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

First, we convert the logarithm with base 4 to base 2:
\(3\log_4 p = 3 \times \frac{\log_2 p}{\log_2 4} = \frac{3}{2}\log_2 p = \log_2(p^{3/2})\).

Now substitute this back into the original equation:
\(\log_2(p^{3/2}) + \log_2 q = 6\)
\(\log_2(p^{3/2} q) = 6\).

Express this in exponential form:
\(p^{3/2} q = 2^6\)
\(p^{3/2} q = 64\).

Square both sides to find the value of \(p^3 q^2\):
\((p^{3/2} q)^2 = 64^2\)
\(p^3 q^2 = 4096\).

M1: Use the change of base formula to show \(3\log_4 p = \frac{3}{2}\log_2 p\) or equivalent.
M1: Use laws of logarithms to obtain \(p^{3/2} q = 64\).
A1: Square both sides correctly to get the final answer 4096.

Using the product rule for differentiation:
\(\frac{\text{d}y}{\text{d}x} = \frac{\text{d}}{\text{d}x}(x^2) \cdot e^{3x} + x^2 \cdot \frac{\text{d}}{\text{d}x}(e^{3x})\)
\(\frac{\text{d}y}{\text{d}x} = 2x e^{3x} + x^2 (3e^{3x}) = e^{3x}(2x + 3x^2)\).

Substitute \(x = \frac{1}{3}\) into the derivative:
\(\frac{\text{d}y}{\text{d}x} = e^{3(1/3)} \left(2\left(\frac{1}{3}\right) + 3\left(\frac{1}{3}\right)^2\right)\)
\(\frac{\text{d}y}{\text{d}x} = e^{1} \left(\frac{2}{3} + \frac{3}{9}\right)\)
\(\frac{\text{d}y}{\text{d}x} = e \left(\frac{2}{3} + \frac{1}{3}\right) = e(1) = e\).

M1: For a correct attempt at differentiation using the product rule.
A1: For a correct derivative expression, e.g., \(2x e^{3x} + 3x^2 e^{3x}\).
A1: For correct substitution and simplification to obtain the exact value \(e\).

Since the terms form a geometric progression, the common ratio \(r\) is constant:
\(\frac{k}{k+4} = \frac{k-3}{k}\)
\(k^2 = (k+4)(k-3)\)
\(k^2 = k^2 + k - 12\)
\(k = 12\).

Now substitute \(k = 12\) to find the first term \(a\) and the common ratio \(r\):
First term, \(a = k + 4 = 12 + 4 = 16\).
Common ratio, \(r = \frac{k}{k+4} = \frac{12}{16} = \frac{3}{4}\).

Since \(|r| < 1\), the sum to infinity \(S_{\infty}\) exists:
\(S_{\infty} = \frac{a}{1 - r} = \frac{16}{1 - 3/4} = \frac{16}{1/4} = 64\).

M1: Set up the ratio equation and solve to find \(k = 12\).
M1: Determine the first term \(a = 16\) and common ratio \(r = \frac{3}{4}\).
A1: Use the sum to infinity formula correctly to obtain \(64\).

Using the trigonometric identity \
\(\sin^2 \theta = 1 - \cos^2 \theta\):
\(3(1 - \cos^2 \theta) + 5\cos \theta - 5 = 0\)
\(3 - 3\cos^2 \theta + 5\cos \theta - 5 = 0\)
\(3\cos^2 \theta - 5\cos \theta + 2 = 0\).

Factorising this quadratic equation in \(\cos \theta\):
\((3\cos \theta - 2)(\cos \theta - 1) = 0\).

This yields two possible values:
1) \(\cos \theta = 1\), which gives \(\theta = 0^\circ\).
2) \(\cos \theta = \frac{2}{3}\), which gives \(\theta = 48.2^\circ\) (correct to 1 decimal place).

M1: Substitute \(1 - \cos^2 \theta\) for \(\sin^2 \theta\) and write as a 3-term quadratic equation.
M1: Solve or factorise the quadratic equation to find \(\cos \theta = 1\) and \(\cos \theta = \frac{2}{3}\).
A1: Provide both correct angles \(\theta = 0^\circ\) and \(\theta = 48.2^\circ\) (accept \(48.2\)).

The letters of the word JOURNEY are J, O, U, R, N, E, Y. The vowels are O, U, and E.
Treat the group of vowels {O, U, E} as a single entity.
This leaves us with 5 entities to arrange: the group {O, U, E} and the individual consonants J, R, N, and Y.

The number of ways to arrange these 5 entities is:
\(5! = 120\) ways.

Within the group, the 3 vowels can be arranged amongst themselves in:
\(3! = 6\) ways.

Therefore, the total number of arrangements is:
\(120 \times 6 = 720\).

M1: Treat vowels as a single unit and calculate \(5!\) or 120.
M1: Calculate the internal arrangements of the vowels as \(3!\) or 6.
A1: Multiply the outcomes to obtain 720.

(a) (i) Using the product rule:
\( \frac{\text{d}y}{\text{d}x} = 3e^{-2x} + (3x+1)(-2e^{-2x}) \)
\( \frac{\text{d}y}{\text{d}x} = (3 - 6x - 2)e^{-2x} = (1 - 6x)e^{-2x} \)

(ii) At the stationary point, \( \frac{\text{d}y}{\text{d}x} = 0 \):
\( (1 - 6x)e^{-2x} = 0 \)
Since \( e^{-2x} \ne 0 \), \( 1 - 6x = 0 \Rightarrow x = \frac{1}{6} \).
Substitute \( x = \frac{1}{6} \) into the curve equation:
\( y = \left(3\left(\frac{1}{6}\right)+1\right)e^{-2\left(\frac{1}{6}\right)} = \frac{3}{2}e^{-\frac{1}{3}} \).
Stationary point: \( \left(\frac{1}{6}, \frac{3}{2}e^{-\frac{1}{3}}\right) \).

(b) When \( x = 0 \), \( y = (0+1)e^0 = 1 \).
The gradient of the tangent at \( x = 0 \) is \( m_t = (1 - 0)e^0 = 1 \).
The gradient of the normal is \( m_n = -\frac{1}{m_t} = -1 \).
The equation of the normal is:
\( y - 1 = -1(x - 0) \Rightarrow y = -x + 1 \).
To find the coordinates of \(P\), set \( y = 0 \):
\( 0 = -x + 1 \Rightarrow x = 1 \).
So, \( P \) has coordinates \( (1, 0) \).

(a) (i)
- **M1**: Applying product rule correctly.
- **A1**: Finding \( 3e^{-2x} \) and \( -2e^{-2x} \).
- **A1**: Obtaining simplified derivative \( (1-6x)e^{-2x} \).

(ii)
- **M1**: Setting \( \frac{\text{d}y}{\text{d}x} = 0 \).
- **A1**: Finding \( x = \frac{1}{6} \).
- **A1**: Finding exact \( y = \frac{3}{2}e^{-\frac{1}{3}} \).

(b)
- **B1**: Finding the coordinates \( (0, 1) \) when \( x = 0 \).
- **M1**: Finding the gradient of the tangent and normal at \( x = 0 \).
- **A1**: Finding the normal gradient \( = -1 \).
- **M1**: Setting up the equation of the normal and substituting \( y = 0 \).
- **A1**: For \( P(1, 0) \).

(a) LHS:
\( \frac{\sin^2 \theta + (1 - \cos \theta)^2}{\sin \theta (1 - \cos \theta)} \)
\( = \frac{\sin^2 \theta + 1 - 2\cos \theta + \cos^2 \theta}{\sin \theta (1 - \cos \theta)} \)
Using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( = \frac{2 - 2\cos \theta}{\sin \theta (1 - \cos \theta)} \)
\( = \frac{2(1 - \cos \theta)}{\sin \theta (1 - \cos \theta)} = \frac{2}{\sin \theta} = 2\text{cosec }\theta \).

(b) Using the identity from part (a) with \( \theta = 2x \):
\( 2\text{cosec }2x = 3\text{sec }2x \)
\( \frac{2}{\sin 2x} = \frac{3}{\cos 2x} \Rightarrow \tan 2x = \frac{2}{3} \).
Since \( 0^\circ \le x \le 180^\circ \), we have \( 0^\circ \le 2x \le 360^\circ \).
\( 2x = \tan^{-1}\left(\frac{2}{3}\right) \approx 33.69^\circ \) or \( 2x = 180^\circ + 33.69^\circ = 213.69^\circ \).
\( x \approx 16.8^\circ \) or \( x \approx 106.8^\circ \).

(a)
- **M1**: Finding a common denominator and writing as a single fraction.
- **M1**: Expanding the numerator and using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
- **A1**: Factoring the numerator to \( 2(1 - \cos \theta) \).
- **A1**: Obtaining the final expression \( 2\text{cosec }\theta \) clearly.

(b)
- **M1**: Utilizing the identity to rewrite the equation as \( 2\text{cosec }2x = 3\text{sec }2x \).
- **M1**: Expressing in terms of sine and cosine.
- **A1**: Deducing \( \tan 2x = \frac{2}{3} \).
- **B1**: Identifying the range for \( 2x \) is \( 0^\circ \le 2x \le 360^\circ \).
- **M1**: Solving for the principal value of \( 2x \).
- **A1**: Finding \( x \approx 16.8^\circ \).
- **A1**: Finding \( x \approx 106.8^\circ \) (deduct 1 mark if there are extra incorrect values inside the range).

(a) From the arithmetic progression formulas:
\( u_3 = a + 2d = 14 \) (Equation 1)
\( S_{10} = \frac{10}{2}(2a + 9d) = 5(2a + 9d) = 215 \Rightarrow 2a + 9d = 43 \) (Equation 2)
From Equation 1, \( a = 14 - 2d \).
Substitute into Equation 2:
\( 2(14 - 2d) + 9d = 43 \)
\( 28 - 4d + 9d = 43 \Rightarrow 5d = 15 \Rightarrow d = 3 \).
Substitute \( d = 3 \) back:
\( a = 14 - 2(3) = 8 \).
Thus, \( a = 8 \) and \( d = 3 \).

(b) For the geometric progression:
First term \( A = a = 8 \).
Sum to infinity \( S_\infty = \frac{A}{1-R} = \frac{8}{1-R} \).
We are given \( S_\infty = 4 \times u_3 = 4 \times 14 = 56 \).
Thus,
\( \frac{8}{1-R} = 56 \)
\( 1-R = \frac{8}{56} = \frac{1}{7} \)
\( R = 1 - \frac{1}{7} = \frac{6}{7} \).

(a)
- **B1**: For the equation \( a + 2d = 14 \).
- **M1**: For using the formula \( S_n = \frac{n}{2}(2a + (n-1)d) \) with \( n = 10 \).
- **A1**: For obtaining \( 2a + 9d = 43 \).
- **M1**: For attempting to solve the simultaneous equations.
- **A1**: For finding both \( a = 8 \) and \( d = 3 \).

(b)
- **B1**: For identifying that the first term of the GP is \( A = 8 \).
- **B1**: For identifying the target sum to infinity is \( 56 \).
- **M1**: For setting up the sum to infinity equation \( \frac{8}{1-R} = 56 \).
- **M1**: For algebraic steps to isolate \( R \).
- **A2**: For obtaining the correct exact value \( R = \frac{6}{7} \) (A1 if left as unsimplified decimal/fraction).

(a) The perimeter of the sector is given by:
\( P = 2r + r\theta = 30 \).
\( r\theta = 30 - 2r \Rightarrow \theta = \frac{30 - 2r}{r} = \frac{30}{r} - 2 \).

(b) The area \( A \) of the sector is given by:
\( A = \frac{1}{2}r^2\theta \).
Substitute the expression for \( \theta \) from part (a):
\( A = \frac{1}{2}r^2\left(\frac{30 - 2r}{r}\right) \)
\( A = \frac{1}{2}r(30 - 2r) = 15r - r^2 \).

(c) To find the maximum area, we can differentiate \( A \) with respect to \( r \):
\( \frac{\text{d}A}{\text{d}r} = 15 - 2r \).
Set \( \frac{\text{d}A}{\text{d}r} = 0 \) for stationary points:
\( 15 - 2r = 0 \Rightarrow r = 7.5 \).
Since \( \frac{\text{d}^2A}{\text{d}r^2} = -2 < 0 \), this gives a maximum value.
Maximum area: \( A_{\text{max}} = 15(7.5) - (7.5)^2 = 112.5 - 56.25 = 56.25 \text{ cm}^2 \).
Corresponding value of \( \theta \):
\( \theta = \frac{30}{7.5} - 2 = 4 - 2 = 2 \) radians.

(a)
- **M1**: For the perimeter expression \( 2r + r\theta \).
- **A1**: For equation \( 2r + r\theta = 30 \).
- **A1**: For isolating \( \theta = \frac{30-2r}{r} \) or equivalent.

(b)
- **M1**: For using the area formula \( A = \frac{1}{2}r^2\theta \).
- **M1**: Substituting their \( \theta \) from part (a) into the area formula.
- **A1**: Clearly demonstrating the simplification to obtain \( A = 15r - r^2 \).

(c)
- **M1**: Attempting differentiation of \( A \) with respect to \( r \) (or completing the square).
- **A1**: Finding \( r = 7.5 \).
- **M1**: Substituting \( r = 7.5 \) to find the maximum area.
- **A1**: For \( A_{\text{max}} = 56.25 \).
- **A1**: For \( \theta = 2 \).

(a) Since there are no restrictions, we choose 5 people out of 12:
\( \binom{12}{5} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792 \).

(b) To have more women than men, the combinations of (Women, Men) on the committee of 5 can be:
1. 3 Women and 2 Men:
\( \binom{6}{3} \times \binom{6}{2} = 20 \times 15 = 300 \)

2. 4 Women and 1 Man:
\( \binom{6}{4} \times \binom{6}{1} = 15 \times 6 = 90 \)

3. 5 Women and 0 Men:
\( \binom{6}{5} \times \binom{6}{0} = 6 \times 1 = 6 \)

Total ways: \( 300 + 90 + 6 = 396 \).

(c) There are two alternate configurations of seats:
Configuration 1: M W M W M W M W M W M W
Configuration 2: W M W M W M W M W M W M
For each configuration, the 6 men can be arranged in the 6 chosen male seats in \( 6! \) ways, and the 6 women in \( 6! \) ways.
Total arrangements: \( 2 \times 6! \times 6! = 2 \times 720 \times 720 = 1,036,800 \).

(a)
- **M1**: For writing \( \binom{12}{5} \).
- **A1**: For 792.

(b)
- **M1**: For identifying the three valid cases: (3W, 2M), (4W, 1M), (5W, 0M).
- **M1**: Calculating at least one correct combination product.
- **A1**: Finding all three case counts correctly (300, 90, 6).
- **M1**: Summing the three valid cases.
- **A1**: For 396.

(c)
- **B1**: For identifying the multiplier of 2 (for both configurations starting with M or W).
- **M1**: For using \( 6! \) for either men or women arrangement.
- **A1**: For the calculation structure \( 2 \times 6! \times 6! \).
- **A1**: For 1,036,800.

(a) Using the logarithm quotient property:
\( \log_3\left(\frac{2x+1}{x-1}\right) = 2 \).
Convert to exponential form:
\( \frac{2x+1}{x-1} = 3^2 = 9 \).
Solve the equation:
\( 2x + 1 = 9(x - 1) \)
\( 2x + 1 = 9x - 9 \)
\( 7x = 10 \Rightarrow x = \frac{10}{7} \).
Check domain restrictions: Since \( x = 10/7 > 1 \), both original logarithm arguments are positive. The solution is valid.

(b) Rewrite the equation:
\( 3 \cdot (3^y)^2 - 10(3^y) + 3 = 0 \).
Let \( u = 3^y \). This gives the quadratic equation:
\( 3u^2 - 10u + 3 = 0 \).
Factor the quadratic:
\( (3u - 1)(u - 3) = 0 \).
So, \( u = \frac{1}{3} \) or \( u = 3 \).
Case 1: \( 3^y = \frac{1}{3} = 3^{-1} \Rightarrow y = -1 \).
Case 2: \( 3^y = 3^1 \Rightarrow y = 1 \).
Both values of \( y \) are valid.

(a)
- **M1**: For applying the subtraction rule of logarithms to obtain \( \log_3\left(\frac{2x+1}{x-1}\right) \).
- **M1**: For converting the logarithmic equation to exponential form \( 3^2 \) or 9.
- **A1**: For obtaining the correct linear equation \( 2x+1 = 9(x-1) \).
- **A1**: For \( x = \frac{10}{7} \) (or equivalent simplified fraction/decimal \( \approx 1.43 \)).

(b)
- **M1**: For recognizing \( 3^{2y+1} \) can be written as \( 3 \cdot 3^{2y} \) or \( 3 \cdot (3^y)^2 \).
- **M1**: Setting up a quadratic in terms of \( 3^y \) (or equivalent substitution \( u = 3^y \)).
- **A1**: For the correct quadratic equation \( 3u^2 - 10u + 3 = 0 \).
- **M1**: For factorising or solving the quadratic equation.
- **A1**: For finding \( u = 3 \) and \( u = \frac{1}{3} \).
- **M1**: For equating back \( 3^y = 3 \) and \( 3^y = \frac{1}{3} \) and solving for \( y \).
- **A1**: For both solutions \( y = 1 \) and \( y = -1 \).