2026 Cambridge IAS-Level Mathematics (9709) 模拟试题及答案详解

Rewrite the derivative in index form:
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^{\frac{1}{2}} - 4x^{-\frac{1}{2}}\]

Integrate to find \(y\):
\[y = \int \left(3x^{\frac{1}{2}} - 4x^{-\frac{1}{2}}\right) \mathrm{d}x = \frac{3x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{4x^{\frac{1}{2}}}{\frac{1}{2}} + c\]
\[y = 2x^{\frac{3}{2}} - 8x^{\frac{1}{2}} + c\]

Substitute the coordinates of the given point \((4, 6)\) to find the constant \(c\):
\[6 = 2(4)^{\frac{3}{2}} - 8(4)^{\frac{1}{2}} + c\]
\[6 = 2(8) - 8(2) + c\]
\[6 = 16 - 16 + c \implies c = 6\]

Hence, the equation of the curve is:
\[y = 2x^{\frac{3}{2}} - 8x^{\frac{1}{2}} + 6\quad \left(\text{or } y = 2\sqrt{x^3} - 8\sqrt{x} + 6\right)\]

M1: Attempt integration with at least one power increased by 1 (e.g. obtaining terms in \(x^{\frac{3}{2}}\) or \(x^{\frac{1}{2}}\)).
A1: Correct integrated terms \(2x^{\frac{3}{2}} - 8x^{\frac{1}{2}}\) (with or without \(+ c\)).
M1: Substitute \(x = 4\) and \(y = 6\) into an integrated expression containing a constant of integration \(c\).
A1: Correct equation \(y = 2x^{\frac{3}{2}} - 8x^{\frac{1}{2}} + 6\) or exact equivalent.

Equating the equations of the line and the curve:
\[x^2 - 4x + 7 = 2x + k\]
Rearranging into standard quadratic form \(ax^2 + bx + c = 0\):
\[x^2 - 6x + (7 - k) = 0\]

Since the line and the curve do not intersect, the discriminant must be strictly negative (\(b^2 - 4ac < 0\)):
\[(-6)^2 - 4(1)(7 - k) < 0\]
\[36 - 28 + 4k < 0\]
\[8 + 4k < 0\]
\[4k < -8\]
\[k < -2\]

M1: Equate the linear and quadratic equations and rearrange to form a 3-term quadratic in \(x\).
M1: Apply the discriminant condition \(b^2 - 4ac < 0\) to their 3-term quadratic.
A1: Correct inequality in terms of \(k\), e.g. \(36 - 4(7 - k) < 0\) or \(8 + 4k < 0\).
A1: Fully correct final inequality \(k < -2\).

(a) Completing the square:
\[\mathrm{f}(x) = 2(x^2 - 6x) + 13\]
\[= 2\left[(x - 3)^2 - 9\right] + 13\]
\[= 2(x - 3)^2 - 18 + 13\]
\[= 2(x - 3)^2 - 5\]

(b) Let \(y = 2(x - 3)^2 - 5\):
\[y + 5 = 2(x - 3)^2\]
\[(x - 3)^2 = \frac{y + 5}{2}\]

Since the domain is \(x \leqslant 3\), \(x - 3 \leqslant 0\), so we take the negative square root:
\[x - 3 = -\sqrt{\frac{y + 5}{2}}\]
\[x = 3 - \sqrt{\frac{y + 5}{2}}\]

Hence:
\[\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x + 5}{2}}\]

(a)
M1: Complete the square to obtain the form \(2(x - 3)^2 + c\).
A1: Correct form \(2(x - 3)^2 - 5\).

(b)
M1: Correct algebraic process to isolate \((x - 3)^2\) and take the square root from their completed-square form.
A1: Correct expression \(\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x + 5}{2}}\) (with correct choice of negative root and expressed in terms of \(x\)).

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

Now multiply by \((1 + ax)\):
\[(1 + ax)(32 - 80x + 80x^2 + \dots)\]

The terms giving \(x^2\) are:
\[1 \times (80x^2) + ax \times (-80x) = (80 - 80a)x^2\]

Equating the coefficient to \(-40\):
\[80 - 80a = -40\]
\[-80a = -120\]
\[a = \frac{120}{80} = \frac{3}{2}\]

M1: Correct attempt to expand \((2 - x)^5\) up to the \(x^2\) term, using correct binomial coefficients.
A1: Obtain the correct terms \(-80x\) and \(80x^2\).
M1: Form an equation for the coefficient of \(x^2\) in the product: \(80 - 80a = -40\).
A1: Correct final value \(a = \frac{3}{2}\) or \(1.5\).

Write the gradient function as \(\frac{\mathrm{d}y}{\mathrm{d}x} = 12(2x+1)^{-\frac{1}{2}} - 1\).

Integrate with respect to \(x\):
\[ y = \int \left(12(2x+1)^{-\frac{1}{2}} - 1\right) \mathrm{d}x \]
\[ y = \frac{12(2x+1)^{\frac{1}{2}}}{\frac{1}{2} \times 2} - x + c \]
\[ y = 12\sqrt{2x+1} - x + c \]

Substitute the given point \((4, 15)\) to find \(c\):
\[ 15 = 12\sqrt{2(4)+1} - 4 + c \]
\[ 15 = 12(3) - 4 + c \]
\[ 15 = 32 + c \implies c = -17 \]

Hence, the equation of the curve is \(y = 12\sqrt{2x+1} - x - 17\).

M1: Attempt to integrate of the form \(k(2x+1)^{\frac{1}{2}} - x\)
A1: Correct integration \(12\sqrt{2x+1} - x (+ c)\)
M1: Substitute \(x = 4, y = 15\) into an integrated expression containing a constant \(c\)
A1: Correct equation \(y = 12\sqrt{2x+1} - x - 17\) or equivalent

(a) Since the terms are in geometric progression,
\[ \frac{p+6}{p} = \frac{3p+6}{p+6} \]
\[ (p + 6)^2 = p(3p + 6) \]
\[ p^2 + 12p + 36 = 3p^2 + 6p \]
\[ 2p^2 - 6p - 36 = 0 \implies p^2 - 3p - 18 = 0 \]
\[ (p - 6)(p + 3) = 0 \]
Since \(p > 0\), \(p = 6\).
The common ratio is \(r = \frac{6 + 6}{6} = 2\).

(b) The first three terms of the GP are \(6, 12, 24\).
For the arithmetic progression:
1st term: \(a = 6\)
5th term: \(a + 4d = 12 \implies 6 + 4d = 12 \implies 4d = 6 \implies d = 1.5\).
Check 13th term: \(a + 12d = 6 + 12(1.5) = 24\), which matches.
Thus, \(a = 6\) and \(d = 1.5\).

(c) The sum of the first 20 terms is:
\[ S_{20} = \frac{20}{2}\left[2(6) + (20 - 1)(1.5)\right] = 10[12 + 28.5] = 10(40.5) = 405 \]

(a) M1: Setting up equation using constant ratio \((p+6)^2 = p(3p+6)\) and expanding.
A1: Correct quadratic and obtaining \(p = 6\) (rejecting \(p = -3\)).
A1: Correct common ratio \(r = 2\).

(b) B1: Correct value \(a = 6\).
B1: Using \(a + 4d = 12\) to find \(d = 1.5\) (or \(\frac{3}{2}\)).

(c) M1: Correct use of AP sum formula \(S_n = \frac{n}{2}[2a + (n-1)d]\) with \(n=20\) and candidate's \(a, d\).
A1: 405

(a) Completing the square:
\[ 2x^2 - 16x + 35 = 2(x^2 - 8x) + 35 = 2\left[(x - 4)^2 - 16\right] + 35 = 2(x - 4)^2 + 3 \]
So \(a = 2\), \(h = 4\), \(k = 3\).
Since the vertex is at \((4, 3)\) and the domain is \(x \le 4\), the minimum value is \(3\).
Range: \(\mathrm{f}(x) \ge 3\) (or \(y \ge 3\)).

(b) Let \(y = 2(x - 4)^2 + 3\):
\[ 2(x - 4)^2 = y - 3 \implies (x - 4)^2 = \frac{y - 3}{2} \]
Since the domain is \(x \le 4\), \(x - 4 \le 0\), so we take the negative square root:
\[ x - 4 = -\sqrt{\frac{y - 3}{2}} \implies x = 4 - \sqrt{\frac{y - 3}{2}} \]
Thus, \(\mathrm{f}^{-1}(x) = 4 - \sqrt{\frac{x - 3}{2}}\).

(c) \(\mathrm{gf}(x) = \frac{8}{2\mathrm{f}(x) - 1} = \frac{8}{21}\)
\[ 2\mathrm{f}(x) - 1 = 21 \implies 2\mathrm{f}(x) = 22 \implies \mathrm{f}(x) = 11 \]
\[ 2(x - 4)^2 + 3 = 11 \implies 2(x - 4)^2 = 8 \implies (x - 4)^2 = 4 \]
Since \(x \le 4\), \(x - 4 = -2 \implies x = 2\).

(a) B1: Correctly completing the square to get \(2(x - 4)^2 + 3\).
B1: Stating correct range \(\mathrm{f}(x) \ge 3\) or \(y \ge 3\).

(b) M1: Correct algebraic steps to make \(x\) the subject of \(y = 2(x-4)^2 + 3\).
A1: \(\mathrm{f}^{-1}(x) = 4 - \sqrt{\frac{x - 3}{2}}\) (must select negative square root).

(c) M1: Setting \(\frac{8}{2\mathrm{f}(x)-1} = \frac{8}{21}\) to find \(\mathrm{f}(x) = 11\) and solving \(2(x-4)^2 + 3 = 11\).
A1: \(x = 2\) (rejecting \(x = 6\) due to domain restriction).

(a) Completing the square for the circle equation:
\[ (x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0 \]
\[ (x - 3)^2 + (y + 2)^2 = 25 \]
Centre is \((3, -2)\), radius is \(\sqrt{25} = 5\).

(b) Substitute \(y = 2x + k\) into the circle equation:
\[ (x - 3)^2 + (2x + k + 2)^2 = 25 \]
\[ x^2 - 6x + 9 + 4x^2 + 4(k + 2)x + (k + 2)^2 = 25 \]
\[ 5x^2 + (4k + 2)x + (k^2 + 4k - 12) = 0 \]
For two distinct intersections, the discriminant \(\Delta > 0\):
\[ (4k + 2)^2 - 4(5)(k^2 + 4k - 12) > 0 \]
\[ 16k^2 + 16k + 4 - 20k^2 - 80k + 240 > 0 \]
\[ -4k^2 - 64k + 244 > 0 \implies k^2 + 16k - 61 < 0 \]
Solving \(k^2 + 16k - 61 = 0\):
\[ k = \frac{-16 \pm \sqrt{16^2 - 4(1)(-61)}}{2} = \frac{-16 \pm \sqrt{500}}{2} = -8 \pm 5\sqrt{5} \]
Thus, \(-8 - 5\sqrt{5} < k < -8 + 5\sqrt{5}\).

(c) When \(k = -3\), the line equation is \(2x - y - 3 = 0\).
The perpendicular distance \(d\) from the centre \((3, -2)\) to the line is:
\[ d = \frac{|2(3) - (-2) - 3|}{\sqrt{2^2 + (-1)^2}} = \frac{|6 + 2 - 3|}{\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5} \]
Using Pythagoras' theorem on the triangle formed by the centre, chord midpoint, and an endpoint:
\[ \left(\frac{1}{2}AB\right)^2 = r^2 - d^2 = 5^2 - (\sqrt{5})^2 = 25 - 5 = 20 \]
\[ \frac{1}{2}AB = \sqrt{20} = 2\sqrt{5} \implies AB = 4\sqrt{5} \]

(a) B1: Centre \((3, -2)\).
B1: Radius \(r = 5\).

(b) M1: Substituting \(y = 2x + k\) into the circle equation to obtain a 3-term quadratic in \(x\).
M1: Applying discriminant \(b^2 - 4ac > 0\).
A1: Correct quadratic inequality \(k^2 + 16k - 61 < 0\) (or equivalent).
A1: \(-8 - 5\sqrt{5} < k < -8 + 5\sqrt{5}\).

(c) M1: Finding perpendicular distance from centre to line or finding coordinates of intersection points \(A\) and \(B\).
A1: \(4\sqrt{5}\) (exact form required).

(a) To find the coordinates of \( A \), set \( y = 0 \):
\[ 6\sqrt{x} - x\sqrt{x} = 0 \implies \sqrt{x}(6 - x) = 0 \]
Since \( A \) lies on the positive \( x \)-axis, \( x > 0 \), so \( x = 6 \).
Hence, \( A \) is \( (6, 0) \).

(b) Rewrite the equation of the curve in index form:
\[ y = 6x^{\frac{1}{2}} - x^{\frac{3}{2}} \]
Differentiating with respect to \( x \):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = 3x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}} \]
At the stationary point \( M \), \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \):
\[ \frac{3}{\sqrt{x}} - \frac{3}{2}\sqrt{x} = 0 \implies 3 = \frac{3}{2}x \implies x = 2 \]
Substitute \( x = 2 \) into the curve equation:
\[ y = 6\sqrt{2} - 2\sqrt{2} = 4\sqrt{2} \]
Thus, \( M \) has coordinates \( (2, 4\sqrt{2}) \).

To determine the nature, find the second derivative:
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{3}{2}x^{-\frac{3}{2}} - \frac{3}{4}x^{-\frac{1}{2}} \]
At \( x = 2 \):
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{3}{2(2\sqrt{2})} - \frac{3}{4\sqrt{2}} = -\frac{3}{4\sqrt{2}} - \frac{3}{4\sqrt{2}} = -\frac{6}{4\sqrt{2}} < 0 \]
Since \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0 \), the stationary point is a maximum.

(c) The region enclosed by the curve and the positive \( x \)-axis lies between \( x = 0 \) and \( x = 6 \):
\[ \text{Area} = \int_0^6 (6x^{\frac{1}{2}} - x^{\frac{3}{2}})\,\mathrm{d}x = \left[ 4x^{\frac{3}{2}} - \frac{2}{5}x^{\frac{5}{2}} \right]_0^6 \]
Evaluating at the limits:
\[ = \left( 4(6)^{\frac{3}{2}} - \frac{2}{5}(6)^{\frac{5}{2}} \right) - 0 = 4(6\sqrt{6}) - \frac{2}{5}(36\sqrt{6}) = 24\sqrt{6} - \frac{72}{5}\sqrt{6} = \frac{48\sqrt{6}}{5} \]

(d) The volume of revolution about the \( x \)-axis is given by:
\[ V = \pi \int_0^4 y^2\,\mathrm{d}x \]
First, expand \( y^2 \):
\[ y^2 = (6x^{\frac{1}{2}} - x^{\frac{3}{2}})^2 = 36x - 12x^2 + x^3 \]
Now integrate:
\[ V = \pi \int_0^4 (36x - 12x^2 + x^3)\,\mathrm{d}x = \pi \left[ 18x^2 - 4x^3 + \frac{1}{4}x^4 \right]_0^4 \]
Evaluating at \( x = 4 \):
\[ = \pi \left( 18(16) - 4(64) + \frac{1}{4}(256) \right) = \pi (288 - 256 + 64) = 96\pi \]

(a)
B1: State \( (6, 0) \) or \( x = 6, y = 0 \).

(b)
M1: Differentiate to obtain \( ax^{-\frac{1}{2}} + bx^{\frac{1}{2}} \).
A1: Obtain correct derivative \( 3x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}} \), equate to 0 and solve for \( x = 2 \).
A1: Obtain \( y = 4\sqrt{2} \) (or exact equivalent).
B1: Evaluate \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \) (or consider gradient change) and correctly deduce Maximum.

(c)
M1: Integrate \( y \) with at least one power increased by 1.
A1: Correct integrated expression \( 4x^{\frac{3}{2}} - \frac{2}{5}x^{\frac{5}{2}} \).
A1: Substitute limits 0 and 6 to obtain \( \frac{48\sqrt{6}}{5} \) (or exact equivalent surd form).

(d)
M1: Attempt \( \pi \int y^2\,\mathrm{d}x \) with correct expansion of \( y^2 \).
A1: Correct integration \( \left[ 18x^2 - 4x^3 + \frac{1}{4}x^4 \right] \).
A1: Substitute limits 0 and 4 to obtain \( 96\pi \).

(a) Rewrite the curve equation:
\[ y = 2(x - 3)^{\frac{1}{2}} \]
Differentiating with respect to \( x \):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = 2 \cdot \frac{1}{2}(x - 3)^{-\frac{1}{2}} = \frac{1}{\sqrt{x - 3}} \]
At \( x = 7 \):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{7 - 3}} = \frac{1}{2} \]
The equation of the tangent at \( A(7, 4) \) is:
\[ y - 4 = \frac{1}{2}(x - 7) \implies y = \frac{1}{2}x + \frac{1}{2} \quad \text{(or } x - 2y + 1 = 0\text{)} \]

(b) At the \( y \)-axis, \( x = 0 \):
\[ y = \frac{1}{2}(0) + \frac{1}{2} = \frac{1}{2} \]
Hence, \( B \) has coordinates \( \left(0, \frac{1}{2}\right) \).

(c) For rotation about the \( y \)-axis, express \( x \) in terms of \( y \):
\[ y = 2\sqrt{x - 3} \implies y^2 = 4(x - 3) \implies x - 3 = \frac{1}{4}y^2 \implies x = 3 + \frac{1}{4}y^2 \]
The volume is rotated between \( y = 0 \) and \( y = 4 \):
\[ V = \pi \int_0^4 x^2\,\mathrm{d}y = \pi \int_0^4 \left( 3 + \frac{1}{4}y^2 \right)^2\,\mathrm{d}y \]
Expand the integrand:
\[ x^2 = 9 + \frac{3}{2}y^2 + \frac{1}{16}y^4 \]
Integrate with respect to \( y \):
\[ V = \pi \left[ 9y + \frac{1}{2}y^3 + \frac{1}{80}y^5 \right]_0^4 \]
Evaluate at \( y = 4 \):
\[ V = \pi \left( 9(4) + \frac{1}{2}(64) + \frac{1024}{80} \right) = \pi \left( 36 + 32 + \frac{64}{5} \right) = \pi \left( 68 + \frac{64}{5} \right) = \frac{404\pi}{5} \]

(d) The tangent line \( y = \frac{1}{2}x + \frac{1}{2} \) crosses the \( x \)-axis when \( y = 0 \implies x = -1 \).
The area under the tangent from \( x = -1 \) to \( x = 7 \) is the area of a right-angled triangle with base \( 7 - (-1) = 8 \) and height \( 4 \):
\[ \text{Area of triangle} = \frac{1}{2} \times 8 \times 4 = 16 \]
The curve meets the \( x \)-axis at \( x = 3 \). The area under the curve from \( x = 3 \) to \( x = 7 \) is:
\[ \int_3^7 2(x - 3)^{\frac{1}{2}}\,\mathrm{d}x = \left[ \frac{4}{3}(x - 3)^{\frac{3}{2}} \right]_3^7 = \frac{4}{3}(7 - 3)^{\frac{3}{2}} - 0 = \frac{4}{3}(4^{\frac{3}{2}}) = \frac{4}{3}(8) = \frac{32}{3} \]
The area of the enclosed region is:
\[ \text{Area} = 16 - \frac{32}{3} = \frac{16}{3} \]

(a)
M1: Differentiate to obtain \( k(x - 3)^{-\frac{1}{2}} \).
A1: Obtain gradient \( \frac{1}{2} \) at \( x = 7 \).
A1: Obtain tangent equation \( y = \frac{1}{2}x + \frac{1}{2} \) or equivalent.

(b)
B1: State \( \left(0, \frac{1}{2}\right) \).

(c)
M1: Rearrange to express \( x \) in terms of \( y \) and square to form \( \pi \int x^2\,\mathrm{d}y \).
A1: Correct integrated expression \( \left[ 9y + \frac{1}{2}y^3 + \frac{1}{80}y^5 \right] \).
A1: Correct substitution of limits 0 and 4 to obtain \( \frac{404\pi}{5} \) (or \( 80.8\pi \)).

(d)
M1: Find area under tangent line from \( x = -1 \) to \( x = 7 \) (as triangle area \( 16 \) or by integration).
M1: Integrate curve \( \int_3^7 2(x-3)^{\frac{1}{2}}\,\mathrm{d}x \) to obtain \( \frac{32}{3} \).
A1: Subtract to obtain exact area \( \frac{16}{3} \).

(a) Write \(y = 16(2x + 1)^{-2}\).
Using the chain rule:
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 16 \cdot (-2)(2x + 1)^{-3} \cdot 2 = -64(2x + 1)^{-3} = -\frac{64}{(2x + 1)^3}\]

(b) At \(x = \frac{1}{2}\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{64}{\left(2\left(\frac{1}{2}\right) + 1\right)^3} = -\frac{64}{2^3} = -\frac{64}{8} = -8\]
The equation of the tangent at \(P\left(\frac{1}{2}, 4\right)\) is:
\[y - 4 = -8\left(x - \frac{1}{2}\right)\]
\[y - 4 = -8x + 4 \implies y = -8x + 8\]

(c) At the point \(Q\) where the tangent crosses the \(x\)-axis, \(y = 0\):
\[0 = -8x + 8 \implies x = 1\]
So \(Q\) is \((1, 0)\).

The required region lies between \(x = \frac{1}{2}\) and \(x = 2\). The total area under the curve from \(x = \frac{1}{2}\) to \(x = 2\) is:
\[\int_{1/2}^2 16(2x + 1)^{-2} \,\mathrm{d}x = \left[ 16 \cdot \frac{(2x + 1)^{-1}}{-1 \cdot 2} \right]_{1/2}^2 = \left[ -\frac{8}{2x + 1} \right]_{1/2}^2\]
\[= \left(-\frac{8}{2(2) + 1}\right) - \left(-\frac{8}{2\left(\frac{1}{2}\right) + 1}\right) = -\frac{8}{5} - (-4) = 4 - \frac{8}{5} = \frac{12}{5}\]

The area under the tangent line from \(x = \frac{1}{2}\) to \(x = 1\) forms a right-angled triangle with base \(1 - \frac{1}{2} = \frac{1}{2}\) and height \(4\):
\[\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{1}{2} \times 4 = 1\]

Therefore, the area of the region is:
\[\text{Area} = \frac{12}{5} - 1 = \frac{7}{5} = 1.4\]

(a)
M1: Applies chain rule to differentiate \(k(2x+1)^{-2}\) obtaining \(a(2x+1)^{-3}\).
A1: Obtains correct derivative \(-64(2x+1)^{-3}\) or \(-\frac{64}{(2x+1)^3}\).

(b)
M1: Evaluates their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) at \(x = \frac{1}{2}\) to find the numerical gradient.
M1: Forms equation of line using their gradient and point \(\left(\frac{1}{2}, 4\right)\).
A1: Obtains correct tangent equation \(y = -8x + 8\) or \(8x + y - 8 = 0\).

(c)
B1: Finds \(x = 1\) for the \(x\)-intercept of the tangent.
M1: Integrates \(16(2x+1)^{-2}\) to obtain form \(k(2x+1)^{-1}\).
A1: Obtains correct integral \(-\frac{8}{2x+1}\) and evaluates between \(x = \frac{1}{2}\) and \(x = 2\) to get \(\frac{12}{5}\).
M1: Calculates triangle area \(1\) (or integrates tangent from \(\frac{1}{2}\) to \(1\)) and subtracts from curve area.
A1: Obtains exact area \(\frac{7}{5}\) or \(1.4\).

To solve the inequality \(|3x - 5| < 2|x + 1|\), square both sides since both sides are non-negative:
\[ (3x - 5)^2 < 4(x + 1)^2 \]

Expand both expressions:
\[ 9x^2 - 30x + 25 < 4(x^2 + 2x + 1) \]
\[ 9x^2 - 30x + 25 < 4x^2 + 8x + 4 \]

Rearrange into standard quadratic form:
\[ 5x^2 - 38x + 21 < 0 \]

Find the critical values by solving the corresponding quadratic equation:
\[ (5x - 3)(x - 7) = 0 \]
\[ x = \frac{3}{5} \quad \text{or} \quad x = 7 \]

Since the inequality is \(< 0\), the solution is the region between the roots:
\[ \frac{3}{5} < x < 7 \]

M1: Valid method to solve the inequality, e.g. squaring both sides or considering separate linear cases.
A1: Correct 3-term quadratic inequality \(5x^2 - 38x + 21 < 0\) or linear critical equations.
A1: Correct critical values \(x = \frac{3}{5}\) (or \(0.6\)) and \(x = 7\).
A1: Correct final inequality \(\frac{3}{5} < x < 7\) (accept \(0.6 < x < 7\)).

Take the natural logarithm (or logarithm to base 10) of both sides:
\[ \ln\left(5^{2x - 1}\right) = \ln\left(3^{x + 2}\right) \]

Apply the power law of logarithms:
\[ (2x - 1)\ln 5 = (x + 2)\ln 3 \]

Expand the brackets:
\[ 2x\ln 5 - \ln 5 = x\ln 3 + 2\ln 3 \]

Rearrange to collect terms in \(x\) on one side:
\[ 2x\ln 5 - x\ln 3 = \ln 5 + 2\ln 3 \]
\[ x(2\ln 5 - \ln 3) = \ln 5 + 2\ln 3 \]

Make \(x\) the subject:
\[ x = \frac{\ln 5 + 2\ln 3}{2\ln 5 - \ln 3} = \frac{\ln 5 + \ln 9}{\ln 25 - \ln 3} = \frac{\ln 45}{\ln\left(\frac{25}{3}\right)} \]

Evaluate using a calculator:
\[ x = \frac{3.80666...}{2.12026...} \approx 1.79537... \]

Rounding to 3 significant figures gives \(x = 1.80\).

M1: Take logarithms of both sides and apply the power law \(\ln(a^b) = b\ln a\).
M1: Expand correctly and collect like terms in \(x\) on one side.
A1: Obtain a correct unsimplified exact expression for \(x\), e.g. \(x = \frac{\ln 5 + 2\ln 3}{2\ln 5 - \ln 3}\).
A1: Obtain \(1.80\) (must be given to 3 s.f., accept \(1.8\) only if trailing zero is clear from working).

(a) Using \(x_1 = 1.5\):
\[ x_2 = \frac{2(1.5)^3 + 15}{3(1.5)^2 + 4} = \frac{2(3.375) + 15}{3(2.25) + 4} = \frac{21.75}{10.75} \approx 2.0233 \]

Using \(x_2 = 2.023256\dots\):
\[ x_3 = \frac{2(2.023256)^3 + 15}{3(2.023256)^2 + 4} \approx \frac{31.5649}{16.2807} \approx 1.9388 \]

Using \(x_3 = 1.938794\dots\):
\[ x_4 = \frac{2(1.938794)^3 + 15}{3(1.938794)^2 + 4} \approx \frac{29.5754}{15.2767} \approx 1.9360 \]

(b) Since subsequent iterates round to \(1.94\), we have \(\alpha = 1.94\) correct to 2 decimal places.

(a) M1: Substitute \(x_1 = 1.5\) into the iterative formula to calculate \(x_2\).
A1: Obtain \(x_2 = 2.0233\).
A1: Obtain \(x_3 = 1.9388\) and \(x_4 = 1.9360\).

(b) B1: State \(\alpha = 1.94\) (or show sufficient convergence to justify 2 d.p.).

(a) Expanding \(R \cos(\theta - \alpha) = R \cos \theta \cos \alpha + R \sin \theta \sin \alpha\).
Equating coefficients:
\(R \cos \alpha = 6\) and \(R \sin \alpha = 8\).
\(R = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10\).
\(\tan \alpha = \frac{8}{6} = \frac{4}{3} \implies \alpha = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ = 53.1^\circ\).
Thus, \(6 \cos \theta + 8 \sin \theta = 10 \cos(\theta - 53.1^\circ)\).

(b) Using the result from part (a):
\(10 \cos(\theta - 53.13^\circ) = 5 \implies \cos(\theta - 53.13^\circ) = 0.5\).
\(\theta - 53.13^\circ = 60^\circ\) or \(\theta - 53.13^\circ = 360^\circ - 60^\circ = 300^\circ\) (or \(-60^\circ\)).
- \(\theta = 60^\circ + 53.13^\circ = 113.13^\circ \approx 113.1^\circ\)
- \(\theta = 300^\circ + 53.13^\circ = 353.13^\circ \approx 353.1^\circ\)

(c) The expression is \(\frac{1}{10 \cos(\theta - 53.13^\circ) + 15}\).
The fraction is maximised when the denominator is minimised and positive.
The minimum value of \(\cos(\theta - 53.13^\circ)\) is \(-1\).
Minimum denominator \(= 10(-1) + 15 = 5\).
Hence, the exact maximum value is \(\frac{1}{5}\).
This occurs when \(\cos(\theta - 53.13^\circ) = -1\):
\(\theta - 53.13^\circ = 180^\circ \implies \theta = 233.13^\circ \approx 233.1^\circ\).

(a)
- B1: State \(R = 10\).
- M1: Use \(\tan \alpha = \frac{8}{6}\) (or equivalent with correct trigonometric ratio).
- A1: Obtain \(\alpha = 53.1^\circ\) (accept 53.13).

(b)
- M1: Form the equation \(\cos(\theta - 53.1^\circ) = 0.5\) and find one correct angle (e.g. \(60^\circ\)).
- A1: Obtain \(\theta = 113.1^\circ\).
- A1: Obtain \(\theta = 353.1^\circ\) and no other values in the interval.

(c)
- M1: Recognise that the maximum occurs when \(\cos(\theta - \alpha) = -1\) (evaluating denominator \(= 15 - R\)).
- A1: Obtain exact maximum value \(\frac{1}{5}\).
- B1 (0.5 mark): Obtain \(\theta = 233.1^\circ\) (allow FT from their \(\alpha\)).

(a) Using the Factor Theorem, \(P(2) = 0\):
\(2(2)^3 + a(2)^2 + b(2) + 24 = 0\)
\(16 + 4a + 2b + 24 = 0 \implies 4a + 2b = -40 \implies 2a + b = -20\) --- (1)

Using the Remainder Theorem, \(P(-1) = 45\):
\(2(-1)^3 + a(-1)^2 + b(-1) + 24 = 45\)
\(-2 + a - b + 24 = 45 \implies a - b = 23\) --- (2)

Adding equations (1) and (2):
\(3a = 3 \implies a = 1\).
Substituting \(a = 1\) into (2):
\(1 - b = 23 \implies b = -22\).

(b) \(P(x) = 2x^3 + x^2 - 22x + 24\).
Since \((x - 2)\) is a factor, divide \(P(x)\) by \((x - 2)\):
\(2x^3 + x^2 - 22x + 24 = (x - 2)(2x^2 + 5x - 12)\).
Factorising the quadratic factor:
\(2x^2 + 5x - 12 = (2x - 3)(x + 4)\).
Therefore, \(P(x) = (x - 2)(2x - 3)(x + 4)\).

(c) The given equation is \(P(\tan \theta) = 0\).
From the factorisation:
\((\tan \theta - 2)(2\tan \theta - 3)(\tan \theta + 4) = 0\)
This gives:
1. \(\tan \theta = 2 \implies \theta = \tan^{-1}(2) \approx 1.107 \approx 1.11\text{ rad}\)
2. \(\tan \theta = \frac{3}{2} = 1.5 \implies \theta = \tan^{-1}(1.5) \approx 0.9828 \approx 0.983\text{ rad}\)
3. \(\tan \theta = -4 \implies \theta = \pi + \tan^{-1}(-4) = \pi - 1.3258 \approx 1.816 \approx 1.82\text{ rad}\)

All solutions in \(0 \leqslant \theta \leqslant \pi\) are \(\theta = 0.983, 1.11, 1.82\).

(a)
- M1: Substitute \(x = 2\) to set \(P(2) = 0\) and substitute \(x = -1\) to set \(P(-1) = 45\).
- M1: Solve the resulting simultaneous linear equations in \(a\) and \(b\).
- A1: Obtain \(a = 1\) and \(b = -22\).

(b)
- M1: Divide by \((x - 2)\) to find a quadratic factor \(2x^2 + kx + c\).
- A1: Obtain quadratic factor \(2x^2 + 5x - 12\).
- A0.5: Complete factorisation as \((x - 2)(2x - 3)(x + 4)\) (or \(2(x - 2)(x - 1.5)(x + 4)\)).

(c)
- B1: Identify \(\tan \theta = 2, 1.5, -4\) from linear factors.
- M1: Apply inverse tangent function and correctly handle angles within \([0, \pi]\) (especially the negative tangent value \(\pi - \tan^{-1}(4)\)).
- A1: Obtain all three correct values: \(0.983, 1.11, 1.82\) (allow 3 s.f. or better; penalise extra solutions in range).

(a) Differentiating $x$ and $y$ with respect to $\theta$:
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = 2 - 2\cos 2\theta = 2(1 - \cos 2\theta) \]
Using the double-angle identity $1 - \cos 2\theta = 2\sin^2 \theta$:
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = 2(2\sin^2 \theta) = 4\sin^2 \theta \]
Differentiating $y$ using the chain rule:
\[ \frac{\mathrm{d}y}{\mathrm{d}\theta} = 4 \times 2\sin \theta \cos \theta = 8\sin \theta \cos \theta \]
Using the chain rule for parametric differentiation:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}} = \frac{8\sin \theta \cos \theta}{4\sin^2 \theta} = \frac{2\cos \theta}{\sin \theta} = 2\cot \theta \]

(b) At $\theta = \frac{1}{6}\pi$:
\[ x = 2\left(\frac{\pi}{6}\right) - \sin\left(\frac{\pi}{3}\right) = \frac{\pi}{3} - \frac{\sqrt{3}}{2} \]
\[ y = 4\sin^2\left(\frac{\pi}{6}\right) = 4\left(\frac{1}{2}\right)^2 = 1 \]
Gradient of the tangent:
\[ m = 2\cot\left(\frac{\pi}{6}\right) = 2\left(\frac{\cos(\pi/6)}{\sin(\pi/6)}\right) = 2\left(\frac{\sqrt{3}/2}{1/2}\right) = 2\sqrt{3} \]
Equation of the tangent:
\[ y - 1 = 2\sqrt{3}\left(x - \left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right)\right) \]
\[ y - 1 = 2\sqrt{3}x - \frac{2\sqrt{3}}{3}\pi + 3 \]
\[ y = 2\sqrt{3}x + 4 - \frac{2\sqrt{3}}{3}\pi \]

(c) Using the identity $\cos^2\left(\frac{1}{2}x\right) = \frac{1 + \cos x}{2}$:
\[ y = 8\left(\frac{1 + \cos x}{2}\right) = 4 + 4\cos x \]
The area of the region is:
\[ \text{Area} = \int_{0}^{\frac{1}{2}\pi} (4 + 4\cos x)\,\mathrm{d}x \]
\[ = \left[ 4x + 4\sin x \right]_{0}^{\frac{1}{2}\pi} \]
\[ = \left( 4\left(\frac{\pi}{2}\right) + 4\sin\left(\frac{\pi}{2}\right) \right) - \left( 0 + 4\sin 0 \right) \]
\[ = 2\pi + 4(1) - 0 = 2\pi + 4 \]

(a)
B1: Correct differentiation of $x$, obtaining $2 - 2\cos 2\theta$, and use of $1 - \cos 2\theta = 2\sin^2 \theta$ to reach $4\sin^2 \theta$.
B1: Correct differentiation of $y$, obtaining $8\sin\theta\cos\theta$.
M1: Application of $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}\theta}{\mathrm{d}x/\mathrm{d}\theta}$.
A0.5: Clear algebraic simplification to the given result $2\cot\theta$.

(b)
B1: Correct evaluation of $x = \frac{\pi}{3} - \frac{\sqrt{3}}{2}$, $y = 1$, and gradient $m = 2\sqrt{3}$.
M1: Correct method to form the equation of a straight line using their coordinates and gradient.
A1: Correct exact equation in the required form $y = 2\sqrt{3}x + 4 - \frac{2\sqrt{3}}{3}\pi$ (or $y = 2\sqrt{3}x + 4 - \frac{2\pi}{\sqrt{3}}$).

(c)
B1: Correctly rewrites the integrand as $4 + 4\cos x$.
M1: Correct integration to obtain $4x + 4\sin x$ and substitution of limits $0$ and $\frac{1}{2}\pi$.
A1: Correct exact answer $2\pi + 4$.

(a) Differentiating $x$ and $y$ with respect to $\theta$:
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = 2\sec^2 \theta = \frac{2}{\cos^2 \theta} \]
\[ \frac{\mathrm{d}y}{\mathrm{d}\theta} = 12\sin\theta\cos\theta \]
Using the parametric differentiation rule:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}} = \frac{12\sin\theta\cos\theta}{2\sec^2 \theta} = 6\sin\theta\cos\theta \cdot \cos^2 \theta = 6\sin\theta\cos^3 \theta \]

(b) The gradient of the normal is given by $m_N = -\frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}}$.
Setting $m_N = -\frac{8}{9\sqrt{3}}$:
\[ -\frac{1}{6\sin\theta\cos^3 \theta} = -\frac{8}{9\sqrt{3}} \implies 6\sin\theta\cos^3 \theta = \frac{9\sqrt{3}}{8} \]
\[ \sin\theta\cos^3 \theta = \frac{3\sqrt{3}}{16} \]
For $\theta = \frac{1}{6}\pi$:
\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}, \quad \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \implies \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)^3 = \frac{3\sqrt{3}}{16} \]
Hence $\theta = \frac{1}{6}\pi$.
Finding the coordinates:
\[ x = 2\tan\left(\frac{\pi}{6}\right) = 2\left(\frac{1}{\sqrt{3}}\right) = \frac{2\sqrt{3}}{3} \]
\[ y = 6\sin^2\left(\frac{\pi}{6}\right) = 6\left(\frac{1}{2}\right)^2 = \frac{3}{2} \]
The exact coordinates are $\left(\frac{2\sqrt{3}}{3}, \frac{3}{2}\right)$.

(c)(i) Using the identity $\sin^2(2x) = \frac{1 - \cos 4x}{2}$:
\[ 12\sin^2(2x) = 12\left(\frac{1 - \cos 4x}{2}\right) = 6 - 6\cos 4x \]
Thus $a = 6$ and $b = -6$.

(c)(ii) The area is:
\[ \text{Area} = \int_{0}^{\frac{1}{4}\pi} (6 - 6\cos 4x)\,\mathrm{d}x = \left[ 6x - \frac{6}{4}\sin 4x \right]_{0}^{\frac{1}{4}\pi} = \left[ 6x - \frac{3}{2}\sin 4x \right]_{0}^{\frac{1}{4}\pi} \]
Evaluating at the limits:
\[ = \left( 6\left(\frac{\pi}{4}\right) - \frac{3}{2}\sin\pi \right) - (0 - 0) = \frac{3\pi}{2} - 0 = \frac{3}{2}\pi \]

(a)
B1: Correct derivative $\frac{\mathrm{d}x}{\mathrm{d}\theta} = 2\sec^2\theta$.
B1: Correct derivative $\frac{\mathrm{d}y}{\mathrm{d}\theta} = 12\sin\theta\cos\theta$.
M1: Correct division $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}\theta}{\mathrm{d}x/\mathrm{d}\theta}$ leading fully to $6\sin\theta\cos^3\theta$.

(b)
M1: Forms equation for normal gradient $-\frac{1}{\mathrm{d}y/\mathrm{d}x} = -\frac{8}{9\sqrt{3}}$ to get $\sin\theta\cos^3\theta = \frac{3\sqrt{3}}{16}$.
A1: Deduces $\theta = \frac{1}{6}\pi$ (or $\frac{\pi}{6}$).
M1: Substitutes their $\theta$ into the parametric expressions for $x$ and $y$.
A0.5: Obtains exact coordinates $\left(\frac{2\sqrt{3}}{3}, \frac{3}{2}\right)$ or $\left(\frac{2}{\sqrt{3}}, \frac{3}{2}\right)$.

(c)(i)
B1: States $a = 6$ and $b = -6$ (or writes $6 - 6\cos 4x$).

(c)(ii)
M1: Integrates their $a + b\cos 4x$ to obtain $ax + \frac{b}{4}\sin 4x$ and applies limits $0$ to $\frac{1}{4}\pi$.
A1: Obtains exact area $\frac{3}{2}\pi$ (or $1.5\pi$).