2026 Cambridge IAS-Level Mathematics (9709) 模擬試題連答案詳解

Let \( u = x^{\frac{1}{4}} \), so \( u^2 = x^{\frac{1}{2}} \).

The equation becomes:
\[ 2u^2 - 7u + 3 = 0 \]

Factorising the quadratic equation:
\[ (2u - 1)(u - 3) = 0 \]

This gives:
\[ u = \frac{1}{2} \quad \text{or} \quad u = 3 \]

Now substitute back \( u = x^{\frac{1}{4}} \):
\[ x^{\frac{1}{4}} = \frac{1}{2} \implies x = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]
\[ x^{\frac{1}{4}} = 3 \implies x = 3^4 = 81 \]

Thus, the solutions are \( x = \frac{1}{16} \) and \( x = 81 \).

M1: Recognising the disguised quadratic and attempting to factorise or solve a 3-term quadratic in \( x^{\frac{1}{4}} \) (or in \( u \) where \( u = x^{\frac{1}{4}} \)).
A1: Finding correct values for the intermediate variable: \( x^{\frac{1}{4}} = \frac{1}{2} \) and \( x^{\frac{1}{4}} = 3 \).
A1: Both correct final answers \( x = \frac{1}{16} \) (or 0.0625) and \( x = 81 \) obtained from valid working.

(a)
\[ \mathrm{f}(x) = 2(x^2 - 6x) + 13 = 2\left((x - 3)^2 - 9\right) + 13 = 2(x - 3)^2 - 18 + 13 = 2(x - 3)^2 - 5 \]
Thus, \( a = 2 \), \( h = 3 \), and \( k = -5 \).

(b)
Applying a stretch parallel to the \( y \)-axis with scale factor 3 to \( y = \mathrm{f}(x) \):
\[ y_1 = 3\mathrm{f}(x) = 3\left(2(x - 3)^2 - 5\right) = 6(x - 3)^2 - 15 \]
Then, applying a translation by the vector \( \begin{pmatrix} -1 \\ 4 \end{pmatrix} \) replaces \( x \) with \( x + 1 \) and adds 4 to the expression:
\[ \mathrm{g}(x) = 6\left((x + 1) - 3\right)^2 - 15 + 4 = 6(x - 2)^2 - 11 \]
Expanding to the form \( px^2 + qx + r \):
\[ \mathrm{g}(x) = 6(x^2 - 4x + 4) - 11 = 6x^2 - 24x + 24 - 11 = 6x^2 - 24x + 13 \]

(c)
Starting from \( y = x^2 \) to reach \( y = 6(x - 2)^2 - 11 \):
1. Stretch parallel to the \( y \)-axis with scale factor 6 (gives \( y = 6x^2 \)).
2. Translation by the vector \( \begin{pmatrix} 2 \\ -11 \end{pmatrix} \) (gives \( y = 6(x - 2)^2 - 11 \)).

Alternatively:
1. Translation by the vector \( \begin{pmatrix} 2 \\ -\frac{11}{6} \end{pmatrix} \), followed by a stretch parallel to the \( y \)-axis with scale factor 6.

(a)
- B1: For \( 2(x - 3)^2 \) or identifying \( a = 2 \) and \( h = 3 \).
- B1: For \( -5 \) (or \( k = -5 \)).

(b)
- M1: For correctly applying the stretch and translation in order, e.g. \( 3\mathrm{f}(x + 1) + 4 \) or replacing \( x \to x + 1 \) and \( y \to 3y + 4 \) on vertex/standard form.
- A1: For \( \mathrm{g}(x) = 6x^2 - 24x + 13 \) (accept \( p = 6, q = -24, r = 13 \)).

(c)
- B1: Stretch parallel to the \( y \)-axis with scale factor 6 (or stretch in the \( y \)-direction with scale factor 6).
- B1: Translation by the vector \( \begin{pmatrix} 2 \\ -11 \end{pmatrix} \) (or translation of 2 units in the positive \( x \)-direction and 11 units in the negative \( y \)-direction).
- B1: Correct order clearly stated (must correspond to the transformations given, e.g. stretch then translation).

(a)
First, find the midpoint \(M\) of the line segment \(AB\):
\[M = \left(\frac{-2 + 4}{2}, \frac{7 + (-1)}{2}\right) = (1, 3)\]

Next, calculate the gradient of the line segment \(AB\):
\[m_{AB} = \frac{-1 - 7}{4 - (-2)} = \frac{-8}{6} = -\frac{4}{3}\]

The gradient of the perpendicular bisector is the negative reciprocal:
\[m = -\frac{1}{m_{AB}} = \frac{3}{4}\]

Using the point-slope form with \(M(1, 3)\):
\[y - 3 = \frac{3}{4}(x - 1)\]
\[y = \frac{3}{4}x - \frac{3}{4} + 3\]
\[y = \frac{3}{4}x + \frac{9}{4}\]

(b)
The centre of any circle passing through \(A\) and \(B\) must lie on the perpendicular bisector of \(AB\).

To find the centre, solve the simultaneous equations:
\[y = \frac{3}{4}x + \frac{9}{4}\]
\[y = 2x - 4\]

Equating the two expressions for \(y\):
\[2x - 4 = \frac{3}{4}x + \frac{9}{4}\]
\[\frac{5}{4}x = \frac{25}{4} \implies x = 5\]

Substitute \(x = 5\) into \(y = 2x - 4\):
\[y = 2(5) - 4 = 6\]
So the centre of the circle is \((5, 6)\).

Now, calculate the square of the radius \(r^2\) using point \(A(-2, 7)\):
\[r^2 = (5 - (-2))^2 + (6 - 7)^2 = 7^2 + (-1)^2 = 49 + 1 = 50\]

Thus, the equation of the circle is:
\[(x - 5)^2 + (y - 6)^2 = 50\]

(a)
- M1: For finding the midpoint \((1, 3)\) and gradient of \(AB\) \(\left(-\frac{4}{3}\right)\).
- M1: For using the perpendicular gradient rule \(m_{\perp} = -\frac{1}{m}\) and forming the line equation through their midpoint.
- A1: For correct equation \(y = \frac{3}{4}x + \frac{9}{4}\) or \(y = 0.75x + 2.25\).

(b)
- M1: For solving simultaneously the perpendicular bisector from (a) with \(y = 2x - 4\) to find the coordinates of the centre.
- M1: For a correct method to find the radius or \(r^2\) using their centre and point \(A\) or \(B\).
- A1: For correct equation \((x - 5)^2 + (y - 6)^2 = 50\) (or equivalent expanded form \(x^2 + y^2 - 10x - 12y + 11 = 0\)).

(a) The general term in the binomial expansion of \(\left(2x - \dfrac{a}{x}\right)^6\) is:
\[ \binom{6}{r} (2x)^{6-r} \left(-\dfrac{a}{x}\right)^r = \binom{6}{r} 2^{6-r} (-a)^r x^{6-2r} \]
For the term independent of \(x\), set the power of \(x\) to zero:
\[ 6 - 2r = 0 \implies r = 3 \]
Evaluating the term:
\[ \binom{6}{3} (2x)^3 \left(-\dfrac{a}{x}\right)^3 = 20 \times 8x^3 \times \left(-\dfrac{a^3}{x^3}\right) = -160a^3 \]

(b) In the expansion of \(\left(1 + \dfrac{3}{x^2}\right)\left(2x - \dfrac{a}{x}\right)^6\), the terms that contribute to the constant term are:
1. \(1 \times (\text{term independent of } x \text{ in } (2x - \frac{a}{x})^6) = 1 \times (-160a^3) = -160a^3\)
2. \(\dfrac{3}{x^2} \times (\text{term in } x^2 \text{ in } (2x - \frac{a}{x})^6)\)

To find the term in \(x^2\), set \(6 - 2r = 2 \implies 2r = 4 \implies r = 2\):
\[ \binom{6}{2} (2x)^4 \left(-\dfrac{a}{x}\right)^2 = 15 \times 16x^4 \times \dfrac{a^2}{x^2} = 240a^2 x^2 \]
Multiplying by \(\dfrac{3}{x^2}\):
\[ \dfrac{3}{x^2} \times 240a^2 x^2 = 720a^2 \]
Thus, the term independent of \(x\) is:
\[ -160a^3 + 720a^2 \]
Setting this equal to \(2160\):
\[ -160a^3 + 720a^2 = 2160 \]
Divide through by \(80\):
\[ -2a^3 + 9a^2 = 27 \implies 2a^3 - 9a^2 + 27 = 0 \]
Factorising the cubic:
\[ (a - 3)(2a^2 - 3a - 9) = 0 \]
\[ (a - 3)(a - 3)(2a + 3) = 0 \implies (a - 3)^2(2a + 3) = 0 \]
Hence, the possible values of \(a\) are:
\[ a = 3 \quad \text{or} \quad a = -\dfrac{3}{2} \]

(a)
- M1: Identifies the correct term using \(r = 3\) or sets up the general term correctly with powers adding to 6.
- M1: Correct evaluation of binomial coefficient and powers: \(\binom{6}{3} \times 2^3 \times (-a)^3\) or \(20 \times 8 \times (-a^3)\).
- A1: \(-160a^3\) (must include the negative sign).

(b)
- B1: Finds the correct term in \(x^2\) for the single expansion: \(240a^2 x^2\) or coefficient \(240a^2\).
- M1: Correctly combines the two relevant product terms to form the constant term: \(1 \times (-160a^3) + 3 \times 240a^2\) (ft their part (a)).
- M1: Sets their expression equal to \(2160\) and demonstrates a valid method to solve the resulting cubic equation (e.g. testing factors and factorising).
- A1: Obtains \(a = 3\) and \(a = -\dfrac{3}{2}\) (or \(-1.5\)) with no extra incorrect real roots.

(a)
Integrate \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) with respect to \(x\):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \int \left(12x + 4x^{-3}\right) \mathrm{d}x = 6x^2 - 2x^{-2} + c \]
Since \(P(1, 5)\) is a stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) when \(x = 1\):
\[ 6(1)^2 - 2(1)^{-2} + c = 0 \implies 6 - 2 + c = 0 \implies c = -4 \]
Thus, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - \frac{2}{x^2} - 4\).

Now integrate \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find \(y\):
\[ y = \int \left(6x^2 - 2x^{-2} - 4\right) \mathrm{d}x = 2x^3 + 2x^{-1} - 4x + d \]
Substitute the coordinates of \(P(1, 5)\) into the equation:
\[ 5 = 2(1)^3 + \frac{2}{1} - 4(1) + d \]
\[ 5 = 2 + 2 - 4 + d \implies 5 = d \]
Therefore, the equation of the curve is:
\[ y = 2x^3 + \frac{2}{x} - 4x + 5 \]

(b)
Evaluate \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) at \(x = 1\):
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(1) + \frac{4}{1^3} = 12 + 4 = 16 \]
Since \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0\), the stationary point \(P\) is a minimum.

(a)
- M1: Integrate \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) to find an expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\) with at least one term correctly integrated (power increased by 1).
- A1: Correct expression \(\frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 2x^{-2} + c\) (constant \(c\) not required for this mark).
- M1: Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) with \(x = 1\) and solve to find \(c = -4\).
- M1: Integrate their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to obtain \(y\), including a second constant of integration \(d\), and substitute \((1, 5)\) to find \(d\).
- A1: Correct final equation \(y = 2x^3 + \frac{2}{x} - 4x + 5\) (or exact equivalent).

(b)
- M1: Substitute \(x = 1\) into the given expression for \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\).
- A1: Obtain 16 (or state \(> 0\)) and conclude that \(P\) is a minimum (WWW).

Use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\):

\[6(1 - \sin^2 \theta) - \sin \theta - 4 = 0\]
\[6 - 6\sin^2 \theta - \sin \theta - 4 = 0\]
\[6\sin^2 \theta + \sin \theta - 2 = 0\]

Factorise the quadratic equation in \(\sin \theta\):

\[(2\sin \theta - 1)(3\sin \theta + 2) = 0\]

This gives:

\[\sin \theta = \frac{1}{2} \quad \text{or} \quad \sin \theta = -\frac{2}{3}\]

For \(\sin \theta = \frac{1}{2}\):
\[\theta = 30^\circ, \quad 180^\circ - 30^\circ = 150^\circ\]

For \(\sin \theta = -\frac{2}{3}\):
Basic angle \(\alpha = \sin^{-1}\left(\frac{2}{3}\right) \approx 41.81^\circ\)
\[\theta = 180^\circ + 41.81^\circ = 221.8^\circ\]
\[\theta = 360^\circ - 41.81^\circ = 318.2^\circ\]

Therefore, the solutions in the interval \(0^\circ \leqslant \theta \leqslant 360^\circ\) are \(\theta = 30^\circ, 150^\circ, 221.8^\circ, 318.2^\circ\).

M1: Uses \(\cos^2 \theta = 1 - \sin^2 \theta\) to obtain a 3-term quadratic in \(\sin \theta\).
M1: Solves their 3-term quadratic equation in \(\sin \theta\) by factorisation, formula, or completing the square to find values for \(\sin \theta\).
A1: Obtains \(\theta = 30^\circ, 150^\circ\).
A1: Obtains \(\theta = 221.8^\circ, 318.2^\circ\) (accept 1 d.p.) and no other solutions in the given interval.

(a) (i) The 1st, 4th, and 13th terms of the arithmetic progression are:
\(u_1 = a\)
\(u_4 = a + 3d\)
\(u_{13} = a + 12d\)

Since these form consecutive terms of a geometric progression:
\(\frac{a + 3d}{a} = \frac{a + 12d}{a + 3d}\)
\((a + 3d)^2 = a(a + 12d)\)
\(a^2 + 6ad + 9d^2 = a^2 + 12ad\)
\(9d^2 - 6ad = 0\)
\(3d(3d - 2a) = 0\)

Since \(d \neq 0\), we have \(3d - 2a = 0 \implies d = \frac{2}{3}a\).

(ii) The common ratio is:
\(r = \frac{a + 3d}{a} = \frac{a + 3\left(\frac{2}{3}a\right)}{a} = \frac{3a}{a} = 3\).

(b) The formula for the sum of the first \(n\) terms of an arithmetic progression is:
\(S_n = \frac{n}{2}[2a + (n-1)d]\)

For \(n = 20\):
\(S_{20} = \frac{20}{2}[2a + 19d] = 10[2a + 19d]\)

Substituting \(d = \frac{2}{3}a\):
\(S_{20} = 10\left[2a + 19\left(\frac{2}{3}a\right)\right] = 10\left(\frac{44}{3}a\right) = \frac{440}{3}a\)

Given \(S_{20} = 440\):
\(\frac{440}{3}a = 440 \implies a = 3\)

Then \(d = \frac{2}{3}(3) = 2\).

(c) (i) For the second geometric progression:
\(S_\infty = \frac{b}{1-k} = 440\)

The sum of the first two terms is:
\(b + bk = b(1+k) = 330\)

Dividing the two equations:
\(\frac{b(1+k)}{\frac{b}{1-k}} = (1+k)(1-k) = 1 - k^2\)
\(1 - k^2 = \frac{330}{440} = \frac{3}{4}\)
\(k^2 = \frac{1}{4}\)

Since \(0 < k < 1\), \(k = \frac{1}{2}\).

Substituting \(k = \frac{1}{2}\) into \(S_\infty\):
\(\frac{b}{1 - \frac{1}{2}} = 440 \implies b = 220\).

(ii) The sum of the first \(n\) terms is:
\(S_n = \frac{b(1-k^n)}{1-k} = \frac{220\left(1 - \left(\frac{1}{2}\right)^n\right)}{1 - \frac{1}{2}} = 440\left(1 - \left(\frac{1}{2}\right)^n\right)\)

We require \(S_n > 439\):
\(440\left(1 - \left(\frac{1}{2}\right)^n\right) > 439\)
\(1 - \left(\frac{1}{2}\right)^n > \frac{439}{440}\)
\(\left(\frac{1}{2}\right)^n < \frac{1}{440}\)
\(2^n > 440\)

Since \(2^8 = 256\) and \(2^9 = 512\), the least integer value is \(n = 9\).

(a)(i) M1: Equating ratios \(\frac{a+3d}{a} = \frac{a+12d}{a+3d}\) or using \((a+3d)^2 = a(a+12d)\)
A1: Correct expansion and simplifying to obtain given result \(d = \frac{2}{3}a\) WWW AG

(a)(ii) B1: Correct common ratio \(r = 3\)

(b) M1: Uses correct sum formula \(S_{20} = \frac{20}{2}(2a + 19d)\) and equates to 440
M1: Substitutes \(d = \frac{2}{3}a\) into their equation to obtain a single equation in \(a\)
A1: Correct values: \(a = 3\) and \(d = 2\)

(c)(i) M1: Sets up simultaneous equations \(\frac{b}{1-k} = 440\) and \(b(1+k) = 330\), eliminating \(b\) or \(k\)
A1: Both correct values \(k = \frac{1}{2}\) (or 0.5) and \(b = 220\)

(c)(ii) M1: Sets up inequality/equation \(\frac{220(1 - 0.5^n)}{1 - 0.5} > 439\) and solves for \(n\) using logs or powers of 2
A1: Correct integer \(n = 9\)

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

(b) Rearranging the line equation gives $x = 2y - 9$.
Substitute this into the circle equation:
\[ (2y - 9 - 2)^2 + (y - 3)^2 = 25 \]
\[ (2y - 11)^2 + (y - 3)^2 = 25 \]
\[ (4y^2 - 44y + 121) + (y^2 - 6y + 9) = 25 \]
\[ 5y^2 - 50y + 105 = 0 \]
\[ y^2 - 10y + 21 = 0 \]
\[ (y - 3)(y - 7) = 0 \]
So $y = 3$ or $y = 7$.

When $y = 3$, $x = 2(3) - 9 = -3$, giving the point $A(-3, 3)$.
When $y = 7$, $x = 2(7) - 9 = 5$, giving the point $B(5, 7)$.

The length of the chord $AB$ is:
\[ AB = \sqrt{(5 - (-3))^2 + (7 - 3)^2} = \sqrt{8^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5} \]

(c) Let the centre of the circle be $M(2, 3)$.

For the tangent at $A(-3, 3)$:
The radius $MA$ is horizontal as both points have $y$-coordinate $3$ (gradient $= 0$).
Therefore, the tangent at $A$ is a vertical line with equation $x = -3$.

For the tangent at $B(5, 7)$:
Gradient of radius $MB = \frac{7 - 3}{5 - 2} = \frac{4}{3}$.
Hence, the gradient of the tangent at $B$ is $-\frac{3}{4}$.
The equation of the tangent at $B$ is:
\[ y - 7 = -\frac{3}{4}(x - 5) \implies 4y - 28 = -3x + 15 \implies 3x + 4y = 43 \]

To find the intersection point $T$, substitute $x = -3$ into the tangent equation:
\[ 3(-3) + 4y = 43 \implies -9 + 4y = 43 \implies 4y = 52 \implies y = 13 \]

Thus, the coordinates of $T$ are $(-3, 13)$.

(a)
B1: State centre $(2, 3)$
B1: State radius $r = 5$

(b)
M1: Substitute linear expression into circle equation to obtain a 3-term quadratic in one variable
A1: Correct 3-term quadratic, e.g. $y^2 - 10y + 21 = 0$ or $5x^2 - 10x - 75 = 0$
A1: Correct coordinates for both points $A(-3, 3)$ and $B(5, 7)$ (in either order)
M1: Apply correct distance formula $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ to candidate coordinates
A1: Correct exact chord length $4\sqrt{5}$ or $\sqrt{80}$

(c)
M1: Find gradient of at least one radius and state/use perpendicular gradient $m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}}$ (or state vertical tangent $x = -3$)
M1: Form equation of tangent at $B$: $y - 7 = -\frac{3}{4}(x - 5)$ (or equivalent)
M1: Solve equations of the two tangents simultaneously
A1: Correct coordinates $T(-3, 13)$

(a) Write the curve as \(y = 12(2x + 1)^{-\frac{1}{2}}\).
Differentiating with respect to \(x\) using the chain rule:
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 12 \times \left(-\frac{1}{2}\right)(2x + 1)^{-\frac{3}{2}} \times 2 = -12(2x + 1)^{-\frac{3}{2}}\]
At \(x = 4\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = -12(2(4) + 1)^{-\frac{3}{2}} = -12(9)^{-\frac{3}{2}} = -12 \times \frac{1}{27} = -\frac{4}{9}\]
The equation of the tangent at \(P(4, 4)\) is:
\[y - 4 = -\frac{4}{9}(x - 4)\]
\[y = -\frac{4}{9}x + \frac{16}{9} + 4 \implies y = -\frac{4}{9}x + \frac{52}{9} \quad \text{or} \quad 4x + 9y = 52\]

(b) Integrating \(12(2x + 1)^{-\frac{1}{2}}\) with respect to \(x\):
\[\int 12(2x + 1)^{-\frac{1}{2}} \, \mathrm{d}x = \frac{12(2x + 1)^{\frac{1}{2}}}{\frac{1}{2} \times 2} + c = 12(2x + 1)^{\frac{1}{2}} + c = 12\sqrt{2x + 1} + c\]

(c) The enclosed region is bounded by \(x = 0\) and \(x = 4\).
First, find the area under the curve from \(x = 0\) to \(x = 4\):
\[\int_{0}^{4} \frac{12}{\sqrt{2x + 1}} \, \mathrm{d}x = \left[ 12\sqrt{2x + 1} \right]_{0}^{4} = 12\sqrt{9} - 12\sqrt{1} = 36 - 12 = 24\]
Next, find the area under the tangent from \(x = 0\) to \(x = 4\):
\[\int_{0}^{4} \left(-\frac{4}{9}x + \frac{52}{9}\right) \, \mathrm{d}x = \left[ -\frac{2}{9}x^2 + \frac{52}{9}x \right]_{0}^{4} = -\frac{2(16)}{9} + \frac{52(4)}{9} = -\frac{32}{9} + \frac{208}{9} = \frac{176}{9}\]
(Alternatively, using the area of a trapezium with parallel vertical heights \(y(0) = \frac{52}{9}\) and \(y(4) = 4\):
\[\text{Area} = \frac{1}{2}\left(\frac{52}{9} + 4\right) \times 4 = 2 \times \frac{88}{9} = \frac{176}{9}\)) The area of the enclosed region is: \[\text{Area} = 24 - \frac{176}{9} = \frac{216 - 176}{9} = \frac{40}{9}\]

(a)
M1: Differentiates to obtain \(k(2x + 1)^{-\frac{3}{2}}\) where \(k\) is a constant.
A1: Correct derivative \(-12(2x + 1)^{-\frac{3}{2}}\) (or unsimplified equivalent).
M1: Substitutes \(x = 4\) into their derivative to find gradient and forms equation of line through \((4, 4)\).
A1: Correct equation in any equivalent form, e.g. \(y = -\frac{4}{9}x + \frac{52}{9}\) or \(4x + 9y - 52 = 0\).

(b)
M1: Integrates to obtain \(A(2x + 1)^{\frac{1}{2}}\) where \(A\) is a constant.
A1: \(12\sqrt{2x + 1} + c\) or \(12(2x + 1)^{\frac{1}{2}} + c\) (omission of \(+ c\) is condoned).

(c)
M1: Applies limits \(0\) and \(4\) correctly to their integral from part (b).
A1: Correct area under curve \(= 24\).
M1: Finds area under tangent between \(x = 0\) and \(x = 4\) by integration or trapezium formula.
M1: (Dependent on previous M marks) Subtracts area under tangent from area under curve (or evaluates \(\int_0^4 (y_{\text{curve}} - y_{\text{tangent}})\,\mathrm{d}x\)).
A1: Correct exact area \(\frac{40}{9}\) (or \(4\frac{4}{9}\)).

(a) Completing the square:
\[ 2x^2 - 12x + 13 = 2(x^2 - 6x) + 13 = 2\left[(x - 3)^2 - 9\right] + 13 = 2(x - 3)^2 - 18 + 13 = 2(x - 3)^2 - 5 \]
So \(p = 3\) and \(q = -5\).
The vertex of the parabola is at \(x = 3\). For \(\mathrm{f}\) to be a one-one function on \(x \ge c\), the least value of \(c\) is \(3\).

(b) Let \(y = 2(x - 3)^2 - 5\) for \(x \ge 3\).
\[ y + 5 = 2(x - 3)^2 \]
\[ (x - 3)^2 = \frac{y + 5}{2} \]
Since \(x \ge 3\), take the positive square root:
\[ x - 3 = \sqrt{\frac{y + 5}{2}} \]
\[ x = 3 + \sqrt{\frac{y + 5}{2}} \]
Therefore, \(\mathrm{f}^{-1}(x) = 3 + \sqrt{\frac{x + 5}{2}}\) (or \(3 + \sqrt{\frac{1}{2}(x + 5)}\)).
The domain of \(\mathrm{f}^{-1}\) is the range of \(\mathrm{f}\), which is \(x \ge -5\).

(c) Set \(\mathrm{f}(x) = \mathrm{g}(x)\):
\[ 2x^2 - 12x + 13 = 2x + k \]
\[ 2x^2 - 14x + (13 - k) = 0 \]
For the equation to have no real roots, the discriminant must be strictly less than 0:
\[ b^2 - 4ac < 0 \]
\[ (-14)^2 - 4(2)(13 - k) < 0 \]
\[ 196 - 8(13 - k) < 0 \]
\[ 196 - 104 + 8k < 0 \]
\[ 92 + 8k < 0 \]
\[ 8k < -92 \]
\[ k < -\frac{92}{8} = -\frac{23}{2} \quad (\text{or } k < -11.5) \]

(a)
- M1: Attempt to complete the square, obtaining \(2(x - 3)^2 + q\) or finding \(p = 3, q = -5\).
- A1: Fully correct form \(2(x - 3)^2 - 5\).
- B1: State \(c = 3\) (allow \(c \ge 3\) if \(c = 3\) is identified as the least value).

(b)
- M1: Set \(y = 2(x - 3)^2 - 5\) and correctly make \((x - 3)^2\) or \(x - 3\) the subject.
- A1: Obtain \(\mathrm{f}^{-1}(x) = 3 + \sqrt{\frac{x + 5}{2}}\) (must be in terms of \(x\) with the positive root selected).
- B1: State domain \(x \ge -5\) or \([-5, \infty)\) (must use \(x\) or state domain clearly, not \(y\)).

(c)
- M1: Equate \(2x^2 - 12x + 13 = 2x + k\) and form a 3-term quadratic equation \(2x^2 - 14x + (13 - k) = 0\).
- M1: Use discriminant \(b^2 - 4ac < 0\) on their quadratic equation in \(x\).
- A1: Obtain correct strict inequality \(k < -\frac{23}{2}\) (or \(k < -11.5\)).

Integrate each term with respect to \(x\):
\[ \int \left( 3\mathrm{e}^{2x} - 2\mathrm{e}^{-2x} \right) \mathrm{d}x = \left[ \frac{3}{2}\mathrm{e}^{2x} + \mathrm{e}^{-2x} \right]_{0}^{\ln 2} \]

Substitute the upper limit \(x = \ln 2\):
\[ \frac{3}{2}\mathrm{e}^{2\ln 2} + \mathrm{e}^{-2\ln 2} = \frac{3}{2}\mathrm{e}^{\ln 4} + \mathrm{e}^{\ln(1/4)} = \frac{3}{2}(4) + \frac{1}{4} = 6 + \frac{1}{4} = \frac{25}{4} \]

Substitute the lower limit \(x = 0\):
\[ \frac{3}{2}\mathrm{e}^{0} + \mathrm{e}^{0} = \frac{3}{2}(1) + 1 = \frac{5}{2} \]

Subtract the lower limit value from the upper limit value:
\[ \frac{25}{4} - \frac{5}{2} = \frac{25}{4} - \frac{10}{4} = \frac{15}{4} \]

B1: Obtain \(\frac{3}{2}\mathrm{e}^{2x}\) from integrating \(3\mathrm{e}^{2x}\)
B1: Obtain \(+\mathrm{e}^{-2x}\) (or \(-\frac{2}{-2}\mathrm{e}^{-2x}\)) from integrating \(-2\mathrm{e}^{-2x}\)
M1: Substitute limits \(0\) and \(\ln 2\) into an expression of the form \(a\mathrm{e}^{2x} + b\mathrm{e}^{-2x}\) and apply power/log laws (e.g. \(\mathrm{e}^{2\ln 2} = 4\), \(\mathrm{e}^{-2\ln 2} = \frac{1}{4}\))
A1: Obtain exact value \(\frac{15}{4}\) (or \(3\frac{3}{4}\) or \(3.75\))

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

\[ 2(1 + \tan^2 \theta) - 5\tan \theta = 5 \]
\[ 2 + 2\tan^2 \theta - 5\tan \theta - 5 = 0 \]
\[ 2\tan^2 \theta - 5\tan \theta - 3 = 0 \]

Factorising the quadratic equation in \(\tan \theta\):

\[ (2\tan \theta + 1)(\tan \theta - 3) = 0 \]

This gives:
\[ \tan \theta = 3 \quad \text{or} \quad \tan \theta = -\frac{1}{2} \]

For \(\tan \theta = 3\):
\[ \theta = \tan^{-1}(3) \approx 71.565^\circ \implies \theta = 71.6^\circ \]
\[ \theta = 180^\circ + 71.565^\circ = 251.565^\circ \implies \theta = 251.6^\circ \]

For \(\tan \theta = -\frac{1}{2}\):
\[ \text{Basic angle } \alpha = \tan^{-1}(0.5) \approx 26.565^\circ \]
\[ \theta = 180^\circ - 26.565^\circ = 153.435^\circ \implies \theta = 153.4^\circ \]
\[ \theta = 360^\circ - 26.565^\circ = 333.435^\circ \implies \theta = 333.4^\circ \]

Therefore, the solutions in the given interval are:
\[ \theta = 71.6^\circ, 153.4^\circ, 251.6^\circ, 333.4^\circ \]

M1: Use \(\sec^2 \theta = 1 + \tan^2 \theta\) to form a 3-term quadratic equation in \(\tan \theta\).
A1: Obtain \(2\tan^2 \theta - 5\tan \theta - 3 = 0\) and obtain correct values \(\tan \theta = 3\) and \(\tan \theta = -\frac{1}{2}\).
A1: Obtain \(71.6^\circ\) and \(251.6^\circ\) (allow 1 d.p. rounding from \(71.57^\circ\) and \(251.57^\circ\)).
A1: Obtain \(153.4^\circ\) and \(333.4^\circ\) (allow 1 d.p. rounding from \(153.43^\circ\) and \(333.43^\circ\)) and no extra solutions in the range \(0^\circ \leqslant \theta \leqslant 360^\circ\).

(a)
Taking natural logarithms of both sides of \(y = A x^k\):
\[\ln y = \ln(A x^k) = \ln A + k \ln x\]
This is in linear form \(Y = m X + c\), where \(Y = \ln y\), \(X = \ln x\), gradient \(m = k\), and vertical intercept \(c = \ln A\).

Calculate the gradient \(k\):
\[k = \frac{0.54 - 3.24}{2.00 - 0.50} = \frac{-2.70}{1.50} = -1.8\]

Using the point \((2.00, 0.54)\) to find \(\ln A\):
\[0.54 = \ln A + (-1.8)(2.00)\]
\[\ln A = 0.54 + 3.60 = 4.14\]
\[A = \mathrm{e}^{4.14} \approx 62.8030... = 62.8\text{ (to 3 s.f.)}\]

(b)
When \(x = 4.5\):
\[\ln y = 4.14 - 1.8 \ln(4.5)\]
\[\ln y = 4.14 - 1.8(1.50408...) = 4.14 - 2.70734... = 1.43266...\]
\[y = \mathrm{e}^{1.43266...} \approx 4.19\text{ (to 3 s.f.)}\]

(a)
- M1: For expressing \(\ln y = \ln A + k \ln x\) and attempting to calculate the gradient using the given coordinates.
- A1: For \(k = -1.8\) (or \(-\frac{9}{5}\)).
- A1: For finding \(\ln A = 4.14\) and evaluating \(A = 62.8\) (accept \(62.80\)).

(b)
- M1: For substituting \(x = 4.5\) into \(\ln y = \ln A + k \ln x\) or \(y = A x^k\) using their values from part (a).
- A1: For obtaining \(y = 4.19\) (allow \(4.18\) to \(4.20\) due to intermediate rounding).

(a)
Method 1: Squaring both sides
\((2x + 1)^2 \leqslant (x - 5)^2\)
\(4x^2 + 4x + 1 \leqslant x^2 - 10x + 25\)
\(3x^2 + 14x - 24 \leqslant 0\)

Factorising to find the critical values:
\((3x - 4)(x + 6) = 0\)
\(x = \frac{4}{3}\) or \(x = -6\)

Since the inequality is \(\leqslant 0\), the solution is the region between the roots:
\(-6 \leqslant x \leqslant \frac{4}{3}\)

Method 2: Solving linear equations
\(2x + 1 = x - 5 \implies x = -6\)
\(2x + 1 = -(x - 5) \implies 3x = 4 \implies x = \frac{4}{3}\)
Testing regions gives \(-6 \leqslant x \leqslant \frac{4}{3}\).

(b)
Let \(x = \ln(3y)\).
Using the result from part (a):
\(-6 \leqslant \ln(3y) \leqslant \frac{4}{3}\)

Taking exponentials of each part:
\(\mathrm{e}^{-6} \leqslant 3y \leqslant \mathrm{e}^{\frac{4}{3}}\)

Dividing by 3:
\(\frac{1}{3}\mathrm{e}^{-6} \leqslant y \leqslant \frac{1}{3}\mathrm{e}^{\frac{4}{3}}\) (or \(\frac{1}{3\mathrm{e}^6} \leqslant y \leqslant \frac{1}{3}\mathrm{e}^{\frac{4}{3}}\))

(a)
M1: State a correct method to find critical values, e.g. squaring both sides to form a 3-term quadratic or setting up and solving two linear equations \(2x + 1 = x - 5\) and \(2x + 1 = -(x - 5)\).
A1: Obtain critical values \(-6\) and \(\frac{4}{3}\) (or \(1\frac{1}{3}\)).
M1: Identify the correct inside region for the inequality using their critical values.
A1: Obtain correct final inequality \(-6 \leqslant x \leqslant \frac{4}{3}\) (accept equivalent notation such as \([-6, \frac{4}{3}]\)).

(b)
M1: Relate part (b) to part (a) by setting \(x = \ln(3y)\) to state \(-6 \leqslant \ln(3y) \leqslant \frac{4}{3}\) (allow follow-through on their critical values from (a)).
M1: Correctly apply the exponential function to undo the natural logarithm and solve for \(y\).
A1: Obtain the correct exact range \(\frac{1}{3}\mathrm{e}^{-6} \leqslant y \leqslant \frac{1}{3}\mathrm{e}^{\frac{4}{3}}\) (or \(\frac{1}{3\mathrm{e}^6} \leqslant y \leqslant \frac{1}{3}\mathrm{e}^{\frac{4}{3}}\)).

(a)
Using the Factor Theorem, \( p(2) = 0 \):
\[ 2^3 + a(2^2) - 4(2) + b = 0 \implies 8 + 4a - 8 + b = 0 \implies 4a + b = 0 \quad \text{--- (1)} \]

Using the Remainder Theorem, \( p(-1) = 9 \):
\[ (-1)^3 + a(-1)^2 - 4(-1) + b = 9 \implies -1 + a + 4 + b = 9 \implies a + b = 6 \quad \text{--- (2)} \]

Subtracting equation (2) from equation (1):
\[ 3a = -6 \implies a = -2 \]

Substituting \( a = -2 \) into equation (2):
\[ -2 + b = 6 \implies b = 8 \]

---

(b)
With \( a = -2 \) and \( b = 8 \), the polynomial is \( p(x) = x^3 - 2x^2 - 4x + 8 \).
The equation \( p(x) = 15 \) becomes:
\[ x^3 - 2x^2 - 4x + 8 = 15 \]
\[ x^3 = 2x^2 + 4x + 7 \]
\[ x = \sqrt[3]{2x^2 + 4x + 7} \]

---

(c)
Let \( f(x) = x^3 - 2x^2 - 4x - 7 \).
Evaluating at \( x = 3 \) and \( x = 4 \):
\[ f(3) = 3^3 - 2(3^2) - 4(3) - 7 = 27 - 18 - 12 - 7 = -10 < 0 \]
\[ f(4) = 4^3 - 2(4^2) - 4(4) - 7 = 64 - 32 - 16 - 7 = 9 > 0 \]

Since there is a sign change and \( f(x) \) is continuous, the equation \( p(x) = 15 \) has a root between \( 3 \) and \( 4 \).

---

(d)
Using the iterative formula:
\[ x_{n+1} = \sqrt[3]{2x_n^2 + 4x_n + 7} \]

Taking an initial value \( x_1 = 3.5 \):
- \( x_2 = \sqrt[3]{2(3.5)^2 + 4(3.5) + 7} = 3.5700 \)
- \( x_3 = 3.6028 \)
- \( x_4 = 3.6183 \)
- \( x_5 = 3.6256 \)
- \( x_6 = 3.6290 \)
- \( x_7 = 3.6306 \)
- \( x_8 = 3.6313 \)

Thus, the root correct to 2 decimal places is \( 3.63 \).

(a)
- M1: Substitute \( x = 2 \) and equate to \( 0 \), OR substitute \( x = -1 \) and equate to \( 9 \) (or equivalent algebraic division).
- A1: Obtain two correct linear equations in \( a \) and \( b \), e.g. \( 4a + b = 0 \) and \( a + b = 6 \).
- A1: Solve simultaneously to find \( a = -2 \) and \( b = 8 \).

(b)
- B1: Set \( x^3 - 2x^2 - 4x + 8 = 15 \) and show sufficient intermediate working to reach the given result \( x = \sqrt[3]{2x^2 + 4x + 7} \) with no errors.

(c)
- B1: Evaluate values of a relevant continuous function at \( 3 \) and \( 4 \) (e.g. \( -10 \) and \( 9 \)) and state a conclusion referring to the change of sign.

(d)
- M1: Use the iterative formula \( x_{n+1} = \sqrt[3]{2x_n^2 + 4x_n + 7} \) correctly at least once.
- A1: Obtain values of iterations to at least 4 decimal places showing convergence towards \( 3.63 \).
- A1: Obtain final value \( 3.63 \) correct to 2 decimal places with sufficient iterations shown to justify accuracy.

(a) Expanding the bracket gives:
\[(3\sin x + 2\cos x)^2 = 9\sin^2 x + 12\sin x\cos x + 4\cos^2 x\]
Using the identities \(\sin 2x = 2\sin x\cos x\), \(\sin^2 x = \frac{1 - \cos 2x}{2}\), and \(\cos^2 x = \frac{1 + \cos 2x}{2}\):
\[9\left(\frac{1 - \cos 2x}{2}\right) + 6(2\sin x\cos x) + 4\left(\frac{1 + \cos 2x}{2}\right)\]
\[= \frac{9}{2} - \frac{9}{2}\cos 2x + 6\sin 2x + 2 + 2\cos 2x\]
\[= \frac{13}{2} + 6\sin 2x - \frac{5}{2}\cos 2x\]
Thus, \(a = \frac{13}{2}\), \(b = 6\), and \(c = -\frac{5}{2}\).

(b) Using the result from part (a):
\[\int_0^{\frac{1}{4}\pi} (3\sin x + 2\cos x)^2\,\mathrm{d}x = \int_0^{\frac{1}{4}\pi} \left(\frac{13}{2} + 6\sin 2x - \frac{5}{2}\cos 2x\right)\,\mathrm{d}x\]
\[= \left[ \frac{13}{2}x - 3\cos 2x - \frac{5}{4}\sin 2x \right]_0^{\frac{1}{4}\pi}\]
Substituting the upper limit \(x = \frac{1}{4}\pi\):
\[\frac{13}{2}\left(\frac{\pi}{4}\right) - 3\cos\left(\frac{\pi}{2}\right) - \frac{5}{4}\sin\left(\frac{\pi}{2}\right) = \frac{13\pi}{8} - 0 - \frac{5}{4} = \frac{13\pi}{8} - \frac{5}{4}\]
Substituting the lower limit \(x = 0\):
\[\frac{13}{2}(0) - 3\cos(0) - \frac{5}{4}\sin(0) = 0 - 3 - 0 = -3\]
Subtracting the lower limit value from the upper limit value:
\[\left(\frac{13\pi}{8} - \frac{5}{4}\right) - (-3) = \frac{13\pi}{8} - \frac{5}{4} + 3 = \frac{13}{8}\pi + \frac{7}{4} = \frac{13\pi + 14}{8}\]

(a)
- M1: Expand bracket to obtain three terms and attempt to use at least one double-angle formula.
- M1: Use both \(\sin 2x = 2\sin x\cos x\) and \(\cos 2x = 1 - 2\sin^2 x\) or \(2\cos^2 x - 1\) correctly.
- A1: Obtain \(\frac{13}{2} + 6\sin 2x - \frac{5}{2}\cos 2x\) or state correct values for \(a, b, c\).

(b)
- M1: Integrate expression of the form \(a + b\sin 2x + c\cos 2x\) to obtain \(ax - \frac{1}{2}b\cos 2x + \frac{1}{2}c\sin 2x\).
- A1ft: Obtain correct unsimplified or simplified integral \(\frac{13}{2}x - 3\cos 2x - \frac{5}{4}\sin 2x\) (follow through on their non-zero constants \(a, b, c\)).
- M1: Substitute limits \(0\) and \(\frac{1}{4}\pi\) correctly into an integrated expression.
- A1: Obtain \(\frac{13}{8}\pi + \frac{7}{4}\) or \(\frac{13\pi + 14}{8}\) or equivalent exact form.

(a) Differentiating \(x\) and \(y\) with respect to \(\theta\):
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = 2\sec^2\theta = \frac{2}{\cos^2\theta} \]
\[ \frac{\mathrm{d}y}{\mathrm{d}\theta} = -4\sin 2\theta + 4\cos\theta \]
Using the double-angle identity \(\sin 2\theta = 2\sin\theta\cos\theta\):
\[ \frac{\mathrm{d}y}{\mathrm{d}\theta} = -8\sin\theta\cos\theta + 4\cos\theta = 4\cos\theta(1 - 2\sin\theta) \]
Applying the chain rule for parametric differentiation:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}} = \frac{4\cos\theta(1 - 2\sin\theta)}{\frac{2}{\cos^2\theta}} = 2\cos^3\theta(1 - 2\sin\theta) \]

(b) At the stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[ 2\cos^3\theta(1 - 2\sin\theta) = 0 \]
For \(0 < \theta < \frac{1}{2}\pi\), \(\cos\theta \neq 0\), so:
\[ 1 - 2\sin\theta = 0 \implies \sin\theta = \frac{1}{2} \implies \theta = \frac{1}{6}\pi \]
Substituting \(\theta = \frac{1}{6}\pi\) into the parametric equations:
\[ x = 2\tan\left(\frac{1}{6}\pi\right) = 2\left(\frac{1}{\sqrt{3}}\right) = \frac{2\sqrt{3}}{3} \]
\[ y = 2\cos\left(\frac{1}{3}\pi\right) + 4\sin\left(\frac{1}{6}\pi\right) = 2\left(\frac{1}{2}\right) + 4\left(\frac{1}{2}\right) = 1 + 2 = 3 \]
Hence, the exact coordinates are \(\left(\frac{2\sqrt{3}}{3}, 3\right)\).

(c) When \(\theta = \frac{1}{4}\pi\):
\[ x = 2\tan\left(\frac{1}{4}\pi\right) = 2(1) = 2 \]
\[ y = 2\cos\left(\frac{1}{2}\pi\right) + 4\sin\left(\frac{1}{4}\pi\right) = 0 + 4\left(\frac{\sqrt{2}}{2}\right) = 2\sqrt{2} \]
The gradient of the tangent is:
\[ m_T = 2\cos^3\left(\frac{1}{4}\pi\right)\left(1 - 2\sin\left(\frac{1}{4}\pi\right)\right) = 2\left(\frac{\sqrt{2}}{2}\right)^3\left(1 - 2\left(\frac{\sqrt{2}}{2}\right)\right) = \frac{\sqrt{2}}{2}(1 - \sqrt{2}) = \frac{\sqrt{2} - 2}{2} \]
The gradient of the normal is:
\[ m_N = -\frac{1}{m_T} = -\frac{2}{\sqrt{2} - 2} = \frac{2}{2 - \sqrt{2}} = \frac{2(2 + \sqrt{2})}{(2 - \sqrt{2})(2 + \sqrt{2})} = \frac{2(2 + \sqrt{2})}{2} = 2 + \sqrt{2} \]
The equation of the normal is:
\[ y - 2\sqrt{2} = (2 + \sqrt{2})(x - 2) \]
\[ y = (2 + \sqrt{2})x - 4 - 2\sqrt{2} + 2\sqrt{2} \implies y = (2 + \sqrt{2})x - 4 \]

(a)
B1: Correct derivative \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 2\sec^2\theta\)
M1: Differentiate \(y\) to obtain form \(a\sin 2\theta + b\cos\theta\) and apply identity \(\sin 2\theta = 2\sin\theta\cos\theta\)
M1: Use \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}\theta}{\mathrm{d}x/\mathrm{d}\theta}\) with \(\sec^2\theta = \frac{1}{\cos^2\theta}\)
A1 (AG): Fully correct intermediate steps leading convincingly to \(2\cos^3\theta(1 - 2\sin\theta)\)

(b)
M1: Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) to find \(\sin\theta = \frac{1}{2}\) and hence \(\theta = \frac{1}{6}\pi\)
A1: Obtain exact coordinates \(\left(\frac{2\sqrt{3}}{3}, 3\right)\) or \(\left(\frac{2}{\sqrt{3}}, 3\right)\)

(c)
M1: Substitute \(\theta = \frac{1}{4}\pi\) to find \(x = 2\), \(y = 2\sqrt{2}\), and evaluate gradient of tangent \(m_T = \frac{\sqrt{2}-2}{2}\)
M1: Use \(m_N = -\frac{1}{m_T}\) and form the equation of the normal \(y - y_1 = m_N(x - x_1)\)
A1: Obtain \(y = (2 + \sqrt{2})x - 4\) (or \(y = (\sqrt{2} + 2)x - 4\))