2026 Cambridge IAS-Level Mathematics (9709) Practice Paper with Answers

(a)
Equate the equations of the curve and the line:
\[2x^2 + kx + 3 = 3x - k\]
Rearrange into standard quadratic form \(ax^2 + bx + c = 0\):
\[2x^2 + (k - 3)x + (k + 3) = 0\]
For the line and the curve not to intersect, the discriminant must be strictly negative (\(b^2 - 4ac < 0\)):
\[(k - 3)^2 - 4(2)(k + 3) < 0\]
\[k^2 - 6k + 9 - 8k - 24 < 0\]
\[k^2 - 14k - 15 < 0\]
Factorise the quadratic expression:
\[(k - 15)(k + 1) < 0\]
The critical values are \(k = -1\) and \(k = 15\).
Since the inequality is \(< 0\), the solution is:
\[-1 < k < 15\]

(b)
Substitute \(k = -3\) into both equations:
Curve: \(y = 2x^2 - 3x + 3\)
Line: \(y = 3x - (-3) = 3x + 3\)

The curve lies strictly above the line when:
\[2x^2 - 3x + 3 > 3x + 3\]
\[2x^2 - 6x > 0\]
\[2x(x - 3) > 0\]
The critical values are \(x = 0\) and \(x = 3\).
Since the inequality is \(> 0\), the set of values of \(x\) is:
\[x < 0 \text{ or } x > 3\]

(a)
* M1: Sets line equation equal to curve equation and forms a 3-term quadratic in \(x\), then applies \(b^2 - 4ac < 0\) (or \(= 0\) / \(\leqslant 0\)).
* A1: Obtains correct 3-term quadratic in \(k\): \(k^2 - 14k - 15 < 0\) (or \(= 0\)) with correct critical values \(k = -1, 15\).
* A1: Fully correct strict inequality \(-1 < k < 15\) (accept equivalent interval notation \((-1, 15)\); do not accept \(k > -1 \text{ and } k < 15\) if written as two separate unconnected inequalities without link).

(b)
* M1: Substitutes \(k = -3\) to form the inequality \(2x^2 - 6x > 0\) (or \(= 0\)) and finds critical values \(x = 0, 3\).
* A1: Fully correct solution: \(x < 0 \text{ or } x > 3\) (accept \(x < 0, x > 3\) or \((-\infty, 0) \cup (3, \infty)\); reject \(0 > x > 3\)).

(a)
Complete the square on the quadratic in \(x^2\):
\[\mathrm{f}(x) = 3\left(x^4 - \frac{14}{3}x^2\right) + 8\]
\[= 3\left[\left(x^2 - \frac{7}{3}\right)^2 - \left(\frac{7}{3}\right)^2\right] + 8\]
\[= 3\left(x^2 - \frac{7}{3}\right)^2 - 3\left(\frac{49}{9}\right) + 8\]
\[= 3\left(x^2 - \frac{7}{3}\right)^2 - \frac{49}{3} + \frac{24}{3}\]
\[= 3\left(x^2 - \frac{7}{3}\right)^2 - \frac{25}{3}\]

Since \(\left(x^2 - \frac{7}{3}\right)^2 \geqslant 0\) for all real \(x\), the minimum value of \(\mathrm{f}(x)\) is \(-\frac{25}{3}\).
This minimum occurs when:
\[x^2 - \frac{7}{3} = 0 \implies x^2 = \frac{7}{3} \implies x = \pm \sqrt{\frac{7}{3}} = \pm \frac{\sqrt{21}}{3}\]

(b)
Solve \(3x^4 - 14x^2 + 8 = 0\):
Let \(u = x^2\):
\[3u^2 - 14u + 8 = 0\]
Factorise the quadratic:
\[(3u - 2)(u - 4) = 0\]
Thus \(u = \frac{2}{3}\) or \(u = 4\).

Substituting back \(u = x^2\):
\[x^2 = 4 \implies x = \pm 2\]
\[x^2 = \frac{2}{3} \implies x = \pm \sqrt{\frac{2}{3}} = \pm \frac{\sqrt{6}}{3}\]

Hence the solutions are \(x = \pm 2\) and \(x = \pm \frac{\sqrt{6}}{3}\).

(a)
* B1: Obtains \(a = 3\) and \(b = -\frac{7}{3}\) in \(3\left(x^2 - \frac{7}{3}\right)^2 + c\).
* B1: Obtains \(c = -\frac{25}{3}\) and correctly states the minimum value as \(-\frac{25}{3}\).
* B1: Correctly finds the exact values of \(x\): \(x = \pm \sqrt{\frac{7}{3}}\) or \(x = \pm \frac{\sqrt{21}}{3}\) (both positive and negative roots required).

(b)
* M1: Treats equation as a quadratic in \(x^2\) and factorises or uses quadratic formula to find values for \(x^2\) (\(x^2 = 4\) and \(x^2 = \frac{2}{3}\)).
* A1: Obtains all 4 exact solutions: \(x = \pm 2\) and \(x = \pm \sqrt{\frac{2}{3}}\) (or \(\pm \frac{\sqrt{6}}{3}\)).

(a)
Starting with the left-hand side:
\[ \text{LHS} = \frac{\sin\theta}{1 - \cos\theta} - \frac{1}{\tan\theta} \]
Using \(\tan\theta = \frac{\sin\theta}{\cos\theta}\):
\[ = \frac{\sin\theta}{1 - \cos\theta} - \frac{\cos\theta}{\sin\theta} \]
Putting over a common denominator:
\[ = \frac{\sin^2\theta - \cos\theta(1 - \cos\theta)}{\sin\theta(1 - \cos\theta)} \]
\[ = \frac{\sin^2\theta - \cos\theta + \cos^2\theta}{\sin\theta(1 - \cos\theta)} \]
Using \(\sin^2\theta + \cos^2\theta = 1\):
\[ = \frac{1 - \cos\theta}{\sin\theta(1 - \cos\theta)} \]
Cancelling the factor \((1 - \cos\theta)\):
\[ = \frac{1}{\sin\theta} = \text{RHS} \]

(b)
Using the identity from part (a) with angle \(2\theta\):
\[ \frac{1}{\sin 2\theta} = \sin 2\theta + \frac{3}{2} \]
Multiply both sides by \(2\sin 2\theta\):
\[ 2 = 2\sin^2 2\theta + 3\sin 2\theta \]
\[ 2\sin^2 2\theta + 3\sin 2\theta - 2 = 0 \]
Factorising the quadratic equation:
\[ (2\sin 2\theta - 1)(\sin 2\theta + 2) = 0 \]
Since \(\sin 2\theta + 2 = 0\) has no real solutions (as \(-1 \le \sin 2\theta \le 1\)), we have:
\[ \sin 2\theta = \frac{1}{2} \]
For \(0^\circ < \theta < 180^\circ\), the interval for \(2\theta\) is \(0^\circ < 2\theta < 360^\circ\).
\[ 2\theta = 30^\circ \quad \text{or} \quad 2\theta = 180^\circ - 30^\circ = 150^\circ \]
Dividing by 2 gives:
\[ \theta = 15^\circ, \quad \theta = 75^\circ \]

(a)
* M1: Express \(\tan\theta\) as \(\frac{\sin\theta}{\cos\theta}\) and combine fractions over a common denominator \(\sin\theta(1-\cos\theta)\).
* M1: Use the identity \(\sin^2\theta + \cos^2\theta = 1\) correctly in the numerator.
* A1: Simplify correctly to obtain \(\frac{1}{\sin\theta}\) with no algebraic errors.

(b)
* M1: Substitute \(\frac{1}{\sin 2\theta}\) for the LHS to form an equation in \(\sin 2\theta\).
* M1: Form a three-term quadratic in \(\sin 2\theta\) and attempt to solve by factorisation or formula.
* A1: Obtain \(\sin 2\theta = \frac{1}{2}\) and reject or discard \(\sin 2\theta = -2\).
* A1: Both correct values \(\theta = 15^\circ\) and \(\theta = 75^\circ\) obtained, with no additional solutions in the specified range.

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

(b)
When \(x = 1\):
\[y = 4(3(1) + 1)^{\frac{1}{2}} - 1 = 4\sqrt{4} - 1 = 8 - 1 = 7\]
So \(P\) is the point \((1, 7)\).

The gradient of the tangent at \(x = 1\) is:
\[m_{\text{tangent}} = 6(3(1) + 1)^{-\frac{1}{2}} - 1 = \frac{6}{2} - 1 = 2\]
Therefore, the gradient of the normal is:
\[m_{\text{normal}} = -\frac{1}{2}\]
The equation of the normal is:
\[y - 7 = -\frac{1}{2}(x - 1)\]
\[y = -\frac{1}{2}x + \frac{1}{2} + 7\]
\[y = -\frac{1}{2}x + \frac{15}{2}\]

(c)
Finding the intercepts of the normal line \(y = -\frac{1}{2}x + \frac{15}{2}\):
- When \(x = 0\), \(y = \frac{15}{2}\), so \(A = \left(0, \frac{15}{2}\right)\).
- When \(y = 0\), \(0 = -\frac{1}{2}x + \frac{15}{2} \implies x = 15\), so \(B = (15, 0)\).

The area of triangle \(OAB\) is:
\[\text{Area} = \frac{1}{2} \times OA \times OB = \frac{1}{2} \times 15 \times \frac{15}{2} = \frac{225}{4} = 56.25\]

(a)
* M1: Differentiates to obtain \(k(3x + 1)^{-\frac{1}{2}} - 1\) (where \(k\) is a constant, demonstrating the chain rule).
* A1: Correct derivative \(6(3x + 1)^{-\frac{1}{2}} - 1\) or \(\frac{6}{\sqrt{3x + 1}} - 1\).

(b)
* B1: Obtains \(y = 7\) at \(x = 1\).
* M1: Substitutes \(x = 1\) into their derivative to find tangent gradient, then uses \(m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}\) and forms a straight-line equation with their \((1, y)\).
* A1: Correct equation in the required form: \(y = -\frac{1}{2}x + \frac{15}{2}\) (or \(y = -0.5x + 7.5\)).

(c)
* M1: Sets \(x = 0\) and \(y = 0\) in their normal equation to find both axis intercepts, and applies \(\frac{1}{2} \times \text{base} \times \text{height}\).
* A1: Correct area \(\frac{225}{4}\) or \(56.25\) (or \(56\frac{1}{4}\)).

(a)
Using the binomial theorem for \((2 - ax)^5\):
\[ (2 - ax)^5 = 2^5 + \binom{5}{1} 2^4 (-ax) + \binom{5}{2} 2^3 (-ax)^2 + \dots \]
\[ = 32 + 5(16)(-ax) + 10(8)(a^2x^2) + \dots \]
\[ = 32 - 80ax + 80a^2x^2 \]

(b)
Consider the product \((1 + 4x)(2 - ax)^5 = (1 + 4x)(32 - 80ax + 80a^2x^2 + \dots)\).
The term in \(x^2\) is obtained from:
\[ 1 \times (80a^2x^2) + 4x \times (-80ax) = (80a^2 - 320a)x^2 \]
Equating the coefficient to \(400\):
\[ 80a^2 - 320a = 400 \]
Dividing through by \(80\):
\[ a^2 - 4a - 5 = 0 \]
\[ (a - 5)(a + 1) = 0 \]
\[ a = 5 \quad \text{or} \quad a = -1 \]

(c)
When \(a = -1\), \((2 - ax)^5 = (2 + x)^5\).
We need the term in \(x^2\) from the product \(\left(3 - \frac{2}{x}\right)(2 + x)^5\).
The expansion of \((2 + x)^5\) up to the term in \(x^3\) is:
\[ (2 + x)^5 = 32 + 80x + 80x^2 + \binom{5}{3}(2)^2(x)^3 + \dots = 32 + 80x + 80x^2 + 40x^3 + \dots \]
The term in \(x^2\) in the expansion of \(\left(3 - \frac{2}{x}\right)(2 + x)^5\) comes from:
\[ 3 \times (80x^2) + \left(-\frac{2}{x}\right) \times (40x^3) = 240x^2 - 80x^2 = 160x^2 \]
Therefore, the coefficient of \(x^2\) is \(160\).

(a)
* B1: For constant term \(32\).
* B1: For \(-80ax\).
* B1: For \(+80a^2x^2\) (allow unsimplified coefficients if fully evaluated correctly to this stage).

(b)
* M1: For correctly identifying and summing the two terms in \(x^2\): \(1 \times (80a^2) + 4 \times (-80a)\) (allow FT from part (a)).
* M1: For setting their expression equal to \(400\) and attempting to solve a 3-term quadratic in \(a\).
* A1: For obtaining both correct values \(a = 5\) and \(a = -1\) with no extra values.

(c)
* M1: For finding the term in \(x^3\) of \((2+x)^5\) (i.e. \(40x^3\)) and combining \(3 \times (80) - 2 \times (40)\).
* A1: For the correct coefficient \(160\).

(a)
Rewrite $\mathrm{f}(x)$ as $\mathrm{f}(x) = 4x + 9(x-1)^{-1}$.

Differentiating with respect to $x$:
$$\mathrm{f}'(x) = 4 - 9(x-1)^{-2}$$

At a stationary point, $\mathrm{f}'(x) = 0$:
$$4 - \frac{9}{(x-1)^2} = 0$$
$$(x-1)^2 = \frac{9}{4}$$

Since $x > 1$, we take the positive square root:
$$x - 1 = \frac{3}{2} \implies x = \frac{5}{2} = 2.5$$

Substitute $x = \frac{5}{2}$ into $\mathrm{f}(x)$:
$$y = 4\left(\frac{5}{2}\right) + \frac{9}{\frac{5}{2} - 1} = 10 + \frac{9}{\frac{3}{2}} = 10 + 6 = 16$$

To determine the nature of the stationary point, find the second derivative:
$$\mathrm{f}''(x) = 18(x-1)^{-3} = \frac{18}{(x-1)^3}$$

At $x = \frac{5}{2}$:
$$\mathrm{f}''\left(\frac{5}{2}\right) = \frac{18}{\left(\frac{3}{2}\right)^3} = \frac{18}{\frac{27}{8}} = \frac{16}{3} > 0$$

Since $\mathrm{f}''\left(\frac{5}{2}\right) > 0$, the stationary point is a local minimum.

Coordinates: $\left(\frac{5}{2}, 16\right)$, Minimum.

---

(b)
Applying the reflection in the $y$-axis gives $y = \mathrm{f}(-x)$.
Applying the translation by $\begin{pmatrix} 2 \\ 5 \end{pmatrix}$ replaces $x$ by $x - 2$ and adds $5$:
$$\mathrm{g}(x) = \mathrm{f}(-(x - 2)) + 5 = \mathrm{f}(2 - x) + 5$$

Domain:
The original domain is $x > 1$. Replacing $x$ with $2 - x$ gives:
$$2 - x > 1 \implies x < 1$$

Expression for $\mathrm{g}(x)$:
$$\mathrm{g}(x) = 4(2 - x) + \frac{9}{(2 - x) - 1} + 5$$
$$\mathrm{g}(x) = 8 - 4x + \frac{9}{1 - x} + 5$$
$$\mathrm{g}(x) = 13 - 4x + \frac{9}{1 - x} \quad \left(\text{or } 13 - 4x - \frac{9}{x - 1}\right)$$

(a)
- M1: Differentiating to obtain at least one term correct (e.g. $4$ or $k(x-1)^{-2}$).
- A1: Correct first derivative $\mathrm{f}'(x) = 4 - 9(x-1)^{-2}$.
- A1: Setting $\mathrm{f}'(x) = 0$ and finding both coordinates correctly: $\left(\frac{5}{2}, 16\right)$ or $(2.5, 16)$. Reject $x = -\frac{1}{2}$ as it lies outside the domain $x > 1$.
- B1: Valid method for determining nature (e.g. evaluating $\mathrm{f}''\left(\frac{5}{2}\right) = \frac{16}{3} > 0$ or testing gradients either side) and concluding minimum.

(b)
- B1: Stating correct domain $x < 1$ (or equivalent inequality/interval notation).
- M1: Valid algebraic application of both transformations, obtaining $\mathrm{f}(2-x) + 5$.
- A1: Fully correct simplified expression $\mathrm{g}(x) = 13 - 4x + \frac{9}{1-x}$ or $\mathrm{g}(x) = 13 - 4x - \frac{9}{x-1}$.

(a)
First, find the arc length \(AB\):
\[\text{Arc } AB = r\theta = 6 \times \frac{1}{3}\pi = 2\pi\text{ cm}\]

In the right-angled triangle \(OAC\), \(\angle OAC = \frac{1}{2}\pi\) and \(\angle AOC = \frac{1}{3}\pi\):
\[\tan\left(\frac{1}{3}\pi\right) = \frac{AC}{OA} \implies AC = 6\tan\left(\frac{1}{3}\pi\right) = 6\sqrt{3}\text{ cm}\]
\[\cos\left(\frac{1}{3}\pi\right) = \frac{OA}{OC} \implies OC = \frac{6}{\cos\left(\frac{1}{3}\pi\right)} = \frac{6}{\frac{1}{2}} = 12\text{ cm}\]

Since \(B\) lies on \(OC\) and \(OB = 6\text{ cm}\):
\[BC = OC - OB = 12 - 6 = 6\text{ cm}\]

Therefore, the perimeter of the shaded region is:
\[\text{Perimeter} = \text{Arc } AB + AC + BC = 2\pi + 6\sqrt{3} + 6 = 6 + 6\sqrt{3} + 2\pi\text{ cm}\]

(b)
The area of the right-angled triangle \(OAC\) is:
\[\text{Area of } \triangle OAC = \frac{1}{2} \times OA \times AC = \frac{1}{2} \times 6 \times 6\sqrt{3} = 18\sqrt{3}\text{ cm}^2\]

The area of the sector \(OAB\) is:
\[\text{Area of sector } OAB = \frac{1}{2}r^2\theta = \frac{1}{2} \times 6^2 \times \frac{1}{3}\pi = \frac{1}{2} \times 36 \times \frac{1}{3}\pi = 6\pi\text{ cm}^2\]

The area of the shaded region is:
\[\text{Area} = \text{Area of } \triangle OAC - \text{Area of sector } OAB = 18\sqrt{3} - 6\pi\text{ cm}^2\]
Thus, \(a = 18\) and \(b = 6\).

(a)
* M1: Uses \(s = r\theta\) with \(r = 6\) and \(\theta = \frac{\pi}{3}\), and uses correct right-angled trigonometry to find \(AC\) or \(OC\).
* A1: Obtains \(\text{Arc } AB = 2\pi\), \(AC = 6\sqrt{3}\), and \(BC = 6\) (all three lengths correct).
* A1: Fully correct exact perimeter \(6 + 6\sqrt{3} + 2\pi\) or \(2(3 + 3\sqrt{3} + \pi)\).

(b)
* M1: Uses \(\frac{1}{2} \times \text{base} \times \text{height}\) for the area of \(\triangle OAC\).
* M1: Uses \(\frac{1}{2}r^2\theta\) with \(r = 6\) and \(\theta = \frac{\pi}{3}\) for the area of sector \(OAB\).
* A1: Obtains \(18\sqrt{3} - 6\pi\) (or identifies \(a = 18, b = 6\)).

(a)
Rewrite the integrand in index form:
\[ \int_0^k 12(2x + 3)^{-2} \,\mathrm{d}x \]
Integrate with respect to \(x\):
\[ \left[ \frac{12(2x + 3)^{-1}}{-1 \times 2} \right]_0^k = \left[ -\frac{6}{2x + 3} \right]_0^k \]
Substitute the limits \(x = k\) and \(x = 0\):
\[ \left(-\frac{6}{2k + 3}\right) - \left(-\frac{6}{2(0) + 3}\right) = -\frac{6}{2k + 3} + \frac{6}{3} = 2 - \frac{6}{2k + 3} \]
Alternatively expressed as a single fraction:
\[ \frac{2(2k + 3) - 6}{2k + 3} = \frac{4k}{2k + 3} \]

(b)
As \(k \to \infty\), \(\frac{6}{2k + 3} \to 0\).
Therefore, the area approaches:
\[ 2 - 0 = 2 \]

(c)
Set the expression for the area equal to \(1.8\):
\[ 2 - \frac{6}{2k + 3} = 1.8 \]
\[ \frac{6}{2k + 3} = 0.2 \]
\[ 2k + 3 = \frac{6}{0.2} = 30 \]
\[ 2k = 27 \implies k = 13.5 \quad \left(\text{or } \frac{27}{2}\right) \]

(a)
* M1: Integrate to obtain the form \(a(2x + 3)^{-1}\), where \(a\) is a non-zero constant.
* A1: Correct integral \(-\frac{6}{2x + 3}\) (or \(\frac{12}{-2(2x + 3)}\)).
* A1: Correct substitution of limits \(0\) and \(k\) to obtain \(2 - \frac{6}{2k + 3}\) or \(\frac{4k}{2k + 3}\).

(b)
* B1: State \(2\) (or exact equivalent) from evaluating the limit as \(k \to \infty\).

(c)
* B1: Correct value of \(k = 13.5\) or \(\frac{27}{2}\).

(a) Let \(r\) be the radius of the water surface when the depth is \(h\).
By similar triangles:
\[ \frac{r}{h} = \frac{8}{20} \implies r = \frac{2}{5}h \]
The volume of a cone is \(V = \frac{1}{3}\pi r^2 h\).
Substituting \(r = \frac{2}{5}h\):
\[ V = \frac{1}{3}\pi \left(\frac{2}{5}h\right)^2 h = \frac{1}{3}\pi \left(\frac{4}{25}h^2\right) h = \frac{4}{75}\pi h^3 \]

(b) Differentiating \(V\) with respect to \(h\):
\[ \frac{\mathrm{d}V}{\mathrm{d}h} = 3 \times \frac{4}{75}\pi h^2 = \frac{4}{25}\pi h^2 \]
When \(h = 10\):
\[ \frac{\mathrm{d}V}{\mathrm{d}h} = \frac{4}{25}\pi (10)^2 = 16\pi \]
Using the chain rule with \(\frac{\mathrm{d}V}{\mathrm{d}t} = 12\):
\[ \frac{\mathrm{d}h}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}t} \div \frac{\mathrm{d}V}{\mathrm{d}h} = \frac{12}{16\pi} = \frac{3}{4\pi}\text{ cm s}^{-1} \]

(c) The area of the circular water surface is \(A = \pi r^2\).
Since \(r = \frac{2}{5}h\):
\[ A = \pi \left(\frac{2}{5}h\right)^2 = \frac{4}{25}\pi h^2 \]
Differentiating with respect to \(h\):
\[ \frac{\mathrm{d}A}{\mathrm{d}h} = \frac{8}{25}\pi h \]
When \(h = 10\):
\[ \frac{\mathrm{d}A}{\mathrm{d}h} = \frac{8}{25}\pi (10) = \frac{16}{5}\pi \]
Using the chain rule:
\[ \frac{\mathrm{d}A}{\mathrm{d}t} = \frac{\mathrm{d}A}{\mathrm{d}h} \times \frac{\mathrm{d}h}{\mathrm{d}t} = \left(\frac{16}{5}\pi\right) \times \left(\frac{3}{4\pi}\right) = \frac{12}{5} = 2.4\text{ cm}^2\text{ s}^{-1} \]

(a)
* M1: Uses similar triangles to express \(r\) in terms of \(h\) (i.e. \(r = \frac{2}{5}h\) or \(\frac{r}{h} = \frac{8}{20}\)) and substitutes into \(V = \frac{1}{3}\pi r^2 h\).
* A1: Completes the algebraic working correctly to show \(V = \frac{4}{75}\pi h^3\) (AG).

(b)
* B1: Differentiates to obtain \(\frac{\mathrm{d}V}{\mathrm{d}h} = \frac{4}{25}\pi h^2\) (or equivalent).
* M1: Correctly applies the chain rule \(\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}t} \div \frac{\mathrm{d}V}{\mathrm{d}h}\) with \(\frac{\mathrm{d}V}{\mathrm{d}t} = 12\) and their \(\frac{\mathrm{d}V}{\mathrm{d}h}\) evaluated at \(h = 10\).
* A1: Obtains \(\frac{3}{4\pi}\) (or \(\frac{0.75}{\pi}\)).

(c)
* M1: Expresses \(A\) in terms of \(h\) (or \(r\)) and applies a correct chain rule to find \(\frac{\mathrm{d}A}{\mathrm{d}t}\).
* A1: Obtains \(2.4\) or \(\frac{12}{5}\).

(a)
The first, fifth and seventh terms of the arithmetic progression are \(a\), \(a + 4d\) and \(a + 6d\) respectively.

Since these form the first three terms of a geometric progression:
\[ \frac{a + 4d}{a} = \frac{a + 6d}{a + 4d} \]
\[ (a + 4d)^2 = a(a + 6d) \]
\[ a^2 + 8ad + 16d^2 = a^2 + 6ad \]
\[ 2ad + 16d^2 = 0 \]
\[ 2d(a + 8d) = 0 \]
Since \(d \neq 0\), it follows that \(a + 8d = 0\), so \(a = -8d\).

(b)(i)
The common ratio \(r\) of the geometric progression is:
\[ r = \frac{a + 4d}{a} = \frac{-8d + 4d}{-8d} = \frac{-4d}{-8d} = \frac{1}{2} \]
The sum to infinity of the geometric progression is given by:
\[ S_\infty = \frac{a}{1 - r} = \frac{a}{1 - \frac{1}{2}} = 2a \]
Given that \(S_\infty = 48\):
\[ 2a = 48 \implies a = 24 \]
Using \(a = -8d\):
\[ 24 = -8d \implies d = -3 \]

(b)(ii)
The sum of the first \(n\) terms of an arithmetic progression is:
\[ S_n = \frac{n}{2}\left[2a + (n - 1)d\right] \]
For \(n = 30\), with \(a = 24\) and \(d = -3\):
\[ S_{30} = \frac{30}{2}\left[2(24) + (30 - 1)(-3)\right] \]
\[ S_{30} = 15\left[48 + 29(-3)\right] = 15(48 - 87) = 15(-39) = -585 \]

(a)
* M1: Sets up a correct equation using the geometric progression property, e.g. \((a + 4d)^2 = a(a + 6d)\) or \(\frac{a+4d}{a} = \frac{a+6d}{a+4d}\).
* A1: Expands and simplifies correctly with full working shown to establish \(a = -8d\) (AG).

(b)(i)
* M1: Uses \(a = -8d\) to find \(r = \frac{1}{2}\) and sets \(\frac{a}{1 - r} = 48\).
* A1: Obtains \(a = 24\) and \(d = -3\).

(b)(ii)
* M1: Applies the AP sum formula \(S_n = \frac{n}{2}[2a + (n-1)d]\) with \(n = 30\).
* M1: Correct substitution of their values of \(a\) and \(d\) into the AP sum formula.
* A1: Obtains \(-585\).

(a)
Completing the square on the equation of the circle:
\((x - 4)^2 - 16 + (y - 2)^2 - 4 - 80 = 0\)
\((x - 4)^2 + (y - 2)^2 = 100\)

Therefore, the centre is \(C(4, 2)\) and the radius is \(r = \sqrt{100} = 10\).

(b) (i)
Gradient of radius \(CA\):
\(m_{CA} = \frac{8 - 2}{12 - 4} = \frac{6}{8} = \frac{3}{4}\)

Since the tangent is perpendicular to the radius, the gradient of the tangent at \(A\) is \(-\frac{4}{3}\).
Equation of tangent at \(A\):
\(y - 8 = -\frac{4}{3}(x - 12) \implies 3y - 24 = -4x + 48 \implies 4x + 3y = 72\) (or \(y = -\frac{4}{3}x + 24\))

Gradient of radius \(CB\):
\(m_{CB} = \frac{10 - 2}{-2 - 4} = \frac{8}{-6} = -\frac{4}{3}\)

The gradient of the tangent at \(B\) is \(\frac{3}{4}\).
Equation of tangent at \(B\):
\(y - 10 = \frac{3}{4}(x - (-2)) \implies 4y - 40 = 3x + 6 \implies 3x - 4y = -46\) (or \(y = \frac{3}{4}x + 11.5\))

(b) (ii)
Solve the simultaneous equations:
\(4x + 3y = 72\) --- (1)
\(3x - 4y = -46\) --- (2)

Multiply (1) by 4 and (2) by 3:
\(16x + 12y = 288\)
\(9x - 12y = -138\)

Adding the two equations:
\(25x = 150 \implies x = 6\)

Substitute \(x = 6\) into (1):
\(4(6) + 3y = 72 \implies 24 + 3y = 72 \implies 3y = 48 \implies y = 16\)

So \(T = (6, 16)\).

(c) (i)
- \(m_{CA} \times m_{CB} = \left(\frac{3}{4}\right) \times \left(-\frac{4}{3}\right) = -1\), so \(CA \perp CB\), which means \(\angle ACB = 90^\circ\).
- Tangents are perpendicular to radii at the points of contact, so \(\angle CAT = 90^\circ\) and \(\angle CBT = 90^\circ\).
- The sum of angles in quadrilateral \(ACBT\) is \(360^\circ\), so \(\angle ATB = 360^\circ - 3(90^\circ) = 90^\circ\). Thus, all four interior angles are right angles, so \(ACBT\) is a rectangle.
- Adjacent sides are equal in length: \(CA = CB = r = 10\).
A rectangle with adjacent equal sides is a square. Hence, \(ACBT\) is a square.

(c) (ii)
Area of square \(ACBT = 10 \times 10 = 100\).
Angle of minor sector \(CAB = \frac{\pi}{2}\) radians.
Area of sector \(CAB = \frac{1}{2} r^2 \theta = \frac{1}{2} (10^2) \left(\frac{\pi}{2}\right) = 25\pi\).

Required enclosed area \(= \text{Area of square} - \text{Area of sector} = 100 - 25\pi\).

(a)
* M1: For attempting to complete the square on both \(x\) and \(y\) terms.
* A1: For correct centre \((4, 2)\) and correct radius \(10\).

(b) (i)
* M1: For calculating the gradient of \(CA\) or \(CB\) and taking the negative reciprocal to find the gradient of the tangent.
* A1: For correct tangent equation at \(A\): \(4x + 3y = 72\) or \(y = -\frac{4}{3}x + 24\) (or any equivalent unsimplified form).
* M1: For finding the gradient of the other radius and its corresponding perpendicular tangent gradient.
* A1: For correct tangent equation at \(B\): \(3x - 4y = -46\) or \(y = \frac{3}{4}x + \frac{23}{2}\) (or equivalent).

(b) (ii)
* M1: For an algebraic method to solve their two linear equations simultaneously.
* A1: For \(x = 6, y = 16\) (or coordinates \((6, 16)\)).

(c) (i)
* B1: For showing \(m_{CA} \times m_{CB} = -1\) (or equivalent vector/distance method) to deduce \(\angle ACB = 90^\circ\).
* B1: For complete and rigorous argument: stating \(\angle CAT = \angle CBT = 90^\circ\) and \(CA = CB = 10\) to conclude that \(ACBT\) is a square.

(c) (ii)
* M1: For calculating the area of the minor sector using \(\frac{1}{2} r^2 \theta\) with \(r = 10\) and \(\theta = \frac{\pi}{2}\) (or \(90^\circ\)).
* A1: For exact area \(100 - 25\pi\) (must be exact in terms of \(\pi\)).

To evaluate the definite integral, integrate each term with respect to \(x\):

\[ \int (3\mathrm{e}^{3x} - 4\mathrm{e}^{-2x}) \, \mathrm{d}x = \left[ \frac{3}{3}\mathrm{e}^{3x} - \frac{4}{-2}\mathrm{e}^{-2x} \right] = \left[ \mathrm{e}^{3x} + 2\mathrm{e}^{-2x} \right] \]

Now, substitute the limits \(x = 1\) and \(x = 0\):

At the upper limit \(x = 1\):
\[ \mathrm{e}^{3(1)} + 2\mathrm{e}^{-2(1)} = \mathrm{e}^3 + 2\mathrm{e}^{-2} \]

At the lower limit \(x = 0\):
\[ \mathrm{e}^{0} + 2\mathrm{e}^{0} = 1 + 2(1) = 3 \]

Subtracting the value at the lower limit from the value at the upper limit:
\[ (\mathrm{e}^3 + 2\mathrm{e}^{-2}) - 3 = \mathrm{e}^3 + 2\mathrm{e}^{-2} - 3 \]

M1: For integrating at least one term correctly to the form \(k\mathrm{e}^{ax}\), giving either \(\mathrm{e}^{3x}\) or \(2\mathrm{e}^{-2x}\).
A1: For fully correct integral \(\mathrm{e}^{3x} + 2\mathrm{e}^{-2x}\).
DM1: For correctly substituting limits \(0\) and \(1\) into an integrated expression involving exponentials (dependent on previous M1).
A1: For obtaining the correct exact value \(\mathrm{e}^3 + 2\mathrm{e}^{-2} - 3\) or \(\mathrm{e}^3 + \frac{2}{\mathrm{e}^2} - 3\) or \(\frac{\mathrm{e}^5 - 3\mathrm{e}^2 + 2}{\mathrm{e}^2}\).

Use the identity \( \operatorname{cosec}^2 \theta = 1 + \cot^2 \theta \):
\[ 2(1 + \cot^2 \theta) + \cot \theta = 8 \]
\[ 2\cot^2 \theta + \cot \theta - 6 = 0 \]

Factorise the quadratic equation:
\[ (2\cot \theta - 3)(\cot \theta + 2) = 0 \]

This gives:
\[ \cot \theta = \frac{3}{2} \quad \text{or} \quad \cot \theta = -2 \]

Convert to \( \tan \theta \):
\[ \tan \theta = \frac{2}{3} \quad \text{or} \quad \tan \theta = -\frac{1}{2} \]

For \( \tan \theta = \frac{2}{3} \) in the interval \( 0^\circ < \theta < 180^\circ \):
\[ \theta = \tan^{-1}\left(\frac{2}{3}\right) \approx 33.69^\circ = 33.7^\circ \text{ (1 d.p.)} \]

For \( \tan \theta = -\frac{1}{2} \) in the interval \( 0^\circ < \theta < 180^\circ \):
\[ \theta = 180^\circ - \tan^{-1}\left(\frac{1}{2}\right) = 180^\circ - 26.565^\circ \approx 153.4^\circ \text{ (1 d.p.)} \]

Thus, the solutions are \( \theta = 33.7^\circ \) and \( \theta = 153.4^\circ \).

M1: Use the identity \( \operatorname{cosec}^2 \theta = 1 + \cot^2 \theta \) to obtain a quadratic equation in \( \cot \theta \) or \( \tan \theta \).
A1: Obtain correct 3-term quadratic equation \( 2\cot^2 \theta + \cot \theta - 6 = 0 \) (or \( 6\tan^2 \theta - \tan \theta - 2 = 0 \)).
M1: Solve a 3-term quadratic equation in \( \cot \theta \) or \( \tan \theta \) and attempt at least one angle in the given range.
A1: Obtain \( \theta = 33.7^\circ \) and \( \theta = 153.4^\circ \) (allow 33.7 and 153.4), and no other solutions in the range \( 0^\circ < \theta < 180^\circ \).

(a)
Taking natural logarithms of both sides of \(y = ax^b\):
\[ \ln y = \ln(ax^b) = \ln a + b \ln x \]

This is in the linear form \(Y = mX + c\), where \(Y = \ln y\), \(X = \ln x\), gradient \(m = b\), and vertical intercept \(c = \ln a\).

The gradient of the line is:
\[ b = \frac{5.74 - 2.14}{2.40 - 0.80} = \frac{3.60}{1.60} = 2.25 \]

Using the point \((0.80, 2.14)\) to find the intercept:
\[ \ln a = 2.14 - b(0.80) = 2.14 - 2.25(0.80) = 2.14 - 1.80 = 0.34 \]

Hence:
\[ a = \mathrm{e}^{0.34} \approx 1.4049... \approx 1.40 \]

So, \(a = 1.40\) and \(b = 2.25\).

(b)
When \(x = 4\):
\[ \ln x = \ln 4 \approx 1.3863 \]
\[ \ln y = 0.34 + 2.25\ln 4 = 0.34 + 2.25(1.38629) = 3.45916 \]
\[ y = \mathrm{e}^{3.45916} \approx 31.79... \approx 31.8 \]

*(Alternatively, \(y = ax^b = 1.4049 \times 4^{2.25} = 1.4049 \times 22.6274 = 31.8\).)*

(a)
* M1: Correctly applies laws of logarithms to express \(y = ax^b\) as \(\ln y = \ln a + b \ln x\) (or using \(\lg\)).
* M1: Uses given coordinates to calculate the gradient: \(\frac{5.74 - 2.14}{2.40 - 0.80}\).
* A1: Obtains \(b = 2.25\) (or \(\frac{9}{4}\)).
* A1: Evaluates \(\ln a = 0.34\) and calculates \(a = \mathrm{e}^{0.34} = 1.40\) (allow \(1.40\) or \(1.41\), to 3 s.f.).

(b)
* B1: Obtains \(y = 31.8\) (allow answers in the range \(31.7\) to \(31.9\)).

(a)

Method 1: Squaring both sides
\[ (3x - 5)^2 < (2x + 7)^2 \]
\[ 9x^2 - 30x + 25 < 4x^2 + 28x + 49 \]
\[ 5x^2 - 58x - 24 < 0 \]

Factorising the quadratic expression:
\[ (5x + 2)(x - 12) < 0 \]

The critical values are \(x = -\frac{2}{5}\) and \(x = 12\).

Since the inequality is \(< 0\), the solution is:
\[ -\frac{2}{5} < x < 12 \quad \text{or} \quad -0.4 < x < 12 \]

---

(b)

Notice that \(3^{y+1} = 3(3^y)\).

Let \(x = 3^y\). The inequality becomes:
\[ |3x - 5| < |2x + 7| \]

Using the result from part (a):
\[ -\frac{2}{5} < 3^y < 12 \]

Since \(3^y > 0\) for all real \(y\), the lower bound is automatically satisfied for all real \(y\). Thus:
\[ 3^y < 12 \]

Taking natural logarithms on both sides:
\[ y \ln 3 < \ln 12 \]
\[ y < \frac{\ln 12}{\ln 3} \approx 2.26186... \]

Therefore, the greatest integer value of \(y\) is \(2\).

(a)
* M1: Attempt to solve by squaring both sides to obtain a 3-term quadratic, or by solving the linear equations \(3x - 5 = 2x + 7\) and \(3x - 5 = -(2x + 7)\).
* A1: Obtain critical values \(-\frac{2}{5}\) (or \(-0.4\)) and \(12\).
* M1: Choose the interior region for their two critical values.
* A1: Obtain correct final inequality \(-\frac{2}{5} < x < 12\) (or \(-0.4 < x < 12\)).

(b)
* M1: Use the substitution \(x = 3^y\) with the result from part (a) to set up \(3^y < 12\) (or \(-\frac{2}{5} < 3^y < 12\)).
* M1: Apply logarithms correctly to solve \(3^y < 12\) to find the upper bound (e.g. \(y < \frac{\ln 12}{\ln 3}\) or \(y < 2.26\)).
* A1: State the final correct integer value \(y = 2\) (or greatest integer is \(2\)).

(a)
Since \((2x - 1)\) is a factor, by the Factor Theorem \(\mathrm{p}\left(\frac{1}{2}\right) = 0\):
\[ 2\left(\frac{1}{2}\right)^3 - 5\left(\frac{1}{2}\right)^2 + a\left(\frac{1}{2}\right) + b = 0 \]
\[ 2\left(\frac{1}{8}\right) - 5\left(\frac{1}{4}\right) + \frac{1}{2}a + b = 0 \]
\[ \frac{1}{4} - \frac{5}{4} + \frac{1}{2}a + b = 0 \implies -1 + \frac{1}{2}a + b = 0 \implies a + 2b = 2 \quad \text{--- (1)} \]

By the Remainder Theorem, \(\mathrm{p}(-2) = -45\):
\[ 2(-2)^3 - 5(-2)^2 + a(-2) + b = -45 \]
\[ 2(-8) - 5(4) - 2a + b = -45 \]
\[ -16 - 20 - 2a + b = -45 \implies -36 - 2a + b = -45 \implies 2a - b = 9 \quad \text{--- (2)} \]

Multiplying (1) by 2 gives \(2a + 4b = 4\).
Subtracting (2) from this:
\[ (2a + 4b) - (2a - b) = 4 - 9 \implies 5b = -5 \implies b = -1 \]

Substituting \(b = -1\) into (1):
\[ a + 2(-1) = 2 \implies a = 4 \]

So \(a = 4\) and \(b = -1\).

(b)(i)
With \(a = 4\) and \(b = -1\), \(\mathrm{p}(x) = 2x^3 - 5x^2 + 4x - 1\).
The equation \(\mathrm{p}(x) = 5\) can be written as \(f(x) = 2x^3 - 5x^2 + 4x - 6 = 0\).
Evaluating \(f(x)\) at the interval endpoints:
\[ f(2) = 2(2)^3 - 5(2)^2 + 4(2) - 6 = 16 - 20 + 8 - 6 = -2 < 0 \]
\[ f(3) = 2(3)^3 - 5(3)^2 + 4(3) - 6 = 54 - 45 + 12 - 6 = 15 > 0 \]
Since there is a change of sign and \(f(x)\) is continuous, the equation \(\mathrm{p}(x) = 5\) has a root between \(x = 2\) and \(x = 3\).

(b)(ii)
Using \(x_{n+1} = \sqrt[3]{\frac{5x_n^2 - 4x_n + 6}{2}}\) with \(x_1 = 2.2\):
\[ x_2 = \sqrt[3]{\frac{5(2.2)^2 - 4(2.2) + 6}{2}} = \sqrt[3]{\frac{24.2 - 8.8 + 6}{2}} = \sqrt[3]{10.7} \approx 2.2037 \]
\[ x_3 = \sqrt[3]{\frac{5(2.2037)^2 - 4(2.2037) + 6}{2}} \approx 2.2060 \]
\[ x_4 = \sqrt[3]{\frac{5(2.2060)^2 - 4(2.2060) + 6}{2}} \approx 2.2074 \]
\[ x_5 = \sqrt[3]{\frac{5(2.2074)^2 - 4(2.2074) + 6}{2}} \approx 2.2083 \]
\[ x_6 = \sqrt[3]{\frac{5(2.2083)^2 - 4(2.2083) + 6}{2}} \approx 2.2088 \]

Since successive iterates round to \(2.21\), the root correct to 2 decimal places is \(x = 2.21\).

(a)
* M1: Use Factor Theorem to write \(\mathrm{p}\left(\frac{1}{2}\right) = 0\) and form an equation in \(a\) and \(b\).
* M1: Use Remainder Theorem to write \(\mathrm{p}(-2) = -45\) and form a second linear equation in \(a\) and \(b\).
* M1: Solve the pair of simultaneous linear equations for \(a\) or \(b\).
* A1: Both correct values obtained: \(a = 4\) and \(b = -1\).

(b)(i)
* B1: Evaluate \(f(2)\) and \(f(3)\) (where \(f(x) = 2x^3 - 5x^2 + 4x - 6\) or \(\mathrm{p}(x) - 5\)) obtaining \(-2\) and \(15\), noting the sign change and concluding a root exists in \((2, 3)\).

(b)(ii)
* M1: Apply the iteration formula correctly at least once using \(x_1 = 2.2\).
* A1: Obtain \(x_2 = 2.2037\) and \(x_3 = 2.2060\) (accept values rounded or truncated to 4 decimal places).
* A1: Obtain the final answer \(x = 2.21\) with sufficient evidence of convergence shown to 2 decimal places (must show at least up to \(x_4\) or \(x_5\)).

(a)
Using the double angle identity:
\[\cos 2x = 2\cos^2 x - 1\]
Rearranging for \(\cos^2 x\):
\[\cos^2 x = \frac{1}{2}(1 + \cos 2x)\]
Integrating with respect to \(x\):
\[\int \cos^2 x \,\mathrm{d}x = \int \frac{1}{2}(1 + \cos 2x)\,\mathrm{d}x = \frac{1}{2}\left(x + \frac{1}{2}\sin 2x\right) + c = \frac{1}{2}x + \frac{1}{4}\sin 2x + c\]

(b)
The area of region \(R\) is given by the definite integral:
\[\text{Area} = \int_{0}^{\frac{1}{6}\pi} (8\cos^2 x + 6\sin 2x)\,\mathrm{d}x\]
Using the result from part (a):
\[\int 8\cos^2 x \,\mathrm{d}x = 8\left(\frac{1}{2}x + \frac{1}{4}\sin 2x\right) = 4x + 2\sin 2x\]
Integrating the second term:
\[\int 6\sin 2x \,\mathrm{d}x = -3\cos 2x\]
Thus, the integrated expression is:
\[\left[ 4x + 2\sin 2x - 3\cos 2x \right]_{0}^{\frac{1}{6}\pi}\]
Substitute the upper limit \(x = \frac{1}{6}\pi\):
\[4\left(\frac{1}{6}\pi\right) + 2\sin\left(\frac{1}{3}\pi\right) - 3\cos\left(\frac{1}{3}\pi\right) = \frac{2}{3}\pi + 2\left(\frac{\sqrt{3}}{2}\right) - 3\left(\frac{1}{2}\right) = \frac{2}{3}\pi + \sqrt{3} - \frac{3}{2}\]
Substitute the lower limit \(x = 0\):
\[4(0) + 2\sin 0 - 3\cos 0 = 0 + 0 - 3(1) = -3\]
Subtract the lower limit from the upper limit:
\[\text{Area} = \left(\frac{2}{3}\pi + \sqrt{3} - \frac{3}{2}\right) - (-3) = \frac{2}{3}\pi + \sqrt{3} + \frac{3}{2}\]

(a)
* B1: State \(\cos 2x = 2\cos^2 x - 1\) or equivalent rearrangement \(\cos^2 x = \frac{1}{2}(1 + \cos 2x)\).
* B1: Complete the integration correctly to obtain \(\frac{1}{2}x + \frac{1}{4}\sin 2x + c\) with all intermediate steps clearly shown.

(b)
* M1: State or use \(\int (8\cos^2 x + 6\sin 2x)\,\mathrm{d}x\) and apply the result from (a) to obtain a term of the form \(4x + 2\sin 2x\).
* A1: Correct integration of \(6\sin 2x\) to \(-3\cos 2x\), giving complete anti-derivative \(4x + 2\sin 2x - 3\cos 2x\).
* M1: Substitute both limits \(x = \frac{1}{6}\pi\) and \(x = 0\) into an integrated expression of the form \(px + q\sin 2x + r\cos 2x\) (must evaluate at \(x = 0\)).
* M1: Correctly evaluate exact trigonometric values: \(\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\), \(\cos(\frac{\pi}{3}) = \frac{1}{2}\), and \(\cos 0 = 1\).
* A1: Obtain exact final answer \(\frac{2}{3}\pi + \sqrt{3} + \frac{3}{2}\) (or \(\frac{4\pi + 6\sqrt{3} + 9}{6}\)) with \(a = \frac{2}{3}\), \(b = 1\), \(c = \frac{3}{2}\).

(a)
Differentiating \(x\) with respect to \(\theta\):
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = -6\sin 2\theta = -6(2\sin\theta\cos\theta) = -12\sin\theta\cos\theta \]

Differentiating \(y\) with respect to \(\theta\) using the chain rule:
\[ \frac{\mathrm{d}y}{\mathrm{d}\theta} = 24\sin^2\theta\cos\theta - 6\cos\theta = 6\cos\theta(4\sin^2\theta - 1) \]

Using the parametric differentiation rule \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}}\):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{6\cos\theta(4\sin^2\theta - 1)}{-12\sin\theta\cos\theta} = -\frac{4\sin^2\theta - 1}{2\sin\theta} = \frac{1 - 4\sin^2\theta}{2\sin\theta} \]

(b)
At a stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[ 1 - 4\sin^2\theta = 0 \implies \sin^2\theta = \frac{1}{4} \]

Since \(0 < \theta < \frac{1}{2}\pi\), we have \(\sin\theta = \frac{1}{2}\), which gives \(\theta = \frac{1}{6}\pi\).

Substituting \(\theta = \frac{1}{6}\pi\) into the parametric equations:
\[ x = 3\cos\left(2 \cdot \frac{\pi}{6}\right) = 3\cos\left(\frac{\pi}{3}\right) = 3 \cdot \frac{1}{2} = \frac{3}{2} \]
\[ y = 8\left(\sin\frac{\pi}{6}\right)^3 - 6\left(\sin\frac{\pi}{6}\right) = 8\left(\frac{1}{2}\right)^3 - 6\left(\frac{1}{2}\right) = 8\left(\frac{1}{8}\right) - 3 = 1 - 3 = -2 \]

Thus, the exact coordinates of the stationary point are \(\left(\frac{3}{2}, -2\right)\).

(c)
When \(\theta = \frac{1}{4}\pi\):
\[ \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \implies \sin^2\left(\frac{\pi}{4}\right) = \frac{1}{2} \]

The gradient of the tangent is:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1 - 4\left(\frac{1}{2}\right)}{2\left(\frac{1}{\sqrt{2}}\right)} = \frac{1 - 2}{\sqrt{2}} = -\frac{1}{\sqrt{2}} \]

The gradient of the normal is the negative reciprocal of the tangent gradient:
\[ m_{\text{normal}} = -\frac{1}{-\frac{1}{\sqrt{2}}} = \sqrt{2} \]

(a)
- B1: Correct derivative \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = -6\sin 2\theta\) (or \(-12\sin\theta\cos\theta\)).
- M1: Differentiate \(y\) using chain rule to obtain an expression of the form \(a\sin^2\theta\cos\theta + b\cos\theta\).
- A1: Correct derivative \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 24\sin^2\theta\cos\theta - 6\cos\theta\).
- A1 (AG): Correct application of \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}\theta}{\mathrm{d}x/\mathrm{d}\theta}\), use of double-angle identity \(\sin 2\theta = 2\sin\theta\cos\theta\), and convincing simplification to the given result with no errors seen.

(b)
- M1: Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and solve for \(\sin\theta\) or \(\theta\).
- A1: Obtain \(\theta = \frac{1}{6}\pi\) (or \(\sin\theta = \frac{1}{2}\)).
- A1: Obtain correct exact coordinates \(\left(\frac{3}{2}, -2\right)\) (both \(x\) and \(y\) correct).

(c)
- M1: Substitute \(\theta = \frac{1}{4}\pi\) into their expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\) and use \(m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}\).
- A1: Obtain exact gradient \(\sqrt{2}\).