2026 Edexcel IAL Pure Mathematics (YPM01) 模擬試題連答案詳解

First, write the equation of the curve in index form:
\[y = 3x^2 - 4x^{-\frac{1}{2}} + 5\]

Differentiate with respect to \(x\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 2 \times 3x^{2-1} - \left(-\frac{1}{2}\right) \times 4x^{-\frac{1}{2}-1} + 0\]
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 6x + 2x^{-\frac{3}{2}}\]

Substitute \(x = 4\) into the expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\):
\[\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=4} = 6(4) + 2(4)^{-\frac{3}{2}}\]

Since \(4^{-\frac{3}{2}} = \frac{1}{(\sqrt{4})^3} = \frac{1}{2^3} = \frac{1}{8}\), we have:
\[\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=4} = 24 + 2 \times \frac{1}{8} = 24 + \frac{1}{4} = \frac{97}{4}\]

M1: Attempts to differentiate with at least one term correct (e.g. \(6x\) or \(k x^{-\frac{3}{2}}\)).
A1: Correct derivative \(6x + 2x^{-\frac{3}{2}}\) (or equivalent).
A1: Correct final value of \(\frac{97}{4}\) (or \(24.25\) or \(24\frac{1}{4}\)).

To find the equation of the curve, integrate \(\frac{\mathrm{d}y}{\mathrm{d}x}\) with respect to \(x\):
\[y = \int \left(3x^2 - 4x + x^{-2}\right) \mathrm{d}x\]

Integrate each term:
\[y = \frac{3x^3}{3} - \frac{4x^2}{2} + \frac{x^{-1}}{-1} + C\]
\[y = x^3 - 2x^2 - \frac{1}{x} + C\]
where \(C\) is the constant of integration.

Use the given point \((1, 5)\) to find \(C\). Substitute \(x = 1\) and \(y = 5\):
\[5 = (1)^3 - 2(1)^2 - \frac{1}{1} + C\]
\[5 = 1 - 2 - 1 + C\]
\[5 = -2 + C\]
\[C = 7\]

Thus, the equation of the curve is:
\[y = x^3 - 2x^2 - \frac{1}{x} + 7\]

M1: Attempts to integrate \(\frac{\mathrm{d}y}{\mathrm{d}x}\), with at least two terms integrated correctly (raising the power by 1).
M1: Substitutes \(x = 1\) and \(y = 5\) into an integrated expression containing a constant of integration \(C\) to find the value of \(C\).
A1: \(y = x^3 - 2x^2 - \frac{1}{x} + 7\) or equivalent (e.g. \(y = x^3 - 2x^2 - x^{-1} + 7\)).

From the first equation, we get \( y = 3 - 2x \). Substituting this into the second equation gives: \( x^2 + x(3 - 2x) = -4 \) which expands to \( x^2 + 3x - 2x^2 = -4 \). Rearranging terms gives \( -x^2 + 3x + 4 = 0 \), which simplifies to \( x^2 - 3x - 4 = 0 \). Factorising the quadratic yields \( (x - 4)(x + 1) = 0 \), which gives \( x = 4 \) or \( x = -1 \). Substituting these back into the linear equation gives the corresponding \( y \) values: for \( x = 4 \), \( y = 3 - 2(4) = -5 \); for \( x = -1 \), \( y = 3 - 2(-1) = 5 \). Thus, the solutions are \( x = 4, y = -5 \) and \( x = -1, y = 5 \).

M1: Attempts to make \( y \) or \( x \) the subject of the linear equation and substitutes into the quadratic equation to form a 3-term quadratic equation in a single variable. M1: Solves their 3-term quadratic equation to find two values of \( x \) (or \( y \)). A1: Both pairs of solutions correct and correctly paired: \( x = 4, y = -5 \) and \( x = -1, y = 5 \).

From the first equation, we can write \( y = 2x - 5 \). Substituting this into the second equation gives: \( x^2 - (2x - 5)^2 = 3 \). Expanding the bracket gives \( x^2 - (4x^2 - 20x + 25) = 3 \), which simplifies to \( -3x^2 + 20x - 25 = 3 \). Rearranging to set the quadratic to zero gives \( 3x^2 - 20x + 28 = 0 \). Factorising this quadratic equation gives \( (3x - 14)(x - 2) = 0 \), so \( x = 2 \) or \( x = \frac{14}{3} \). Substituting these back into \( y = 2x - 5 \): for \( x = 2 \), \( y = 2(2) - 5 = -1 \); for \( x = \frac{14}{3} \), \( y = 2(\frac{14}{3}) - 5 = \frac{13}{3} \). Thus, the solutions are \( x = 2, y = -1 \) and \( x = \frac{14}{3}, y = \frac{13}{3} \).

M1: Attempts to make \( y \) or \( x \) the subject of the linear equation and substitutes into the quadratic equation to form a 3-term quadratic equation in a single variable. M1: Solves their 3-term quadratic equation to find two values of \( x \) (or \( y \)). A1: Both pairs of solutions correct and correctly paired: \( x = 2, y = -1 \) and \( x = \frac{14}{3}, y = \frac{13}{3} \).

First, solve the linear inequality: \( 2(x + 1) > 5 - x \Rightarrow 2x + 2 > 5 - x \Rightarrow 3x > 3 \Rightarrow x > 1 \). Second, solve the quadratic inequality: \( x^2 - 4x - 12 < 0 \Rightarrow (x - 6)(x + 2) < 0 \). The critical values are \( x = -2 \) and \( x = 6 \). Since we want the quadratic expression to be less than zero, the solution is \( -2 < x < 6 \). To satisfy both inequalities, find the intersection of the two solution sets: \( x > 1 \) and \( -2 < x < 6 \). This yields the set of values \( 1 < x < 6 \).

M1: Attempts to solve the linear inequality to get \( x > c \) (where \( c \) is a constant), or attempts to solve the quadratic equation to find the critical values \( -2 \) and \( 6 \).
A1: Correctly identifies both individual solution sets: \( x > 1 \) and \( -2 < x < 6 \).
A0.5: Correctly finds the intersection of both sets to give \( 1 < x < 6 \) (or equivalent set notation \( \{x : 1 < x < 6\} \)).

For a point to lie within \( R \), there must exist a \( y \)-value satisfying both \( y \ge x^2 - 4 \) and \( y \le 2x - 1 \), which requires \( x^2 - 4 \le 2x - 1 \). Rearranging this gives: \( x^2 - 2x - 3 \le 0 \). Factorising the quadratic expression: \( (x - 3)(x + 1) \le 0 \), which yields the interval \( -1 \le x \le 3 \). Applying the final constraint \( x \ge 0 \), the intersection of \( -1 \le x \le 3 \) and \( x \ge 0 \) gives the range of \( x \)-coordinates as \( 0 \le x \le 3 \).

M1: Sets up the inequality \( x^2 - 4 \le 2x - 1 \) and attempts to solve the resulting quadratic equation to find the boundary values (e.g. by factorising \( x^2 - 2x - 3 = 0 \)).
A1: Correctly identifies the interval \( -1 \le x \le 3 \).
A0.5: Combines this with the constraint \( x \ge 0 \) to give the final range \( 0 \le x \le 3 \) (accept interval notation \( [0, 3] \)).

To find where the line intersects the \(x\)-axis (point \(A\)), we substitute \(y = 0\): \(3x + 4(0) - 24 = 0 \implies 3x = 24 \implies x = 8\). So, the coordinates of \(A\) are \((8, 0)\). To find where the line intersects the \(y\)-axis (point \(B\)), we substitute \(x = 0\): \(3(0) + 4y - 24 = 0 \implies 4y = 24 \implies y = 6\). So, the coordinates of \(B\) are \((0, 6)\). The triangle \(OAB\) is right-angled at the origin \(O(0,0)\), with base \(OA = 8\) and height \(OB = 6\). The area of triangle \(OAB\) is given by: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 6 = 24\).

M1: Attempts to find the coordinates of \(A\) and \(B\) by substituting \(y = 0\) and \(x = 0\) respectively. A1: Correctly identifies the coordinates \(A(8,0)\) and \(B(0,6)\) (or states that the intercepts are 8 and 6). A1: Correctly calculates the area of the triangle to be 24.

The points \(P(-2, 3)\) and \(Q(4, 3)\) lie on the horizontal line \(y = 3\). The length of the base \(PQ\) is \(4 - (-2) = 6\). Let \(h\) be the perpendicular height of the triangle \(PQR\) from the base \(PQ\) to the vertex \(R\). Using the area formula: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \implies 15 = \frac{1}{2} \times 6 \times h \implies 3h = 15 \implies h = 5\). Since the base lies on the line \(y = 3\), the \(y\)-coordinate of \(R\) must be either \(3 + 5 = 8\) or \(3 - 5 = -2\). Since the \(y\)-coordinate of \(R\) is positive, we have \(y = 8\). Substituting \(y = 8\) into the equation of the line \(y = 2x + 1\) gives: \(8 = 2x + 1 \implies 2x = 7 \implies x = 3.5\). Thus, the coordinates of \(R\) are \((3.5, 8)\).

M1: Finds the length of the base \(PQ = 6\) and sets up the area equation \(\frac{1}{2} \times 6 \times h = 15\) to find the height \(h = 5\). M1: Uses the vertical height to find the positive \(y\)-coordinate of \(R\) as \(y = 8\). A1: Correctly calculates \(x = 3.5\) and states the coordinates of \(R\) as \((3.5, 8)\) or \((\frac{7}{2}, 8)\).

First, find the equation of \(l_2\). Since \(l_2\) is perpendicular to \(l_1\) (which has gradient 2), its gradient is \(-\frac{1}{2}\). Since it passes through \((0, 9)\), its equation is \(y = -\frac{1}{2}x + 9\). Next, find the coordinates of the intersection point \(P\) by solving the equations simultaneously: \(2x + 4 = -\frac{1}{2}x + 9 \implies \frac{5}{2}x = 5 \implies x = 2\). Substituting \(x = 2\) into \(y = 2x + 4\) gives \(y = 8\). So, the coordinates of \(P\) are \((2, 8)\). Next, find the coordinates of \(Q\) by setting \(y = 0\) in the equation of \(l_1\): \(0 = 2x + 4 \implies x = -2\). So, the coordinates of \(Q\) are \((-2, 0)\). In triangle \(OPQ\), the vertices are \(O(0,0)\), \(Q(-2, 0)\), and \(P(2, 8)\). Taking \(OQ\) as the base, the base length is \(2\) units, and the perpendicular height is the \(y\)-coordinate of \(P\), which is \(8\). Therefore, the area is: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 8 = 8\).

M1: Correctly determines the equation of \(l_2\) as \(y = -\frac{1}{2}x + 9\) and solves simultaneously with \(l_1\) to find the intersection point \(P(2, 8)\). M1: Finds the \(x\)-intercept \(Q(-2, 0)\) and identifies the base as 2 and the height as 8. A1: Correctly calculates the area of triangle \(OPQ\) to be 8.

We are given two equations:

1. The perimeter of the sector:
\( P = 2r + r\theta = 40 \)

2. The area of the sector:
\( A = \frac{1}{2}r^2\theta = 75 \)

From the perimeter equation, we can express \( r\theta \) in terms of \( r \):
\( r\theta = 40 - 2r \)

Substitute this expression for \( r\theta \) into the area equation:
\( A = \frac{1}{2}r(r\theta) = 75 \)
\( \frac{1}{2}r(40 - 2r) = 75 \)
\( r(20 - r) = 75 \)
\( 20r - r^2 = 75 \)
\( r^2 - 20r + 75 = 0 \)

Solving this quadratic equation by factorisation:
\( (r - 5)(r - 15) = 0 \)

This gives \( r = 5 \) or \( r = 15 \).

Since we are given that \( r > 5 \), we select \( r = 15 \).

M1: Writes down equations for both perimeter and area and eliminates \( \theta \) to form a quadratic equation in \( r \).
M1: Solves the quadratic equation \( r^2 - 20r + 75 = 0 \) to find the two roots \( r = 5 \) and \( r = 15 \).
A1: Chooses the correct value of \( r = 15 \) with a clear reason or reference to the constraint \( r > 5 \).

The area of the triangle \( OAB \) is given by:
\( \text{Area} = \frac{1}{2} r^2 \sin\theta \)

Substitute \( r = 8 \) and \( \text{Area} = 16 \):
\( 16 = \frac{1}{2} (8)^2 \sin\theta \)
\( 16 = 32 \sin\theta \)
\( \sin\theta = \frac{1}{2} \)

Since \( \theta \) is obtuse (\( \frac{\pi}{2} < \theta < \pi \)):
\( \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \text{ radians} \)

Now, the area of the sector is:
\( \text{Area of sector} = \frac{1}{2} r^2 \theta \)
\( \text{Area of sector} = \frac{1}{2} (8)^2 \left(\frac{5\pi}{6}\right) = 32 \times \frac{5\pi}{6} = \frac{160\pi}{6} = \frac{80\pi}{3} \text{ cm}^2 \)

M1: Attempts to use the area of a triangle formula \( \frac{1}{2}r^2\sin\theta = 16 \) to find \( \sin\theta = \frac{1}{2} \).
M1: Identifies the correct obtuse angle \( \theta = \frac{5\pi}{6} \) in radians.
A1: Obtains the correct exact area of \( \frac{80\pi}{3} \) (accept equivalent exact forms, e.g. \( 26\frac{2}{3}\pi \)).

First, find the length of the arc \( AB \):
\( s = r\theta = 10 \times 1.2 = 12\text{ cm} \)

Next, find the length of the chord \( AB \). Using the cosine rule in triangle \( OAB \):
\( AB^2 = OA^2 + OB^2 - 2(OA)(OB)\cos\theta \)
\( AB^2 = 10^2 + 10^2 - 2(10)(10)\cos(1.2) \)
\( AB^2 = 200 - 200\cos(1.2) \)

Since \( \cos(1.2) \approx 0.36236 \):
\( AB^2 \approx 200 - 200(0.36236) = 127.528 \)
\( AB \approx \sqrt{127.528} \approx 11.293\text{ cm} \)

(Alternatively, using trigonometry on the isosceles triangle: \( AB = 2r\sin\left(\frac{\theta}{2}\right) = 20\sin(0.6) \approx 11.293\text{ cm} \))

The perimeter of the segment is the sum of the arc length and the chord length:
\( \text{Perimeter} = s + AB \approx 12 + 11.293 = 23.293\text{ cm} \)

Rounding to 3 significant figures gives \( 23.3\text{ cm} \).

M1: Calculates the arc length \( s = r\theta = 12 \).
M1: Attempts to find the length of the chord \( AB \) using the cosine rule or \( 2r\sin(\theta/2) \) with \( \theta \) in radians.
A1: Adds the arc length and chord length to get \( 23.3 \) (accept \( 23.3 \) or answers rounding to \( 23.3 \)).

To find the points of intersection, equate the equations of the curve and the line:
\( x^3 - 3x^2 - 2x + 5 = 2x + 5 \)

Rearrange the equation to set it equal to zero:
\( x^3 - 3x^2 - 4x = 0 \)

Factor out the common term \( x \):
\( x(x^2 - 3x - 4) = 0 \)

Factorise the quadratic expression:
\( x(x - 4)(x + 1) = 0 \)

This gives the \( x \)-coordinates of the intersection points:
\( x = -1, \, x = 0, \, x = 4 \)

M1: Sets the equation of the curve equal to the equation of the line and simplifies to a cubic equation of the form \( x^3 - 3x^2 - 4x = 0 \) (or equivalent).
M1: Factorises the cubic expression to find the roots, showing at least \( x(x^2 - 3x - 4) = 0 \) or equivalent.
A1: Correctly identifies all three \( x \)-coordinates: \( x = -1, \, x = 0, \, x = 4 \) (any order).

Since the curve intersects the \( y \)-axis at \( (0, 12) \), substitute \( x = 0 \) and \( y = 12 \) into the equation of the curve:
\( 12 = (0 - 2)^2(0 + k) \)
\( 12 = 4k \)
\( k = 3 \)

Thus, the equation of the curve is \( y = (x - 2)^2(x + 3) \).

To find the intersections with the \( x \)-axis, set \( y = 0 \):
\( (x - 2)^2(x + 3) = 0 \)

This yields \( x = 2 \) (a repeated root) and \( x = -3 \).

The intersection on the negative \( x \)-axis is therefore \( (-3, 0) \).

M1: Substitutes \( x = 0 \) and \( y = 12 \) into the equation of the curve to set up an equation in terms of \( k \).
A1: Solves to find \( k = 3 \).
A1: States the correct coordinates of the negative \( x \)-axis intersection: \( (-3, 0) \). (Accept \( x = -3, y = 0 \); do not accept just \( x = -3 \) unless coordinates are clearly implied).

To find the points of intersection, equate the two curves:
\( x^3 - 2x^2 + 3x = x^2 + 3x \)

Subtract \( x^2 + 3x \) from both sides to form a single cubic equation:
\( x^3 - 3x^2 = 0 \)

Factorise the equation:
\( x^2(x - 3) = 0 \)

This gives the \( x \)-coordinates of the intersection points:
\( x = 0 \) and \( x = 3 \)

Substitute these \( x \)-values back into either curve's equation to find the corresponding \( y \)-coordinates:

For \( x = 0 \): \( y = 0^2 + 3(0) = 0 \), giving the point \( (0, 0) \).

For \( x = 3 \): \( y = 3^2 + 3(3) = 18 \), giving the point \( (3, 18) \).

Therefore, the coordinates of the points of intersection are \( (0, 0) \) and \( (3, 18) \).

M1: Equates the two equations and simplifies to a cubic equation of the form \( x^3 - 3x^2 = 0 \) (or equivalent).
M1: Factorises the equation to find both \( x = 0 \) and \( x = 3 \), and attempts to find at least one corresponding \( y \)-coordinate.
A1: Correctly states both coordinates: \( (0, 0) \) and \( (3, 18) \).

The curve is given by \(y = 2x^2 - 8x^{\frac{3}{2}} + 12\). Differentiating with respect to \(x\) gives \(\frac{dy}{dx} = 4x - 12x^{\frac{1}{2}}\). The line \(y + 8x = 5\) has gradient \(-8\). For the tangent to be parallel to this line, we set \(\frac{dy}{dx} = -8\), so \(4x - 12\sqrt{x} = -8\). Dividing by 4 gives \(x - 3\sqrt{x} + 2 = 0\). This is a quadratic in \(\sqrt{x}\), which factorises to \((\sqrt{x} - 1)(\sqrt{x} - 2) = 0\). Thus, \(\sqrt{x} = 1\) or \(\sqrt{x} = 2\), which gives \(x = 1\) or \(x = 4\). The larger of these \(x\)-coordinates is 4.

M1: Attempts to differentiate the equation of the curve \(C\) with at least one term correct. A1: Correct derivative \(\frac{dy}{dx} = 4x - 12\sqrt{x}\). M1: Sets their derivative equal to \(-8\) (the gradient of the line) and attempts to solve the resulting quadratic in \(\sqrt{x}\). A1: Obtains the correct larger \(x\)-coordinate of 4.

Let the height of the box be \(h\text{ cm}\). The volume is given by \(V = 2x^2 h = 36\), which gives \(h = \frac{18}{x^2}\). The surface area \(S\) of the open-topped box is given by the area of the base plus the four sides: \(S = 2x^2 + 2xh + 2(2xh) = 2x^2 + 6xh\). Substituting \(h = \frac{18}{x^2}\) into the surface area formula gives \(S = 2x^2 + 6x\left(\frac{18}{x^2}\right) = 2x^2 + \frac{108}{x}\). To find the minimum surface area, we differentiate with respect to \(x\): \(\frac{dS}{dx} = 4x - \frac{108}{x^2}\). Setting \(\frac{dS}{dx} = 0\) gives \(4x^3 = 108\), so \(x^3 = 27\), which yields \(x = 3\). The minimum surface area is found by substituting \(x = 3\) back into the expression for \(S\): \(S = 2(3)^2 + \frac{108}{3} = 18 + 36 = 54\text{ cm}^2\).

M1: Expresses the height \(h\) in terms of \(x\) using the volume formula. M1: Writes a correct expression for the surface area \(S\) in terms of \(x\). M1: Differentiates \(S\) with respect to \(x\) and sets \(\frac{dS}{dx} = 0\) to find the critical value of \(x\). A1: Finds \(x = 3\) and correctly calculates the minimum surface area to be 54.

First, find the gradient of the curve by differentiating: \(\frac{dy}{dx} = 3x^2 - 6x - 9\). Setting \(\frac{dy}{dx} = 0\) to find stationary points: \(3(x^2 - 2x - 3) = 0 \implies 3(x-3)(x+1) = 0\). This gives stationary points at \(x = 3\) and \(x = -1\). The second derivative is \(\frac{d^2y}{dx^2} = 6x - 6\). At \(x = 3\), \(\frac{d^2y}{dx^2} = 12 > 0\), so the local minimum occurs at \(x = 3\). The \(y\)-coordinate of the local minimum is \(y = 3^3 - 3(3)^2 - 9(3) + 15 = -12\). Thus, the horizontal tangent at the local minimum is the line with equation \(y = -12\). Next, find the tangent to \(C\) at \(x = 0\). At \(x = 0\), the \(y\)-coordinate is 15 and the gradient is \(\frac{dy}{dx} = -9\). The equation of this tangent is \(y - 15 = -9(x - 0) \implies y = -9x + 15\). To find the intersection of this tangent with the horizontal line \(y = -12\), set \(-9x + 15 = -12 \implies -9x = -27 \implies x = 3\).

M1: Differentiates the equation of \(C\) and sets the derivative to 0 to find the \(x\)-coordinates of the stationary points. A1: Correctly identifies \(x=3\) as the local minimum and finds the horizontal tangent equation \(y = -12\). M1: Finds the equation of the tangent at \(x=0\) as \(y = -9x + 15\). A1: Equates the tangent equation to \(-12\) and solves to find \(x = 3\).

First, rewrite the integrand by dividing each term in the numerator by the denominator: \(\frac{3x^4 - 2x^{1/2}}{x^2} = 3x^2 - 2x^{-3/2}\). Next, integrate term by term: \(\int (3x^2 - 2x^{-3/2}) \, dx = \frac{3x^3}{3} - \frac{2x^{-1/2}}{-1/2} + C\). Simplifying the coefficients gives: \(x^3 + 4x^{-1/2} + C\), which can also be written as \(x^3 + \frac{4}{\sqrt{x}} + C\).

M1: Divides to express the integrand as two terms in index form: \(3x^2 - 2x^{-3/2}\). A1: Integrates at least one term correctly, e.g., obtaining \(x^3\) or \(4x^{-1/2}\). A1: Fully correct integrated expression: \(x^3 + 4x^{-1/2}\) (or equivalent). B1: Correctly includes the constant of integration \(+ C\).

First, write the curve's equation as \(y = 2x^2 - 8x^{-1}\). Differentiating with respect to \(x\) gives \(\frac{dy}{dx} = 4x + 8x^{-2}\). At the point \(P(2, 4)\), substitute \(x = 2\) to find the gradient of the tangent: \(\frac{dy}{dx} = 4(2) + 8(2)^{-2} = 8 + 2 = 10\). The gradient of the normal is the negative reciprocal of the tangent gradient: \(m = -\frac{1}{10}\). Use the line equation \(y - y_1 = m(x - x_1)\) with point \((2, 4)\): \(y - 4 = -\frac{1}{10}(x - 2)\). Multiply by 10 to clear the fraction: \(10y - 40 = -x + 2\). Rearrange into the form \(ax + by + c = 0\): \(x + 10y - 42 = 0\).

M1: Differentiates the curve's equation correctly to find \(\frac{dy}{dx} = 4x + 8x^{-2}\). A1: Substitutes \(x = 2\) to find the tangent gradient is 10. M1: Finds the negative reciprocal gradient \(m = -\frac{1}{10}\) and uses the line formula with the point \((2, 4)\). A1: Obtains the correct equation in the specified form: \(x + 10y - 42 = 0\) (or any integer multiple thereof, e.g., \(-x - 10y + 42 = 0\)).

Rewrite the gradient function as \(\frac{dy}{dx} = 6x^2 - 5 + 4x^{-3}\). Integrate with respect to \(x\) to find \(y\): \(y = \int (6x^2 - 5 + 4x^{-3}) \, dx = \frac{6x^3}{3} - 5x + \frac{4x^{-2}}{-2} + C = 2x^3 - 5x - 2x^{-2} + C\). To find \(C\), substitute the coordinates of the point \((1, 8)\): \(8 = 2(1)^3 - 5(1) - 2(1)^{-2} + C\), which simplifies to \(8 = 2 - 5 - 2 + C \Rightarrow 8 = -5 + C \Rightarrow C = 13\). Therefore, the equation of the curve is \(y = 2x^3 - 5x - \frac{2}{x^2} + 13\).

M1: Integrates the gradient function, attempting to raise the power of at least two terms by 1. A1: Obtains a correct integrated expression with constant of integration, e.g., \(2x^3 - 5x - 2x^{-2} + C\). M1: Substitutes \(x = 1\) and \(y = 8\) into their integrated equation to solve for \(C\). A1: Correct final equation: \(y = 2x^3 - 5x - \frac{2}{x^2} + 13\) (or equivalent).

The maximum value of the function \(y = -5\cos(3x)\) occurs when \(\cos(3x) = -1\). Since the maximum value of \(\cos(3x)\) is \(1\) and its minimum value is \(-1\), the maximum value of \(y\) is \(-5 \times (-1) = 5\). We solve \(\cos(3x) = -1\) for the interval \(0^\circ \le 3x \le 540^\circ\): \(3x = 180^\circ, 540^\circ\). Dividing by 3 gives \(x = 60^\circ, 180^\circ\). Thus, the maximum points are \((60^\circ, 5)\) and \((180^\circ, 5)\).

M1: Identifies that the maximum y-coordinate is 5, or sets \(\cos(3x) = -1\). M1: Solves to find at least one correct value of \(x\) in the interval (e.g., \(x = 60^\circ\) or \(x = 180^\circ\)). A1: Gives both correct coordinate pairs \((60^\circ, 5)\) and \((180^\circ, 5)\) (accept coordinates written without degree symbols if clear).

The period of \(\cos(bx)\) is given by \(\frac{360^\circ}{b}\). Since the period is \(120^\circ\), we have \(\frac{360^\circ}{b} = 120^\circ \implies b = 3\). The maximum value of the curve is \(a + c = 8\) and the minimum value of the curve is \(-a + c = 2\) (since \(a > 0\)). Solving these simultaneous equations by adding them gives \(2c = 10 \implies c = 5\). Substituting \(c = 5\) into the first equation gives \(a + 5 = 8 \implies a = 3\).

B1: Correctly finds \(b = 3\). M1: Sets up the equations \(a + c = 8\) and \(-a + c = 2\) (or equivalent reasoning to find amplitude and vertical shift). A1: Correctly finds both \(a = 3\) and \(c = 5\).

Translating the curve \(y = \tan(x)\) by the vector \(\begin{pmatrix} 30^\circ \\ 2 \end{pmatrix}\) results in the equation for \(C_2\): \(y = \tan(x - 30^\circ) + 2\). The point \(P\) is the \(y\)-intercept, which occurs where \(x = 0\). Substituting \(x = 0\) gives \(y = \tan(-30^\circ) + 2\). Since \(\tan(-30^\circ) = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}\), the exact \(y\)-coordinate of \(P\) is \(2 - \frac{\sqrt{3}}{3}\).

M1: Identifies the equation of the translated curve as \(y = \tan(x - 30^\circ) + 2\). M1: Substitutes \(x = 0\) to find the \(y\)-intercept. A1: Obtains the exact value \(2 - \frac{\sqrt{3}}{3}\) (or equivalent exact forms such as \(\frac{6 - \sqrt{3}}{3}\) or \(2 - \frac{1}{\sqrt{3}}\)).

The general term in the expansion of \((2 + kx)^6\) is given by \(\binom{6}{r} 2^{6-r} (kx)^r\). For the term in \(x^3\), we set \(r = 3\). This gives the term: \(\binom{6}{3} 2^3 (kx)^3 = 20 \times 8 \times k^3 x^3 = 160k^3 x^3\). We are given that the coefficient of \(x^3\) is 1280, so: \(160k^3 = 1280\), which simplifies to \(k^3 = 8\). Since \(k\) is a positive constant, we have \(k = 2\).

M1: Attempts to find the term in \(x^3\), showing a term of the form \(\binom{6}{3} \times 2^3 \times (kx)^3\) with at least two of the three parts correct. A1: Obtains the correct equation \(160k^3 = 1280\) or equivalent. A1: Correctly finds \(k = 2\) (accept with no other values, or if other values are explicitly rejected).

First, expand \((1 + 3x)^5\) up to the term in \(x^2\): \((1 + 3x)^5 = 1 + \binom{5}{1}(3x) + \binom{5}{2}(3x)^2 + \dots = 1 + 15x + 90x^2 + \dots\). Next, consider the product \((x - 2)(1 + 15x + 90x^2 + \dots)\). The terms in \(x^2\) are obtained by multiplying: \(x \times (15x) = 15x^2\) and \(-2 \times (90x^2) = -180x^2\). Combining these terms gives: \(15x^2 - 180x^2 = -165x^2\). Thus, the coefficient of \(x^2\) is \(-165\).

M1: Attempts to expand \((1 + 3x)^5\) up to the term in \(x^2\), showing binomial coefficients and correct powers of 3. Accept \(1 + 15x + 90x^2\). M1: Uses a complete method to find the coefficient of \(x^2\) in the product, i.e., \(1 \times A - 2 \times B\) where \(A\) is their coefficient of \(x\) and \(B\) is their coefficient of \(x^2\) from the expansion of \((1 + 3x)^5\). A1: Obtains \(-165\).

Using the linearity of the trapezium rule:

\(T(2\mathrm{f}(x) - 3) = 2 \times T(\mathrm{f}(x)) - T(3)\)

Since the trapezium rule evaluates linear and constant functions exactly:

\(T(3) = \int_{2}^{8} 3 \, \mathrm{d}x = 3 \times (8 - 2) = 18\)

Therefore, the new approximation is:

\(T_{\text{new}} = 2(15.4) - 18 = 30.8 - 18 = 12.8\)

M1: Recognising that the new estimate is given by \(2 \times 15.4 - C\) where \(C\) is the exact integral of the constant 3 over the interval \([2, 8]\) (i.e. \(3 \times 6 = 18\)), or setting up the corresponding coordinate/ordinate formula to show this expansion.

A1: Correct answer of \(12.8\) (or exact equivalent).

The trapezium rule is linear, meaning that the approximation scales directly with the function's transformation.

Let \(T(g(x)) = 5.8\) be the original estimate.

For the constant term, the trapezium rule is exact over the interval \([0, 2]\):

\(T(4) = 4 \times (2 - 0) = 8\)

Thus, the new estimate is:

\(T_{\text{new}} = 3 \times T(g(x)) + T(4) = 3(5.8) + 8 = 17.4 + 8 = 25.4\)

M1: For establishing the method \(3(5.8) + \int_{0}^{2} 4 \, \mathrm{d}x\) or using the ordinate formula to write \(T_{\text{new}} = 3(5.8) + 8\).

A1: For \(25.4\) (or exact equivalent).

Let the original trapezium rule step size be \(h_{\text{old}}\):

\(h_{\text{old}} = \frac{6 - 2}{4} = 1\)

The original estimate is:

\(T_{\text{old}} = \frac{1}{2} [ \mathrm{f}(2) + 2(\mathrm{f}(3) + \mathrm{f}(4) + \mathrm{f}(5)) + \mathrm{f}(6) ] = 8.4\)

For the new integral \(\int_{1}^{3} \mathrm{f}(2x) \, \mathrm{d}x\), the new step size is:

\(h_{\text{new}} = \frac{3 - 1}{4} = 0.5\)

The 5 ordinates for \(g(x) = \mathrm{f}(2x)\) are evaluated at \(x = 1, 1.5, 2, 2.5, 3\), giving:

\(g(1) = \mathrm{f}(2), \ g(1.5) = \mathrm{f}(3), \ g(2) = \mathrm{f}(4), \ g(2.5) = \mathrm{f}(5), \ g(3) = \mathrm{f}(6)\)

Since these ordinates are identical to the original ones, the new estimate is:

\(T_{\text{new}} = \frac{h_{\text{new}}}{2} [ \mathrm{f}(2) + 2(\mathrm{f}(3) + \mathrm{f}(4) + \mathrm{f}(5)) + \mathrm{f}(6) ]\)

\(T_{\text{new}} = \frac{h_{\text{new}}}{h_{\text{old}}} \times T_{\text{old}} = 0.5 \times 8.4 = 4.2\)

M1: For identifying that the ordinates are identical for both integrals and that the new step size is half of the original step size (\(h_{\text{new}} = 0.5\), \(h_{\text{old}} = 1\)), hence \(T_{\text{new}} = 0.5 \times 8.4\).

A1: For \(4.2\) (or exact equivalent).

Since \((x - 2)\) is a factor of \(\mathrm{f}(x)\), by the factor theorem: \(\mathrm{f}(2) = 0\). Substituting \(x = 2\) into the polynomial: \(2(2)^3 + a(2)^2 + b(2) + 6 = 0 \implies 16 + 4a + 2b + 6 = 0 \implies 4a + 2b = -22 \implies 2a + b = -11\) (Equation 1). Since the remainder when \(\mathrm{f}(x)\) is divided by \((x + 1)\) is \(-12\), by the remainder theorem: \(\mathrm{f}(-1) = -12\). Substituting \(x = -1\) into the polynomial: \(2(-1)^3 + a(-1)^2 + b(-1) + 6 = -12 \implies -2 + a - b + 6 = -12 \implies a - b = -16\) (Equation 2). Solving Equations 1 and 2 simultaneously, from Equation 2 we have \(b = a + 16\). Substituting this into Equation 1 gives: \(2a + (a + 16) = -11 \implies 3a = -27 \implies a = -9\). Substituting \(a = -9\) back into the expression for \(b\) gives: \(b = -9 + 16 = 7\). Thus, \(a = -9\) and \(b = 7\).

M1: Attempts to apply the factor theorem or remainder theorem by setting \(\mathrm{f}(2) = 0\) or \(\mathrm{f}(-1) = -12\). A1: Obtains two correct linear equations in terms of \(a\) and \(b\), e.g., \(2a + b = -11\) and \(a - b = -16\) (or equivalent). M1: Uses a valid algebraic method to solve their simultaneous linear equations to find at least one variable. A0.5: Obtains correct values of both \(a = -9\) and \(b = 7\).

Since \((2x - 1)\) is a factor of \(\mathrm{g}(x)\), by the factor theorem: \(\mathrm{g}\left(\frac{1}{2}\right) = 0\). Substituting \(x = \frac{1}{2}\) into the expression: \(k\left(\frac{1}{2}\right)^3 - 5\left(\frac{1}{2}\right)^2 - 11\left(\frac{1}{2}\right) + 6 = 0 \implies \frac{k}{8} - \frac{5}{4} - \frac{11}{2} + 6 = 0\). Multiplying the entire equation by 8 yields: \(k - 10 - 44 + 48 = 0 \implies k - 6 = 0 \implies k = 6\). Thus, the polynomial is \(\mathrm{g}(x) = 6x^3 - 5x^2 - 11x + 6\). To find the remainder when divided by \((x + 2)\), we calculate \(\mathrm{g}(-2)\) using the remainder theorem: \(\text{Remainder} = \mathrm{g}(-2) = 6(-2)^3 - 5(-2)^2 - 11(-2) + 6 = 6(-8) - 5(4) + 22 + 6 = -48 - 20 + 22 + 6 = -40\).

M1: Applies the factor theorem by substituting \(x = \frac{1}{2}\) into \(\mathrm{g}(x)\) and setting the expression equal to 0. A1: Obtains \(k = 6\). M1: Uses the remainder theorem by substituting \(x = -2\) into their \(\mathrm{g}(x)\) with their value of \(k\). A0.5: Obtains a final remainder of \(-40\).

First, write the integrand by dividing each term in the numerator by the denominator:

\[\frac{3x^2 - 2x + 1}{2\sqrt{x}} = \frac{3x^2}{2x^{1/2}} - \frac{2x}{2x^{1/2}} + \frac{1}{2x^{1/2}}\]

\[= \frac{3}{2}x^{3/2} - x^{1/2} + \frac{1}{2}x^{-1/2}\]

Next, integrate each term with respect to \(x\):

\[\int \left( \frac{3}{2}x^{3/2} - x^{1/2} + \frac{1}{2}x^{-1/2} \right) \text{d}x = \left[ \frac{\frac{3}{2}}{\frac{5}{2}}x^{5/2} - \frac{1}{\frac{3}{2}}x^{3/2} + \frac{\frac{1}{2}}{\frac{1}{2}}x^{1/2} \right]_{1}^{4}\]

\[= \left[ \frac{3}{5}x^{5/2} - \frac{2}{3}x^{3/2} + x^{1/2} \right]_{1}^{4}\]

Now substitute the upper limit \(x = 4\):

\[\frac{3}{5}(4)^{5/2} - \frac{2}{3}(4)^{3/2} + (4)^{1/2} = \frac{3}{5}(32) - \frac{2}{3}(8) + 2\]

\[= \frac{96}{5} - \frac{16}{3} + 2 = \frac{288 - 80 + 30}{15} = \frac{238}{15}\]

Substitute the lower limit \(x = 1\):

\[\frac{3}{5}(1)^{5/2} - \frac{2}{3}(1)^{3/2} + (1)^{1/2} = \frac{3}{5} - \frac{2}{3} + 1\]

\[= \frac{9 - 10 + 15}{15} = \frac{14}{15}\]

Finally, subtract the value at the lower limit from the value at the upper limit:

\[\frac{238}{15} - \frac{14}{15} = \frac{224}{15}\]

**M1**: For attempting to divide at least two terms of the numerator by the denominator to express the integrand as a sum of terms in powers of \(x\).
**A1**: Correct simplified expression: \(\frac{3}{2}x^{3/2} - x^{1/2} + \frac{1}{2}x^{-1/2}\) (or equivalent).
**M1**: For attempting to integrate their expression, with the power increased by 1 in at least two terms.
**A1**: Correct integrated expression: \(\frac{3}{5}x^{5/2} - \frac{2}{3}x^{3/2} + x^{1/2}\).
**A1**: For substituting both limits \(4\) and \(1\) into their integrated expression, subtracting the correct way round, and simplifying to the exact value of \(\frac{224}{15}\) (or \(14\frac{14}{15}\)).

First, complete the square for the equation of \( C_1 \):
\( (x - 2)^2 - 4 + (y - 5)^2 - 25 + 20 = 0 \)
\( (x - 2)^2 + (y - 5)^2 = 9 \)

From this, we find that \( C_1 \) has center \( (2, 5) \) and radius \( r_1 = 3 \).

The distance \( d \) between the center of \( C_1 \), which is \( (2, 5) \), and the center of \( C_2 \), which is \( (10, 11) \), is given by:
\( d = \sqrt{(10 - 2)^2 + (11 - 5)^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10 \)

Since the circles touch externally, the distance between their centres is equal to the sum of their radii:
\( d = r_1 + R \)
\( 10 = 3 + R \implies R = 7 \)

M1: Attempts to complete the square for the equation of \( C_1 \) to find either its center or its radius. (Center \( (2, 5) \) or radius \( 3 \) seen or implied).

M1: Uses the distance formula to find the distance between the centres of the two circles, obtaining \( 10 \), and sets up the equation \( r_1 + R = d \).

A1: Correctly obtains \( R = 7 \).

First, complete the square for the equation of \( C_1 \):
\( (x - 1)^2 - 1 + (y - 2)^2 - 4 - 59 = 0 \)
\( (x - 1)^2 + (y - 2)^2 = 64 \)

This shows that \( C_1 \) has center \( (1, 2) \) and radius \( R_1 = 8 \).

The distance \( d \) between the center of \( C_1 \), which is \( (1, 2) \), and the center of \( C_2 \), which is \( (4, 6) \), is given by:
\( d = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \)

Since \( r < 8 \) and the circles touch internally, the circle \( C_2 \) lies inside \( C_1 \). The distance between their centres is equal to the difference between their radii:
\( d = R_1 - r \)
\( 5 = 8 - r \implies r = 3 \)

M1: Attempts to complete the square for the equation of \( C_1 \) to find either its center or its radius. (Center \( (1, 2) \) or radius \( 8 \) seen or implied).

M1: Uses the distance formula to find the distance between the centres of the two circles, obtaining \( 5 \), and sets up the equation \( R_1 - r = d \).

A1: Correctly obtains \( r = 3 \).

To prove the inequality, we begin by rearranging all terms to one side of the inequality: \(2x^2 + y^2 - 2xy + 2x + 1 \ge 0\). We can split the term \(2x^2\) into \(x^2 + x^2\) and group the terms as follows: \((x^2 - 2xy + y^2) + (x^2 + 2x + 1) \ge 0\). Factoring both groups gives: \((x-y)^2 + (x+1)^2 \ge 0\). Since the square of any real number is non-negative, we know that \((x-y)^2 \ge 0\) and \((x+1)^2 \ge 0\) for all real numbers \(x\) and \(y\). The sum of two non-negative numbers is also non-negative, hence \((x-y)^2 + (x+1)^2 \ge 0\) is always true. Thus, the original inequality \(2x^2 + y^2 \ge 2x(y - 1) - 1\) holds for all real numbers \(x\) and \(y\).

M1: Attempts to rearrange the inequality by moving all terms to one side to obtain \(2x^2 + y^2 - 2xy + 2x + 1 \ge 0\) or equivalent. M1: Attempts to group terms to form at least one completed square, such as \((x-y)^2\) or \((x+1)^2\). A1: Correctly expresses the inequality in the form \((x-y)^2 + (x+1)^2 \ge 0\). A1*: Completes the proof with a clear concluding statement explaining why the sum of squares of real numbers must be non-negative.

We test each integer \(n\) in the interval \(1 \le n \le 5\) individually: For \(n = 1\), \(f(1) = 1^3 - 1 + 7 = 7\), which is prime because its only positive divisors are 1 and 7. For \(n = 2\), \(f(2) = 2^3 - 2 + 7 = 13\), which is prime because its only positive divisors are 1 and 13. For \(n = 3\), \(f(3) = 3^3 - 3 + 7 = 31\), which is prime because its only positive divisors are 1 and 31. For \(n = 4\), \(f(4) = 4^3 - 4 + 7 = 67\), which is prime because its only positive divisors are 1 and 67. For \(n = 5\), \(f(5) = 5^3 - 5 + 7 = 127\), which is prime because its only positive divisors are 1 and 127. Since \(f(n)\) is prime for every integer in the specified interval, the statement has been proven by exhaustion.

M1: Attempts to evaluate \(f(n)\) for at least three integer values of \(n\) in the given interval. A1: Correctly calculates \(f(1) = 7\), \(f(2) = 13\), and \(f(3) = 31\), stating or showing that they are prime numbers. A1: Correctly calculates \(f(4) = 67\) and \(f(5) = 127\), stating or showing that they are prime numbers. A1*: Evaluates all five cases correctly and provides a concluding statement that the proof is complete by exhaustion.

Let \( y = 2^x \).

Using the laws of indices, we can rewrite the term \( 2^{2x+1} \) as:
\[ 2^{2x+1} = 2^1 \times 2^{2x} = 2(2^x)^2 = 2y^2 \]

Substituting this into the original equation gives a quadratic in terms of \( y \):
\[ 2y^2 - 9y + 7 = 0 \]

Factorizing the quadratic equation:
\[ (2y - 7)(y - 1) = 0 \]

This gives two possible values for \( y \):
\[ y = 3.5 \quad \text{or} \quad y = 1 \]

Now, substitute back \( y = 2^x \):

1. For \( 2^x = 1 \):
\[ x = 0 \]

2. For \( 2^x = 3.5 \):
\[ \ln(2^x) = \ln(3.5) \implies x \ln 2 = \ln(3.5) \implies x = \frac{\ln(3.5)}{\ln 2} \approx 1.81 \text{ (to 3 s.f.)} \]

Thus, the solutions are \( x = 0 \) and \( x \approx 1.81 \).

M1: Expresses the equation as a quadratic in \( 2^x \) or a substituted variable (e.g., \( y = 2^x \)), writing the first term as \( 2(2^x)^2 \) or \( 2y^2 \) to obtain \( 2y^2 - 9y + 7 = 0 \).


M1: Solves their 3-term quadratic equation in \( y \) (or \( 2^x \)) by factorizing, using the quadratic formula, or completing the square to find two values for \( y \) (or \( 2^x \)).


A1: Obtains correct values of \( 2^x = 1 \) and \( 2^x = 3.5 \) (or equivalent fractions).


A1: Obtains the exact solution \( x = 0 \).


A0.5: Obtains the second solution \( x \approx 1.81 \) (accept 1.81, or the exact equivalent \( \log_2(3.5) \) or \( \frac{\ln 3.5}{\ln 2} \)).

Using the laws of logarithms:

First, apply the power law to the term \( 2\log_3(x + 1) \):
\[ 2\log_3(x + 1) = \log_3(x + 1)^2 \]

The equation becomes:
\[ \log_3(2x - 1) = \log_3(x + 1)^2 - \log_3(4) \]

Next, apply the subtraction law of logarithms (or rearrange and apply the addition law):
\[ \log_3(2x - 1) + \log_3(4) = \log_3(x + 1)^2 \]
\[ \log_3(4(2x - 1)) = \log_3(x + 1)^2 \]

Since the bases of the logarithms are the same, we can equate the arguments:
\[ 4(2x - 1) = (x + 1)^2 \]

Expand both sides:
\[ 8x - 4 = x^2 + 2x + 1 \]

Rearrange into a standard quadratic equation:
\[ x^2 - 6x + 5 = 0 \]

Factorize the quadratic:
\[ (x - 1)(x - 5) = 0 \]

This gives:
\[ x = 1 \quad \text{or} \quad x = 5 \]

We must check if both solutions are valid in the original logarithmic equation:
- For \( x = 1 \), the terms inside the logarithms are \( 2x - 1 = 1 > 0 \) and \( x + 1 = 2 > 0 \). Thus, \( x = 1 \) is valid.
- For \( x = 5 \), the terms inside the logarithms are \( 2x - 1 = 9 > 0 \) and \( x + 1 = 6 > 0 \). Thus, \( x = 5 \) is valid.

Therefore, the solutions are \( x = 1 \) and \( x = 5 \).

M1: Correctly applies the power law of logarithms to rewrite \( 2\log_3(x + 1) \) as \( \log_3(x + 1)^2 \).


M1: Correctly applies the addition or subtraction law of logarithms to combine terms, e.g., obtaining \( \log_3\left(\frac{(x+1)^2}{4}\right) \) or \( \log_3(4(2x-1)) \).


M1: Removes logarithms correctly to form a quadratic equation, e.g., \( 4(2x - 1) = (x + 1)^2 \), and simplifies to a standard 3-term quadratic form \( x^2 - 6x + 5 = 0 \).


A1: Solves the quadratic equation to find both \( x = 1 \) and \( x = 5 \).


A0.5: Shows or states that both solutions are valid because they result in positive arguments for the original logarithmic functions (i.e., \( 2x - 1 > 0 \) and \( x + 1 > 0 \) for both values).

We use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to rewrite the equation in terms of \(\sin \theta\) only: \(6(1 - \sin^2 \theta) - \sin \theta - 5 = 0\). Expanding and rearranging gives: \(6 - 6 \sin^2 \theta - \sin \theta - 5 = 0 \implies 6 \sin^2 \theta + \sin \theta - 1 = 0\). This quadratic equation can be factorised as: \((3\sin \theta - 1)(2\sin \theta + 1) = 0\). This yields two possible values for \(\sin \theta\): \(\sin \theta = \frac{1}{3}\) or \(\sin \theta = -\frac{1}{2}\). For \(\sin \theta = \frac{1}{3}\) in the interval \(0 \le \theta < 360^\circ\): \(\theta = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.47^\circ \approx 19.5^\circ\) and \(\theta = 180^\circ - 19.47^\circ \approx 160.5^\circ\). For \(\sin \theta = -\frac{1}{2}\) in the interval \(0 \le \theta < 360^\circ\): \(\theta = 180^\circ - (-30^\circ) = 210^\circ\) and \(\theta = 360^\circ - 30^\circ = 330^\circ\). Thus, the solutions are \(\theta = 19.5^\circ, 160.5^\circ, 210^\circ, 330^\circ\).

M1: Uses the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to form a quadratic equation in \(\sin \theta\). M1: Correctly factorises or solves their quadratic equation to find two values for \(\sin \theta\). A1: Finds any two correct angles (e.g., \(210^\circ\) and \(330^\circ\)). A1: Finds a third correct angle (e.g., \(19.5^\circ\) or \(160.5^\circ\)). A1: Finds all four correct solutions: \(19.5^\circ, 160.5^\circ, 210^\circ, 330^\circ\) and no extras in the range.

Divide both sides of the equation by 2: \(\sin\left(2x + \frac{\pi}{4}\right) = \frac{\sqrt{3}}{2}\). Let \(\phi = 2x + \frac{\pi}{4}\). Since \(0 \le x < 2\pi\), the interval for \(\phi\) is \(\frac{\pi}{4} \le \phi < \frac{17\pi}{4}\). Solving \(\sin\phi = \frac{\sqrt{3}}{2}\) in this interval: the principal value is \(\phi = \frac{\pi}{3}\). The other solutions within the range are \(\phi = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\), \(\phi = 2\pi + \frac{\pi}{3} = \frac{7\pi}{3}\), and \(\phi = 3\pi - \frac{\pi}{3} = \frac{8\pi}{3}\). Now solve for \(x\) using \(2x = \phi - \frac{\pi}{4}\): for \(\phi = \frac{\pi}{3} \implies 2x = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12} \implies x = \frac{\pi}{24}\); for \(\phi = \frac{2\pi}{3} \implies 2x = \frac{2\pi}{3} - \frac{\pi}{4} = \frac{5\pi}{12} \implies x = \frac{5\pi}{24}\); for \(\phi = \frac{7\pi}{3} \implies 2x = \frac{7\pi}{3} - \frac{\pi}{4} = \frac{25\pi}{12} \implies x = \frac{25\pi}{24}\); for \(\phi = \frac{8\pi}{3} \implies 2x = \frac{8\pi}{3} - \frac{\pi}{4} = \frac{29\pi}{12} \implies x = \frac{29\pi}{24}\). These are all the solutions within the range.

M1: Divides by 2 and finds the principal value of \(\arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}\) or \(60^\circ\). M1: Sets up the correct interval for the substituted variable, e.g., \(\frac{\pi}{4} \le 2x + \frac{\pi}{4} < \frac{17\pi}{4}\), and finds at least two correct values for \(2x + \frac{\pi}{4}\). A1: Finds any two correct solutions for \(x\). A1: Finds a third correct solution for \(x\). A1: Finds all four correct solutions: \(x = \frac{\pi}{24}, \frac{5\pi}{24}, \frac{25\pi}{24}, \frac{29\pi}{24}\) with no extra values in the range.

The sum of the first \(n\) terms of an arithmetic series is given by \(S_n = \frac{n}{2}[2a + (n-1)d]\). Given \(S_{10} = 210\), we have: \(\frac{10}{2}[2a + 9d] = 210 \implies 5[2a + 9d] = 210 \implies 2a + 9d = 42\) (Equation 1). The \(n\)th term of an arithmetic series is given by \(u_n = a + (n-1)d\). Given \(u_{11} = 43\), we have: \(a + 10d = 43\) (Equation 2). Multiplying Equation 2 by 2 gives: \(2a + 20d = 86\) (Equation 3). Subtracting Equation 1 from Equation 3: \(11d = 44 \implies d = 4\). Substituting \(d = 4\) into Equation 2: \(a + 10(4) = 43 \implies a = 3\).

M1: Set up two correct equations in \(a\) and \(d\) using the sum and term formulas. A1: Solves the simultaneous equations to find \(d = 4\) or \(a = 3\). A0.5: Finds both correct values: \(a = 3\) and \(d = 4\).

The \(n\)th term of a geometric series is \(u_n = a r^{n-1}\). Given \(u_3 = 18\), we have \(a r^2 = 18\) (Equation 1). Given \(u_5 = 40.5\), we have \(a r^4 = 40.5\) (Equation 2). Dividing Equation 2 by Equation 1: \(\frac{a r^4}{a r^2} = \frac{40.5}{18} \implies r^2 = 2.25\). Since \(r > 0\), we take the positive square root: \(r = \sqrt{2.25} = 1.5\). Substituting \(r = 1.5\) back into Equation 1: \(a (1.5)^2 = 18 \implies 2.25a = 18 \implies a = 8\).

M1: Attempt to divide the fifth term by the third term to set up an equation for \(r^2\). A1: Correctly solve for \(r = 1.5\) (accept \(3/2\)), rejecting the negative root. A0.5: Correctly find \(a = 8\).

The given series is a geometric series: \(\sum_{n=1}^{\infty} 6 \left(-\frac{1}{3}\right)^n = 6\left(-\frac{1}{3}\right) + 6\left(-\frac{1}{3}\right)^2 + \dots\). The first term \(a\) is \(6\left(-\frac{1}{3}\right) = -2\). The common ratio \(r\) is \(-\frac{1}{3}\). Since \(|r| < 1\), the sum to infinity exists and is given by \(S_{\infty} = \frac{a}{1 - r}\). Substituting the values: \(S_{\infty} = \frac{-2}{1 - (-1/3)} = \frac{-2}{4/3} = -2 \times \frac{3}{4} = -1.5\).

M1: Identify the first term \(a = -2\) and common ratio \(r = -1/3\), and substitute into the sum to infinity formula. A1: Correctly evaluate the denominator as \(4/3\) or equivalent. A0.5: Obtain the correct final exact value of \(-1.5\) (or \(-3/2\)).

Since the terms are in arithmetic progression, the difference between consecutive terms is constant: \((4x + 2) - 3x = (6x - 2) - (4x + 2)\). Simplifying both sides: \(x + 2 = 2x - 4\). Rearranging to solve for \(x\): \(2x - x = 2 + 4 \implies x = 6\). Using \(x = 6\), the common difference is \(d = x + 2 = 6 + 2 = 8\).

M1: Set up a correct equation using the common difference property, i.e., \(u_2 - u_1 = u_3 - u_2\). A1: Correctly solve the linear equation to find \(x = 6\). A0.5: Find the correct common difference \(d = 8\).

First, find the points of intersection by setting the curve and line equal to each other:
\[3x^2 - 10x + 8 = x + 2\]
\[3x^2 - 11x + 6 = 0\]
\[(3x - 2)(x - 3) = 0\]
Thus, the limits of integration are \(x = \frac{2}{3}\) and \(x = 3\).

Now, set up the definite integral to find the area of the region:
\[\text{Area} = \int_{\frac{2}{3}}^{3} \left( (x + 2) - (3x^2 - 10x + 8) \right) \text{d}x\]
\[= \int_{\frac{2}{3}}^{3} \left( -3x^2 + 11x - 6 \right) \text{d}x\]
\[= \left[ -x^3 + \frac{11}{2}x^2 - 6x \right]_{\frac{2}{3}}^{3}\]

Evaluate at the upper limit \(x = 3\):
\[-(3)^3 + \frac{11}{2}(3)^2 - 6(3) = -27 + 49.5 - 18 = 4.5 = \frac{9}{2}\]

Evaluate at the lower limit \(x = \frac{2}{3}\):
\[-\left(\frac{2}{3}\right)^3 + \frac{11}{2}\left(\frac{2}{3}\right)^2 - 6\left(\frac{2}{3}\right) = -\frac{8}{27} + \frac{22}{9} - 4 = -\frac{50}{27}\]

Subtract the lower limit from the upper limit:
\[\text{Area} = \frac{9}{2} - \left(-\frac{50}{27}\right) = \frac{243 + 100}{54} = \frac{343}{54}\]

M1: Attempts to find the x-coordinates of the intersection points by equating \(3x^2 - 10x + 8 = x + 2\) and solving the quadratic equation.
A1: Identifies the correct limits of integration, \(x = \frac{2}{3}\) and \(x = 3\).
M1: Integrates \(-3x^2 + 11x - 6\) with at least two terms integrated correctly (power increased by 1).
A1: Correct substitution of limits into the fully correct integrated function to obtain \(\frac{343}{54}\) (or exact equivalent fraction).

To find the stationary point, we first differentiate \(y\) with respect to \(x\):
\[\frac{\text{d}y}{\text{d}x} = 2 \cdot \frac{3}{2}x^{\frac{1}{2}} - 9 = 3x^{\frac{1}{2}} - 9\]
Set \rac{\text{d}y}{\text{d}x} = 0\ to find the stationary point:
\[3x^{\frac{1}{2}} - 9 = 0\]
\[x^{\frac{1}{2}} = 3 \implies x = 9\]
Substitute \(x = 9\) back into the equation of the curve to find the y-coordinate:
\[y = 2(9)^{\frac{3}{2}} - 9(9) + 4 = 2(27) - 81 + 4 = 54 - 81 + 4 = -23\]
So the coordinates of the stationary point are \((9, -23)\).

To determine the nature of this stationary point, we find the second derivative:
\[\frac{\text{d}^2y}{\text{d}x^2} = \frac{d}{\text{d}x}(3x^{\frac{1}{2}} - 9) = \frac{3}{2}x^{-\frac{1}{2}}\]
Evaluate \(\frac{\text{d}^2y}{\text{d}x^2}\) at \(x = 9\):
\[\frac{\text{d}^2y}{\text{d}x^2} = \frac{3}{2\sqrt{9}} = \frac{1}{2}\]
Since \(\frac{\text{d}^2y}{\text{d}x^2} > 0\), the stationary point is a local minimum.

M1: Differentiates the given curve equation to find \(\frac{\text{d}y}{\text{d}x}\) (at least one term with a fractional power integrated/differentiated correctly).
A1: Obtains \(\frac{\text{d}y}{\text{d}x} = 3x^{\frac{1}{2}} - 9\), sets it to 0, and solves to find \(x = 9\).
A1: Substitutes \(x = 9\) to find \(y = -23\) to obtain the coordinate \((9, -23)\).
M1: Finds the second derivative \(\frac{\text{d}^2y}{\text{d}x^2} = \frac{3}{2}x^{-\frac{1}{2}}\), evaluates it at \(x = 9\) to get \(\frac{1}{2}\), and correctly concludes it is a local minimum because this value is positive.

Substitute \(x_0 = 1.5\) into the iterative formula to find \(x_1\): \(x_1 = \ln(3(1.5) + 2) + 0.5 = \ln(6.5) + 0.5 \approx 1.871802 + 0.5 = 2.371802\). To 3 decimal places, \(x_1 \approx 2.372\). Now substitute \(x_1 \approx 2.371802\) into the formula to find \(x_2\): \(x_2 = \ln(3(2.371802) + 2) + 0.5 = \ln(9.115407) + 0.5 \approx 2.210015 + 0.5 = 2.710015\). To 3 decimal places, \(x_2 \approx 2.710\).

M1: Attempts to calculate \(x_1\) by substituting \(x_0 = 1.5\) into the formula. A1: Both \(x_1 \approx 2.372\) and \(x_2 \approx 2.710\) correct to 3 decimal places.

Let \(f(x) = \text{e}^{-x} - \sin(x)\). Evaluate the function at the boundaries of the interval: \(f(0.5) = \text{e}^{-0.5} - \sin(0.5) \approx 0.60653 - 0.47943 = 0.1271\) (which is greater than 0). \(f(0.6) = \text{e}^{-0.6} - \sin(0.6) \approx 0.54881 - 0.56464 = -0.0158\) (which is less than 0). There is a change of sign in the interval \([0.5, 0.6]\) and \(f(x)\) is a continuous function, which confirms there is at least one root \(\alpha\) in this interval.

M1: Attempts to evaluate both \(f(0.5)\) and \(f(0.6)\) with at least one correct calculation to 2 significant figures. A1: Both values calculated correctly (accept \(0.13\) and \(-0.016\) or more accurate) with a concluding statement mentioning 'change of sign' and 'continuous function' (or equivalent).

We rearrange the equation to express \(x\) in terms of the given form: \(2x^3 - x^2 - 4 = 0 \implies 2x^3 - x^2 = 4\). Factor out \(x^2\): \(x^2(2x - 1) = 4\). Divide by \((2x - 1)\): \(x^2 = \frac{4}{2x - 1}\). Take the square root of both sides: \(x = \sqrt{\frac{4}{2x - 1}}\). Matching this to the iterative form \(x_{n+1} = \sqrt{\frac{a}{2x_n - b}}\), we identify the constants as \(a = 4\) and \(b = 1\).

M1: Factorizes \(2x^3 - x^2\) to \(x^2(2x - 1)\) or attempts to make \(x^2\) the subject of the formula. A1: Deduces \(a = 4\) and \(b = 1\) (both correct).

To find the stationary point of the curve \(C\), we differentiate \(y = (2x - 3)\mathrm{e}^{4x}\) with respect to \(x\) using the product rule. Let \(u = 2x - 3\) and \(v = \mathrm{e}^{4x}\). Then \(\frac{\mathrm{d}u}{\mathrm{d}x} = 2\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = 4\mathrm{e}^{4x}\). Applying the product rule: \(\frac{\mathrm{d}y}{\mathrm{d}x} = u\frac{\mathrm{d}v}{\mathrm{d}x} + v\frac{\mathrm{d}u}{\mathrm{d}x} = (2x - 3)(4\mathrm{e}^{4x}) + 2\mathrm{e}^{4x} = (8x - 12 + 2)\mathrm{e}^{4x} = (8x - 10)\mathrm{e}^{4x}\). At a stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\). Since \(\mathrm{e}^{4x} \neq 0\) for all real \(x\), we have \(8x - 10 = 0 \implies x = \frac{5}{4}\). Substituting \(x = \frac{5}{4}\) back into the equation of \(C\) gives \(y = \left(2\left(\frac{5}{4}\right) - 3\right)\mathrm{e}^{4\left(\frac{5}{4}\right)} = \left(\frac{5}{2} - 3\right)\mathrm{e}^5 = -\frac{1}{2}\mathrm{e}^5\). Thus, the exact coordinates of the stationary point of \(C\) are \(\left(\frac{5}{4}, -\frac{1}{2}\mathrm{e}^5\right)\).

M1: Attempts to differentiate using the product rule to obtain an expression of the form \(A\mathrm{e}^{4x} + B(2x-3)\mathrm{e}^{4x}\) where \(A, B\) are non-zero constants. A1: Correct derivative \(\frac{\mathrm{d}y}{\mathrm{d}x} = (8x - 10)\mathrm{e}^{4x}\) or equivalent. A1: Correct exact coordinates \(\left(\frac{5}{4}, -\frac{1}{2}\mathrm{e}^5\right)\) or equivalent exact forms.

To find the stationary point, we differentiate \(y = \frac{\ln x}{x^3}\) with respect to \(x\) using the quotient rule. Let \(u = \ln x\) and \(v = x^3\). Then \(\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x}\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = 3x^2\). Applying the quotient rule: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v\frac{\mathrm{d}u}{\mathrm{d}x} - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} = \frac{x^3\left(\frac{1}{x}\right) - 3x^2\ln x}{(x^3)^2} = \frac{x^2 - 3x^2\ln x}{x^6} = \frac{1 - 3\ln x}{x^4}\). At a stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \implies 1 - 3\ln x = 0\), which gives \(\ln x = \frac{1}{3}\). Solving this yields the exact \(x\)-coordinate: \(x = \mathrm{e}^{\frac{1}{3}}\).

M1: Attempts to differentiate using the quotient rule to obtain an expression of the form \(\frac{a x^2 - b x^2 \ln x}{x^6}\) where \(a, b\) are non-zero constants (or uses product rule on \(x^{-3}\ln x\) to obtain a correct form). A1: Correct derivative \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1 - 3\ln x}{x^4}\) or equivalent. A1: Correct exact \(x\)-coordinate \(x = \mathrm{e}^{\frac{1}{3}}\) or \(\sqrt[3]{\mathrm{e}}\).

First, express the integrand in partial fractions:

\[\frac{2x+1}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}\]

Multiplying by the denominator gives:

\[2x + 1 = A(x-2) + B(x-1)\]

Let \(x = 2\):

\[2(2) + 1 = B(2-1) \implies B = 5\]

Let \(x = 1\):

\[2(1) + 1 = A(1-2) \implies A = -3\]

Thus, the integral becomes:

\[\int_{3}^{4} \left( \frac{5}{x-2} - \frac{3}{x-1} \right) dx\]

Integrating each term:

\[\left[ 5\ln|x-2| - 3\ln|x-1| \right]_3^4\]

Substitute the upper limit \(x = 4\):

\[5\ln(2) - 3\ln(3)\]

Substitute the lower limit \(x = 3\):

\[5\ln(1) - 3\ln(2) = -3\ln(2)\]

Subtract the lower limit from the upper limit:

\[(5\ln 2 - 3\ln 3) - (-3\ln 2) = 8\ln 2 - 3\ln 3\]

Evaluating this numerically:

\[8\ln 2 - 3\ln 3 \approx 5.54518 - 3.29584 = 2.24934\]

To 3 significant figures, the answer is \(2.25\).

M1: Writing the algebraic fraction in partial fraction form with correct coefficients \(A = -3\) and \(B = 5\) (1.0 mark)
M1: Integrating to obtain terms in the form \(P\ln|x-2| + Q\ln|x-1|\) (1.0 mark)
A1: Substituting the limits correctly and obtaining the correct final answer to 3 significant figures: \(2.25\) (1.3 marks)

First, decompose the integrand into partial fractions:

\[\frac{5x-3}{x(2x-1)} = \frac{A}{x} + \frac{B}{2x-1}\]

Multiply through by the common denominator:

\[5x - 3 = A(2x-1) + Bx\]

Let \(x = 0\):

\[-3 = A(-1) \implies A = 3\]

Let \(x = \frac{1}{2}\):

\[5\left(\frac{1}{2}\right) - 3 = B\left(\frac{1}{2}\right) \implies -\frac{1}{2} = \frac{B}{2} \implies B = -1\]

Now rewrite the integral:

\[\int_{1}^{2} \left( \frac{3}{x} - \frac{1}{2x-1} \right) dx\]

Integrate each term, noting the chain rule factor of \(\frac{1}{2}\) for the second term:

\[\left[ 3\ln|x| - \frac{1}{2}\ln|2x-1| \right]_1^2\]

Substitute the limits:

At \(x = 2\):

\[3\ln(2) - \frac{1}{2}\ln(3)\]

At \(x = 1\):

\[3\ln(1) - \frac{1}{2}\ln(1) = 0\]

Subtracting the lower limit from the upper limit:

\[3\ln 2 - 0.5\ln 3 \approx 2.07944 - 0.54931 = 1.53013\]

To 3 significant figures, the value is \(1.53\).

M1: Correctly splitting into partial fractions to find \(A = 3\) and \(B = -1\) (1.0 mark)
M1: Integrating to obtain \(3\ln|x| - \frac{1}{2}\ln|2x-1|\) (accept missing modulus signs) (1.0 mark)
A1: Correctly substituting limits and evaluating to 3 significant figures: \(1.53\) (1.3 marks)

Factorise the denominator of the integrand:

\[x^2 + 3x + 2 = (x+1)(x+2)\]

Express in partial fractions:

\[\frac{4}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}\]

Multiply through by \((x+1)(x+2)\):

\[4 = A(x+2) + B(x+1)\]

Let \(x = -1\):

\[4 = A(1) \implies A = 4\]

Let \(x = -2\):

\[4 = B(-1) \implies B = -4\]

Now integrate:

\[\int_{0}^{2} \left( \frac{4}{x+1} - \frac{4}{x+2} \right) dx = \left[ 4\ln|x+1| - 4\ln|x+2| \right]_0^2\]

This can be simplified using logarithm laws:

\[\left[ 4\ln\left|\frac{x+1}{x+2}\right| \right]_0^2\]

Substitute the limits:

At \(x = 2\):

\[4\ln\left(\frac{3}{4}\right)\]

At \(x = 0\):

\[4\ln\left(\frac{1}{2}\right)\]

Calculate the difference:

\[4\ln\left(\frac{3}{4}\right) - 4\ln\left(\frac{1}{2}\right) = 4\ln\left(\frac{3/4}{1/2}\right) = 4\ln\left(\frac{3}{2}\right)\]

Evaluate numerically:

\[4\ln(1.5) \approx 4 \times 0.405465 = 1.62186\]

To 3 significant figures, the value is \(1.62\).

M1: Factorising the denominator and writing as partial fractions to get \(A = 4\) and \(B = -4\) (1.0 mark)
M1: Integrating to get \(4\ln|x+1| - 4\ln|x+2|\) (1.0 mark)
A1: Correct substitution of limits and evaluating to 3 significant figures: \(1.62\) (1.3 marks)

The initial mass is at \(t = 0\), which gives \(M_0 = 120\). When \(t = 8\), the mass has halved, so \(M = 60\). Substituting these values into the formula gives \(60 = 120 a^8\), which simplifies to \(a^8 = 0.5\). Solving for \(a\), we get \(a = 0.5^{1/8} \approx 0.917004\). Thus, \(a = 0.917\) to 3 significant figures.

M1: Set up a correct equation in \(a\), such as \(a^8 = 0.5\) or \(120 a^8 = 60\). A1: Correct value of \(a = 0.917\) to 3 significant figures.

Equating the two expressions for \(\theta\) gives \(\theta_0 e^{-kt} = \theta_0 10^{-ct}\), which simplifies to \(e^{-kt} = 10^{-ct}\). Taking the natural logarithm of both sides gives \(-kt = -ct \ln(10)\). Dividing by \(-t\) (since \(t \neq 0\)) gives \(k = c \ln(10)\). Therefore, \(c = \frac{k}{\ln(10)}\). Substituting \(k = 0.084\), we find \(c = \frac{0.084}{\ln(10)} \approx 0.0364808\). To 3 significant figures, \(c = 0.0365\).

M1: Equating the exponential terms or taking logarithms to relate \(k\) and \(c\), for example showing \(e^{-k} = 10^{-c}\) or \(k = c \ln(10)\). A1: Correct value of \(c = 0.0365\) to 3 significant figures.

Equating the two expressions for \(V\) gives \(5 \times 3^{2t} = 5 e^{pt}\). Dividing both sides by 5 yields \(3^{2t} = e^{pt}\). Since \(3^{2t} = (3^2)^t = 9^t\), we have \(9^t = (e^p)^t\), which implies \(e^p = 9\). Taking the natural logarithm of both sides gives \(p = \ln(9) \approx 2.19722\). To 3 significant figures, \(p = 2.20\).

M1: Equating the exponential expressions to obtain \(e^p = 9\) or taking logarithms on both sides to get \(2t \ln(3) = pt\). A1: Correct value of \(p = 2.20\) (accept 2.2) to 3 significant figures.

First, evaluate \( g(1) \):
\( g(1) = \mathrm{e}^{2(1)} - 3 = \mathrm{e}^2 - 3 \)

Next, substitute this into \( f(x) \):
\( fg(1) = f(g(1)) = f(\mathrm{e}^2 - 3) = \ln((\mathrm{e}^2 - 3) + 2) = \ln(\mathrm{e}^2 - 1) \)

Calculate the numerical value:
\( fg(1) = \ln(\mathrm{e}^2 - 1) \approx 1.8546... \)

To 3 significant figures, this is \( 1.85 \).

M1: Attempts to find \( g(1) \) by substituting \( x = 1 \) into \( g(x) \).
M1: Correctly applies the composite function order to find \( f(g(1)) \) and obtains an expression in terms of \( \mathrm{e} \).
A1: Correct answer to 3 significant figures: \( 1.85 \).

To find \( f^{-1}(5) \), we set \( f(x) = 5 \) and solve for \( x \):
\( \frac{3\mathrm{e}^x + 1}{\mathrm{e}^x - 2} = 5 \)

Multiply both sides by \( \mathrm{e}^x - 2 \):
\( 3\mathrm{e}^x + 1 = 5(\mathrm{e}^x - 2) \)
\( 3\mathrm{e}^x + 1 = 5\mathrm{e}^x - 10 \)

Rearrange to solve for \( \mathrm{e}^x \):
\( 2\mathrm{e}^x = 11 \)
\( \mathrm{e}^x = 5.5 \)

Take the natural logarithm of both sides:
\( x = \ln(5.5) \approx 1.7047... \)

Since \( 1.7047... > \ln 2 \), this is a valid solution in the domain.
To 3 significant figures, \( f^{-1}(5) = 1.70 \).

M1: Sets \( f(x) = 5 \) or attempts to find the general inverse function \( f^{-1}(x) \) by changing the subject.
M1: Solves the resulting equation to find a value for \( \mathrm{e}^x \).
A1: Correct value to 3 significant figures: \( 1.70 \).

First, find \( g(3) \):
\( g(3) = 3^2 - 2 = 7 \)

Next, evaluate \( f(7) \):
\( fg(3) = f(7) = \frac{2(7) + 3}{7 - 1} = \frac{17}{6} \)

Convert to a decimal and round to 3 significant figures:
\( \frac{17}{6} \approx 2.8333... \approx 2.83 \)

M1: Evaluates \( g(3) \) correctly to get \( 7 \).
M1: Substitutes their value of \( g(3) \) into \( f(x) \).
A1: Correct answer to 3 significant figures: \( 2.83 \).

We start with the Left-Hand Side (LHS) of the identity: LHS = \(\frac{\cos \theta}{\sec \theta - \tan \theta}\). Since \(\sec \theta = \frac{1}{\cos \theta}\) and \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we can substitute these into the expression: LHS = \(\frac{\cos \theta}{\frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta}}\). Combining the terms in the denominator gives: LHS = \(\frac{\cos \theta}{\frac{1 - \sin \theta}{\cos \theta}}\). Multiplying the numerator by the reciprocal of the denominator, we get: LHS = \(\frac{\cos^2 \theta}{1 - \sin \theta}\). Using the Pythagorean identity \(\cos^2 \theta = 1 - \sin^2 \theta\), we substitute this in: LHS = \(\frac{1 - \sin^2 \theta}{1 - \sin \theta}\). Factorising the numerator as a difference of squares: LHS = \(\frac{(1 - \sin \theta)(1 + \sin \theta)}{1 - \sin \theta}\). Cancelling the common factor of \(1 - \sin \theta\) in the numerator and denominator yields: LHS = \(1 + \sin \theta\), which is equal to the Right-Hand Side (RHS). Hence, the identity is proved.

M1: Substitute \(\sec \theta = \frac{1}{\cos \theta}\) and \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) into the denominator. M1: Simplify the complex fraction to obtain \(\frac{\cos^2 \theta}{1 - \sin \theta}\). M1: Use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to express the numerator in terms of sine. A1*: Factorise the numerator as \((1 - \sin \theta)(1 + \sin \theta)\) and cancel the common factor correctly to show the final result with no errors in working.

Using the identity \(\cot^2 x = \csc^2 x - 1\), we substitute this into the given equation: \(2(\csc^2 x - 1) + 3\csc x = 0\). Expanding and simplifying, we obtain the quadratic equation: \(2\csc^2 x + 3\csc x - 2 = 0\). Factorising this quadratic equation gives: \((2\csc x - 1)(\csc x + 2) = 0\). This yields two possible cases: 1) \(\csc x = \frac{1}{2}\), which means \(\sin x = 2\). Since the range of the sine function is \([-1, 1]\), there are no real solutions for this case. 2) \(\csc x = -2\), which means \(\sin x = -\frac{1}{2}\). For \(0 \le x < 2\pi\), the solutions for \(\sin x = -\frac{1}{2}\) lie in the third and fourth quadrants. The reference angle is \(\frac{\pi}{6}\). Thus, \(x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}\) and \(x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}\). Therefore, the solutions are \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\).

M1: Applies the identity \(\cot^2 x = \csc^2 x - 1\) to form a quadratic equation in \(\csc x\). M1: Correctly factorises or solves their quadratic equation to find values for \(\csc x\). M1: Recognises that \(\csc x = \frac{1}{2} \implies \sin x = 2\) has no solutions, and identifies \(\csc x = -2 \implies \sin x = -\frac{1}{2}\). A1: Finds both correct values of \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\) and no others in the given range.

To solve \(|2x - 5| = x + 1\), we consider two cases. Case 1: \(2x - 5 = x + 1\) which gives \(x = 6\). Case 2: \(-(2x - 5) = x + 1\), which simplifies to \(-2x + 5 = x + 1\), leading to \(3x = 4\) and thus \(x = \frac{4}{3}\). Checking both solutions in the original equation: for \(x = 6\), LHS = \(|12 - 5| = 7\) and RHS = \(6 + 1 = 7\) (Valid). For \(x = \frac{4}{3}\), LHS = \(|2(4/3) - 5| = 7/3\) and RHS = \(4/3 + 1 = 7/3\) (Valid). Therefore, the solutions are \(x = \frac{4}{3}\) and \(x = 6\).

M1: Attempts to solve one correct linear equation, e.g., \(2x - 5 = x + 1\) or \(-(2x - 5) = x + 1\). A1: Obtains one correct value of \(x\) (either \(x = 6\) or \(x = \frac{4}{3}\)). A1 (0.5 mark): Obtains both correct values of \(x\) and no extra incorrect values.

First, find the critical values by solving the corresponding equation \(|3x - 2| = 2x + 1\). Case 1: \(3x - 2 = 2x + 1\), which gives \(x = 3\). Case 2: \(-(3x - 2) = 2x + 1\), which simplifies to \(-3x + 2 = 2x + 1\), leading to \(5x = 1\) and thus \(x = 0.2\). Next, test the regions defined by these critical values. For \(x < 0.2\) (e.g. \(x = 0\)), \(3(0) - 2| = 2 > 2(0) + 1 = 1\) (True). For \(0.2 < x < 3\) (e.g. \(x = 1\)), \(|3(1) - 2| = 1 > 2(1) + 1 = 3\) (False). For \(x > 3\) (e.g. \(x = 4\)), \(|3(4) - 2| = 10 > 2(4) + 1 = 9\) (True). Therefore, the solution to the inequality is \(x < 0.2\) or \(x > 3\).

M1: Formulates and attempts to solve the boundary equations \(3x - 2 = 2x + 1\) or \(-(3x - 2) = 2x + 1\). A1: Obtains both critical values \(x = 0.2\) and \(x = 3\). A1 (0.5 mark): Correctly identifies the outside regions, writing the final inequality as \(x < 0.2\) or \(x > 3\).

The modulus term \(|2x + 3|\) has its minimum value of 0 when \(2x + 3 = 0\), which gives \(x = -1.5\). At this point, the value of \(y\) is \(4 - 0 = 4\). Therefore, the coordinates of the vertex of the graph of \(y = 4 - |2x + 3|\) are \((-1.5, 4)\). Since the coefficient of the modulus term is negative, the graph opens downwards, meaning \(y = 4\) is the maximum value of the function. For any horizontal line \(y = k\) to intersect the graph at exactly two distinct points, the line must lie strictly below the maximum point. Thus, we must have \(k < 4\).

B1: Correctly identifies the coordinates of the vertex as \((-1.5, 4)\). M1: Deduces that the graph has a maximum at the vertex and that two distinct solutions occur when \(k\) is strictly less than this maximum y-value. A1 (0.5 mark): States the correct range \(k < 4\).

The curve \(C: y = |2x - 3|\) has two branches: \(y = 2x - 3\) for \(x \ge 1.5\) (gradient 2) and \(y = -2x + 3\) for \(x < 1.5\) (gradient -2). Both branches meet at the vertex \((1.5, 0)\). The line \(l: y = mx + 5\) passes through the y-intercept \((0, 5)\). If \(m = 2\), \(l\) is parallel to the right branch of \(C\). Since \(5 > -3\), \(l\) lies entirely above the right branch and does not intersect it, but it intersects the left branch exactly once. Thus, \(m = 2\) gives exactly one intersection. By symmetry, if \(m = -2\), \(l\) is parallel to the left branch of \(C\) and lies entirely above it, but intersects the right branch exactly once. Thus, \(m = -2\) gives exactly one intersection. If the gradient is steeper than these branches, i.e., \(m > 2\) or \(m < -2\), the line will only intersect one of the branches. For \(-2 < m < 2\), the line is less steep than the branches of the modulus graph, resulting in two intersection points. Therefore, for exactly one intersection, we require \(m \le -2\) or \(m \ge 2\) (or \(|m| \ge 2\)).

M1: Identifies the critical gradients by setting \(m = 2\) and \(m = -2\) (the gradients of the two branches of the modulus graph). A1: Correctly analyzes the behavior for steeper gradients, showing that \(m \ge 2\) or \(m \le -2\) yields exactly one intersection. A1 (0.5 mark): Expresses the final range correctly as \(m \le -2\) or \(m \ge 2\).

To find the stationary points, we differentiate \(y\) with respect to \(x\). Using the chain rule, \(\frac{d}{dx}(3^{2x}) = 2 \cdot 3^{2x} \ln 3\). Thus, the derivative of the curve is: \(\frac{dy}{dx} = 2 \cdot 3^{2x} \ln 3 - 18\). Setting \(\frac{dy}{dx} = 0\) for the stationary point: \(2 \cdot 3^{2x} \ln 3 - 18 = 0\), which simplifies to \(3^{2x} \ln 3 = 9\), so \(3^{2x} = \frac{9}{\ln 3}\). Taking the natural logarithm of both sides gives: \(\ln(3^{2x}) = \ln\left(\frac{9}{\ln 3}\right)\), which simplifies to \(2x \ln 3 = \ln 9 - \ln(\ln 3)\). Since \(\ln 9 = \ln(3^2) = 2 \ln 3\), we can write this as: \(2x \ln 3 = 2 \ln 3 - \ln(\ln 3)\). Dividing both sides by \(2 \ln 3\) yields: \(x = 1 - \frac{\ln(\ln 3)}{2 \ln 3}\). Using the law of logarithms \(2 \ln 3 = \ln(3^2) = \ln 9\), we get: \(x = 1 - \frac{\ln(\ln 3)}{\ln 9}\). This is in the form \(a - \frac{\ln(\ln 3)}{\ln b}\) where \(a = 1\) and \(b = 9\).

M1: Attempts to differentiate \(3^{2x}\) to obtain a form \(k \cdot 3^{2x} \ln 3\), where \(k\) is a constant. A1: Correct derivative \(\frac{dy}{dx} = 2 \cdot 3^{2x} \ln 3 - 18\) (or equivalent). M1: Sets their \(\frac{dy}{dx} = 0\) and applies correct logarithm laws to solve for \(x\). A1: Correct final expression \(1 - \frac{\ln(\ln 3)}{\ln 9}\) (or clearly identifies \(a = 1\) and \(b = 9\)).

We want to express \( 5\sin\theta - 12\cos\theta \) in the form \( R\sin(\theta - \alpha) \). Expanding the form using the compound angle identity: \( R\sin(\theta - \alpha) = R\sin\theta\cos\alpha - R\cos\theta\sin\alpha \). Comparing coefficients of \( \sin\theta \) and \( \cos\theta \): \( R\cos\alpha = 5 \) and \( R\sin\alpha = 12 \). To find \( R \): \( R = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \). To find \( \alpha \): \( \tan\alpha = \frac{12}{5} = 2.4 \), which gives \( \alpha = \arctan(2.4) \approx 67.380^\circ \). To 1 decimal place, \( \alpha = 67.4^\circ \). Thus, the expression is \( 13\sin(\theta - 67.4^\circ) \).

M1: Attempts to find \( R \) using Pythagoras' theorem \( R = \sqrt{5^2 + 12^2} \). M1: Attempts to find \( \alpha \) using \( \tan\alpha = \frac{12}{5} \). A1: Correctly expresses the final answer as \( 13\sin(\theta - 67.4^\circ) \) or clearly states \( R = 13 \) and \( \alpha = 67.4^\circ \).

Compare \( \sqrt{3}\sin x + \cos x \) with \( R\sin(x+\alpha) = R\sin x\cos\alpha + R\cos x\sin\alpha \). We have: \( R\cos\alpha = \sqrt{3} \) and \( R\sin\alpha = 1 \). Therefore: \( R = \sqrt{(\sqrt{3})^2 + 1^2} = 2 \) and \( \tan\alpha = \frac{1}{\sqrt{3}} \implies \alpha = \frac{\pi}{6} \). The expression is \( 2\sin(x + \frac{\pi}{6}) \). The maximum value of this expression is \( 2 \). This maximum occurs when \( \sin(x + \frac{\pi}{6}) = 1 \). The smallest positive value is given by: \( x + \frac{\pi}{6} = \frac{\pi}{2} \implies x = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} \).

M1: Correctly finds \( R = 2 \) and \( \alpha = \frac{\pi}{6} \) to write \( 2\sin(x + \frac{\pi}{6}) \). A1: Correctly identifies the maximum value as 2. A1: Correctly determines the smallest positive value of \( x \) to be \( \frac{\pi}{3} \).

First express \( \sin\theta + \cos\theta \) in the form \( R\sin(\theta + \alpha) \): \( R = \sqrt{1^2 + 1^2} = \sqrt{2} \). \( \tan\alpha = \frac{1}{1} = 1 \implies \alpha = \frac{\pi}{4} \). Thus, the equation becomes: \( \sqrt{2}\sin(\theta + \frac{\pi}{4}) = 1 \implies \sin(\theta + \frac{\pi}{4}) = \frac{1}{\sqrt{2}} \). Given \( 0 \le \theta < 2\pi \), the interval for \( \theta + \frac{\pi}{4} \) is \( \frac{\pi}{4} \le \theta + \frac{\pi}{4} < \frac{9\pi}{4} \). Solving within this range: \( \theta + \frac{\pi}{4} = \frac{\pi}{4} \implies \theta = 0 \). \( \theta + \frac{\pi}{4} = \frac{3\pi}{4} \implies \theta = \frac{\pi}{2} \).

M1: Writes the equation in the form \( \sqrt{2}\sin(\theta + \frac{\pi}{4}) = 1 \) or equivalent. M1: Solves to find at least one correct value for \( \theta + \frac{\pi}{4} \) (e.g., \( \frac{\pi}{4} \) or \( \frac{3\pi}{4} \)). A1: Obtains both correct values \( \theta = 0 \) and \( \theta = \frac{\pi}{2} \) and no others in the range.

We need to find the rate of increase, which is the derivative \( \frac{\mathrm{d}P}{\mathrm{d}t} \), evaluated at \( t = 10 \).

Rewrite the equation using index notation:
\[ P = 1200(1 + 3e^{-0.2t})^{-1} \]

Differentiating with respect to \( t \) using the chain rule:
\[ \frac{\mathrm{d}P}{\mathrm{d}t} = -1200(1 + 3e^{-0.2t})^{-2} \times \left(-0.6e^{-0.2t}\right) \]
\[ \frac{\mathrm{d}P}{\mathrm{d}t} = \frac{720e^{-0.2t}}{(1 + 3e^{-0.2t})^2} \]

Substitute \( t = 10 \):
\[ \frac{\mathrm{d}P}{\mathrm{d}t} = \frac{720e^{-2}}{(1 + 3e^{-2})^2} \]

Using \( e^{-2} \approx 0.135335 \):
\[ \frac{\mathrm{d}P}{\mathrm{d}t} \approx \frac{720 \times 0.135335}{(1 + 3 \times 0.135335)^2} \approx \frac{97.441}{(1.406)^2} \approx 49.3 \]

- **M1**: Attempts to differentiate \( P \) with respect to \( t \). Look for a form of \( A e^{-0.2t} (1 + 3e^{-0.2t})^{-2} \).
- **A1**: Correct derivative: \( \frac{\mathrm{d}P}{\mathrm{d}t} = \frac{720e^{-0.2t}}{(1 + 3e^{-0.2t})^2} \) or equivalent.
- **A1**: Correct final rate of \( 49.3 \) (accept \( 49.3 \) or \( 49 \) following correct working).

Using the given information:

At \( t = 2 \), \( 150 = A e^{2k} \) --- (Equation 1)

At \( t = 5 \), \( 1350 = A e^{5k} \) --- (Equation 2)

Dividing Equation 2 by Equation 1:
\[ \frac{1350}{150} = e^{3k} \implies e^{3k} = 9 \]

This gives:
\[ k = \frac{1}{3}\ln 9 = \ln(9^{1/3}) \approx 0.73241 \]

We need to find the rate of increase, \( \frac{\mathrm{d}C}{\mathrm{d}t} \), when \( t = 4 \).

Differentiating \( C \) with respect to \( t \):
\[ \frac{\mathrm{d}C}{\mathrm{d}t} = k A e^{kt} = k C \]

First, find the population \( C \) when \( t = 4 \):
\[ C(4) = A e^{4k} = (A e^{2k}) \times e^{2k} = 150 e^{2k} \]

Since \( e^{3k} = 9 \), we have \( e^{2k} = 9^{2/3} \approx 4.32675 \).

Thus:
\[ C(4) = 150 \times 9^{2/3} \approx 649.01 \]

Now, calculate the rate of increase:
\[ \frac{\mathrm{d}C}{\mathrm{d}t} = k C(4) \approx 0.73241 \times 649.01 \approx 475.3 \]

To 3 significant figures, the rate of increase is 475 cells per hour.

- **M1**: Attempts to find the growth constant by setting up \( e^{3k} = 9 \) or equivalent to find \( k = \frac{1}{3}\ln 9 \) (or the base multiplier \( b = 9^{1/3} \approx 2.08 \)).
- **M1**: Correct method to find the rate of growth at \( t = 4 \), by finding \( C(4) \approx 649 \) and using \( \frac{\mathrm{d}C}{\mathrm{d}t} = k C \), or by differentiating \( C = A e^{kt} \) directly and substituting \( t = 4 \).
- **A1**: Correct rate of \( 475 \) (accept \( 475 \) or \( 475.3 \)).

The general solution to the differential equation \( \frac{\mathrm{d}P}{\mathrm{d}t} = 0.08 P \) with an initial population \( P(0) = 500 \) is:
\[ P = 500 e^{0.08t} \]

We want to find the time \( t \) when the rate of increase \( \frac{\mathrm{d}P}{\mathrm{d}t} = 100 \).

Substituting \( \frac{\mathrm{d}P}{\mathrm{d}t} = 100 \) into the differential equation:
\[ 100 = 0.08 P \implies P = 1250 \]

Now, substitute \( P = 1250 \) into our population equation:
\[ 1250 = 500 e^{0.08t} \]
\[ e^{0.08t} = 2.5 \]
\[ 0.08t = \ln(2.5) \]
\[ t = \frac{\ln(2.5)}{0.08} \approx 11.458 \]

To 3 significant figures, the time taken is 11.5 days.

- **M1**: Formulates the population equation as \( P = 500 e^{0.08t} \) or establishes that the population must reach \( P = 1250 \) when the rate is 100.
- **M1**: Sets up a correct exponential equation to solve for \( t \), e.g., \( 500 e^{0.08t} = 1250 \) or \( 40 e^{0.08t} = 100 \).
- **A1**: Correct time of \( 11.5 \) (accept \( 11.5 \) or \( 11.46 \)).

First, express the term in index form: \(\frac{1}{\sqrt{4 - 3x}} = (4 - 3x)^{-\frac{1}{2}}\). Factoring out the 4 gives: \(4^{-\frac{1}{2}} \left(1 - \frac{3}{4}x\right)^{-\frac{1}{2}} = \frac{1}{2} \left(1 - \frac{3}{4}x\right)^{-\frac{1}{2}}\). Applying the binomial expansion formula \((1 + y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \dots\) with \(y = -\frac{3}{4}x\) and \(n = -\frac{1}{2}\): \(\left(1 - \frac{3}{4}x\right)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)\left(-\frac{3}{4}x\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(-\frac{3}{4}x\right)^2 + \dots = 1 + \frac{3}{8}x + \frac{27}{128}x^2 + \dots\). Multiplying through by the scale factor \(\frac{1}{2}\): \(\frac{1}{2} \left(1 + \frac{3}{8}x + \frac{27}{128}x^2\right) = \frac{1}{2} + \frac{3}{16}x + \frac{27}{256}x^2\).

M1: For factoring out 4 to obtain a leading term of \(4^{-\frac{1}{2}}\) or \(\frac{1}{2}\). M1: For applying the binomial expansion formula with correct structure to \((1 - \frac{3}{4}x)^{-\frac{1}{2}}\), having at least two terms correct. A0.6: For simplifying terms to obtain the final exact expansion: \(\frac{1}{2} + \frac{3}{16}x + \frac{27}{256}x^2\).

We use the binomial expansion \((1 + y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \dots\) with \(y = 5x\) and \(n = -2\). Substitute these values: \((1 + 5x)^{-2} = 1 + (-2)(5x) + \frac{(-2)(-3)}{2!}(5x)^2 + \dots = 1 - 10x + 3(25x^2) = 1 - 10x + 75x^2\). The expansion is valid when \(|5x| < 1\), which simplifies to \(|x| < \frac{1}{5}\).

M1: For attempting the binomial expansion with correct structure for at least two terms of \((1 + 5x)^{-2}\). A1: For obtaining the correct expansion \(1 - 10x + 75x^2\). B0.6: For identifying the correct range of validity, \(|x| < \frac{1}{5}\) or equivalent.

Rewrite the expression by factoring out 9: \((9 - x)^{\frac{1}{2}} = 9^{\frac{1}{2}} \left(1 - \frac{1}{9}x\right)^{\frac{1}{2}} = 3 \left(1 - \frac{1}{9}x\right)^{\frac{1}{2}}\). Expand \(\left(1 - \frac{1}{9}x\right)^{\frac{1}{2}}\) using the formula with \(n = \frac{1}{2}\) and \(y = -\frac{1}{9}x\): \(\left(1 - \frac{1}{9}x\right)^{\frac{1}{2}} = 1 + \left(\frac{1}{2}\right)\left(-\frac{1}{9}x\right) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2}\left(-\frac{1}{9}x\right)^2 + \dots = 1 - \frac{1}{18}x - \frac{1}{648}x^2 + \dots\). Multiplying by the factor of 3: \(3 \left(1 - \frac{1}{18}x - \frac{1}{648}x^2\right) = 3 - \frac{1}{6}x - \frac{1}{216}x^2\).

M1: For factoring out 9 to obtain \(3\) as a multiplier. M1: For correct binomial expansion structure applied to \((1 - \frac{1}{9}x)^{\frac{1}{2}}\). A0.6: For obtaining the final simplified expansion \(3 - \frac{1}{6}x - \frac{1}{216}x^2\).

Let \(I = \int_{1}^{4} \frac{\ln(2x)}{\sqrt{x}} \, \mathrm{d}x\). We use integration by parts: \(\int u \frac{\mathrm{d}v}{\mathrm{d}x} \, \mathrm{d}x = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x} \, \mathrm{d}x\). Let \(u = \ln(2x)\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = x^{-1/2}\). Differentiating \(u\) gives \(\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x}\). Integrating \(\frac{\mathrm{d}v}{\mathrm{d}x}\) gives \(v = 2x^{1/2}\). Applying the integration by parts formula: \(I = \left[ 2x^{1/2}\ln(2x) \right]_{1}^{4} - \int_{1}^{4} 2x^{1/2} \cdot \frac{1}{x} \, \mathrm{d}x = \left[ 2\sqrt{x}\ln(2x) \right]_{1}^{4} - \int_{1}^{4} 2x^{-1/2} \, \mathrm{d}x = \left[ 2\sqrt{x}\ln(2x) - 4\sqrt{x} \right]_{1}^{4}\). Substituting the upper limit \(x = 4\): \(2\sqrt{4}\ln(8) - 4\sqrt{4} = 4(3\ln 2) - 8 = 12\ln 2 - 8\). Substituting the lower limit \(x = 1\): \(2\sqrt{1}\ln(2) - 4\sqrt{1} = 2\ln 2 - 4\). Subtracting the lower limit from the upper limit: \(I = (12\ln 2 - 8) - (2\ln 2 - 4) = 10\ln 2 - 4\). Thus, \(a = 10\) and \(b = -4\).

M1: Attempts integration by parts with \(u = \ln(2x)\) and \(v' = x^{-1/2}\) to find \(u' = \frac{1}{x}\) and \(v = k x^{1/2}\) (where \(k > 0\)).
A1: Obtains a correct intermediate expression for the integration by parts, e.g., \(2\sqrt{x}\ln(2x) - \int 2x^{-1/2} \, \mathrm{d}x\).
A1: Correctly integrates to obtain the full integrated expression \(2\sqrt{x}\ln(2x) - 4\sqrt{x}\).
M1: Substitutes limits of 4 and 1 into an expression of the form \(A\sqrt{x}\ln(2x) - B\sqrt{x}\) and subtracts.
A1: Obtains the correct final exact value \(10\ln 2 - 4\) (or clearly states \(a = 10\) and \(b = -4\)).

Assume that there exists an integer \(n\) such that \(n^2 + 5\) is an odd integer and \(n\) is an odd integer.

Since \(n\) is odd, we can write \(n = 2k + 1\) for some integer \(k \in \mathbb{Z}\).

Substitute this expression into \(n^2 + 5\):
\(n^2 + 5 = (2k + 1)^2 + 5\)
\(n^2 + 5 = 4k^2 + 4k + 1 + 5\)
\(n^2 + 5 = 4k^2 + 4k + 6\)
\(n^2 + 5 = 2(2k^2 + 2k + 3)\)

Since \(k\) is an integer, \(2k^2 + 2k + 3\) is also an integer. Therefore, \(n^2 + 5\) is an even integer.

This contradicts the assumption that \(n^2 + 5\) is an odd integer. Hence, the assumption that \(n\) is odd must be false, which proves that \(n\) is an even integer.

M1: States the negation of the statement, assuming that there exists an integer \(n\) such that \(n^2 + 5\) is odd and \(n\) is odd.
M1: Writes \(n = 2k + 1\) (or equivalent) and substitutes this into \(n^2 + 5\) to obtain a factor of 2, i.e., \(2(2k^2 + 2k + 3)\).
A1: Explains clearly that this expression is even, which contradicts the assumption that \(n^2 + 5\) is odd, and concludes the proof.

Assume that there exist positive integers \(a\) and \(b\) such that \(a^2 - 4b^2 = 15\).

We can factorise the left-hand side of the equation as a difference of two squares:
\((a - 2b)(a + 2b) = 15\)

Since \(a\) and \(b\) are positive integers, both \(a - 2b\) and \(a + 2b\) must be integers. Furthermore, since \(a, b \ge 1\), we have \(a + 2b > a - 2b\) and \(a + 2b > 0\), which implies that both factors must be positive integers.

The only pairs of positive integer factors of 15 are \((15, 1)\) and \((5, 3)\).

Case 1: \(a + 2b = 15\) and \(a - 2b = 1\)
Subtracting the second equation from the first gives:
\(4b = 14 \implies b = 3.5\)
This contradicts the assumption that \(b\) is an integer.

Case 2: \(a + 2b = 5\) and \(a - 2b = 3\)
Subtracting the second equation from the first gives:
\(4b = 2 \implies b = 0.5\)
This also contradicts the assumption that \(b\) is an integer.

Since both possible cases lead to a contradiction, our assumption must be false. Therefore, there are no positive integers \(a\) and \(b\) such that \(a^2 - 4b^2 = 15\).

M1: States the negation of the statement, assuming that there exist positive integers \(a\) and \(b\) such that \(a^2 - 4b^2 = 15\).
M1: Factorises to \((a-2b)(a+2b) = 15\) and identifies the possible positive integer factor pairs of 15, which are \((15, 1)\) and \((5, 3)\).
A1: Solves for \(b\) in both cases to show that \(b\) is not an integer (i.e., \(b = 3.5\) and \(b = 0.5\)), and concludes that this contradicts the assumption that \(a\) and \(b\) are integers.

Let \(u = \mathrm{e}^x\). Then \(\frac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^x\), which gives \(\mathrm{d}x = \frac{\mathrm{d}u}{u}\). Changing the limits of integration: When \(x = 0\), \(u = \mathrm{e}^0 = 1\). When \(x = \ln 2\), \(u = \mathrm{e}^{\ln 2} = 2\). Substituting these into the integral gives: \(\int_{0}^{\ln 2} \frac{\mathrm{e}^{3x}}{\mathrm{e}^{2x} + 3} \, \mathrm{d}x = \int_{1}^{2} \frac{u^3}{u^2 + 3} \frac{\mathrm{d}u}{u} = \int_{1}^{2} \frac{u^2}{u^2 + 3} \, \mathrm{d}u\). Using algebraic division or matching terms: \(\frac{u^2}{u^2 + 3} = \frac{u^2 + 3 - 3}{u^2 + 3} = 1 - \frac{3}{u^2 + 3}\). Now integrate with respect to \(u\): \(\int_{1}^{2} \left(1 - \frac{3}{u^2 + 3}\right) \mathrm{d}u = \left[ u - \frac{3}{\sqrt{3}}\arctan\left(\frac{u}{\sqrt{3}}\right) \right]_{1}^{2} = \left[ u - \sqrt{3}\arctan\left(\frac{u}{\sqrt{3}}\right) \right]_{1}^{2}\). Evaluating this at the limits: \(= \left(2 - \sqrt{3}\arctan\left(\frac{2}{\sqrt{3}}\right)\right) - \left(1 - \sqrt{3}\arctan\left(\frac{1}{\sqrt{3}}\right)\right)\). Using the known value \(\arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\) and rationalising the denominator \(\frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\), we obtain: \(= 1 + \frac{1}{6}\sqrt{3}\pi - \sqrt{3}\arctan\left(\frac{2\sqrt{3}}{3}\right)\). Comparing this with the required form, we find the constants: \(A = 1\), \(B = \frac{1}{6}\), and \(C = -1\).

M1: Differentiates the substitution to find \(\mathrm{d}x = \frac{\mathrm{d}u}{u}\) or equivalent, and attempts to change the limits to \(u = 1\) and \(u = 2\). A1: Obtains the correct simplified integral in terms of \(u\): \(\int_{1}^{2} \frac{u^2}{u^2 + 3} \, \mathrm{d}u\). M1: Rewrites the integrand in the form \(1 - \frac{c}{u^2 + 3}\) and integrates to the form \(u - k\arctan\left(\frac{u}{\sqrt{3}}\right)\). A1: Correctly integrates to obtain \(u - \sqrt{3}\arctan\left(\frac{u}{\sqrt{3}}\right)\). B0.5: Evaluates the integral using the limits to find the exact values, correctly identifying \(A = 1\), \(B = \frac{1}{6}\), and \(C = -1\).

Let \(x = 2\sin\theta\). Then \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 2\cos\theta\), which gives \(\mathrm{d}x = 2\cos\theta \, \mathrm{d}\theta\). Now change the limits: When \(x = 0\), \(2\sin\theta = 0 \implies \theta = 0\). When \(x = 1\), \(2\sin\theta = 1 \implies \sin\theta = \frac{1}{2} \implies \theta = \frac{\pi}{6}\). Substituting these into the integral: \(\int_{0}^{1} \frac{x^2}{\sqrt{4-x^2}} \, \mathrm{d}x = \int_{0}^{\pi/6} \frac{4\sin^2\theta}{\sqrt{4 - 4\sin^2\theta}} (2\cos\theta) \, \mathrm{d}\theta\). Since \(\sqrt{4 - 4\sin^2\theta} = \sqrt{4\cos^2\theta} = 2\cos\theta\) (as \(\cos\theta \ge 0\) for \(0 \le \theta \le \frac{\pi}{6}\)), the integral simplifies to: \(\int_{0}^{\pi/6} \frac{4\sin^2\theta}{2\cos\theta} (2\cos\theta) \, \mathrm{d}\theta = \int_{0}^{\pi/6} 4\sin^2\theta \, \mathrm{d}\theta\). Apply the double-angle identity \(\cos 2\theta = 1 - 2\sin^2\theta \implies 2\sin^2\theta = 1 - \cos 2\theta\): \(\int_{0}^{\pi/6} 4\sin^2\theta \, \mathrm{d}\theta = \int_{0}^{\pi/6} 2(1 - \cos 2\theta) \, \mathrm{d}\theta\). Integrate with respect to \(\theta\): \(= \left[ 2\theta - \sin 2\theta \right]_{0}^{\pi/6}\). Substitute the limits: \(= \left( 2\left(\frac{\pi}{6}\right) - \sin\left(\frac{\pi}{3}\right) \right) - (0) = \frac{\pi}{3} - \frac{\sqrt{3}}{2}\). Comparing this with the form \(a\pi - b\sqrt{3}\), we obtain \(a = \frac{1}{3}\) and \(b = \frac{1}{2}\).

M1: Correctly differentiates to find \(\mathrm{d}x = 2\cos\theta \, \mathrm{d}\theta\) and converts limits to \(0\) and \(\frac{\pi}{6\pi}\). M1: Substitutes expressions into the integrand and correctly simplifies \(\sqrt{4 - x^2}\) to \(2\cos\theta\). A1: Obtains the correct simplified integral: \(\int_{0}^{\pi/6} 4\sin^2\theta \, \mathrm{d}\theta\). M1: Uses the double angle identity \(\cos 2\theta = 1 - 2\sin^2\theta\) to rewrite the integrand and integrates to the form \(p\theta - q\sin 2\theta\). A0.5: Evaluates the limits correctly to obtain the exact expression and identifies \(a = \frac{1}{3}\) and \(b = \frac{1}{2}\).

To find the gradient of the tangent, we use parametric differentiation:
\[\frac{dx}{dt} = 4t\]
\[\frac{dy}{dt} = 3t^2 - 1\]
Using the chain rule, the gradient \(\frac{dy}{dx}\) is given by:
\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2 - 1}{4t}\]
At the point where \(t = 2\):
\[\frac{dy}{dx} = \frac{3(2)^2 - 1}{4(2)} = \frac{12 - 1}{8} = \frac{11}{8}\]

M1: Attempts to differentiate both \(x\) and \(y\) with respect to \(t\) to find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
M1: Applies the chain rule \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\) and substitutes \(t = 2\).
A0.75: Correct exact gradient of \(\frac{11}{8}\) or equivalent.

First, find the coordinates of the point where \(t = \frac{\pi}{6}\):
\[x = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\]
\[y = \cos\left(2 \cdot \frac{\pi}{6}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\]
Next, differentiate \(x\) and \(y\) with respect to \(t\):
\[\frac{dx}{dt} = \cos(t)\]
\[\frac{dy}{dt} = -2\sin(2t)\]
Using the chain rule, the gradient of the tangent is:
\[\frac{dy}{dx} = \frac{-2\sin(2t)}{\cos(t)}\]
At \(t = \frac{\pi}{6}\):
\[\frac{dy}{dx} = \frac{-2\sin(\frac{\pi}{3})}{\cos(\frac{\pi}{6})} = \frac{-2 \left(\frac{\sqrt{3}}{2}\right)}{\frac{\sqrt{3}}{2}} = -2\]
The equation of the tangent is:
\[y - y_1 = m(x - x_1)\]
\[y - \frac{1}{2} = -2\left(x - \frac{1}{2}\right)\]
\[y - \frac{1}{2} = -2x + 1 \implies y = -2x + \frac{3}{2}\]

M1: Finds the coordinates of the point at \(t = \frac{\pi}{6}\) and attempts to differentiate to find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
M1: Uses the chain rule to find \(\frac{dy}{dx}\) and evaluates this gradient at \(t = \frac{\pi}{6}\).
A0.75: Correct equation of the tangent in the required form, e.g., \(y = -2x + 1.5\) or \(y = -2x + \frac{3}{2}\).

Differentiating \(x\) with respect to \(t\):
\[\frac{dx}{dt} = 2e^{2t}\]
Differentiating \(y\) with respect to \(t\) using the product rule:
\[\frac{dy}{dt} = 1 \cdot e^t + t \cdot e^t = (1+t)e^t\]
Using the chain rule, the gradient is:
\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{(1+t)e^t}{2e^{2t}} = \frac{1+t}{2e^t}\]
At \(t = 0\):
\[\frac{dy}{dx} = \frac{1+0}{2e^0} = \frac{1}{2}\]

M1: Differentiates \(x\) and \(y\) with respect to \(t\), correctly applying the product rule for \(y\).
M1: Applies the chain rule to obtain an expression for \(\frac{dy}{dx}\) in terms of \(t\).
A0.75: Correctly evaluates \(\frac{dy}{dx} = \frac{1}{2}\) at \(t = 0\).

Differentiating both parametric equations with respect to \(t\):
\[\frac{dx}{dt} = \frac{1}{t+1}\]
\[\frac{dy}{dt} = 2t\]
Applying the chain rule:
\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t}{\frac{1}{t+1}} = 2t(t+1) = 2t^2 + 2t\]
The tangent is parallel to the line \(y = 4x - 5\), so its gradient must be \(4\):
\[2t^2 + 2t = 4 \implies 2t^2 + 2t - 4 = 0 \implies t^2 + t - 2 = 0\]
Factoring the quadratic:
\[(t+2)(t-1) = 0\]
Since we are given \(t > -1\), we choose \(t = 1\).
Now we find the coordinates at \(t = 1\):
\[x = \ln(1 + 1) = \ln(2)\]
\[y = 1^2 + 2 = 3\]
Thus, the exact coordinates of the point are \((\ln(2), 3)\).

M1: Correctly differentiates \(x\) and \(y\) with respect to \(t\) to find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
M1: Equates their expression for \(\frac{dy}{dx}\) to 4, solves the quadratic equation, and selects the correct value \(t = 1\) based on the domain.
A0.75: Substitutes \(t = 1\) to find the correct exact coordinates \((\ln(2), 3)\).

Let \( X \) be a general point on the line \( l \). The coordinates of \( X \) are \( (2 + \lambda, 1 - \lambda, -1 + 2\lambda) \).
The vector \( \vec{PX} \) is given by:
\[ \vec{PX} = \vec{OX} - \vec{OP} = \begin{pmatrix} 2 + \lambda \\ 1 - \lambda \\ -1 + 2\lambda \end{pmatrix} - \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 + \lambda \\ 3 - \lambda \\ -5 + 2\lambda \end{pmatrix} \]
For the shortest distance from \( P \) to \( l \), the vector \( \vec{PX} \) must be perpendicular to the direction vector of the line, \( \mathbf{d} = \mathbf{i} - \mathbf{j} + 2\mathbf{k} \).
Setting the scalar product \( \vec{PX} \cdot \mathbf{d} = 0 \):
\[ (1 + \lambda)(1) + (3 - \lambda)(-1) + (-5 + 2\lambda)(2) = 0 \]
\[ 1 + \lambda - 3 + \lambda - 10 + 4\lambda = 0 \]
\[ 6\lambda - 12 = 0 \implies \lambda = 2 \]
Substitute \( \lambda = 2 \) back into the vector \( \vec{PX} \):
\[ \vec{PX} = \begin{pmatrix} 1 + 2 \\ 3 - 2 \\ -5 + 4 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix} \]
The shortest distance is the magnitude of this perpendicular vector:
\[ |\vec{PX}| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{9 + 1 + 1} = \sqrt{11} \]

M1: Attempts to express \( \vec{PX} \) in terms of \( \lambda \) and sets up the scalar product equation \( \vec{PX} \cdot \mathbf{d} = 0 \) to find \( \lambda \).
A1: Solves to find \( \lambda = 2 \) and obtains the perpendicular vector \( 3\mathbf{i} + \mathbf{j} - \mathbf{k} \) (or equivalent).
A1: Correctly calculates the magnitude of the perpendicular vector to find the shortest distance is \( \sqrt{11} \).

If the lines intersect, there must exist values of \( \mu \) and \( \nu \) that yield the same position vector for the point of intersection.
Equating the \( x \)- and \( z \)-components of the two lines gives:
1) \( 1 + 2\mu = 2 - \nu \implies 2\mu + \nu = 1 \)
2) \( 3 - \mu = 4 + \nu \implies \mu + \nu = -1 \)

Subtracting the second equation from the first equation:
\[ (2\mu + \nu) - (\mu + \nu) = 1 - (-1) \implies \mu = 2 \]
Substituting \( \mu = 2 \) into the second equation:
\[ 2 + \nu = -1 \implies \nu = -3 \]

Now, equating the \( y \)-components of both lines:
\[ a + \mu = -1 + 3\nu \]
Substitute \( \mu = 2 \) and \( \nu = -3 \):
\[ a + 2 = -1 + 3(-3) \]
\[ a + 2 = -10 \implies a = -12 \]

M1: Equates the \( x \)- and \( z \)-components to set up a system of simultaneous equations in \( \mu \) and \( \nu \).
A1: Solves the equations to find the parameters \( \mu = 2 \) and \( \nu = -3 \).
A1: Equates the \( y \)-components and uses the values of \( \mu \) and \( \nu \) to obtain \( a = -12 \).

The direction vectors of the lines \( l_1 \) and \( l_2 \) are respectively:
\[ \mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} \quad \text{and} \quad \mathbf{d}_2 = \begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix} \]
To find the cosine of the angle between them, we use the scalar product formula:
\[ \cos \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1| |\mathbf{d}_2|} \]

First, calculate the scalar product:
\[ \mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(3) + (-1)(4) + (2)(0) = 6 - 4 = 2 \]

Next, calculate the magnitudes of each vector:
\[ |\mathbf{d}_1| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3 \]
\[ |\mathbf{d}_2| = \sqrt{3^2 + 4^2 + 0^2} = \sqrt{25} = 5 \]

Substitute these values into the formula:
\[ \cos \theta = \frac{2}{3 \times 5} = \frac{2}{15} \]

M1: Identifies the direction vectors and attempts to calculate their scalar product \( \mathbf{d}_1 \cdot \mathbf{d}_2 \).
A1: Correctly calculates both magnitudes: \( |\mathbf{d}_1| = 3 \) and \( |\mathbf{d}_2| = 5 \).
A1: Applies the cosine formula to obtain \( \cos \theta = \frac{2}{15} \).

To find the gradient of the curve, we differentiate the equation implicitly with respect to \(x\):

\[ \frac{d}{dx}(2x^2) + \frac{d}{dx}(3xy) - \frac{d}{dx}(y^2) = \frac{d}{dx}(13) \]

Using the product rule on \(3xy\) and the chain rule on \(y^2\):

\[ 4x + 3\left(y + x\frac{dy}{dx}\right) - 2y\frac{dy}{dx} = 0 \]

Now substitute \(x = 2\) and \(y = 1\):

\[ 4(2) + 3\left(1 + 2\frac{dy}{dx}\right) - 2(1)\frac{dy}{dx} = 0 \]

\[ 8 + 3 + 6\frac{dy}{dx} - 2\frac{dy}{dx} = 0 \]

\[ 11 + 4\frac{dy}{dx} = 0 \]

\[ \frac{dy}{dx} = -\frac{11}{4} \]

M1: Differentiates implicitly with respect to \(x\) to obtain \(4x + 3y + 3x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0\). Allow minor errors, but must see a product rule attempt on \(3xy\) and a correct term for \(-2y\frac{dy}{dx}\).
M1: Substitutes \(x = 2\) and \(y = 1\) into their differentiated expression to find a value for \(\frac{dy}{dx}\).
A1: Correct gradient of \(-\frac{11}{4}\) (or \(-2.75\)).

Differentiating implicitly with respect to \(x\):

\[ 3y^2\frac{dy}{dx} + \left( 6xy + 3x^2\frac{dy}{dx} \right) - 3x^2 = 0 \]

Substitute the coordinates of the point \((1, 2)\), so \(x = 1\) and \(y = 2\):

\[ 3(2)^2\frac{dy}{dx} + 6(1)(2) + 3(1)^2\frac{dy}{dx} - 3(1)^2 = 0 \]

\[ 12\frac{dy}{dx} + 12 + 3\frac{dy}{dx} - 3 = 0 \]

\[ 15\frac{dy}{dx} + 9 = 0 \]

\[ \frac{dy}{dx} = -\frac{9}{15} = -\frac{3}{5} \]

M1: Differentiates implicitly with respect to \(x\) to get \(3y^2\frac{dy}{dx} + 6xy + 3x^2\frac{dy}{dx} - 3x^2 = 0\). Allow M1 if at least two terms are differentiated correctly, with one being a product rule term.
M1: Substitutes \(x = 1\) and \(y = 2\) into their differentiated equation and attempts to solve for \(\frac{dy}{dx}\).
A1: Correct value of \(\frac{dy}{dx} = -\frac{3}{5}\) (or \(-0.6\)).

Differentiating the equation implicitly with respect to \(x\):

\[ \cos(x) - \sin(y)\frac{dy}{dx} = \frac{1}{2}y + \frac{1}{2}x\frac{dy}{dx} \]

Substitute the point \((\pi, 0)\), so \(x = \pi\) and \(y = 0\):

\[ \cos(\pi) - \sin(0)\frac{dy}{dx} = \frac{1}{2}(0) + \frac{1}{2}\pi\frac{dy}{dx} \]

Since \(\cos(\pi) = -1\) and \(\sin(0) = 0\):

\[ -1 - 0 = \frac{\pi}{2}\frac{dy}{dx} \]

\[ \frac{dy}{dx} = -\frac{2}{\pi} \]

M1: Differentiates implicitly with respect to \(x\), showing \(\cos(x)\), \(-\sin(y)\frac{dy}{dx}\), and an application of the product rule on \(\frac{1}{2}xy\).
M1: Substitutes \(x = \pi\) and \(y = 0\) into their differentiated expression to solve for \(\frac{dy}{dx}\).
A1: Correct answer of \(-\frac{2}{\pi}\) or exact equivalent.

Let the volume of the spherical balloon be \(V\) and its radius be \(r\). We are given that \(\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{k}{V}\), where \(k\) is a positive constant. Separating variables gives \(\int V \, \mathrm{d}V = \int k \, \mathrm{d}t\), which integrates to \(\frac{1}{2}V^2 = kt + C\), where \(C\) is a constant of integration. Since the balloon is a sphere, \(V = \frac{4}{3}\pi r^3\). Substituting this into the integrated equation gives \(\frac{1}{2}\left(\frac{4}{3}\pi r^3\right)^2 = kt + C\), which simplifies to \(\frac{8}{9}\pi^2 r^6 = kt + C\). This can be written in the form \(r^6 = \lambda t + \mu\), where \(\lambda = \frac{9k}{8\pi^2}\) and \(\mu = \frac{9C}{8\pi^2}\). Using the initial condition \(t = 0\), \(r = 1\), we get \(1^6 = \lambda(0) + \mu \implies \mu = 1\). Using the condition \(t = 3\), \(r = 2\), we get \(2^6 = 3\lambda + 1 \implies 64 = 3\lambda + 1 \implies \lambda = 21\). Thus, the equation for the radius is \(r^6 = 21t + 1\). When \(t = 9\), we have \(r^6 = 21(9) + 1 = 189 + 1 = 190\). Therefore, the exact radius of the balloon is \(190^{1/6}\text{ cm}\).

M1: Sets up the differential equation and separates variables to obtain \(\int V \mathrm{d}V = \int k \mathrm{d}t\).
M1: Integrates to find \(\frac{1}{2}V^2 = kt + C\) and substitutes \(V = \frac{4}{3}\pi r^3\) to write the relation in terms of \(r\) and \(t\).
M1: Applies the boundary conditions \(t = 0, r = 1\) and \(t = 3, r = 2\) to find the constants of integration.
A1: Obtains the correct relation \(r^6 = 21t + 1\).
A1: Finds the correct exact radius \(190^{1/6}\).

We are given that \(\frac{\mathrm{d}V}{\mathrm{d}t} = k \frac{A}{V}\), where \(k\) is a positive constant. Since \(V = \frac{4}{3}\pi r^3\), the chain rule gives \(\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}r}\frac{\mathrm{d}r}{\mathrm{d}t} = 4\pi r^2 \frac{\mathrm{d}r}{\mathrm{d}t}\). Substituting \(A = 4\pi r^2\) and \(V = \frac{4}{3}\pi r^3\), we get \(4\pi r^2 \frac{\mathrm{d}r}{\mathrm{d}t} = k \frac{4\pi r^2}{\frac{4}{3}\pi r^3} = \frac{3k}{r}\). Dividing both sides by \(4\pi r^2\) (since \(r > 0\)) gives \(\frac{\mathrm{d}r}{\mathrm{d}t} = \frac{3k}{4\pi r^3}\). Separating variables, we get \(\int r^3 \mathrm{d}r = \int \frac{3k}{4\pi} \mathrm{d}t\), which integrates to \(\frac{1}{4}r^4 = \frac{3k}{4\pi}t + C\), where \(C\) is a constant. This can be written as \(r^4 = \lambda t + \mu\), where \(\lambda = \frac{3k}{\pi}\) and \(\mu = 4C\). Using the initial condition \(t = 0\), \(r = 2\), we get \(2^4 = \mu \implies \mu = 16\). Using \(t = 5\), \(r = 3\), we get \(3^4 = 5\lambda + 16 \implies 81 = 5\lambda + 16 \implies 5\lambda = 65 \implies \lambda = 13\). Thus, the relation is \(r^4 = 13t + 16\). When \(t = 13\), we have \(r^4 = 13(13) + 16 = 169 + 16 = 185\). Therefore, the exact radius of the cell is \(185^{1/4}\text{ \mu m}\).

M1: Uses the chain rule \(\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}r}\frac{\mathrm{d}r}{\mathrm{d}t}\) and substitutes the formulas for \(V\) and \(A\) to set up the differential equation in terms of \(r\) and \(t\).
M1: Separates variables and integrates to obtain \(r^4 = \lambda t + \mu\).
M1: Applies the conditions \(t = 0, r = 2\) and \(t = 5, r = 3\) to find the values of the constants.
A1: Finds the correct relation \(r^4 = 13t + 16\).
A1: Obtains the correct exact radius \(185^{1/4}\).

For part (a), we know that \(V = \frac{4}{3}\pi r^3\), so \(\frac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2 = S\). By the chain rule, \(\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}r}\frac{\mathrm{d}r}{\mathrm{d}t} = S \frac{\mathrm{d}r}{\mathrm{d}t\). Substituting this into the given differential equation gives \(S \frac{\mathrm{d}r}{\mathrm{d}t} = k S \mathrm{e}^{-t}\). Since the surface area \(S > 0\), we can divide both sides by \(S\) to obtain \(\frac{\mathrm{d}r}{\mathrm{d}t} = k \mathrm{e}^{-t}\) as required.

For part (b), integrating \(\frac{\mathrm{d}r}{\mathrm{d}t} = k \mathrm{e}^{-t}\) with respect to \(t\) gives \(r = -k\mathrm{e}^{-t} + C\), where \(C\) is a constant of integration. As \(t \to \infty\), \(\mathrm{e}^{-t} \to 0\), so \(r \to C\). Since the limiting value of the radius is \(5\text{ mm}\), we have \(C = 5\). At \(t = 0\), \(r = 1\), so \(1 = -k\mathrm{e}^0 + 5 \implies 1 = -k + 5 \implies k = 4\). The equation for the radius is therefore \(r = 5 - 4\mathrm{e}^{-t}\). When \(t = \ln 2\), \(r = 5 - 4\mathrm{e}^{-\ln 2} = 5 - 4\left(\frac{1}{2}\right) = 5 - 2 = 3\text{ mm}\).

M1: Applies the chain rule \(\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}r}\frac{\mathrm{d}r}{\mathrm{d}t}\) and correctly shows \(\frac{\mathrm{d}r}{\mathrm{d}t} = k \mathrm{e}^{-t}\).
M1: Integrates the differential equation to obtain \(r = -k\mathrm{e}^{-t} + C\).
M1: Applies the limit as \(t \to \infty\) to find \(C = 5\), and the initial condition \(t = 0, r = 1\) to find \(k = 4\).
A1: Finds the correct expression for the radius \(r = 5 - 4\mathrm{e}^{-t}\).
A1: Substitutes \(t = \ln 2\) to find the correct radius \(3\).

To find the volume of revolution about the x-axis, we use the formula:

\[V = \pi \int y^2 \, dx = \pi \int y^2 \frac{dx}{dt} \, dt\]

First, find

\[\frac{dx}{dt} = -2 \sin t\]

Now find the limits of integration for t:
When \(x = 0 \implies 2 \cos t = 0 \implies t = \frac{\pi}{2}\).
When \(x = 2 \implies 2 \cos t = 2 \implies t = 0\).

Substitute \(y = \sin(2t) = 2 \sin t \cos t\) and \(\frac{dx}{dt}\) into the volume formula:

\[V = \pi \int_{\pi/2}^{0} (2 \sin t \cos t)^2 (-2 \sin t) \, dt\]

\[V = \pi \int_{0}^{\pi/2} 8 \sin^3 t \cos^2 t \, dt\]

To integrate this, use the identity \(\sin^2 t = 1 - \cos^2 t\):

\[V = 8\pi \int_{0}^{\pi/2} (1 - \cos^2 t) \cos^2 t \sin t \, dt\]

Let \(u = \cos t\), then \(du = -\sin t \, dt\).
When \(t = 0 \implies u = 1\), and when \(t = \frac{\pi}{2} \implies u = 0\).

\[V = 8\pi \int_{1}^{0} (1 - u^2) u^2 (-du)\]

\[V = 8\pi \int_{0}^{1} (u^2 - u^4) \, du\]

\[V = 8\pi \left[ \frac{u^3}{3} - \frac{u^5}{5} \right]_{0}^{1}\]

\[V = 8\pi \left( \frac{1}{3} - \frac{1}{5} \right) = 8\pi \left( \frac{2}{15} \right) = \frac{16}{15}\pi\]

**M1**: Differentiates x with respect to t to find \(\frac{dx}{dt} = -2\sin t\) and attempts to use the parametric volume formula \(V = \pi \int y^2 \frac{dx}{dt} \, dt\) with appropriate substitution for y.
**M1**: Writes the integrand entirely in terms of \(\sin t\) and \(\cos t\), establishing the correct limits \(t = 0\) and \(t = \frac{\pi}{2}\) to get \(V = 8\pi \int_{0}^{\pi/2} \sin^3 t \cos^2 t \, dt\) (or equivalent).
**M1**: Employs a valid integration method, such as the substitution \(u = \cos t\) to express the integral in the form \(\int (u^2 - u^4) \, du\), and integrates to obtain \(\left[ \frac{u^3}{3} - \frac{u^5}{5} \right]\).
**A1**: Reaches the correct exact answer of \(\frac{16}{15}\pi\).

To find the volume of revolution about the x-axis, we use the formula:

\[V = \pi \int y^2 \, dx = \pi \int y^2 \frac{dx}{d\theta} \, d\theta\]

First, find

\[\frac{dx}{d\theta} = \sec^2 \theta\]

Now find the limits in terms of \(\theta\):
When \(x = 0 \implies \tan \theta = 0 \implies \theta = 0\).
When \(x = 1 \implies \tan \theta = 1 \implies \theta = \frac{\pi}{4}\).

Substitute \(y = \sec \theta\) and \(\frac{dx}{d\theta} = \sec^2 \theta\) into the volume formula:

\[V = \pi \int_{0}^{\pi/4} (\sec \theta)^2 (\sec^2 \theta) \, d\theta = \pi \int_{0}^{\pi/4} \sec^4 \theta \, d\theta\]

Use the trigonometric identity \(\sec^2 \theta = 1 + \tan^2 \theta\):

\[V = \pi \int_{0}^{\pi/4} (1 + \tan^2 \theta) \sec^2 \theta \, d\theta\]

Let \(u = \tan \theta\), then \(du = \sec^2 \theta \, d\theta\).
When \(\theta = 0 \implies u = 0\), and when \(\theta = \frac{\pi}{4} \implies u = 1\).

\[V = \pi \int_{0}^{1} (1 + u^2) \, du\]

\[V = \pi \left[ u + \frac{u^3}{3} \right]_{0}^{1} = \pi \left( 1 + \frac{1}{3} - 0 \right) = \frac{4}{3}\pi\]

**M1**: Differentiates x to get \(\frac{dx}{d\theta} = \sec^2 \theta\) and sets up the parametric volume integral \(V = \pi \int_{0}^{\pi/4} \sec^4 \theta \, d\theta\) with correct limits.
**M1**: Uses the identity \(\sec^2 \theta = 1 + \tan^2 \theta\) to rewrite the integral in a form ready for substitution, i.e., \(\int (1 + \tan^2 \theta) \sec^2 \theta \, d\theta\).
**A1**: Integrates correctly to obtain \(\left[ \tan \theta + \frac{1}{3}\tan^3 \theta \right]\) (or equivalent in terms of u).
**A1**: Substitutes limits correctly to find the exact volume \(\frac{4}{3}\pi\) (or equivalent).