2024 Cambridge IGCSE Mathematics - Additional (0606) 模拟试题及答案详解

For (a): The x-intercepts are found by setting \(y = 0\), giving \(x = -1\), \(x = 2.5\), and \(x = 3\). The y-intercept is found by setting \(x = 0\), giving \(y = -\frac{1}{4}(0-3)(0+1)(2(0)-5) = -3.75\). Since the leading coefficient of the cubic is negative, the graph starts in the second quadrant, crosses the x-axis at \(x = -1\), goes below the x-axis to a local minimum, crosses the y-axis at \((0, -3.75)\), passes through the x-axis at \(x = 2.5\), reaches a local maximum in the first quadrant, crosses the x-axis at \(x = 3\), and continues downwards.
For (b): The inequality is satisfied where the graph lies below the x-axis. Looking at the sketch, this occurs in the intervals \(-1 < x < 2.5\) and \(x > 3\).

1(a): B1 for a correct cubic shape starting in the second quadrant and ending in the fourth quadrant. B1 for x-intercepts correctly marked at \(-1\), \(2.5\), and \(3\). B1 for y-intercept correctly marked at \(-3.75\) (or \(-15/4\)).
1(b): B1 for one correct interval (either \(-1 < x < 2.5\) or \(x > 3\)). B1 for both correct intervals with no extras.

For (a): Performing algebraic long division of \(6x^3 + x^2 - 19x + 18\) by \(2x-3\):
Dividing the leading terms: \(6x^3 / 2x = 3x^2\).
Subtracting \(3x^2(2x-3) = 6x^3 - 9x^2\) leaves \(10x^2 - 19x + 18\).
Dividing leading terms: \(10x^2 / 2x = 5x\).
Subtracting \(5x(2x-3) = 10x^2 - 15x\) leaves \(-4x + 18\).
Dividing leading terms: \(-4x / 2x = -2\).
Subtracting \(-2(2x-3) = -4x + 6\) leaves a remainder of \(12\).
Thus, \(q(x) = 3x^2 + 5x - 2\) and \(r = 12\).
For (b): We have \(p(x) - 12 = (2x-3)(3x^2 + 5x - 2)\). Factoring the quadratic part gives \(3x^2 + 5x - 2 = (3x-1)(x+2)\). Thus, \(p(x) - 12 = (2x-3)(3x-1)(x+2)\).
For (c): The equation \(p(x) = 12\) is equivalent to \(p(x) - 12 = 0\). Using the factored form from (b), we get \((2x-3)(3x-1)(x+2) = 0\), which yields the solutions \(x = 1.5\), \(x = \frac{1}{3}\), and \(x = -2\).

2(a): M1 for a valid division method (long division, synthetic, or equating coefficients). A1 for \(q(x) = 3x^2 + 5x - 2\). A1 for \(r = 12\).
2(b): M1 for factorising their quadratic \(q(x)\) into two linear factors. A1 for the correct product of three linear factors: \((2x-3)(3x-1)(x+2)\).
2(c): B1 for all three correct solutions: \(x = 1.5\), \(x = \frac{1}{3}\), and \(x = -2\) (allow follow-through on incorrect linear factors from part b).

Using the laws of logarithms:

1. Express $$2\log_2(x-3)$$ as $$\log_2(x-3)^2$$.
2. Express the constant $$1$$ as $$\log_2 2$$.
3. Combine the terms on the right-hand side: $$1 + \log_2 3 = \log_2 2 + \log_2 3 = \log_2(2 \times 3) = \log_2 6$$.
4. Combine the terms on the left-hand side: $$\log_2(x-3)^2 - \log_2(x+5) = \log_2\left(\frac{(x-3)^2}{x+5}\right)$$.

Equating the arguments of the logarithms gives:
$$\frac{(x-3)^2}{x+5} = 6$$
$$(x-3)^2 = 6(x+5)$$
$$x^2 - 6x + 9 = 6x + 30$$
$$x^2 - 12x - 21 = 0$$

Solve the quadratic equation using the quadratic formula:
$$x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(-21)}}{2}$$
$$x = \frac{12 \pm \sqrt{144 + 84}}{2}$$
$$x = \frac{12 \pm \sqrt{228}}{2}$$

Since $$228 = 4 \times 57$$, we can simplify $\sqrt{228}$ to $$2\sqrt{57}$$:
$$x = \frac{12 \pm 2\sqrt{57}}{2} = 6 \pm \sqrt{57}$$

Now, check the domain of the original logarithmic terms:
For $$\log_2(x-3)$$ to be defined, we need $$x - 3 > 0 \implies x > 3$$.
For $$\log_2(x+5)$$ to be defined, we need $$x + 5 > 0 \implies x > -5$$.

Since $$6 - \sqrt{57} \approx 6 - 7.55 = -1.55$$, this value is less than $$3$$ and must be rejected.

Therefore, the only valid solution is:
$$x = 6 + \sqrt{57}$$

**M1**: For using the power rule to write $$2\log_2(x-3)$$ as $$\log_2(x-3)^2$$.
**M1**: For writing $$1$$ as $$\log_2 2$$ and combining the RHS using the addition law to get $$\log_2 6$$, or equivalent.
**M1**: For combining the LHS using subtraction law and removing logarithms to obtain a quadratic equation, e.g., $$(x-3)^2 = 6(x+5)$$.
**A1**: For obtaining the correct simplified quadratic equation $$x^2 - 12x - 21 = 0$$.
**M1**: For a valid attempt to solve their quadratic equation using the quadratic formula or completing the square.
**A1**: For identifying that $$x = 6 - \sqrt{57}$$ is invalid and stating the final correct answer $$x = 6 + \sqrt{57}$$ only.

From the first equation, express all terms with base 3:
$$3^x \times (3^2)^{2y} = 3^3 \times 3^{0.5}$$
$$3^x \times 3^{4y} = 3^{3.5}$$
$$3^{x + 4y} = 3^{3.5}$$

Equating the indices yields:
$$x + 4y = 3.5 \quad \text{--- (Equation 1)}$$

From the second equation, apply the laws of logarithms:
$$\log_2(x + 2y) = \log_2(3^2) - \log_2 2$$
$$\log_2(x + 2y) = \log_2 9 - \log_2 2$$
$$\log_2(x + 2y) = \log_2\left(\frac{9}{2}\right)$$

Equating the arguments of the logarithms yields:
$$x + 2y = 4.5 \quad \text{--- (Equation 2)}$$

Now, solve the system of linear equations:
Subtract Equation 2 from Equation 1:
$$(x + 4y) - (x + 2y) = 3.5 - 4.5$$
$$2y = -1$$
$$y = -0.5$$

Substitute $$y = -0.5$$ back into Equation 2:
$$x + 2(-0.5) = 4.5$$
$$x - 1 = 4.5$$
$$x = 5.5$$

Checking for validity inside the logarithm:
$$x + 2y = 5.5 + 2(-0.5) = 4.5 > 0$$, which is positive and valid.

Thus, the solution is $$x = 5.5$$ (or $$\frac{11}{2}$$) and $$y = -0.5$$ (or $$-\frac{1}{2}$$).

**M1**: For expressing $$9^{2y}$$ as $$3^{4y}$$ and $$27\sqrt{3}$$ as $$3^{3.5}$$.
**A1**: For obtaining the correct linear equation $$x + 4y = 3.5$$ (or equivalent).
**M1**: For using the power rule to write $$2\log_2 3$$ as $$\log_2 9$$.
**M1**: For combining the RHS of the log equation to $$\log_2(9/2)$$ (or $$\log_2 4.5$$) and removing logs.
**A1**: For obtaining the correct linear equation $$x + 2y = 4.5$$ (or equivalent).
**M1**: For a valid method to solve the two simultaneous linear equations.
**A1**: For both correct solutions: $$x = 5.5$$ and $$y = -0.5$$.

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

From the first given equation:
\( \tan\theta = 3 - 2x \)

From the second given equation:
\( \sec\theta = \frac{1}{\cos\theta} = \sqrt{y+1} \)
Squaring both sides:
\( \sec^2\theta = y + 1 \)

Substitute these expressions into the identity:
\( y + 1 = 1 + (3 - 2x)^2 \)

Subtract 1 from both sides:
\( y = (3 - 2x)^2 \)

Expand the bracket to get the final simplified form:
\( y = 9 - 12x + 4x^2 \)
\( y = 4x^2 - 12x + 9 \)

B1: Expresses \( \tan\theta \) as \( 3 - 2x \) or \( \tan^2\theta = (3 - 2x)^2 \).
B1: Expresses \( \sec^2\theta = y + 1 \).
M1: Applies the identity \( \sec^2\theta = 1 + \tan^2\theta \) using their expressions.
A1: Correctly simplifies to get the final expanded expression: \( y = 4x^2 - 12x + 9 \).

First, solve for \( \tan\left(2x - \frac{\pi}{4}\right) \):
\( \tan^2\left(2x - \frac{\pi}{4}\right) = \frac{1}{3} \implies \tan\left(2x - \frac{\pi}{4}\right) = \pm\frac{1}{\sqrt{3}} \)

Let \( \theta = 2x - \frac{\pi}{4} \).
Since \( 0 \le x \le \pi \), the range for \( \theta \) is:
\( 0 \le 2x \le 2\pi \implies -\frac{\pi}{4} \le 2x - \frac{\pi}{4} \le \frac{7\pi}{4} \)

Find the solutions for \( \tan\theta = \pm\frac{1}{\sqrt{3}} \) in the interval \( \left[-\frac{\pi}{4}, \frac{7\pi}{4}\right] \):
- For \( \tan\theta = \frac{1}{\sqrt{3}} \):
\( \theta = \frac{\pi}{6}, \frac{7\pi}{6} \)
- For \( \tan\theta = -\frac{1}{\sqrt{3}} \):
\( \theta = -\frac{\pi}{6}, \frac{5\pi}{6} \) (Note: \( \frac{11\pi}{6} \) is outside the range)

Now, solve for \( x \) using \( 2x = \theta + \frac{\pi}{4} \):
1. \( 2x = -\frac{\pi}{6} + \frac{\pi}{4} = \frac{\pi}{12} \implies x = \frac{\pi}{24} \)
2. \( 2x = \frac{\pi}{6} + \frac{\pi}{4} = \frac{5\pi}{12} \implies x = \frac{5\pi}{24} \)
3. \( 2x = \frac{5\pi}{6} + \frac{\pi}{4} = \frac{13\pi}{12} \implies x = \frac{13\pi}{24} \)
4. \( 2x = \frac{7\pi}{6} + \frac{\pi}{4} = \frac{17\pi}{12} \implies x = \frac{17\pi}{24} \)

B1: Obtains \( \tan\left(2x - \frac{\pi}{4}\right) = \pm\frac{1}{\sqrt{3}} \) (condone missing \( \pm \) for this mark).
M1: Finds at least two correct values of \( \theta \) within the interval \( \left[-\frac{\pi}{4}, \frac{7\pi}{4}\right] \).
A2: Correctly identifies all four boundary values for \( \theta \): \( -\frac{\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6} \) (deduct 1 mark for any incorrect or extra values in range).
A1: For all four correct values of \( x \): \( \frac{\pi}{24}, \frac{5\pi}{24}, \frac{13\pi}{24}, \frac{17\pi}{24} \).

Start with the Left Hand Side (LHS):
\( \text{LHS} = \frac{\sin\theta}{1 - \cos\theta} + \frac{1 - \cos\theta}{\sin\theta} \)

Find a common denominator:
\( = \frac{\sin^2\theta + (1 - \cos\theta)^2}{\sin\theta(1 - \cos\theta)} \)

Expand the numerator:
\( = \frac{\sin^2\theta + 1 - 2\cos\theta + \cos^2\theta}{\sin\theta(1 - \cos\theta)} \)

Use the Pythagorean identity \( \sin^2\theta + \cos^2\theta = 1 \):
\( = \frac{1 + 1 - 2\cos\theta}{\sin\theta(1 - \cos\theta)} \)
\( = \frac{2 - 2\cos\theta}{\sin\theta(1 - \cos\theta)} \)

Factor out a 2 in the numerator:
\( = \frac{2(1 - \cos\theta)}{\sin\theta(1 - \cos\theta)} \)

Cancel the common term \( 1 - \cos\theta \):
\( = \frac{2}{\sin\theta} \)

Using the identity \( \csc\theta = \frac{1}{\sin\theta} \):
\( = 2\csc\theta = \text{RHS} \)

M1: For writing the expression over a common denominator \( \sin\theta(1 - \cos\theta) \).
A1: For correct expansion of the numerator to \( \sin^2\theta + 1 - 2\cos\theta + \cos^2\theta \).
M1: For applying \( \sin^2\theta + \cos^2\theta = 1 \) and correctly factorising the numerator as \( 2(1 - \cos\theta) \).
A1: For cancelling the common bracket and completing the proof to reach \( 2\csc\theta \) clearly.

(a)
Using the formula for the sum to infinity of a geometric progression:
\(S_{\infty} = \frac{a}{1-r} = 16 \implies a = 16(1-r)\)

The sum of the first two terms is:
\(a + ar = 12 \implies a(1+r) = 12\)

Substituting the expression for \(a\):
\(16(1-r)(1+r) = 12\)
\(16(1-r^2) = 12\)
\(1-r^2 = \frac{12}{16} = \frac{3}{4}\)
\(r^2 = \frac{1}{4}\)
\(r = \pm \frac{1}{2}\)

So the two possible values of \(r\) are \(0.5\) and \(-0.5\).

(b)
The general term in the binomial expansion is given by:
\(\binom{6}{r} (2x^2)^{6-r} \left(-\frac{1}{4x}\right)^r = \binom{6}{r} 2^{6-r} \left(-\frac{1}{4}\right)^r x^{12-2r-r}\)

For the term independent of \(x\), the power of \(x\) must be 0:
\(12 - 3r = 0 \implies r = 4\)

Substituting \(r = 4\):
\(\binom{6}{4} 2^2 \left(-\frac{1}{4}\right)^4 = 15 \times 4 \times \frac{1}{256} = \frac{60}{256} = \frac{15}{64}\)

(a)
M1: For using the sum to infinity formula to express \(a\) as \(16(1-r)\).
M1: For substituting into \(a(1+r)=12\) to form a quadratic equation in \(r\).
A1: For finding \(r^2 = 0.25\) (or equivalent).
A1: For both \(r = 0.5\) and \(r = -0.5\).

(b)
M1: For setting the exponent of \(x\) to 0 and obtaining \(r = 4\).
M1: For substituting \(r=4\) into the term formula.
A1: For \(\frac{15}{64}\) (or \(0.234\) or exact equivalent).

(a) By using the product rule:
Let \(u = 2x - 3 \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 2\)
Let \(v = \mathrm{e}^{x^2} \implies \frac{\mathrm{d}v}{\mathrm{d}x} = 2x\mathrm{e}^{x^2}\)

\(\frac{\mathrm{d}y}{\mathrm{d}x} = u \frac{\mathrm{d}v}{\mathrm{d}x} + v \frac{\mathrm{d}u}{\mathrm{d}x}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = (2x - 3)(2x\mathrm{e}^{x^2}) + 2\mathrm{e}^{x^2}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{x^2}(4x^2 - 6x + 2) = 2\mathrm{e}^{x^2}(2x^2 - 3x + 1)\)

(b) When \(x = 1\), \(y = (2(1) - 3)\mathrm{e}^{1^2} = -\mathrm{e}\).
Substituting \(x = 1\) into \(\frac{\mathrm{d}y}{\mathrm{d}x}\):
\(m = 2\mathrm{e}^{1^2}(2(1)^2 - 3(1) + 1) = 2\mathrm{e}(0) = 0\).
Equation of the tangent:
\(y - (-\mathrm{e}) = 0(x - 1) \implies y = -\mathrm{e}\).

(a)
M1: Attempt to apply product rule with correct derivative of \(\mathrm{e}^{x^2}\).
A1: Correct derivatives of both terms: \(2\) and \(2x\mathrm{e}^{x^2}\).
M1: Correct factorization/simplification.
A1: Correct final expression: \(\mathrm{e}^{x^2}(4x^2 - 6x + 2)\) or equivalent.

(b)
B1: Find \(y = -\mathrm{e}\) when \(x = 1\).
M1: Substitute \(x = 1\) into their \(\frac{\mathrm{d}y}{\mathrm{d}x}\).
A1: Show that gradient \(m = 0\).
A1: State the final equation \(y = -\mathrm{e}\).

(a) The total surface area of a cylinder is given by:
\(A = 2\pi r^2 + 2\pi r h = 600\pi\)
Dividing by \(2\pi\):
\(r^2 + r h = 300 \implies h = \frac{300 - r^2}{r}\)

The volume of the cylinder is:
\(V = \pi r^2 h = \pi r^2 \left(\frac{300 - r^2}{r}\right)\)
\(V = \pi r (300 - r^2) = 300\pi r - \pi r^3\).

(b) To find the stationary value of \(V\), we calculate \(\frac{\mathrm{d}V}{\mathrm{d}r}\) and set it to 0:
\(\frac{\mathrm{d}V}{\mathrm{d}r} = 300\pi - 3\pi r^2 = 0\)
\(300\pi = 3\pi r^2 \implies r^2 = 100 \implies r = 10\) (since \(r > 0\)).

Checking the second derivative:
\(\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -6\pi r\)
When \(r = 10\), \(\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -60\pi < 0\), which confirms a maximum.

Substituting \(r = 10\) into the volume formula:
\(V = 300\pi(10) - \pi(10)^3 = 3000\pi - 1000\pi = 2000\pi\text{ cm}^3\).

(a)
B1: Set up the correct surface area formula: \(2\pi r^2 + 2\pi r h = 600\pi\).
M1: Express \(h\) in terms of \(r\).
A1: Substitute \(h\) into the volume formula and simplify to obtain \(V = 300\pi r - \pi r^3\).

(b)
M1: Differentiate \(V\) with respect to \(r\).
A1: Set \(\frac{\mathrm{d}V}{\mathrm{d}r} = 0\) and solve to find \(r = 10\).
M1: Check the second derivative or test values to show it is a maximum.
A1: Substitute \(r = 10\) back into the volume formula.
A1: Final correct value of \(2000\pi\text{ cm}^3\).

(a) The particle is instantaneously at rest when \(v = 0\):
\(\frac{12}{(2t+1)^2} - 3 = 0 \implies (2t+1)^2 = 4\)
Since \(t \ge 0\), \(2t + 1 = 2 \implies 2t = 1 \implies t = 0.5\text{ seconds}\).

(b) Acceleration \(a = \frac{\mathrm{d}v}{\mathrm{d}t}\):
\(v = 12(2t+1)^{-2} - 3 \implies a = -24(2t+1)^{-3} \cdot 2 = -\frac{48}{(2t+1)^3}\)
When \(t = 1\):
\(a = -\frac{48}{3^3} = -\frac{48}{27} = -\frac{16}{9} \approx -1.78\text{ m s}^{-2}\).

(c) Displacement \(s = \int v \mathrm{d}t\):
\(s = \int \left(12(2t+1)^{-2} - 3\right) \mathrm{d}t\)
\(s = \frac{12(2t+1)^{-1}}{(-1)(2)} - 3t + c = -\frac{6}{2t+1} - 3t + c\)
When \(t = 0\), \(s = 0\) (since it passes \(O\) at \(t = 0\)):
\(0 = -\frac{6}{1} - 0 + c \implies c = 6\)
So, \(s = 6 - \frac{6}{2t+1} - 3t\)

When \(t = 2\):
\(s = 6 - \frac{6}{5} - 6 = -1.2\text{ m}\).

(a)
M1: Set \(v = 0\) and attempt to solve for \(t\).
A1: Obtain \(t = 0.5\) (or equivalent).

(b)
M1: Attempt to differentiate \(v\) with respect to \(t\) using chain rule.
A1: Correctly obtain \(a = -1.78\text{ m s}^{-2}\) (or \(-\frac{16}{9}\)).

(c)
M1: Attempt to integrate \(v\) with respect to \(t\).
A1: Obtain correct integration: \(-\frac{6}{2t+1} - 3t\).
M1: Use boundary condition \(s = 0\) when \(t = 0\) to find \(c = 6\).
A1: Substitute \(t = 2\) to get final displacement of \(-1.2\text{ m}\).

(a) Using the substitution method or reverse chain rule:
\(\int (4x - 3)^5 \mathrm{d}x = \frac{(4x - 3)^6}{6 \cdot 4} + c = \frac{(4x - 3)^6}{24} + c\).

(b) Notice that the derivative of \(x^3 + 2x\) is \(3x^2 + 2\).
We can rewrite the integrand as:
\(\frac{6x^2 + 4}{x^3 + 2x} = \frac{2(3x^2 + 2)}{x^3 + 2x}\)
Thus, the integral is:
\(2 \int_{1}^{3} \frac{3x^2 + 2}{x^3 + 2x} \mathrm{d}x = 2 \left[ \ln|x^3 + 2x| \right]_1^3\)

Substituting the limits:
\(2 \left[ \ln(3^3 + 2(3)) - \ln(1^3 + 2(1)) \right] = 2 [\ln(33) - \ln(3)]\)
\(= 2 \ln\left(\frac{33}{3}\right) = 2 \ln(11) = \ln(11^2) = \ln 121\).

(a)
M1: Attempt integration by raising power to 6.
A1: Correct division by 4 (the derivative of the inner function).
A1: Correct expression \(\frac{(4x-3)^6}{24} + c\) (must include constant \(c\)).

(b)
M1: Rewrite the integral to expose the derivative of the denominator.
A1: Correct integration to \(2\ln|x^3 + 2x|\).
M1: Correctly substitute limits 3 and 1.
A1: Apply logarithmic subtraction law correctly.
A1: Obtain exact final answer \(\ln 121\) or \(2\ln 11\).

Equating the equations of the line and the curve:
\[kx - 3 = x^2 - 4x + 1\]
Rearranging to form a three-term quadratic equation in \(x\):
\[x^2 - (4 + k)x + 4 = 0\]
For the line and curve not to intersect, the discriminant must be less than zero (\(b^2 - 4ac < 0\)):
\[[-(4 + k)]^2 - 4(1)(4) < 0\]
\[(4 + k)^2 - 16 < 0\]
\[(k + 4 - 4)(k + 4 + 4) < 0\]
\[k(k + 8) < 0\]
Therefore, the critical values are \(k = 0\) and \(k = -8\).
Since the inequality is less than zero, the solution is:
\[-8 < k < 0\]

M1: Equating the line and curve equations and attempting to form a 3-term quadratic equation in \(x\).
A1: Correct quadratic equation \(x^2 - (4+k)x + 4 = 0\) (or equivalent).
M1: Applying \(b^2 - 4ac < 0\) to their quadratic equation.
A1: Correct range \(-8 < k < 0\) (or equivalent notation).

From the first equation, express \(y\) in terms of \(x\):
\[y = 5 - 2x\]
Substitute \(y\) into the second equation:
\[x^2 + x(5 - 2x) + (5 - 2x)^2 = 7\]
Expand the terms:
\[x^2 + 5x - 2x^2 + 25 - 20x + 4x^2 = 7\]
Simplify and form a quadratic equation:
\[3x^2 - 15x + 18 = 0\]
Divide by 3:
\[x^2 - 5x + 6 = 0\]
Factorise the quadratic:
\[(x - 2)(x - 3) = 0\]
This gives critical values of \(x\):
\[x = 2 \quad \text{or} \quad x = 3\]
Substitute back to find corresponding \(y\) values:
When \(x = 2\):
\[y = 5 - 2(2) = 1\]
When \(x = 3\):
\[y = 5 - 2(3) = -1\]
So the solutions are \(x = 2, y = 1\) and \(x = 3, y = -1\).

M1: Substitution of \(y = 5 - 2x\) (or \(x = \frac{5-y}{2}\)) into the quadratic equation.
A1: Correct simplified quadratic equation, e.g., \(x^2 - 5x + 6 = 0\) or \(y^2 - y - 2 = 0\).
M1: Solving their 3-term quadratic equation to find two values for one variable.
A1: Correct pairs of values: \(x = 2, y = 1\) and \(x = 3, y = -1\).

To solve the modulus inequality, we can square both sides since both expressions are non-negative for the valid range of \(x\):
\[(3x - 5)^2 \le (x + 1)^2\]
Expand both sides:
\[9x^2 - 30x + 25 \le x^2 + 2x + 1\]
Rearrange to form a quadratic inequality:
\[8x^2 - 32x + 24 \le 0\]
Divide the entire inequality by 8:
\[x^2 - 4x + 3 \le 0\]
Factorise the quadratic:
\[(x - 1)(x - 3) \le 0\]
The critical values are \(x = 1\) and \(x = 3\).
Since we want the expression to be less than or equal to zero, we choose the interval between the roots:
\[1 \le x \le 3\]

M1: Squaring both sides to form a quadratic inequality, or setting up the two linear boundary equations: \(3x - 5 = x + 1\) and \(3x - 5 = -(x + 1)\).
A1: Finding the correct critical values \(x = 1\) and \(x = 3\).
M1: Attempting to find the correct inside/outside region (e.g. testing points, sketching graphs or factorising their 3-term quadratic inequality).
A1: Correct range \(1 \le x \le 3\) (allow interval notation \([1, 3]\)).

**(a)**
Substitute \(y = 2x - 10\) into the equation of the circle:
\[ (x-3)^2 + (2x - 11)^2 = 20 \]
Expand the brackets:
\[ x^2 - 6x + 9 + 4x^2 - 44x + 121 = 20 \]
\[ 5x^2 - 50x + 130 = 20 \]
\[ 5x^2 - 50x + 110 = 0 \]
Divide through by 5:
\[ x^2 - 10x + 22 = 0 \]
Apply the quadratic formula:
\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(22)}}{2} \]
\[ x = \frac{10 \pm \sqrt{100 - 88}}{2} \]
\[ x = \frac{10 \pm \sqrt{12}}{2} = 5 \pm \sqrt{3} \]

For \(x = 5 + \sqrt{3}\):
\[ y = 2(5 + \sqrt{3}) - 10 = 2\sqrt{3} \]

For \(x = 5 - \sqrt{3}\):
\[ y = 2(5 - \sqrt{3}) - 10 = -2\sqrt{3} \]

The coordinates are \((5 + \sqrt{3}, 2\sqrt{3})\) and \((5 - \sqrt{3}, -2\sqrt{3})\).

**(b)**
Rearrange the linear equation to express \(y\) in terms of \(x\):
\[ y = 10 - 3x \]
Substitute into the equation of the curve:
\[ x(10 - 3x) + 2x^2 = 8 \]
\[ 10x - 3x^2 + 2x^2 = 8 \]
\[ -x^2 + 10x - 8 = 0 \]
\[ x^2 - 10x + 8 = 0 \]
Apply the quadratic formula:
\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(8)}}{2} \]
\[ x = \frac{10 \pm \sqrt{100 - 32}}{2} \]
\[ x = \frac{10 \pm \sqrt{68}}{2} \]
Since \(\sqrt{68} = 2\sqrt{17}\):
\[ x = 5 \pm \sqrt{17} \]

For \(x = 5 + \sqrt{17}\):
\[ y = 10 - 3(5 + \sqrt{17}) = 10 - 15 - 3\sqrt{17} = -5 - 3\sqrt{17} \]

For \(x = 5 - \sqrt{17}\):
\[ y = 10 - 3(5 - \sqrt{17}) = 10 - 15 + 3\sqrt{17} = -5 + 3\sqrt{17} \]

The coordinates are \((5 + \sqrt{17}, -5 - 3\sqrt{17})\) and \((5 - \sqrt{17}, -5 + 3\sqrt{17})\).

**(a)**
* **M1**: Substitute \(y = 2x - 10\) into the equation of the circle and expand brackets.
* **A1**: Obtain simplified 3-term quadratic equation, e.g., \(x^2 - 10x + 22 = 0\) (or \(5x^2 - 50x + 110 = 0\)).
* **M1**: Solve their quadratic equation using the quadratic formula or by completing the square.
* **A1**: Obtain correct final coordinates: \((5 + \sqrt{3}, 2\sqrt{3})\) and \((5 - \sqrt{3}, -2\sqrt{3})\).

**(b)**
* **M1**: Rearrange the linear equation to express \(y\) in terms of \(x\) (or vice versa) and substitute into the curve's equation.
* **A1**: Obtain a correct 3-term quadratic equation, e.g., \(x^2 - 10x + 8 = 0\).
* **M1**: Apply the quadratic formula or complete the square to solve their quadratic equation.
* **A1**: Obtain correct exact \(x\)-values: \(x = 5 \pm \sqrt{17}\).
* **M1**: Substitute their \(x\)-values back into the linear equation to find corresponding \(y\)-values.
* **A1**: Obtain correct final coordinates: \((5 + \sqrt{17}, -5 - 3\sqrt{17})\) and \((5 - \sqrt{17}, -5 + 3\sqrt{17})\).

(a) (i) Two students are placed at the ends of the line. The number of ways to choose and arrange 2 of the 3 students for the ends is \(^{3}P_2 = 6\).
The remaining 10 members can be arranged in the middle in \(10!\) ways.
Total number of arrangements = \(^{3}P_2 \times 10! = 6 \times 3\,628\,800 = 21\,772\,800\).

(ii) Treat the 4 teachers as a single unit. There are now \(8 + 1 = 9\) units to arrange.
These units can be arranged in \(9!\) ways.
The 4 teachers can be arranged among themselves in \(4!\) ways.
Total number of arrangements = \(9! \times 4! = 362\,880 \times 24 = 8\,709\,120\).

(b) (i) To select 5 members with no parent, all 5 must be chosen from the 4 teachers and 3 students (total of 7 people).
Number of selections = \(^{7}C_5 = 21\).

(ii) The committee of 5 can have 3 or 4 teachers.
Case 1: 3 teachers and 2 non-teachers.
Number of ways = \(^{4}C_3 \times ^{8}C_2 = 4 \times 28 = 112\).
Case 2: 4 teachers and 1 non-teacher.
Number of ways = \(^{4}C_4 \times ^{8}C_1 = 1 \times 8 = 8\).
Total number of selections = \(112 + 8 = 120\).

(a) (i)
* **M1** for considering the ends and the remaining members: \(^{3}P_2 \times 10!\) or \(3 \times 2 \times 10!\) oe
* **A1** for \(21\,772\,800\) (accept scientific notation \(2.18 \times 10^7\))

(ii)
* **M1** for treating teachers as a block: \(9! \times 4!\) oe
* **A1** for \(8\,709\,120\) (accept scientific notation \(8.71 \times 10^6\))

(b) (i)
* **M1** for \(^{7}C_5\) oe
* **A1** for \(21\)

(ii)
* **M1** for adding the two scenarios: \(^{4}C_3 \times ^{8}C_2 + ^{4}C_4 \times ^{8}C_1\) oe
* **A1** for \(120\)

(a) For \( x < 0 \), we have \( 2x^2 > 0 \) and \( 3-x > 3 > 0 \). This means that the range of \( f \) is \( f(x) > 0 \). Since the domain of \( g \) is \( x < 0 \), the range of \( f \) does not lie within the domain of \( g \). Hence, the composite function \( gf \) does not exist.

(b) \( fg(x) = f(g(x)) = f\left(\frac{4}{x}\right) = \frac{2(4/x)^2}{3 - 4/x} = \frac{32/x^2}{(3x-4)/x} = \frac{32}{x(3x-4)} \).

(c) Let \( y = \frac{2x^2}{3-x} \).
\( y(3-x) = 2x^2 \implies 3y - xy = 2x^2 \implies 2x^2 + yx - 3y = 0 \).
Using the quadratic formula to solve for \( x \):
\( x = \frac{-y \pm \sqrt{y^2 - 4(2)(-3y)}}{4} = \frac{-y \pm \sqrt{y^2 + 24y}}{4} \).
Since \( x < 0 \) and \( \sqrt{y^2+24y} > y \) for \( y > 0 \), we must select the negative square root to ensure \( x \) remains negative.
\( x = \frac{-y - \sqrt{y^2 + 24y}}{4} \).
Thus, \( f^{-1}(x) = \frac{-x - \sqrt{x^2 + 24x}}{4} \).

(a) B1: For explaining that \( f(x) > 0 \) (or range of \( f \) is positive).
B1: For explaining that this range does not fall within the domain of \( g \) (which is \( x < 0 \)).

(b) M1: For substituting \( g(x) \) into \( f(x) \).
A1: For obtaining the fully simplified expression \( \frac{32}{x(3x-4)} \) or equivalent.

(c) M1: For attempting to form a quadratic equation in \( x \) (or \( y \)).
M1: For correctly applying the quadratic formula to solve for \( x \) and justifying the choice of the negative sign because \( x < 0 \).
A1: For the correct final expression \( f^{-1}(x) = \frac{-x - \sqrt{x^2 + 24x}}{4} \).

(a) Starting with the Left-Hand Side (LHS):
\(\frac{\sin \theta}{1 - \cos \theta} - \frac{1 - \cos \theta}{\sin \theta}\)
Express over a common denominator:
\(= \frac{\sin^2 \theta - (1 - \cos \theta)^2}{\sin \theta(1 - \cos \theta)}\)
Expand the numerator:
\(= \frac{\sin^2 \theta - (1 - 2\cos \theta + \cos^2 \theta)}{\sin \theta(1 - \cos \theta)}\)
\(= \frac{\sin^2 \theta - 1 + 2\cos \theta - \cos^2 \theta}{\sin \theta(1 - \cos \theta)}\)
Since \(\sin^2 \theta - 1 = -\cos^2 \theta\):
\(= \frac{-\cos^2 \theta + 2\cos \theta - \cos^2 \theta}{\sin \theta(1 - \cos \theta)}\)
\(= \frac{2\cos \theta - 2\cos^2 \theta}{\sin \theta(1 - \cos \theta)}\)
Factorise the numerator:
\(= \frac{2\cos \theta(1 - \cos \theta)}{\sin \theta(1 - \cos \theta)}\)
Cancel the common factor of \((1 - \cos \theta)\):
\(= \frac{2\cos \theta}{\sin \theta} = 2\cot \theta\) (as required).

(b) Using the result from part (a), the equation can be written as:
\(2\cot 2\theta = 3\tan 2\theta\)
Since \(\cot 2\theta = \frac{1}{\tan 2\theta}\), this becomes:
\(\frac{2}{\tan 2\theta} = 3\tan 2\theta\)
\(3\tan^2 2\theta = 2\)
\(\tan^2 2\theta = \frac{2}{3}\)
\(\tan 2\theta = \pm\sqrt{\frac{2}{3}} \approx \pm 0.8165\)
For \(0^\circ \leqslant \theta \leqslant 180^\circ\), the range for \(2\theta\) is \(0^\circ \leqslant 2\theta \leqslant 360^\circ\).
The basic angle \(\phi = \tan^{-1}\left(\sqrt{\frac{2}{3}}\right) \approx 39.23^\circ\).
Therefore, the possible values for \(2\theta\) are:
In Quadrant 1: \(2\theta = 39.23^\circ\)
In Quadrant 2: \(2\theta = 180^\circ - 39.23^\circ = 140.77^\circ\)
In Quadrant 3: \(2\theta = 180^\circ + 39.23^\circ = 219.23^\circ\)
In Quadrant 4: \(2\theta = 360^\circ - 39.23^\circ = 320.77^\circ\)
Dividing each by 2 gives:
\(\theta = 19.6^\circ, 70.4^\circ, 109.6^\circ, 160.4^\circ\) (correct to 1 decimal place).

(a)
- **M1**: Correct attempt to combine fractions with a common denominator of \(\sin\theta(1-\cos\theta)\).
- **A1**: Correct expansion of the numerator to get \(\sin^2\theta - 1 + 2\cos\theta - \cos^2\theta\) or equivalent.
- **M1**: Uses identity \(\sin^2\theta + \cos^2\theta = 1\) correctly to simplify the numerator to \(2\cos\theta - 2\cos^2\theta\).
- **A1**: Completes proof fully by factorising numerator to cancel \((1-\cos\theta)\) and obtaining \(2\cot\theta\) with no logical errors.

(b)
- **B1**: Identifies equation as \(2\cot 2\theta = 3\tan 2\theta\).
- **M1**: Uses \(\cot 2\theta = \frac{1}{\tan 2\theta}\) to form a quadratic equation of the form \(a\tan^2 2\theta = b\).
- **M1**: Finds a correct basic angle of approximately \(39.2^\circ\) or finds at least one correct value of \(2\theta\).
- **A1**: All four correct values of \(\theta = 19.6^\circ, 70.4^\circ, 109.6^\circ, 160.4^\circ\) to 1 decimal place, and no extras in the range.

First, find $\frac{dy}{dx}$ using the quotient rule:

$$\frac{dy}{dx} = \frac{x^2 \cdot \frac{2}{2x+1} - 2x \ln(2x+1)}{x^4}$$

$$\frac{dy}{dx} = \frac{2x - 2(2x+1)\ln(2x+1)}{x^3(2x+1)}$$

Now, substitute $x = 2$ into the derivative:

$$\left.\frac{dy}{dx}\right|_{x=2} = \frac{2(2) - 2(2(2)+1)\ln(2(2)+1)}{2^3(2(2)+1)} = \frac{4 - 10\ln 5}{40} = \frac{2 - 5\ln 5}{20} \approx -0.30236$$

Using the small changes formula $\delta y \approx \frac{dy}{dx} \delta x$ with $\delta x = p$:

$$\delta y \approx -0.302p$$

M1: for a correct attempt to differentiate using the product or quotient rule (at least one term correct)
A1: for $\frac{dy}{dx} = \frac{2x - 2(2x+1)\ln(2x+1)}{x^3(2x+1)}$ oe
M1: for substituting $x = 2$ into their derivative
A1: for obtaining $\frac{2 - 5\ln 5}{20}$ or $\approx -0.302$
A1: for final answer $-0.302p$ (allow answers rounding to $-0.302p$)

The area under the curve is given by the definite integral:

$$\text{Area} = \int_{0}^{\ln 2} (e^{3x} + 2e^{-x}) \, dx$$

First, find the indefinite integral:

$$\int (e^{3x} + 2e^{-x}) \, dx = \frac{1}{3}e^{3x} - 2e^{-x}$$

Now evaluate this at the limits $x = \ln 2$ and $x = 0$:

- At $x = \ln 2$:
$$\frac{1}{3}e^{3\ln 2} - 2e^{-\ln 2} = \frac{1}{3}e^{\ln 8} - 2e^{\ln(1/2)} = \frac{8}{3} - 2\left(\frac{1}{2}\right) = \frac{8}{3} - 1 = \frac{5}{3}$$

- At $x = 0$:
$$\frac{1}{3}e^{0} - 2e^{0} = \frac{1}{3} - 2 = -\frac{5}{3}$$

Subtract the lower limit value from the upper limit value:

$$\text{Area} = \frac{5}{3} - \left(-\frac{5}{3}\right) = \frac{10}{3}$$

M1: for setting up the integral $\int_{0}^{\ln 2} (e^{3x} + 2e^{-x}) \, dx$
A1: for correct integration to obtain $\frac{1}{3}e^{3x} - 2e^{-x}$
M1: for correct substitution of $\ln 2$ into their integrated expression, simplifying exponential-log terms correctly
M1: for correct substitution of $0$ and subtraction of terms
A1: for the exact value $\frac{10}{3}$ or $3\frac{1}{3}$

First, express the volume $V$ in terms of $x$ and $h$:
$$V = x^2 h = 18 \implies h = \frac{18}{x^2}$$

Next, write down the formula for the total cost $C$:
- The area of the top and bottom is $2x^2$. Since it costs $$2\text{ per cm}^2\$, the cost is \$2 \times 2x^2 = 4x^2\$.
- The area of the four sides is \$4xh\$. Since it costs $$3\text{ per cm}^2$, the cost is $3 \times 4xh = 12xh$.

Thus, the total cost function is:
$$C = 4x^2 + 12xh$$

Substitute $h = \frac{18}{x^2}$ into the cost function:
$$C = 4x^2 + 12x \left(\frac{18}{x^2}\right) = 4x^2 + \frac{216}{x}$$

To find the minimum cost, differentiate $C$ with respect to $x$ and set the derivative to $0$:
$$\frac{dC}{dx} = 8x - \frac{216}{x^2}$$

Setting $\frac{dC}{dx} = 0$:
$$8x - \frac{216}{x^2} = 0 \implies 8x^3 = 216 \implies x^3 = 27 \implies x = 3$$

To confirm it is a minimum, we find the second derivative:
$$\frac{d^2C}{dx^2} = 8 + \frac{432}{x^3}$$
At $x = 3$, $\frac{d^2C}{dx^2} = 8 + \frac{432}{27} > 0$, so it is indeed a minimum.

The minimum total cost is:
$$C = 4(3)^2 + \frac{216}{3} = 36 + 72 = 108$$

M1: for expressing $h$ in terms of $x$, i.e., $h = \frac{18}{x^2}$
M1: for a correct cost expression in terms of $x$ and $h$, i.e., $C = 4x^2 + 12xh$ oe
A1: for substituting $h$ to get the cost function in one variable, $C = 4x^2 + \frac{216}{x}$ oe
M1: for differentiating and setting their $\frac{dC}{dx} = 0$ to solve for $x$
A1: for $x = 3$
A1: for minimum cost $= 108$

(a) Finding the vector \(\overrightarrow{XN}\):
\(\overrightarrow{XN} = \overrightarrow{XO} + \overrightarrow{ON} = -\mathbf{x} + \frac{1}{2}\mathbf{y}\)
Then,
\(\overrightarrow{OP} = \overrightarrow{OX} + \overrightarrow{XP} = \mathbf{x} + \lambda(-\mathbf{x} + \frac{1}{2}\mathbf{y}) = (1 - \lambda)\mathbf{x} + \frac{1}{2}\lambda\mathbf{y}\)

Finding the vector \(\overrightarrow{YM}\):
\(\overrightarrow{YM} = \overrightarrow{YO} + \overrightarrow{OM} = -\mathbf{y} + \frac{1}{3}\mathbf{x}\)
Then,
\(\overrightarrow{OP} = \overrightarrow{OY} + \overrightarrow{YP} = \mathbf{y} + \mu(-\mathbf{y} + \frac{1}{3}\mathbf{x}) = \frac{1}{3}\mu\mathbf{x} + (1 - \mu)\mathbf{y}\)

Equating the coefficients of \(\mathbf{x}\) and \(\mathbf{y}\):
\(1 - \lambda = \frac{1}{3}\mu\)
\(\frac{1}{2}\lambda = 1 - \mu\)

From the second equation, \(\mu = 1 - \frac{1}{2}\lambda\).
Substituting into the first equation:
\(1 - \lambda = \frac{1}{3}(1 - \frac{1}{2}\lambda) = \frac{1}{3} - \frac{1}{6}\lambda\)
\(\frac{2}{3} = \frac{5}{6}\lambda \implies \lambda = \frac{4}{5}\)
Then \(\mu = 1 - \frac{1}{2}\left(\frac{4}{5}\right) = \frac{3}{5}\).

Since \(\lambda = \frac{4}{5}\), \(\overrightarrow{XP} = \frac{4}{5}\overrightarrow{XN}\), which means \(P\) divides \(XN\) in the ratio \(4 : 1\).
Since \(\mu = \frac{3}{5}\), \(\overrightarrow{YP} = \frac{3}{5}\overrightarrow{YM}\), which means \(P\) divides \(YM\) in the ratio \(3 : 2\).

(b) Substitute \(\lambda = \frac{4}{5}\) into the expression for \(\overrightarrow{OP}\):
\(\overrightarrow{OP} = \left(1 - \frac{4}{5}\right)\mathbf{x} + \frac{1}{2}\left(\frac{4}{5}\right)\mathbf{y} = \frac{1}{5}\mathbf{x} + \frac{2}{5}\mathbf{y} = \frac{1}{5}(\mathbf{x} + 2\mathbf{y})\)
Since \(\overrightarrow{OR} = \mathbf{x} + 2\mathbf{y}\), we have:
\(\overrightarrow{OP} = \frac{1}{5}\overrightarrow{OR}\)
Since \(\overrightarrow{OP}\) is a scalar multiple of \(\overrightarrow{OR}\) and they share the common point \(O\), the points \(O\), \(P\) and \(R\) are collinear.

(a)
B1: For \(\overrightarrow{OP} = (1 - \lambda)\mathbf{x} + \frac{1}{2}\lambda\mathbf{y}\) or equivalent.
B1: For \(\overrightarrow{OP} = \frac{1}{3}\mu\mathbf{x} + (1 - \mu)\mathbf{y}\) or equivalent.
M1: For equating the coefficients of \(\mathbf{x}\) to form an equation in \(\lambda\) and \(\mu\).
M1: For equating the coefficients of \(\mathbf{y}\) to form an equation in \(\lambda\) and \(\mu\).
A1: For finding \(\lambda = \frac{4}{5}\).
A1: For finding \(\mu = \frac{3}{5}\).
A1: For a fully correct conclusion showing that \(P\) divides \(XN\) in the ratio \(4:1\) and \(YM\) in the ratio \(3:2\).

(b)
M1: For writing \(\overrightarrow{OP} = \frac{1}{5}(\mathbf{x} + 2\mathbf{y})\) or equivalent.
A1: For concluding that \(\overrightarrow{OP}\) and \(\overrightarrow{OR}\) are scalar multiples with a common point \(O\), hence collinear.