2023 Cambridge IGCSE Mathematics - Additional (0606) 模拟试题及答案详解

Square both sides of the inequality:
\((3x + 2)^2 < (x - 4)^2\)

Expand both sides:
\(9x^2 + 12x + 4 < x^2 - 8x + 16\)

Rearrange the inequality to form a quadratic expression on one side:
\(8x^2 + 20x - 12 < 0\)

Divide each term by 4 to simplify:
\(2x^2 + 5x - 3 < 0\)

Factorise the quadratic expression:
\((2x - 1)(x + 3) < 0\)

The critical values are \(x = -3\) and \(x = \frac{1}{2}\).

Since the quadratic expression must be less than zero, the solution range lies between the critical values:
\(-3 < x < \frac{1}{2}\).

**M1** for squaring both sides and expanding correctly to obtain a 3-term quadratic.
**A1** for obtaining the correct quadratic \(2x^2 + 5x - 3 < 0\) (or any equivalent form).
**B1** for finding both critical values \(x = -3\) and \(x = \frac{1}{2}\).
**A1** for the correct final inequality range \(-3 < x < \frac{1}{2}\).

Apply the Factor Theorem: Since \(x - 2\) is a factor of \(P(x)\), we have \(P(2) = 0\).
\(P(2) = (2)^3 + a(2)^2 - 7(2) + b = 0\)
\(8 + 4a - 14 + b = 0 \implies 4a + b = 6\) (Equation 1)

Apply the Remainder Theorem: When \(P(x)\) is divided by \(x + 1\), the remainder is \(-9\), so \(P(-1) = -9\).
\(P(-1) = (-1)^3 + a(-1)^2 - 7(-1) + b = -9\)
\(-1 + a + 7 + b = -9 \implies a + b = -15\) (Equation 2)

Subtract Equation 2 from Equation 1:
\((4a + b) - (a + b) = 6 - (-15)\)
\(3a = 21 \implies a = 7\)

Substitute \(a = 7\) back into Equation 2:
\(7 + b = -15 \implies b = -22\).

**M1** for applying the Factor Theorem \(P(2) = 0\) to form a linear equation in \(a\) and \(b\).
**A1** for obtaining the simplified equation \(4a + b = 6\) (or equivalent).
**M1** for applying the Remainder Theorem \(P(-1) = -9\) to form a second linear equation in \(a\) and \(b\).
**A1** for obtaining the simplified equation \(a + b = -15\) (or equivalent).
**A1** for correctly solving the simultaneous equations to find \(a = 7\) and \(b = -22\).

First, find the vector \(\overrightarrow{AB}\):
\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\)
\(\overrightarrow{AB} = (10\mathbf{i} + 12\mathbf{j}) - (2\mathbf{i} - 3\mathbf{j})\)
\(\overrightarrow{AB} = 8\mathbf{i} + 15\mathbf{j}\)

Next, calculate the magnitude of \(\overrightarrow{AB}\):
\(|\overrightarrow{AB}| = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\)

Finally, construct the unit vector in the direction of \(\overrightarrow{AB}\):
\(\hat{u} = \frac{\overrightarrow{AB}}{|\overrightarrow{AB}|} = \frac{8\mathbf{i} + 15\mathbf{j}}{17} = \frac{8}{17}\mathbf{i} + \frac{15}{17}\mathbf{j}\).

**M1** for an attempt to find \(\overrightarrow{AB}\) by subtracting the position vector of \(A\) from \(B\).
**A1** for obtaining \(\overrightarrow{AB} = 8\mathbf{i} + 15\mathbf{j}\).
**M1** for correctly calculating the magnitude of their vector to be 17.
**A1** for the correct unit vector \(\frac{8}{17}\mathbf{i} + \frac{15}{17}\mathbf{j}\) (or equivalent column vector notation).

First, apply the power law of logarithms to the second term:
\(\log_3(5x - 2) - \log_3(x^2) = 1\)

Next, apply the division law of logarithms to combine the terms on the left-hand side:
\(\log_3\left(\frac{5x - 2}{x^2}\right) = 1\)

Convert the equation from logarithmic to exponential form:
\(\frac{5x - 2}{x^2} = 3^1\)

Multiply both sides by \(x^2\) and rearrange to form a quadratic equation:
\(5x - 2 = 3x^2\)
\(3x^2 - 5x + 2 = 0\)

Factorise the quadratic equation:
\((3x - 2)(x - 1) = 0\)
This yields two possible values: \(x = 1\) or \(x = \frac{2}{3}\).

Check for validity against the constraints of original logarithm terms (i.e. \(5x - 2 > 0\) and \(x > 0\)):
For \(x = 1\): \(5(1) - 2 = 3 > 0\) and \(1 > 0\) (valid).
For \(x = \frac{2}{3}\): \(5(\frac{2}{3}) - 2 = \frac{4}{3} > 0\) and \( \frac{2}{3} > 0\) (valid).

Hence, the solutions are \(x = 1\) and \(x = \frac{2}{3}\).

**M1** for applying the power law to obtain \(\log_3(x^2)\).
**M1** for applying the division law to write the left-hand side as a single logarithm.
**M1** for converting the logarithmic equation correctly into exponential form.
**A1** for obtaining the correct quadratic equation \(3x^2 - 5x + 2 = 0\) and solving it to find both values.
**A1** for confirming both answers are valid by checking domain restrictions.

Use the Pythagorean identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to rewrite the equation in terms of \(\sin \theta\):
\(2(1 - \sin^2 \theta) + \sin \theta - 1 = 0\)
\(2 - 2\sin^2 \theta + \sin \theta - 1 = 0\)
\(-2\sin^2 \theta + \sin \theta + 1 = 0\)

Multiply by \(-1\) to make the leading coefficient positive:
\(2\sin^2 \theta - \sin \theta - 1 = 0\)

Factorise the quadratic in terms of \(\sin \theta\):
\((2\sin \theta + 1)(\sin \theta - 1) = 0\)

This gives two scenarios:
1) \(\sin \theta = 1\)
2) \(\sin \theta = -\frac{1}{2}\)

Solve each within the interval \(0^\circ \le \theta \le 360^\circ\):
- From \(\sin \theta = 1\):
\(\theta = 90^\circ\)

- From \(\sin \theta = -\frac{1}{2}\):
The reference angle in the first quadrant is \(30^\circ\). Sine is negative in Quadrants III and IV:
Quadrant III: \(\theta = 180^\circ + 30^\circ = 210^\circ\)
Quadrant IV: \(\theta = 360^\circ - 30^\circ = 330^\circ\)

Thus, the solutions are \(\theta = 90^\circ, 210^\circ, 330^\circ\).

**M1** for using \(\cos^2 \theta = 1 - \sin^2 \theta\) to formulate a quadratic equation purely in terms of \(\sin \theta\).
**A1** for the correct simplified quadratic equation \(2\sin^2 \theta - \sin \theta - 1 = 0\).
**M1** for solving the quadratic equation to find \(\sin \theta = 1\) and \(\sin \theta = -\frac{1}{2}\).
**A1** for identifying all three correct angles \(90^\circ, 210^\circ, 330^\circ\) with no extra angles in the range.

To find the stationary point, we must differentiate \(y\) with respect to \(x\) and set \(\frac{dy}{dx} = 0\).

Apply the quotient rule, where \(u = e^{2x}\) and \(v = 3x + 1\):
\(\frac{du}{dx} = 2e^{2x}\)
\(\frac{dv}{dx} = 3\)

Using the quotient rule formula \(\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\):
\(\frac{dy}{dx} = \frac{(3x + 1)(2e^{2x}) - (e^{2x})(3)}{(3x + 1)^2}\)

Factor out \(e^{2x}\) in the numerator:
\(\frac{dy}{dx} = \frac{e^{2x}[2(3x + 1) - 3]}{(3x + 1)^2}\)
\(\frac{dy}{dx} = \frac{e^{2x}(6x + 2 - 3)}{(3x + 1)^2}\)
\(\frac{dy}{dx} = \frac{e^{2x}(6x - 1)}{(3x + 1)^2}\)

Set \(\frac{dy}{dx} = 0\) to find the stationary point:
\(\frac{e^{2x}(6x - 1)}{(3x + 1)^2} = 0\)

Since \(e^{2x} > 0\) for all real values of \(x\), we solve:
\(6x - 1 = 0 \implies x = \frac{1}{6}\)

This value of \(x\) is in the domain \(x > -\frac{1}{3}\), so the exact \(x\)-coordinate of the stationary point is \(x = \frac{1}{6}\).

**M1** for attempting to apply the quotient rule (or product rule with negative exponent).
**A1** for obtaining the correct derivative of \(e^{2x}\) as \(2e^{2x}\).
**A1** for the correct unsimplified expression of \(\frac{dy}{dx}\).
**M1** for setting \(\frac{dy}{dx} = 0\) and attempting to solve for \(x\).
**A1** for the correct exact value \(x = \frac{1}{6}\).

(a) For the modulus graph \( y = |3x - 4| \):
- It is a V-shaped graph with vertex on the positive \( x \)-axis.
- The vertex is at \( (\frac{4}{3}, 0) \).
- The \( y \)-intercept is at \( (0, 4) \).

For the line \( y = x + 2 \):
- It is a straight line passing through \( (0, 2) \) and \( (-2, 0) \).
- The sketch shows two intersection points in the first quadrant.

(b) To solve \( |3x - 4| > x + 2 \), we first find the critical values by solving the equation:
\( |3x - 4| = x + 2 \)

Case 1: \( 3x - 4 = x + 2 \implies 2x = 6 \implies x = 3 \)
Case 2: \( -(3x - 4) = x + 2 \implies -3x + 4 = x + 2 \implies 4x = 2 \implies x = 0.5 \)

By comparing the graphs or checking regions, the solution to the inequality is \( x < 0.5 \) or \( x > 3 \).

(a)
- **B1**: Correct V-shaped graph for \( y = |3x - 4| \) with vertex on the positive \( x \)-axis.
- **B1**: Vertex marked at \( (4/3, 0) \) and \( y \)-intercept at \( (0, 4) \).
- **B1**: Straight line for \( y = x + 2 \) passing through \( (0, 2) \) and \( (-2, 0) \) with positive gradient.
- **B1**: Correct intersection points shown in the first quadrant.

(b)
- **M1**: Attempt to solve the equation \( |3x - 4| = x + 2 \) by squaring or using two cases.
- **A1**: Finding critical value \( x = 3 \).
- **A1**: Finding critical value \( x = 0.5 \).
- **A1**: Correct inequality range: \( x < 0.5 \) or \( x > 3 \) (or equivalent).

(a) Using the product rule with \( u = 2x - 3 \) and \( v = e^{-x^2} \):
\( \frac{\mathrm{d}u}{\mathrm{d}x} = 2 \) and \( \frac{\mathrm{d}v}{\mathrm{d}x} = -2x e^{-x^2} \)
\( \frac{\mathrm{d}y}{\mathrm{d}x} = 2e^{-x^2} + (2x - 3)(-2x e^{-x^2}) \)
\( \frac{\mathrm{d}y}{\mathrm{d}x} = 2e^{-x^2}(1 + 3x - 2x^2) \)

(b) For stationary points, set \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \):
\( 2x^2 - 3x - 1 = 0 \)
Using the quadratic formula:
\( x = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-1)}}{2(2)} = \frac{3 \pm \sqrt{17}}{4} \)

For the corresponding \( y \)-coordinates:
At \( x = \frac{3+\sqrt{17}}{4} \), we have \( 2x - 3 = \frac{\sqrt{17}-3}{2} \).
\( y = \frac{\sqrt{17}-3}{2} e^{-\left(\frac{3+\sqrt{17}}{4}\right)^2} = \frac{\sqrt{17}-3}{2} e^{-\frac{13+3\sqrt{17}}{8}} \)

At \( x = \frac{3-\sqrt{17}}{4} \), we have \( 2x - 3 = \frac{-\sqrt{17}-3}{2} \).
\( y = -\frac{\sqrt{17}+3}{2} e^{-\left(\frac{3-\sqrt{17}}{4}\right)^2} = -\frac{\sqrt{17}+3}{2} e^{-\frac{13-3\sqrt{17}}{8}} \)

(c) To determine the nature of the positive stationary point at \( x_1 = \frac{3+\sqrt{17}}{4} \approx 1.78 \):
We differentiate \( \frac{\mathrm{d}y}{\mathrm{d}x} = 2e^{-x^2}(1 + 3x - 2x^2) \):
\( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -4x e^{-x^2}(1 + 3x - 2x^2) + 2e^{-x^2}(3 - 4x) \)
Since \( 1 + 3x_1 - 2x_1^2 = 0 \) at the stationary point:
\( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2e^{-x_1^2}(3 - 4x_1) \)
Since \( x_1 \approx 1.78 \), the term \( 3 - 4x_1 < 0 \), meaning \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0 \).
Thus, the stationary point is a local maximum.

(a)
- **M1**: Attempt to use the product rule to differentiate \( y = (2x - 3)e^{-x^2} \).
- **A1**: Correct differentiation of \( e^{-x^2} \) to get \( -2x e^{-x^2} \).
- **A1**: Fully correct expression for \( \frac{\mathrm{d}y}{\mathrm{d}x} = 2e^{-x^2}(1 + 3x - 2x^2) \).

(b)
- **M1**: Setting their derivative to 0 to obtain a quadratic equation in \( x \).
- **A1**: Finding exact \( x \)-coordinates: \( x = \frac{3 \pm \sqrt{17}}{4} \).
- **M1**: Substituting at least one of their \( x \)-values back into the curve equation to find \( y \).
- **A1**: Finding exact coordinate pairs: \( (\frac{3+\sqrt{17}}{4}, \frac{\sqrt{17}-3}{2} e^{-\frac{13+3\sqrt{17}}{8}}) \) and \( (\frac{3-\sqrt{17}}{4}, -\frac{\sqrt{17}+3}{2} e^{-\frac{13-3\sqrt{17}}{8}}) \).

(c)
- **M1**: Finding an expression for the second derivative or testing the gradient on either side of the stationary point.
- **A1**: Valid reasoning showing \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0 \) and concluding it is a maximum.

(a) Combining the fractions on the LHS over a common denominator:
\( \frac{\cos \theta(1 + \sin \theta) - \cos \theta(1 - \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)} = \frac{\cos \theta + \cos \theta \sin \theta - \cos \theta + \cos \theta \sin \theta}{1 - \sin^2 \theta} \)
\( = \frac{2 \cos \theta \sin \theta}{\cos^2 \theta} \) (using \( 1 - \sin^2 \theta = \cos^2 \theta \))
\( = \frac{2 \sin \theta}{\cos \theta} = 2\tan \theta \).

(b) Substituting \( \theta = 2x - \frac{\pi}{4} \), the equation simplifies using part (a) to:
\( 2\tan(2x - \frac{\pi}{4}) = 2 \implies \tan(2x - \frac{\pi}{4}) = 1 \).

For \( 0 < x < \pi \), we have the interval:
\( -\frac{\pi}{4} < 2x - \frac{\pi}{4} < \frac{7\pi}{4} \).

In this interval, \( 2x - \frac{\pi}{4} = \frac{\pi}{4} \) or \( 2x - \frac{\pi}{4} = \frac{5\pi}{4} \).

Solving for \( x \):
1) \( 2x = \frac{\pi}{2} \implies x = \frac{\pi}{4} \)
2) \( 2x = \frac{3\pi}{2} \implies x = \frac{3\pi}{4} \)

(a)
- **M1**: Putting LHS over a common denominator \( (1 - \sin\theta)(1 + \sin\theta) \).
- **A1**: Correct numerator expansion: \( 2\cos\theta\sin\theta \).
- **M1**: Using \( 1 - \sin^2\theta = \cos^2\theta \) to simplify denominator.
- **A1**: Completed proof showing the steps leading to \( 2\tan\theta \).

(b)
- **M1**: Recognizing identity and simplifying equation to \( \tan(2x - \frac{\pi}{4}) = 1 \).
- **M1**: Finding first correct value: \( 2x - \frac{\pi}{4} = \frac{\pi}{4} \).
- **A1**: Finding \( x = \frac{\pi}{4} \).
- **M1**: Finding second correct value: \( 2x - \frac{\pi}{4} = \frac{5\pi}{4} \).
- **A1**: Finding \( x = \frac{3\pi}{4} \) and no other values.

(a) The perimeter of the sector is given by:
\( P = 2r + r\theta = 40 \implies r\theta = 40 - 2r \implies \theta = \frac{40 - 2r}{r} \)

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

Substituting \( \theta \) into the area equation:
\( \frac{1}{2}r^2 \left(\frac{40 - 2r}{r}\right) = 75 \implies \frac{1}{2}r(40 - 2r) = 75 \implies 20r - r^2 = 75 \)
\( r^2 - 20r + 75 = 0 \implies (r - 15)(r - 5) = 0 \)

Thus, the possible values of \( r \) are \( r = 15 \) or \( r = 5 \).

Finding the corresponding values of \( \theta \):
- If \( r = 15 \), \( \theta = \frac{40 - 30}{15} = \frac{2}{3} \) radians.
- If \( r = 5 \), \( \theta = \frac{40 - 10}{5} = 6 \) radians.
Both values of \( \theta \) are in the valid range \( (0, 2\pi) \).

(b) For the larger value of \( r = 15 \), the angle \( \theta = \frac{2}{3} \).
The area of the segment is:
\( A_{\text{segment}} = \frac{1}{2}r^2(\theta - \sin\theta) = \frac{1}{2}(15)^2\left(\frac{2}{3} - \sin\left(\frac{2}{3}\right)\right) \)

Using radians, \( \sin(2/3) \approx 0.61837 \).
\( A_{\text{segment}} = 112.5(0.66667 - 0.61837) = 112.5 \times 0.0483 = 5.43\text{ cm}^2 \) (to 3 significant figures).

(a)
- **B1**: Writing down correct formula equations for perimeter and area.
- **M1**: Eliminating \( \theta \) to form a quadratic equation in \( r \).
- **A1**: Showing \( r^2 - 20r + 75 = 0 \).
- **A1**: Finding \( r = 5 \) and \( r = 15 \).
- **M1**: Substituting \( r \) back to find \( \theta \).
- **A1**: Finding correct pairs: \( (15, 2/3) \) and \( (5, 6) \).

(b)
- **M1**: Using correct segment area formula: \( \frac{1}{2}r^2(\theta - \sin\theta) \) with \( r = 15 \) and \( \theta = 2/3 \).
- **M1**: Evaluating \( \sin(2/3) \approx 0.618 \).
- **A1**: Correctly obtaining \( 5.43\text{ cm}^2 \) (allow 5.43 to 5.44).

(a) The team of 5 must contain more women than men.
Possible scenarios:
- 3 women and 2 men: \( \binom{5}{3} \times \binom{6}{2} = 10 \times 15 = 150 \) ways
- 4 women and 1 man: \( \binom{5}{4} \times \binom{6}{1} = 5 \times 6 = 30 \) ways
- 5 women and 0 men: \( \binom{5}{5} \times \binom{6}{0} = 1 \times 1 = 1 \) way

Total ways = \( 150 + 30 + 1 = 181 \).

(b) The letters in \( \text{ALGEBRA} \) are A, L, G, E, B, R, A. There are 7 letters: 2 A's, and 1 of L, G, E, B, R.
(i) No restrictions:
\( \frac{7!}{2!} = 2520 \).

(ii) A's not together:
Total arrangements - Arrangements where A's are together.
Treating "AA" as a single block, we arrange 6 entities (AA, L, G, E, B, R):
\( 6! = 720 \).

Arrangements with A's not together: \( 2520 - 720 = 1800 \).

(a)
- **M1**: Identifying the 3 distinct cases of (3W, 2M), (4W, 1M), and (5W, 0M).
- **M1**: Calculating combination products for at least one case correctly.
- **A1**: Finding valid counts: 150, 30, 1.
- **A1**: Summing to find the correct total of 181.

(b)(i)
- **M1**: Showing division by 2! to adjust for the two repeated A's.
- **A1**: Finding 2520.

(b)(ii)
- **M1**: Finding arrangements where A's are together (720) or using the gap method.
- **A1**: Correct subtraction/calculation to find 1800.

(a) \( \overrightarrow{OS} = \overrightarrow{OP} + \overrightarrow{PS} = \mathbf{p} + \frac{1}{4}\overrightarrow{PQ} \)
Since \( \overrightarrow{PQ} = \mathbf{q} - \mathbf{p} \):
\( \overrightarrow{OS} = \mathbf{p} + \frac{1}{4}(\mathbf{q} - \mathbf{p}) = \frac{3}{4}\mathbf{p} + \frac{1}{4}\mathbf{q} \).

(b) Using \( \overrightarrow{OT} = \lambda \overrightarrow{OS} \):
\( \overrightarrow{OT} = \frac{3}{4}\lambda\mathbf{p} + \frac{1}{4}\lambda\mathbf{q} \) --- (Equation 1)

Now, express \( \overrightarrow{OT} \) using the route through \( R \):
\( \overrightarrow{OT} = \overrightarrow{OR} + \overrightarrow{RT} \)
Since \( \overrightarrow{OR} = \frac{2}{3}\mathbf{p} \) and \( \overrightarrow{RQ} = \overrightarrow{OQ} - \overrightarrow{OR} = \mathbf{q} - \frac{2}{3}\mathbf{p} \):
\( \overrightarrow{OT} = \frac{2}{3}\mathbf{p} + \mu\left(\mathbf{q} - \frac{2}{3}\mathbf{p}\right) = \left(\frac{2}{3} - \frac{2}{3}\mu\right)\mathbf{p} + \mu\mathbf{q} \) --- (Equation 2)

Equating both representations of \( \overrightarrow{OT} \):
\( \frac{3}{4}\lambda\mathbf{p} + \frac{1}{4}\lambda\mathbf{q} = \left(\frac{2}{3} - \frac{2}{3}\mu\right)\mathbf{p} + \mu\mathbf{q} \)

Comparing coefficients of \( \mathbf{q} \):
\( \frac{1}{4}\lambda = \mu \implies \lambda = 4\mu \)

Comparing coefficients of \( \mathbf{p} \):
\( \frac{3}{4}\lambda = \frac{2}{3} - \frac{2}{3}\mu \)

Substituting \( \lambda = 4\mu \):
\( 3\mu = \frac{2}{3} - \frac{2}{3}\mu \implies 9\mu = 2 - 2\mu \implies 11\mu = 2 \implies \mu = \frac{2}{11} \)

Thus, \( \lambda = 4\mu = \frac{8}{11} \).

(a)
- **M1**: Finding \( \overrightarrow{PQ} = \mathbf{q} - \mathbf{p} \).
- **M1**: Expressing \( \overrightarrow{OS} \) in terms of \( \mathbf{p} \) and \( \overrightarrow{PQ} \).
- **A1**: Correctly obtaining \( \overrightarrow{OS} = \frac{3}{4}\mathbf{p} + \frac{1}{4}\mathbf{q} \).

(b)
- **B1**: Expressing \( \overrightarrow{OT} \) in terms of \( \lambda, \mathbf{p}, \mathbf{q} \).
- **M1**: Expressing \( \overrightarrow{RQ} = \mathbf{q} - \frac{2}{3}\mathbf{p} \).
- **M1**: Expressing second representation of \( \overrightarrow{OT} \) using \( \mu \).
- **A1**: Finding \( \overrightarrow{OT} = \left(\frac{2}{3} - \frac{2}{3}\mu\right)\mathbf{p} + \mu\mathbf{q} \).
- **M1**: Comparing coefficients of \( \mathbf{p} \) and \( \mathbf{q} \) to set up simultaneous equations.
- **A1**: Finding \( \mu = \frac{2}{11} \) and \( \lambda = \frac{8}{11} \) (both correct).

To find the exact value of the integral: \(\int_1^3 \left( \frac{e^{2x} - 2}{e^x} \right) \text{d}x\), first simplify the integrand: \(\frac{e^{2x} - 2}{e^x} = e^x - 2e^{-x}\). Now, integrate each term with respect to \(x\): \(\int (e^x - 2e^{-x}) \text{d}x = e^x + 2e^{-x}\). Apply the limits of integration from \(1\) to \(3\): \(\left[ e^x + 2e^{-x} \right]_1^3 = \left( e^3 + 2e^{-3} \right) - \left( e^1 + 2e^{-1} \right) = e^3 - e + \frac{2}{e^3} - \frac{2}{e}\).

B1: Simplify the integrand to \(e^x - 2e^{-x}\); M1: Attempt to integrate, obtaining of the form \(e^x + k e^{-x}\); A1: Correct integrated expression \(e^x + 2e^{-x}\); M1: Substitute limits 3 and 1; A1: Correct substitution \((e^3 + 2/e^3) - (e + 2/e)\); A1: Exact final answer \(e^3 - e + 2/e^3 - 2/e\) or equivalent.

Let \(u_n = a + (n-1)d\) and \(v_n = a r^{n-1}\) with \(a = 4\). From \(u_3 = v_2\), we have \(4 + 2d = 4r \implies d = 2r - 2\) (Equation 1). From \(u_{11} = v_3\), we have \(4 + 10d = 4r^2 \implies 2 + 5d = 2r^2\) (Equation 2). Substituting Equation 1 into Equation 2: \(2 + 5(2r - 2) = 2r^2 \implies 2r^2 - 10r + 8 = 0 \implies r^2 - 5r + 4 = 0\). Factoring gives \((r - 4)(r - 1) = 0\). Since \(r > 1\), we have \(r = 4\). Substituting \(r = 4\) into Equation 1 gives \(d = 2(4) - 2 = 6\).

B1: Form equation \(4 + 2d = 4r\) or equivalent; B1: Form equation \(4 + 10d = 4r^2\) or equivalent; M1: Eliminate \(d\) to form a quadratic equation in \(r\); A1: Solve quadratic to obtain \(r = 4\) (rejecting \(r = 1\)); M1: Substitute \(r\) to find \(d\); A1: Correct values \(d = 6, r = 4\).

From the second equation, \(2^{x+3} = 2^{2y} \implies x + 3 = 2y \implies x = 2y - 3\) (Equation 1). From the first equation, \(\log_3\left(\frac{x}{y^2}\right) = -1 \implies \frac{x}{y^2} = \frac{1}{3} \implies y^2 = 3x\) (Equation 2). Substitute Equation 1 into Equation 2: \(y^2 = 3(2y - 3) \implies y^2 - 6y + 9 = 0 \implies (y - 3)^2 = 0 \implies y = 3\). Substituting \(y = 3\) into Equation 1 gives \(x = 2(3) - 3 = 3\).

B1: Simplify the index equation to \(x + 3 = 2y\) or equivalent; M1: Apply log laws to combine LHS of the first equation; A1: Convert to exponential form to get \(y^2 = 3x\) or equivalent; M1: Substitute linear equation into quadratic to form a 3-term quadratic in one variable; A1: Solve quadratic to get \(y = 3\); A1: Correctly find \(x = 3\) and verify.

Using the R-formula, let \(\sqrt{3}\sin 2\theta + \cos 2\theta = R\sin(2\theta + \alpha)\). Here, \(R = \sqrt{(\sqrt{3})^2 + 1^2} = 2\) and \(\tan\alpha = \frac{1}{\sqrt{3}} \implies \alpha = \frac{\pi}{6}\). The equation becomes \(2\sin(2\theta + \pi/6) = 1 \implies \sin(2\theta + \pi/6) = 1/2\). Since \(0 \le \theta \le \pi\), we have \(\pi/6 \le 2\theta + \pi/6 \le 13\pi/6\). In this interval, the solutions for \(2\theta + \pi/6\) are \(\pi/6\), \(5\pi/6\), and \(13\pi/6\). This gives \(2\theta = 0\), \(2\pi/3\), and \(2\pi\), which simplifies to \(\theta = 0\), \(\pi/3\), and \(\pi\).

M1: Attempt R-formula or compound angle method; A1: Correctly identify \(R = 2\) and \(\alpha = \pi/6\) or equivalent; M1: Write equation as \(\sin(2\theta + \pi/6) = 1/2\); B1: Determine correct interval for transformed angle \([\pi/6, 13\pi/6]\); M1: Solve for transformed angle to find at least two correct values; A1: Correctly find any two of \(\theta = 0\), \(\pi/3\), \(\pi\); A1: Find all three correct solutions with no extra values in the range.

**(a)** First, find \(g(0)\):
\(g(0) = e^{2(0)} - 3 = 1 - 3 = -2\).

Substitute this value into \(f(x)\):
\(fg(0) = f(-2) = 2(-2)^2 + 3 = 2(4) + 3 = 11\).

**(b)** To find \(gg(x)\):
\(gg(x) = g(g(x)) = e^{2(e^{2x} - 3)} - 3 = e^{2e^{2x} - 6} - 3\).

**(c)** To find \(g^{-1}(x)\), let \(y = e^{2x} - 3\):
\(y + 3 = e^{2x}\)
\(2x = \ln(y + 3)\)
\(x = \frac{1}{2}\ln(y + 3)\)
So, \(g^{-1}(x) = \frac{1}{2}\ln(x + 3)\).

Set \(g^{-1}(x) = \frac{1}{2}\ln 4\):
\(\frac{1}{2}\ln(x + 3) = \frac{1}{2}\ln 4\)
\(\ln(x + 3) = \ln 4\)
\(x + 3 = 4\)
\(x = 1\).

**(a)**
- **B1**: For \(g(0) = -2\) seen or implied
- **B1**: For 11

**(b)**
- **B1**: Correct expression \(e^{2e^{2x} - 6} - 3\) (allow unsimplified \(e^{2(e^{2x}-3)} - 3\))

**(c)**
- **M1**: Rearranging \(y = e^{2x}-3\) to make \(x\) the subject (condone one error)
- **A1**: Finding \(g^{-1}(x) = \frac{1}{2}\ln(x+3)\) oe
- **A1**: For \(x = 1\)

For the quadratic curve to lie completely above the \(x\)-axis, we require the coefficient of \(x^2\) to be positive (which is \(1 > 0\)) and the discriminant to be negative (\(b^2 - 4ac < 0\)).

Identify \(a\), \(b\), and \(c\):
\(a = 1\), \(b = -p\), \(c = 2p + 5\).

Calculate the discriminant:
\(b^2 - 4ac = (-p)^2 - 4(1)(2p + 5) = p^2 - 8p - 20\).

Set the discriminant to be less than zero:
\(p^2 - 8p - 20 < 0\).

Factorise the quadratic expression:
\((p - 10)(p + 2) < 0\).

The critical values are \(p = 10\) and \(p = -2\).
For the product to be negative, \(p\) must lie between the critical values:
\(-2 < p < 10\).

- **M1**: For use of \(b^2 - 4ac < 0\) (accept other inequality signs or \(=\))
- **A1**: Correct quadratic inequality: \(p^2 - 8p - 20 < 0\) oe
- **M1**: For factorising or finding correct critical values \(-2\) and \(10\)
- **A1**: Correct final range: \(-2 < p < 10\)

Let \(u = \log_5 x\) and \(v = \log_5 y\).

The simultaneous equations become:
1) \(2u + v = 5\)
2) \(3u - 2v = 4\)

From equation (1), express \(v\) in terms of \(u\):
\(v = 5 - 2u\).

Substitute this into equation (2):
\(3u - 2(5 - 2u) = 4\)
\(3u - 10 + 4u = 4\)
\(7u = 14\)
\(u = 2\).

Now, find \(v\):
\(v = 5 - 2(2) = 1\).

Convert back to \(x\) and \(y\):
\(\log_5 x = 2 \implies x = 5^2 = 25\).
\(\log_5 y = 1 \implies y = 5^1 = 5\).

Therefore, the solution is \(x = 25\) and \(y = 5\).

- **M1**: For a valid attempt to eliminate one variable (either \(\log_5 x\) or \(\log_5 y\))
- **A1**: For finding \(\log_5 x = 2\) or \(\log_5 y = 1\)
- **A1**: For finding the other logarithmic value correctly
- **A1**: For \(x = 25\) (dep on first M1)
- **A1**: For \(y = 5\) (dep on first M1)

First, expand the numerator of the integrand:
\((2x - 3)^2 = 4x^2 - 12x + 9\).

Now rewrite the integrand by dividing each term by \(x^2\):
\(\frac{4x^2 - 12x + 9}{x^2} = 4 - \frac{12}{x} + 9x^{-2}\).

Integrate each term with respect to \(x\):
\(\int \left(4 - \frac{12}{x} + 9x^{-2}\right) dx = 4x - 12\ln|x| - 9x^{-1} + C = 4x - 12\ln|x| - \frac{9}{x} + C\).

Substitute the limits from 1 to 4:
At \(x = 4\):
\(4(4) - 12\ln 4 - \frac{9}{4} = 16 - 12\ln(2^2) - \frac{9}{4} = 16 - 24\ln 2 - 2.25 = 13.75 - 24\ln 2\).

At \(x = 1\):
\(4(1) - 12\ln 1 - \frac{9}{1} = 4 - 0 - 9 = -5\).

Subtract the value at the lower limit from the value at the upper limit:
\((13.75 - 24\ln 2) - (-5) = 18.75 - 24\ln 2\).

Convert 18.75 to an exact fraction:
\(18.75 = \frac{75}{4}\).

So the exact value is \(\frac{75}{4} - 24\ln 2\).

- **B1**: Correctly expands \((2x-3)^2 = 4x^2 - 12x + 9\)
- **M1**: For dividing each term by \(x^2\) to get separate powers of \(x\)
- **B2**: For correct integration of at least two terms (B1 for one correct term)
- Integration is: \(4x - 12\ln x - \frac{9}{x}\)
- **M1**: For substituting limits of 1 and 4 into their integrated expression
- **A1**: For \(\frac{75}{4} - 24\ln 2\) or exact equivalent (e.g. \(18.75 - 12\ln 4\))

**(a)** Velocity is the first derivative of displacement with respect to time:
\(v = \frac{ds}{dt} = \frac{d}{dt}\left( \ln(t^2 + 5) - \frac{1}{3}t \right) = \frac{2t}{t^2 + 5} - \frac{1}{3}\).

Acceleration is the derivative of velocity with respect to time:
\(a = \frac{dv}{dt} = \frac{d}{dt}\left( \frac{2t}{t^2 + 5} - \frac{1}{3} \right)\).
Using the quotient rule:
\(a = \frac{2(t^2 + 5) - 2t(2t)}{(t^2 + 5)^2} = \frac{2t^2 + 10 - 4t^2}{(t^2 + 5)^2} = \frac{10 - 2t^2}{(t^2 + 5)^2}\).

**(b)** The particle is at rest when \(v = 0\):
\(\frac{2t}{t^2 + 5} - \frac{1}{3} = 0\)
\(\frac{2t}{t^2 + 5} = \frac{1}{3}\)
\(6t = t^2 + 5\)
\(t^2 - 6t + 5 = 0\)
\((t - 1)(t - 5) = 0\).
So, \(t = 1\) and \(t = 5\).

**(c)** Substitute \(t = 1\) and \(t = 5\) into the expression for acceleration:
At \(t = 1\):
\(a = \frac{10 - 2(1)^2}{(1^2 + 5)^2} = \frac{8}{36} = \frac{2}{9}\text{ ms}^{-2}\).

At \(t = 5\):
\(a = \frac{10 - 2(5)^2}{(5^2 + 5)^2} = \frac{10 - 50}{30^2} = \frac{-40}{900} = -\frac{2}{45}\text{ ms}^{-2}\).

**(a)**
- **B1**: For velocity \(v = \frac{2t}{t^2 + 5} - \frac{1}{3}\)
- **M1**: Correct use of quotient rule or product rule on their velocity
- **A1**: For \(\frac{2(t^2 + 5) - 4t^2}{(t^2 + 5)^2}\) oe
- **A1**: Simplifies to \(a = \frac{10 - 2t^2}{(t^2 + 5)^2}\)

**(b)**
- **B1**: Sets \(v = 0\)
- **M1**: Forms a 3-term quadratic in \(t\) and attempts to solve: \(t^2 - 6t + 5 = 0\)
- **A1**: For \(t = 1\) and \(t = 5\) (no other values)

**(c)**
- **M1**: Substitutes their positive values of \(t\) into their acceleration expression
- **A1**: Both \(\frac{2}{9}\) and \(-\frac{2}{45}\) (accept decimal equivalents correct to 3sf: 0.222 and -0.0444)

Let the first term be \(a = 4\).
Let \(d\) be the common difference of the AP.
Let \(r\) be the common ratio of the GP.

From the given properties:
1) The 3rd term of the AP is \(a + 2d = 4 + 2d\).
The 2nd term of the GP is \(ar = 4r\).
Since they are equal:
\(4 + 2d = 4r \implies 2 + d = 2r \implies d = 2r - 2\).

2) The 11th term of the AP is \(a + 10d = 4 + 10d\).
The 3rd term of the GP is \(ar^2 = 4r^2\).
Since they are equal:
\(4 + 10d = 4r^2\).

Substitute the expression for \(d\) into this second equation:
\(4 + 10(2r - 2) = 4r^2\)
\(4 + 20r - 20 = 4r^2\)
\(4r^2 - 20r + 16 = 0\).

Divide the entire equation by 4:
\(r^2 - 5r + 4 = 0\)
\((r - 1)(r - 4) = 0\).

This gives two possible values for \(r\):
\(r = 1\) or \(r = 4\).

Since the common ratio of the GP is greater than 1, we must have \(r = 4\).

Now substitute \(r = 4\) back to find \(d\):
\(d = 2(4) - 2 = 6\).

So, the common difference of the AP is 6 and the common ratio of the GP is 4.

- **B1**: For \(4 + 2d = 4r\) oe
- **B1**: For \(4 + 10d = 4r^2\) oe
- **M1**: For attempting to eliminate one variable to form a quadratic equation (usually in \(r\))
- **A1**: For forming a correct quadratic: \(4r^2 - 20r + 16 = 0\) (or simplified \(r^2 - 5r + 4 = 0\)) or \(2d^2 - 12d = 0\)
- **A1**: For finding the correct value \(r = 4\) (rejecting \(r = 1\) with reason or implicitly)
- **A1**: For finding the correct value \(d = 6\)