2023 Cambridge IGCSE International Mathematics (0607) 模拟试题及答案详解

First, divide both sides of the equation by 3 to get \(\cos(2x) = -0.5\). Since the interval for \(x\) is \(0^\circ \le x \le 180^\circ\), the interval for \(2x\) is \(0^\circ \le 2x \le 360^\circ\). In this interval, the solutions for \(\cos(2x) = -0.5\) are \(2x = 120^\circ\) and \(2x = 240^\circ\). Dividing these by 2 gives the solutions for \(x\) as \(x = 60^\circ\) and \(x = 120^\circ\). The sum of these values is \(60^\circ + 120^\circ = 180^\circ\).

M1 for \(\cos(2x) = -0.5\) or for finding at least one correct solution of \(x = 60^\circ\) or \(x = 120^\circ\). A1 for 180.

Using the given terms, we set up a system of linear equations. For \(n = 2\): \(a(2)^2 + b(2) = 14 \implies 4a + 2b = 14 \implies 2a + b = 7\). For \(n = 3\): \(a(3)^2 + b(3) = 33 \implies 9a + 3b = 33 \implies 3a + b = 11\). Subtracting the first simplified equation from the second gives \((3a + b) - (2a + b) = 11 - 7 \implies a = 4\). Substituting \(a = 4\) into \(2a + b = 7\) gives \(2(4) + b = 7 \implies b = -1\). The sequence formula is therefore \(u_n = 4n^2 - n\). For \(n = 5\), we calculate \(u_5 = 4(5)^2 - 5 = 4(25) - 5 = 95\).

M1 for finding the correct values of \(a = 4\) and \(b = -1\) by setting up and solving simultaneous equations. A1 for 95.

The total number of balls is \(5 + n\). The probability of choosing two red balls without replacement is \(\frac{5}{5+n} \times \frac{4}{4+n} = \frac{2}{9}\). This simplifies to \(\frac{20}{(5+n)(4+n)} = \frac{2}{9}\). Cross-multiplying gives \(180 = 2(5+n)(4+n)\), which simplifies to \(90 = n^2 + 9n + 20\). Rearranging into standard quadratic form gives \(n^2 + 9n - 70 = 0\). Factoring the quadratic gives \((n - 5)(n + 14) = 0\). Since the number of balls \(n\) must be positive, we find \(n = 5\).

M1 for formulating the probability equation \(\frac{20}{(5+n)(4+n)} = \frac{2}{9}\) or the quadratic equation \(n^2 + 9n - 70 = 0\). A1 for 5.

When a fair six-sided die is rolled twice, there are \(6 \times 6 = 36\) equally likely outcomes. The sum of the two numbers ranges from 2 to 12. The prime numbers in this range are 2, 3, 5, 7, and 11. Now we list the successful outcomes for each prime sum: Sum is 2: \((1, 1)\) (1 outcome); Sum is 3: \((1, 2), (2, 1)\) (2 outcomes); Sum is 5: \((1, 4), (2, 3), (3, 2), (4, 1)\) (4 outcomes); Sum is 7: \((1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\) (6 outcomes); Sum is 11: \((5, 6), (6, 5)\) (2 outcomes). Total number of successful outcomes = \(1 + 2 + 4 + 6 + 2 = 15\). The probability is \(\frac{15}{36} = \frac{5}{12}\).

M1 for listing at least 8 correct outcomes or establishing the total space of 36 outcomes. A1 for \(\frac{5}{12}\) (or equivalent fraction, or \(0.417\) to 3 significant figures).

Substitute \(n = 2\) and \(n = 4\) into the given formula: For \(n = 2\): \(a(2)^2 + b(2) = 10 \implies 4a + 2b = 10 \implies 2a + b = 5\) (Equation 1). For \(n = 4\): \(a(4)^2 + b(4) = 36 \implies 16a + 4b = 36 \implies 4a + b = 9\) (Equation 2). Subtracting Equation 1 from Equation 2 gives \(2a = 4\), so \(a = 2\). Substituting \(a = 2\) into Equation 1 gives \(2(2) + b = 5 \implies b = 1\). Thus, the general term is \(u_n = 2n^2 + n\). To find \(u_5\), substitute \(n = 5\): \(u_5 = 2(5)^2 + 5 = 50 + 5 = 55\).

M1 for setting up a correct system of simultaneous equations and solving for \(a\) and \(b\) (finding \(a = 2, b = 1\)). A1 for 55.

Rearranging the equation: \(4 \sin x = 1 - 3\) which simplifies to \(4 \sin x = -2\), so \( \sin x = -0.5\). The basic acute angle (reference angle) is \(\sin^{-1}(0.5) = 30^\circ\). Since the sine value is negative, the angle \(x\) must lie in the third and fourth quadrants. Third quadrant: \(x = 180^\circ + 30^\circ = 210^\circ\). Fourth quadrant: \(x = 360^\circ - 30^\circ = 330^\circ\).

M1 for simplifying to \(\sin x = -0.5\) or finding one correct solution (either \(210^\circ\) or \(330^\circ\)). A1 for both solutions \(210^\circ\) and \(330^\circ\).

First, factorise all the quadratic and linear expressions in the numerator and denominator:
1. \(3x^2 - 10x + 3 = (3x - 1)(x - 3)\)
2. \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)
3. \(4x^2 - 1 = (2x - 1)(2x + 1)\) (difference of two squares)
4. \(3x^2 - x = x(3x - 1)\)

Now substitute these factorised forms back into the expression:
\(\frac{(3x - 1)(x - 3)}{(2x + 1)(x - 3)} \times \frac{(2x - 1)(2x + 1)}{x(3x - 1)}\)

Cancel out the common factors:
- \(x - 3\) cancels out from the first fraction.
- \(3x - 1\) cancels out from the numerator of the first fraction and the denominator of the second.
- \(2x + 1\) cancels out from the denominator of the first fraction and the numerator of the second.

This leaves:
\(\frac{2x - 1}{x}\)

M1 for factorising \(3x^2 - 10x + 3\) to \((3x - 1)(x - 3)\)
M1 for factorising \(2x^2 - 5x - 3\) to \((2x + 1)(x - 3)\)
M1 for factorising \(4x^2 - 1\) to \((2x - 1)(2x + 1)\) and \(3x^2 - x\) to \(x(3x - 1)\)
A1 for final simplified answer \(\frac{2x - 1}{x}\) (or equivalent such as \(2 - \frac{1}{x}\))

Multiply the entire equation by the common denominator \((x-1)(x+2)\) to clear the fractions:
\(4(x+2) + 3(x-1) = 2(x-1)(x+2)\)

Expand both sides:
\(4x + 8 + 3x - 3 = 2(x^2 + x - 2)\)
\(7x + 5 = 2x^2 + 2x - 4\)

Rearrange into standard quadratic form \(ax^2 + bx + c = 0\):
\(2x^2 - 5x - 9 = 0\)

Apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-9)}}{2(2)}\)
\(x = \frac{5 \pm \sqrt{25 + 72}}{4}\)
\(x = \frac{5 \pm \sqrt{97}}{4}\)

Calculating the two solutions:
\(x_1 = \frac{5 + 9.848858}{4} \approx 3.7122 \approx 3.71\)
\(x_2 = \frac{5 - 9.848858}{4} \approx -1.2122 \approx -1.21\)

M1 for multiplying by the common denominator: \(4(x+2) + 3(x-1) = 2(x-1)(x+2)\)
M1 for expanding and simplifying to a three-term quadratic equation, e.g., \(2x^2 - 5x - 9 = 0\) (allow for one arithmetic error)
M1 for applying the quadratic formula correctly to their quadratic equation
A1 for both answers \(3.71\) and \(-1.21\) correct to 2 decimal places

Let the sequence be represented by \(T_n\). Find the differences between consecutive terms:
First differences:
\(11 - 3 = 8\)
\(23 - 11 = 12\)
\(39 - 23 = 16\)

Second differences:
\(12 - 8 = 4\)
\(16 - 12 = 4\)

Since the second difference is constant and equals 4, the sequence is quadratic and has the general form \(T_n = an^2 + bn + c\), where \(a = \frac{\text{second difference}}{2} = \frac{4}{2} = 2\).

Now, subtract \(2n^2\) from each term of the sequence to find the linear part:
For \(n=1\): \(3 - 2(1)^2 = 1\)
For \(n=2\): \(11 - 2(2)^2 = 3\)
For \(n=3\): \(23 - 2(3)^2 = 5\)
For \(n=4\): \(39 - 2(4)^2 = 7\)

The remaining linear sequence is 1, 3, 5, 7, ... which has a first term of 1 and a common difference of 2.
The \(n\)th term of this linear sequence is \(2n - 1\).

Combining the quadratic and linear parts, the overall \(n\)th term is:
\(T_n = 2n^2 + 2n - 1\)

M1 for finding first differences (8, 12, 16) and second differences (4, 4)
M1 for determining the coefficient of \(n^2\) is \(a = 2\)
M1 for setting up equations or subtracting \(2n^2\) to find the linear component (e.g., finding the linear sequence 1, 3, 5, 7)
A1 for the correct expression \(2n^2 + 2n - 1\) (or equivalent)

To simplify the algebraic fraction: 1. Factorise the quadratic expression in the numerator: \(2x^2 + 5x - 3 = (2x - 1)(x + 3)\). 2. Factorise the difference of two squares in the denominator: \(4x^2 - 1 = (2x - 1)(2x + 1)\). 3. Substitute these factored forms back into the fraction: \(\frac{(2x - 1)(x + 3)}{(2x - 1)(2x + 1)}\). 4. Cancel the common factor \((2x - 1)\) to get: \(\frac{x + 3}{2x + 1}\).

M1 for factorising the numerator to \((2x - 1)(x + 3)\) or equivalent. M1 for factorising the denominator to \((2x - 1)(2x + 1)\). A1 for cancelling the common factor \((2x - 1)\). A1 for final simplified fraction \(\frac{x + 3}{2x + 1}\).

1. Substitute \(y = 2x - 3\) into the first equation: \(2x^2 - (2x - 3)^2 = 7\). 2. Expand and simplify the expression: \(2x^2 - (4x^2 - 12x + 9) = 7\), which gives \(-2x^2 + 12x - 9 = 7\). Rearranging this gives \(-2x^2 + 12x - 16 = 0\). Divide by \(-2\) to get: \(x^2 - 6x + 8 = 0\). 3. Factorise the quadratic equation: \((x - 2)(x - 4) = 0\), giving \(x = 2\) or \(x = 4\). 4. Substitute these \(x\)-values back into \(y = 2x - 3\): If \(x = 2\), \(y = 2(2) - 3 = 1\). If \(x = 4\), \(y = 2(4) - 3 = 5\). Thus, the solutions are \(x = 2, y = 1\) and \(x = 4, y = 5\).

M1 for substituting \(y = 2x - 3\) into the quadratic equation to form a single variable equation. A1 for simplifying to standard quadratic form \(x^2 - 6x + 8 = 0\) or equivalent. M1 for solving the quadratic equation to obtain \(x = 2\) and \(x = 4\). A1 for obtaining both correct pairs: \(x = 2, y = 1\) and \(x = 4, y = 5\).

(a) Using the Cosine Rule to find \(AC\):
\(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\)
\(AC^2 = 120^2 + 150^2 - 2 \cdot 120 \cdot 150 \cdot \cos(72^\circ)\)
\(AC^2 = 14400 + 22500 - 36000 \cdot \cos(72^\circ)\)
\(AC^2 \approx 36900 - 11124.61 = 25775.39\)
\(AC \approx \sqrt{25775.39} \approx 160.55\text{ m}\).
To 3 significant figures, \(AC = 161\text{ m}\) (or \(160.5\text{ m}\)).

(b) Using the Sine Rule to find angle \(BAC\):
\(\frac{\sin(BAC)}{BC} = \frac{\sin(ABC)}{AC}\)
\(\frac{\sin(BAC)}{150} = \frac{\sin(72^\circ)}{160.55}\)
\(\sin(BAC) = \frac{150 \cdot \sin(72^\circ)}{160.55}\)
\(\sin(BAC) \approx \frac{142.66}{160.55} \approx 0.8886\)
\(BAC = \arcsin(0.8886) \approx 62.69^\circ\).
To 3 significant figures (or 1 decimal place for angles), \(BAC = 62.7^\circ\).

(a)
M1 for substituting correctly into the Cosine Rule: \(120^2 + 150^2 - 2(120)(150)\cos(72^\circ)\)
A1 for \(AC^2 \approx 25775\) (or equivalent)
A1 for \(161\) or \(160.5\) (accept values in range \(160.5\) to \(161\))

(b)
M1 for substituting correctly into the Sine Rule: \(\frac{\sin(BAC)}{150} = \frac{\sin(72^\circ)}{\text{their } AC}\)
A1 for \(\sin(BAC) \approx 0.888\) to \(0.889\)
A1 for \(62.7^\circ\) (accept \(62.6^\circ\) to \(62.8^\circ\))

(a) Let \(O\) be the center of the square base \(ABCD\). Since \(ABCD\) is a square of side \(10\text{ cm}\), the diagonal \(AC\) is:
\(AC = \sqrt{10^2 + 10^2} = \sqrt{200} \approx 14.142\text{ cm}\).
The distance from a corner of the base to the center \(O\) is:
\(OA = \frac{1}{2} AC = \frac{\sqrt{200}}{2} = \sqrt{50} \approx 7.071\text{ cm}\).
Since \(VO\) is the perpendicular height of the pyramid, triangle \(VOA\) is right-angled at \(O\). Using Pythagoras' theorem:
\(VA^2 = VO^2 + OA^2 = 12^2 + (\sqrt{50})^2 = 144 + 50 = 194\).
\(VA = \sqrt{194} \approx 13.928\text{ cm}\).
To 3 significant figures, \(VA = 13.9\text{ cm}\).

(b) The angle between the sloping edge \(VA\) and the base is the angle \(VAO\). In the right-angled triangle \(VOA\):
\(\tan(VAO) = \frac{VO}{OA} = \frac{12}{\sqrt{50}} \approx 1.6971\)
\(VAO = \arctan(1.6971) \approx 59.49^\circ\).
To 3 significant figures, this is \(59.5^\circ\).

(a)
M1 for finding half the diagonal of the base: \(OA = \sqrt{5^2 + 5^2} = \sqrt{50}\) or \(7.07\)
M1 for substituting into Pythagoras' theorem: \(VA^2 = 12^2 + (\text{their } OA)^2\)
A1 for \(13.9\) or \(13.93\) (accept \(13.9\) to \(14.0\))

(b)
M1 for using a correct trigonometric ratio to find angle \(VAO\), e.g., \(\tan(\theta) = \frac{12}{\text{their } OA}\)
A1 for identifying the correct angle \(VAO\) or substituting values correctly
A1 for \(59.5^\circ\) (accept \(59.4^\circ\) to \(59.6^\circ\))

(a) Using the Cosine Rule on the isosceles triangle \(OAB\) where \(OA = OB = 12\text{ cm}\) and angle \(AOB = 120^\circ\):
\(AB^2 = OA^2 + OB^2 - 2 \cdot OA \cdot OB \cdot \cos(AOB)\)
\(AB^2 = 12^2 + 12^2 - 2 \cdot 12 \cdot 12 \cdot \cos(120^\circ)\)
\(AB^2 = 144 + 144 - 288 \cdot (-0.5)\)
\(AB^2 = 288 + 144 = 432\)
\(AB = \sqrt{432} \approx 20.78\text{ cm}\).
To 3 significant figures, \(AB = 20.8\text{ cm}\).

(b) The area of the segment is found by subtracting the area of triangle \(OAB\) from the area of sector \(OAB\).
Area of sector \(OAB = \frac{120}{360} \cdot \pi \cdot 12^2 = \frac{1}{3} \cdot 144\pi = 48\pi \approx 150.80\text{ cm}^2\).
Area of triangle \(OAB = \frac{1}{2} \cdot OA \cdot OB \cdot \sin(120^\circ) = \frac{1}{2} \cdot 12 \cdot 12 \cdot \sin(120^\circ) = 72 \cdot \sin(120^\circ) \approx 62.35\text{ cm}^2\).
Area of segment \( = 150.80 - 62.35 = 88.45\text{ cm}^2\).
To 3 significant figures, the area is \(88.4\text{ cm}^2\).

(a)
M1 for substituting correctly into the Cosine Rule: \(12^2 + 12^2 - 2(12)(12)\cos(120^\circ)\) or using right-angled trig: \(2 \times 12\sin(60^\circ)\)
A1 for \(AB^2 = 432\) or \(AB = 12\sqrt{3}\)
A1 for \(20.8\) (accept \(20.7\) to \(20.8\))

(b)
M1 for calculating the area of the sector: \(\frac{120}{360} \times \pi \times 12^2\) (or \(48\pi\) or \(150.8\))
M1 for calculating the area of the triangle: \(\frac{1}{2} \times 12^2 \times \sin(120^\circ)\) (or \(36\sqrt{3}\) or \(62.4\) or \(62.35\))
A1 for \(88.4\) (accept \(88.4\) to \(88.5\))

**(a) Find the length \(PR\):**

Using the cosine rule:
\[PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(\angle PQR)\]
\[PR^2 = 15^2 + 10^2 - 2(15)(10)\cos(110^\circ)\]
\[PR^2 = 225 + 100 - 300(-0.342020...)\]
\[PR^2 = 325 + 102.606...\]
\[PR^2 = 427.606...\]
\[PR = \sqrt{427.606...} \approx 20.6786...\text{ cm}\]
To 3 significant figures, \(PR = 20.7\text{ cm}\) (or \(20.68\text{ cm}\) to 4 significant figures).

**(b) Find the shortest distance from \(Q\) to \(PR\):**

First, find the area of triangle \(PQR\):
\[\text{Area} = \frac{1}{2} \times PQ \times QR \times \sin(\angle PQR)\]
\[\text{Area} = \frac{1}{2} \times 15 \times 10 \times \sin(110^\circ) = 75 \times 0.93969... \approx 70.4769...\text{ cm}^2\]

The shortest distance from \(Q\) to \(PR\) is the perpendicular height \(h\) with base \(PR\):
\[\text{Area} = \frac{1}{2} \times PR \times h\]
\[70.4769... = \frac{1}{2} \times 20.6786... \times h\]
\[h = \frac{2 \times 70.4769...}{20.6786...} \approx 6.81639...\text{ cm}\]
To 3 significant figures, the shortest distance is \(6.82\text{ cm}\) (or \(6.816\text{ cm}\) to 4 significant figures).

**(a) [3 marks]**
- **M1** for correct substitution into the cosine rule: \(15^2 + 10^2 - 2(15)(10)\cos(110^\circ)\)
- **A1** for \(PR^2 = 427.6\) to \(428\)
- **A1** for \(20.7\) or \(20.68\) (or \(20.67...\))

**(b) [3 marks]**
- **M1** for finding the area of triangle: \(\frac{1}{2} \times 15 \times 10 \times \sin(110^\circ)\) (\(\approx 70.5\))
- **M1** for equating area to \(\frac{1}{2} \times PR \times h\) using their \(PR\) value
- **A1** for \(6.82\) or \(6.816\) (or \(6.81...\))

(a) To find the stationary points, we first differentiate \(f(x)\): \(f'(x) = 6x^2 - 6x - 12\). Setting \(f'(x) = 0\) gives \(6(x^2 - x - 2) = 0\), which factors to \(6(x - 2)(x + 1) = 0\). Thus, the stationary points occur at \(x = 2\) and \(x = -1\). For \(x = -1\), \(f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 8 = -2 - 3 + 12 + 8 = 15\). For \(x = 2\), \(f(2) = 2(2)^3 - 3(2)^2 - 12(2) + 8 = 16 - 12 - 24 + 8 = -12\). Using the second derivative \(f''(x) = 12x - 6\): \(f''(-1) = -18 < 0\), so (-1, 15) is a local maximum. \(f''(2) = 18 > 0\), so (2, -12) is a local minimum. (b) For the horizontal line \(y = k\) to intersect \(y = f(x)\) at three distinct points, \(k\) must lie strictly between the local minimum y-value and the local maximum y-value. Therefore, \(-12 < k < 15\). (c) The curve crosses the y-axis where \(x = 0\). The y-coordinate is \(f(0) = 8\), so the point is (0, 8). The gradient of the tangent at this point is given by \(f'(0) = 6(0)^2 - 6(0) - 12 = -12\). Using the equation of a line \(y - y_1 = m(x - x_1)\), we get \(y - 8 = -12(x - 0)\), which simplifies to \(y = -12x + 8\).

(a) M1 for differentiating \(f(x)\) to get \(6x^2 - 6x - 12\). M1 for setting their derivative to 0 and finding \(x = 2, x = -1\). A1 for y-coordinates 15 and -12. A1 for identifying (-1, 15) as the local maximum. A1 for identifying (2, -12) as the local minimum. (b) M1 for recognizing that \(k\) must be between the local minimum and maximum values. A1 for correct inequality \(-12 < k < 15\). (c) B1 for finding the point (0, 8). M1 for substituting \(x = 0\) into their derivative to get \(m = -12\). A1 for the final equation \(y = -12x + 8\).

(a) For \(g(x) = \frac{4}{x-2} + 3\), the vertical asymptote occurs where the denominator is zero, so \(x - 2 = 0\) which gives \(x = 2\). As \(x\) approaches infinity, the term \(\frac{4}{x-2}\) approaches 0, so \(g(x)\) approaches 3. Thus, the horizontal asymptote is \(y = 3\). (b) Translating by \(\begin{pmatrix} -3 \\ 1 \end{pmatrix}\) means we replace \(x\) with \(x + 3\) and add 1 to the function: \(h(x) = g(x+3) + 1 = \frac{4}{(x+3)-2} + 3 + 1 = \frac{4}{x+1} + 4\). To write this as a single fraction: \(h(x) = \frac{4 + 4(x+1)}{x+1} = \frac{4x + 8}{x+1}\). (c) Stretching vertically by a factor of 2 gives \(2 \cdot g(x) = \frac{8}{x-2} + 6\). Reflecting this in the \(x\)-axis gives \(p(x) = -(2 \cdot g(x)) = -\frac{8}{x-2} - 6\). To express as a single fraction: \(p(x) = \frac{-8 - 6(x-2)}{x-2} = \frac{-8 - 6x + 12}{x-2} = \frac{4 - 6x}{x-2}\).

(a) B1 for vertical asymptote \(x = 2\). B1 for horizontal asymptote \(y = 3\). (b) M1 for substituting \(x + 3\) into the function. M1 for adding 1 to the function. M1 for finding a common denominator. A1 for the correct simplified fraction \(\frac{4x + 8}{x + 1}\). (c) M1 for multiplying the original function by 2. M1 for multiplying by -1 (reflecting). M1 for combining into a single fraction. A1 for the correct simplified fraction \(\frac{4 - 6x}{x - 2}\).

(a) Substitute the given points into \(f(x) = ax^2 + bx + c\): Point (-1, -6): \(a - b + c = -6\) (Equation 1). Point (2, 9): \(4a + 2b + c = 9\) (Equation 2). Point (4, 29): \(16a + 4b + c = 29\) (Equation 3). Subtract Equation 1 from Equation 2: \(3a + 3b = 15 \implies a + b = 5 \implies b = 5 - a\). Subtract Equation 2 from Equation 3: \(12a + 2b = 20 \implies 6a + b = 10\). Substitute \(b = 5 - a\) into \(6a + b = 10\): \(6a + 5 - a = 10 \implies 5a = 5 \implies a = 1\). Then \(b = 5 - 1 = 4\). Substitute \(a = 1\) and \(b = 4\) into Equation 1: \(1 - 4 + c = -6 \implies c = -3\). This confirms \(a = 1\), \(b = 4\), \(c = -3\). (b) Complete the square for \(f(x) = x^2 + 4x - 3\): \(f(x) = (x + 2)^2 - 2^2 - 3 = (x + 2)^2 - 7\). The vertex is at \((-h, k)\), which gives the coordinates (-2, -7). (c) Set up the inequality \(f(x) \ge 9\): \(x^2 + 4x - 3 \ge 9 \implies x^2 + 4x - 12 \ge 0\). Factorize the quadratic expression: \((x + 6)(x - 2) \ge 0\). The critical values are \(x = -6\) and \(x = 2\). Since the quadratic opens upwards, the solution is \(x \le -6\) or \(x \ge 2\).

(a) M1 for setting up the three equations. M1 for a valid method to eliminate one variable. A1 for finding one correct value. A1 for finding the second correct value. A1 for verifying the third value. (b) M1 for completing the square to get \((x + 2)^2 - 7\). A1 for writing down the correct vertex coordinates (-2, -7). (c) M1 for setting up the inequality \(x^2 + 4x - 12 \ge 0\). M1 for finding critical values \(x = -6\) and \(x = 2\). A1 for the correct solution range \(x \le -6\) or \(x \ge 2\).

(a) The probability of drawing a red ball first is \(\frac{r}{12}\). Since drawing is without replacement, the probability of drawing a second red ball is \(\frac{r-1}{11}\). Thus, \(P(\text{both red}) = \frac{r}{12} \times \frac{r-1}{11} = \frac{r(r-1)}{132}\).

(b) Since \(P(\text{both red}) = \frac{14}{33}\), we set up the equation:
\(\frac{r(r-1)}{132} = \frac{14}{33}\)
Multiply both sides by 132:
\(r(r-1) = \frac{14}{33} \times 132\)
\(r^2 - r = 14 \times 4\)
\(r^2 - r = 56\)
\(r^2 - r - 56 = 0\) (as required).

(c) Factorise the quadratic equation:
\(r^2 - r - 56 = 0 \implies (r - 8)(r + 7) = 0\)
This gives \(r = 8\) or \(r = -7\). Since the number of balls must be positive, \(r = 8\).
The number of yellow balls is \(12 - r = 12 - 8 = 4\).

(d) There are now 8 red balls and 4 yellow balls.
(i) All three balls are the same color (either 3 red or 3 yellow):
\(P(\text{3 Red}) = \frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} = \frac{336}{1320} = \frac{14}{55}\)
\(P(\text{3 Yellow}) = \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} = \frac{24}{1320} = \frac{1}{55}\)
Total probability = \(\frac{14}{55} + \frac{1}{55} = \frac{15}{55} = \frac{3}{11}\) (or approximately 0.273).

(ii) At least one ball of each color is drawn:
This is the complement of drawing three balls of the same color.
\(P(\text{at least one of each}) = 1 - P(\text{all same color}) = 1 - \frac{3}{11} = \frac{8}{11}\) (or approximately 0.727).

(a) M1 for \(\frac{r}{12} \times \frac{r-1}{11}\) or equivalent. A1 for \(\frac{r(r-1)}{132}\) or \(\frac{r^2-r}{132}\).

(b) M1 for setting up \(\frac{r(r-1)}{132} = \frac{14}{33}\). M1 for \(r(r-1) = 56\) or \(r^2 - r = 56\). A1 for reaching the final equation \(r^2 - r - 56 = 0\) with no errors seen.

(c) M1 for factoring as \((r-8)(r+7)\) or applying the quadratic formula. A1 for \(r = 8\) (explicitly discarding \(r = -7\)). A1 for 4 yellow balls.

(d)(i) M1 for \(\frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} + \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10}\) or equivalent sum of products. A1 for \(\frac{3}{11}\) (or exact equivalent fraction, or 0.273 or 27.3%).

(d)(ii) M1 for \(1 - \text{their (d)(i)}\) or for summing \(3 \times \frac{8}{12} \times \frac{7}{11} \times \frac{4}{10} + 3 \times \frac{8}{12} \times \frac{4}{11} \times \frac{3}{10}\). A1 for \(\frac{8}{11}\) (or exact equivalent fraction, or 0.727 or 72.7%).

(a) The total number of students is 80.
\(6 + 14 + x + y + 10 = 80\)
\(30 + x + y = 80\)
\(x + y = 50\) (as required).

(b)(i) The mid-interval values for the classes are 10, 30, 50, 70, and 90.
The estimated mean is the sum of the products of mid-interval values and frequencies, divided by the total frequency:
\(\frac{10(6) + 30(14) + 50x + 70y + 90(10)}{80} = 55\)
\(\frac{60 + 420 + 50x + 70y + 900}{80} = 55\)

(b)(ii) Multiply both sides by 80:
\(1380 + 50x + 70y = 4400\)
\(50x + 70y = 4400 - 1380\)
\(50x + 70y = 3020\)
Divide by 10:
\(5x + 7y = 302\) (as required).

(b)(iii) Solve the simultaneous equations:
1) \(x + y = 50 \implies 5x + 5y = 250\)
2) \(5x + 7y = 302\)
Subtract equation 1 from equation 2:
\(2y = 52 \implies y = 26\)
Substitute \(y = 26\) into \(x + y = 50\):
\(x + 26 = 50 \implies x = 24\).

(c)(i) The median is the 40th value.
Cumulative frequencies:
\(\le 20\): 6
\(\le 40\): \(6 + 14 = 20\)
\(\le 60\): \(20 + 24 = 44\)
Since the 40th value lies between the 20th and 44th values, the median is in the class interval \(40 < m \le 60\).

(c)(ii) Students scoring more than 60 marks are in the intervals \(60 < m \le 80\) and \(80 < m \le 100\).
Number of such students = \(y + 10 = 26 + 10 = 36\).
Two students chosen without replacement:
\(P(\text{both } > 60) = \frac{36}{80} \times \frac{35}{79} = \frac{9}{20} \times \frac{35}{79} = \frac{315}{1580} = \frac{63}{316}\) (or approximately 0.199).

(a) B1 for writing down \(6 + 14 + x + y + 10 = 80\) and showing the simplification to \(x + y = 50\).

(b)(i) M1 for using mid-interval values (10, 30, 50, 70, 90) to find the sum of products. A1 for \(\frac{60 + 420 + 50x + 70y + 900}{80} = 55\) or equivalent.

(b)(ii) M1 for multiplying by 80 and collecting constants: \(50x + 70y = 3020\). A1 for completing the division by 10 to get the given equation \(5x + 7y = 302\).

(b)(iii) M1 for a valid method to solve simultaneous equations (substitution or elimination). A1 for \(y = 26\). A1 for \(x = 24\).

(c)(i) B1 for \(40 < m \le 60\) (or \(40 - 60\)).

(c)(ii) B1 for finding the total number of students scoring > 60 is 36. M1 for multiplying \(\frac{36}{80} \times \frac{35}{79}\) (or \(\frac{\text{their } 36}{80} \times \frac{\text{their } 35}{79}\)). A1 for \(\frac{63}{316}\) (or equivalent fraction, or 0.199 or 19.9%).