at least one of \(\frac{dx}{d\theta} = 2 - k\cos(2\theta)\) or \(\frac{dy}{d\theta} = -k\sin(\theta)\).
- A1: Correct expressions for both: \(\frac{dx}{d\theta} = 2 - 2\cos(2\theta)\) and \(\frac{dy}{d\theta} = -4\sin(\theta)\).
- A1*: Uses \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\) and applies the correct double angle identity \(\cos(2\theta) = 1 - 2\sin^2(\theta)\) to simplify the denominator to \(4\sin^2(\theta)\) and show the given result cleanly.

Part (b):
- M1: Evaluates both \(x\) and \(y\) at \(\theta = \frac{\pi}{3}\).
- A1: Correct coordinates: \(x = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}\) and \(y = 2\).
- M1: Finds the gradient of the tangent at \(\theta = \frac{\pi}{3}\) as \(m = -\frac{2}{\sqrt{3}}\) (or equivalent exact form).
- M1: Substitutes their coordinates and gradient into a valid straight-line equation formula like \(y - y_1 = m(x - x_1)\).
- A1: Correct final equation in the required form: \(y = -\frac{2\sqrt{3}}{3}x + \frac{4\sqrt{3}\pi}{9} + 1\) (or any exact equivalent form, e.g. \(y = -\frac{2}{\sqrt{3}}x + \frac{4\pi}{3\sqrt{3}} + 1\)).