2025 HKDSE Mathematics Answers & Marking Scheme

By the remainder theorem, \(f(1) = -6\) and \(f(-\frac{1}{2}) = -9\).

Since \(f(1) = -6\), we have:
\(2(1)^3 + a(1)^2 + b(1) - 6 = -6\)
\(2 + a + b - 6 = -6\)
\(a + b = -2\) --- (1)

Since \(f(-\frac{1}{2}) = -9\), we have:
\(2(-\frac{1}{2})^3 + a(-\frac{1}{2})^2 + b(-\frac{1}{2}) - 6 = -9\)
\(-\frac{1}{4} + \frac{a}{4} - \frac{b}{2} - 6 = -9\)
Multiply the equation by 4:
\(-1 + a - 2b - 24 = -36\)
\(a - 2b = -11\) --- (2)

Subtracting (2) from (1):
\((a + b) - (a - 2b) = -2 - (-11)\)
\(3b = 9\)
\(b = 3\)

Substitute \(b = 3\) into (1):
\(a + 3 = -2\)
\(a = -5\)

Therefore, the values are \(a = -5\) and \(b = 3\).

- For using remainder theorem to set up equations (1) and (2) [1M]
- For obtaining \(a+b = -2\) and \(a-2b = -11\) (or equivalent) [1A]
- For solving the simultaneous equations [1M]
- For obtaining \(a = -5\) and \(b = 3\) [1A] (Award 1A for both correct answers)

(a) Since \(x-2\) is a factor of \(g(x)\), by the factor theorem, we have \(g(2) = 0\).
\(2(2)^3 + k(2)^2 - 13(2) + 6 = 0\)
\(16 + 4k - 26 + 6 = 0\)
\(4k - 4 = 0\)
\(k = 1\)

(b) From (a), \(g(x) = 2x^3 + x^2 - 13x + 6\).
By long division or comparing coefficients, we have:
\(g(x) = (x-2)(2x^2 + 5x - 3)\)
Factorizing the quadratic part:
\(2x^2 + 5x - 3 = (2x-1)(x+3)\)
Therefore, \(g(x) = (x-2)(2x-1)(x+3)\).

(a)
- For substituting \(x=2\) into \(g(x)\) and setting it to 0 [1M]
- For obtaining \(k = 1\) [1A]

(b)
- For finding the quadratic factor \(2x^2 + 5x - 3\) [1M]
- For completely factorizing into \((x-2)(2x-1)(x+3)\) [1A]

The original graph is \(y = f(x) = x^2 - 4x - 5\).

When the graph is translated horizontally to the right by 3 units, we replace \(x\) by \(x-3\):
\(g(x) = f(x-3) = (x-3)^2 - 4(x-3) - 5\)
\(g(x) = (x^2 - 6x + 9) - 4x + 12 - 5\)
\(g(x) = x^2 - 10x + 16\)

When the graph of \(y = g(x)\) is reflected with respect to the \(x\)-axis, we have:
\(h(x) = -g(x)\)
\(h(x) = -(x^2 - 10x + 16)\)
\(h(x) = -x^2 + 10x - 16\)

Therefore, the equation of the graph \(y = h(x)\) is \(y = -x^2 + 10x - 16\).

- For substituting \(x-3\) into the equation to represent translation [1M]
- For obtaining the equation \(y = x^2 - 10x + 16\) (or equivalent) [1A]
- For multiplying the whole function by \(-1\) to represent reflection across the x-axis [1M]
- For obtaining the final equation \(y = -x^2 + 10x - 16\) [1A]

(a) Using the method of completing the square:
\(f(x) = -2(x^2 - 6x) - 10\)
\(f(x) = -2(x^2 - 6x + 9 - 9) - 10\)
\(f(x) = -2(x-3)^2 + 18 - 10\)
\(f(x) = -2(x-3)^2 + 8\)
Therefore, the coordinates of the vertex of the graph are \((3, 8)\).

(b) Since the vertex of the parabola is a maximum point with coordinates \((3, 8)\) and the coefficient of \(x^2\) is negative, the graph opens downwards.
For the horizontal line \(y = k\) to intersect the graph at two distinct points, the line must lie strictly below the maximum point.
Therefore, the range of values of \(k\) is \(k < 8\).

(Alternative Method for (b)):
For \(y = k\) to intersect \(y = -2x^2 + 12x - 10\) at two distinct points, the equation \(-2x^2 + 12x - 10 = k \implies 2x^2 - 12x + (10+k) = 0\) must have two distinct real roots.
Thus, the discriminant \(\Delta > 0\):
\((-12)^2 - 4(2)(10+k) > 0\)
\(144 - 8(10+k) > 0\)
\(18 - (10+k) > 0\)
\(8 - k > 0\)
\(k < 8\)
Therefore, the range of values is \(k < 8\).

(a)
- For completing the square or using vertex formula \(x = -b/(2a)\) [1M]
- For obtaining the vertex \((3, 8)\) [1A]

(b)
- For using discriminant \(\Delta > 0\) or considering the vertex's y-coordinate [1M]
- For obtaining \(k < 8\) [1A]

(a) The given equation is \(x^2 + y^2 - 6x + 8y + 9 = 0\).
The center of \(C\) is \(\left(-\frac{-6}{2}, -\frac{8}{2}\right) = (3, -4)\).
The radius of \(C\) is \(\sqrt{3^2 + (-4)^2 - 9} = \sqrt{9 + 16 - 9} = \sqrt{16} = 4\).

(b) Let \(P(3, -4)\) be the center of \(C\). The distance from the origin \(O(0,0)\) to \(P\) is:
\(OP = \sqrt{(3-0)^2 + (-4-0)^2} = \sqrt{9 + 16} = 5\).

Let \(T\) be the point of contact of the tangent from \(O\) to \(C\). Since \(PT\) is the radius of \(C\), \(PT = 4\), and \(\angle PTO = 90^\circ\).
By Pythagoras' theorem in \(\triangle PTO\):
\(OT^2 + PT^2 = OP^2\)
\(OT^2 + 4^2 = 5^2\)
\(OT^2 = 25 - 16 = 9\)
\(OT = 3\).

Therefore, the length of the tangent from \(O\) to \(C\) is 3.

(a)
- For finding the center \((3, -4)\) [1A]
- For finding the radius \(4\) [1A]

(b)
- For finding the distance from the origin to the center \(OP = 5\) (or applying the direct formula for tangent length) [1M]
- For obtaining the length of tangent as \(3\) [1A]

(a) The radius \(r\) of circle \(C\) is the distance between \(P(-3, 2)\) and \(Q(1, 5)\):
\(r = \sqrt{(1 - (-3))^2 + (5 - 2)^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5\).
Therefore, the equation of \(C\) is:
\((x+3)^2 + (y-2)^2 = 5^2\)
\((x+3)^2 + (y-2)^2 = 25\) (or \(x^2 + y^2 + 6x - 4y - 12 = 0\)).

(b) The distance between the center \(P(-3, 2)\) and the point \(R(-1, -1)\) is:
\(PR = \sqrt{(-1 - (-3))^2 + (-1 - 2)^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}\).
Since \(PR = \sqrt{13} \approx 3.61 < 5\), the distance from \(R\) to the center is less than the radius of \(C\).
Therefore, the point \(R\) lies inside the circle \(C\).

(a)
- For finding the radius \(r = 5\) [1M]
- For obtaining the equation of the circle \((x+3)^2 + (y-2)^2 = 25\) (or equivalent) [1A]

(b)
- For calculating the distance \(PR = \sqrt{13}\) (or comparing \(PR^2 = 13\) with \(r^2 = 25\)) [1M]
- For concluding that \(R\) lies inside the circle with a valid comparison [1A]

The sum of the 7 known numbers is:
\(11 + 13 + 15 + 15 + 17 + 19 + 21 = 111\).
Since the mean of the 9 numbers is 17, the sum of all 9 numbers is:
\(17 \times 9 = 153\).
Thus, we have:
\(111 + x + y = 153\)
\(x + y = 42\) --- (1)

The known minimum of the given 7 numbers is 11, and the known maximum is 21.
We are given that the range is 12.
Case 1: If \(x\) is the minimum and \(y\) is the maximum of the entire set:
Then the range is \(y - x = 12\).
Combining with (1):
\(x + y = 42\)
\(y - x = 12\)
Adding them gives \(2y = 54 \implies y = 27\), and then \(x = 15\).
However, if \(x = 15\) and \(y = 27\), the minimum of the entire set is actually 11 (since \(11 < 15\)), so the range would be \(27 - 11 = 16 \neq 12\). This is a contradiction.

Case 2: If the minimum of the set is 11, and \(y\) is the maximum of the set:
Then the range is \(y - 11 = 12 \implies y = 23\).
Substitute \(y = 23\) into (1):
\(x + 23 = 42 \implies x = 19\).
The sorted set is: \(11, 13, 15, 15, 17, 19, 19, 21, 23\).
The minimum is 11, the maximum is 23, and the range is \(23 - 11 = 12\), which is correct.
Since \(x = 19 \le y = 23\), this solution is valid.

Case 3: If \(x\) is the minimum of the set, and the maximum is 21:
Then the range is \(21 - x = 12 \implies x = 9\).
Substitute \(x = 9\) into (1):
\(9 + y = 42 \implies y = 33\).
But if \(y = 33\), the maximum of the set is 33, and the range becomes \(33 - 9 = 24 \neq 12\). This is a contradiction.

Therefore, the only valid solution is \(x = 19\) and \(y = 23\).

- For using the mean to set up the equation \(x + y = 42\) [1M]
- For considering possible cases of range [1M]
- For finding \(y = 23\) [1A]
- For finding \(x = 19\) [1A]

(a) Since the mean of the 8 data is 8, the sum of all data is \(8 \times 8 = 64\).
\(4 + 6 + 7 + 8 + 9 + 10 + 11 + a = 64\)
\(55 + a = 64\)
\(a = 9\).

(b) The 8 data are: \(4, 6, 7, 8, 9, 9, 10, 11\).
Using the formula for variance \(\sigma^2 = \frac{1}{N}\sum(x_i - \mu)^2\):
\(\sum(x_i - 8)^2 = (4-8)^2 + (6-8)^2 + (7-8)^2 + (8-8)^2 + (9-8)^2 + (9-8)^2 + (10-8)^2 + (11-8)^2\)
\(= (-4)^2 + (-2)^2 + (-1)^2 + 0^2 + 1^2 + 1^2 + 2^2 + 3^2\)
\(= 16 + 4 + 1 + 0 + 1 + 1 + 4 + 9 = 36\).

Variance \(\sigma^2 = \frac{36}{8} = 4.5\).
Standard deviation \(\sigma = \sqrt{4.5} \approx 2.1213\).

Therefore, the standard deviation is \(2.12\) (correct to 2 decimal places).

(a)
- For setting up the equation using mean: \(4 + 6 + 7 + 8 + 9 + 10 + 11 + a = 64\) [1M]
- For obtaining \(a = 9\) [1A]

(b)
- For calculating the sum of squares of differences (36) or applying standard deviation formula/using calculator [1M]
- For obtaining the standard deviation \(2.12\) [1A] (Accept \(2.12\) only)

(a) Let \(a\) be the first term and \(d\) be the common difference of the arithmetic sequence.
Using the formula for the general term \(T(n) = a + (n-1)d\):
\(T(3) = a + 2d = 13\) --- (1)
\(T(8) = a + 7d = 33\) --- (2)

Subtracting (1) from (2):
\((a + 7d) - (a + 2d) = 33 - 13\)
\(5d = 20\)
\(d = 4\)

Substitute \(d = 4\) into (1):
\(a + 2(4) = 13\)
\(a = 5\)
Therefore, the first term of the sequence is 5 and the common difference is 4.

(b) The sum of the first \(n\) terms is given by \(S(n) = \frac{n}{2}[2a + (n-1)d]\).
For \(n = 20\):
\(S(20) = \frac{20}{2}[2(5) + (20-1)(4)]\)
\(S(20) = 10[10 + 19(4)]\)
\(S(20) = 10[10 + 76]\)
\(S(20) = 10 \times 86 = 860\).

Therefore, the sum of the first 20 terms of the sequence is 860.

(a)
- For setting up the system of linear equations [1M]
- For finding \(a = 5\) and \(d = 4\) [1A] (Award 1A for both correct answers)

(b)
- For using the summation formula of arithmetic sequence [1M]
- For obtaining \(860\) [1A]

(a) Radius of \(C\) is \(r = GP = \sqrt{(3-0)^2 + (4-0)^2} = 5\).
Equation of \(C\): \((x-3)^2 + (y-4)^2 = 5^2\), which is \((x-3)^2 + (y-4)^2 = 25\) (or \(x^2 + y^2 - 6x - 8y = 0\)).
Slope of \(GP = \frac{4-0}{3-0} = \frac{4}{3}\).
Since \(L\) is tangent to \(C\) at \(P\), \(L \perp GP\).
Slope of \(L = -\frac{1}{4/3} = -\frac{3}{4}\).
Equation of \(L\): \(y - 0 = -\frac{3}{4}(x - 0)\), which is \(3x + 4y = 0\).

(b) The center of \(C'\) is \(G'(3-d, 4)\).
Since \(C'\) is tangent to \(L\), the perpendicular distance from \(G'\) to \(L\) is equal to its radius \(5\).
\(\frac{|3(3-d) + 4(4)|}{\sqrt{3^2 + 4^2}} = 5\)
\(\frac{|25 - 3d|}{5} = 5\)
\(|25 - 3d| = 25\)
Since \(d > 0\), we have \(25 - 3d = -25\)
\(3d = 50\)
\(d = \frac{50}{3}\)

(a)
- Finding the radius of \(C\) (1M)
- Finding the equation of \(C\) (1A)
- Finding the slope of \(L\) (1M)
- Finding the equation of \(L\) (1A)
(b)
- Writing the coordinates of \(G'\) as \((3-d, 4)\) (1M)
- Setting up the distance equation equal to 5 (1M)
- Finding \(d = 50/3\) (rejecting \(d=0\)) (1A)

(a) Since \(x-2\) is a factor of \(f(x)\), by Factor Theorem:
\(f(2) = 0\)
\(2(2)^3 + a(2)^2 + b(2) - 12 = 0\)
\(16 + 4a + 2b - 12 = 0\)
\(4a + 2b = -4 \Rightarrow 2a + b = -2\) --- (1)

Since \(f(x)\) divided by \(x+1\) yields a remainder of \(-15\), by Remainder Theorem:
\(f(-1) = -15\)
\(2(-1)^3 + a(-1)^2 + b(-1) - 12 = -15\)
\(-2 + a - b - 12 = -15\)
\(a - b = -1\) --- (2)

Solving (1) and (2) simultaneously:
Adding (1) and (2):
\(3a = -3 \Rightarrow a = -1\)
Substitute \(a = -1\) into (2):
\(-1 - b = -1 \Rightarrow b = 0\).
So, \(a = -1\) and \(b = 0\).

(b) With \(a = -1\) and \(b = 0\), we have \(f(x) = 2x^3 - x^2 - 12\).
Since \(x-2\) is a factor of \(f(x)\), we perform algebraic division:
\(2x^3 - x^2 - 12 = (x-2)(2x^2 + 3x + 6)\).
To find the roots of \(f(x) = 0\), we solve:
\((x-2)(2x^2 + 3x + 6) = 0\)
So \(x = 2\) or \(2x^2 + 3x + 6 = 0\).
For the quadratic equation \(2x^2 + 3x + 6 = 0\):
Discriminant \(\Delta = 3^2 - 4(2)(6) = 9 - 48 = -39 < 0\).
Since \(\Delta < 0\), the equation \(2x^2 + 3x + 6 = 0\) has no real roots.
Therefore, not all roots of the equation \(f(x) = 0\) are real numbers.
The claim is disagreed.

(a)
- Using \(f(2) = 0\) to establish equation (1) (1M)
- Using \(f(-1) = -15\) to establish equation (2) (1M)
- Solving for \(a = -1\) and \(b = 0\) (1A)
(b)
- Performing division to get \((x-2)(2x^2 + 3x + 6)\) (1M for division method, 1A for the quotient)
- Calculating discriminant \(\Delta = -39\) (1M)
- Concluding that not all roots are real and explaining why (1A)

(a) The 15 data points are: 22, 25, 25, 28, 30, 33, 33, 35, 37, 38, 42, 44, 44, 50, 54.
Mean \(= \frac{22+25+25+28+30+33+33+35+37+38+42+44+44+50+54}{15} = \frac{540}{15} = 36\).
Median \(= \text{the 8th datum} = 35\).
Lower quartile \(Q_1 = \text{the 4th datum} = 28\).
Upper quartile \(Q_3 = \text{the 12th datum} = 44\).
Interquartile range (IQR) \(= Q_3 - Q_1 = 44 - 28 = 16\).

(b) (i) Since the mean of the 17 employees is still 36:
New total sum of hourly wages \(= 36 \times 17 = 612\).
Original sum of hourly wages \(= 540\).
Therefore, \(W_1 + W_2 = 612 - 540 = 72\).

(ii) Original range \(= 54 - 22 = 32\).
New range \(= 32 + 6 = 38\).
To keep the median at 35, we must add one datum \(W_1 \le 35\) and one datum \(W_2 \ge 35\).
To make the range 38, we have three cases:
- Case 1: Only the minimum changes. New minimum \(= 22 - 6 = 16\). So \(W_1 = 16\). For the range to be 38, the maximum remains 54, so \(W_2 \le 54\). Thus, any pair with \(W_1 = 16\) and \(35 \le W_2 \le 54\) works, e.g., \(W_1 = 16, W_2 = 40\).
- Case 2: Only the maximum changes. New maximum \(= 54 + 6 = 60\). So \(W_2 = 60\). For the range to be 38, the minimum remains 22, so \(W_1 \ge 22\). Thus, any pair with \(W_2 = 60\) and \(22 \le W_1 \le 35\) works, e.g., \(W_1 = 25, W_2 = 60\).
- Case 3: Both minimum and maximum change. Here, \(W_2 - W_1 = 38\) with \(W_1 < 22\) and \(W_2 > 54\), e.g., \(W_1 = 18, W_2 = 56\).
(Any one valid pair of values is accepted).

(a)
- Mean = 36 (1A)
- Median = 35 (1A)
- IQR = 16 (1A)
(b)
- (i) Finding total sum of 17 employees' wages is 612 (1M)
- (i) Finding \(W_1 + W_2 = 72\) (1A)
- (ii) Analyzing conditions for median and range (1M)
- (ii) Writing down any one valid pair of \(W_1\) and \(W_2\) (1A)

(a) The 1st term, the 3rd term, and the 9th term of \(A\) are:
\(T_1 = a\)
\(T_3 = a + 2d\)
\(T_9 = a + 8d\)
Since they form a geometric sequence in that order, we have:
\((a + 2d)^2 = a(a + 8d)\)
\(a^2 + 4ad + 4d^2 = a^2 + 8ad\)
\(4d^2 = 4ad\)
Since \(d \neq 0\), dividing both sides by \(4d\) yields:
\(d = a\), which is \(a = d\).

(b) (i) The sum of the first 10 terms of \(A\) is 110:
\(S_{10} = \frac{10}{2}[2a + 9d] = 110\)
\(5[2a + 9d] = 110\)
\(2a + 9d = 22\)
Since \(a = d\), we substitute \(d = a\) into the equation:
\(2a + 9a = 22 \Rightarrow 11a = 22 \Rightarrow a = 2\).
Thus, \(d = 2\).

The first term of \(G\) is \(T_1 = a = 2\).
The second term of \(G\) is \(T_3 = a + 2d = 2 + 4 = 6\).
The common ratio of \(G\) is \(r = \frac{6}{2} = 3\).
So, the first term of \(G\) is 2 and the common ratio is 3.

(ii) The sum of the first 8 terms of \(G\) is:
\(S_8 = \frac{2(3^8 - 1)}{3 - 1} = 3^8 - 1 = 6561 - 1 = 6560\).

(a)
- Expressing terms in \(a\) and \(d\) (1M)
- Setting up the geometric progression equation (1M)
- Showing \(a = d\) with a clear explanation of dividing by \(d \neq 0\) (1A)
(b)
- (i) Setting up the arithmetic sum equation and finding \(a = 2, d = 2\) (1M)
- (i) Finding the first term of \(G\) is 2 and the common ratio is 3 (1A)
- (ii) Setting up the sum formula for \(G\) (1M)
- (ii) Finding the sum \(6560\) (1A)

(a) Completing the square:
\(f(x) = x^2 - 4x + 4 - 4 + 3\)
\(f(x) = (x-2)^2 - 1\)
So, the coordinates of the vertex of \(U\) are \((2, -1)\).

(b) (i) First, translating \(U\) vertically downwards by 5 units gives the graph:
\(y = f(x) - 5 = x^2 - 4x + 3 - 5 = x^2 - 4x - 2\).
Next, reflecting the resulting graph with respect to the \(x\)-axis gives the graph \(V\):
\(y = -(x^2 - 4x - 2) = -x^2 + 4x + 2\).
So, the equation of \(V\) is \(y = -x^2 + 4x + 2\).

(ii) Let's find the vertex \(W\) of \(V\):
We can transform the vertex of \(U\), which is \((2, -1)\):
- After translating downwards by 5 units, it becomes \((2, -6)\).
- After reflecting with respect to the \(x\)-axis, it becomes \((2, 6)\).
So, the vertex \(W\) of \(V\) is \((2, 6)\).
Alternatively, we can find it by rewriting \(V\): \(y = -(x^2 - 4x) + 2 = -((x-2)^2 - 4) + 2 = -(x-2)^2 + 6\), which gives vertex \((2, 6)\).

To find the \(x\)-intercepts of \(V\), set \(y = 0\):
\(-x^2 + 4x + 2 = 0\)
\(x^2 - 4x - 2 = 0\)
By quadratic formula:
\(x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-2)}}{2} = \frac{4 \pm \sqrt{16+8}}{2} = \frac{4 \pm \sqrt{24}}{2} = 2 \pm \sqrt{6}\).
So the coordinates of \(A\) and \(B\) are \((2 - \sqrt{6}, 0)\) and \((2 + \sqrt{6}, 0)\).
Distance \(AB = (2 + \sqrt{6}) - (2 - \sqrt{6}) = 2\sqrt{6}\).

The height of triangle \(WAB\) with base \(AB\) on the \(x\)-axis is the \(y\)-coordinate of \(W\), which is \(6\).
Area of triangle \(WAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2\sqrt{6} \times 6 = 6\sqrt{6}\).

(a)
- Completing the square correctly to vertex form (1M)
- Finding the vertex \((2, -1)\) (1A)
(b)
- (i) Writing the translated equation \(y = x^2 - 4x - 2\) (1M)
- (i) Writing the final reflected equation \(y = -x^2 + 4x + 2\) (1A)
- (ii) Finding the coordinates of \(W(2, 6)\) (1M)
- (ii) Setting \(y=0\) and solving for \(x\) to find \(AB = 2\sqrt{6}\) (1M)
- (ii) Finding the area of triangle \(WAB\) as \(6\sqrt{6}\) (1A)

(a) Grouping terms: \((x - 2)^2 - 4 + (y - 6)^2 - 36 + 31 = 0\) \(\implies (x - 2)^2 + (y - 6)^2 = 9\). Hence, center \(G = (2, 6)\) and radius \(r = 3\). (b)(i) The equation of \(L\) is \(y - 1 = m(x - 2) \implies mx - y + (1 - 2m) = 0\). Since \(L\) is tangent to \(C\), the distance from \(G(2, 6)\) to \(L\) is equal to \(r\): \(\frac{|2m - 6 + 1 - 2m|}{\sqrt{m^2+1}} = 3 \implies \frac{5}{\sqrt{m^2+1}} = 3 \implies 25 = 9(m^2 + 1) \implies 9m^2 = 16 \implies m = \pm \frac{4}{3}\). (b)(ii) The distance between \(P(2,1)\) and \(G(2,6)\) is \(5\). Since \(\angle PAG = \angle PBG = 90^\circ\), the tangents \(PA\) and \(PB\) have length \(PA = PB = \sqrt{PG^2 - r^2} = \sqrt{5^2 - 3^2} = 4\). The area of the quadrilateral \(PAGB\) is \(2 \times \text{Area of } \triangle PAG = 2 \times \left(\frac{1}{2} \times PA \times GA\right) = 4 \times 3 = 12\).

(a) 1M for completing square or using formulas; 1A for both correct center and radius. (b)(i) 1M for equation of tangent line; 1M for setting distance = radius; 1A for both values of m. (b)(ii) 1M for finding tangent length PA = 4; 1A for correct area = 12.

(a) Sum of scores = \(10 \times 64 = 640\). Since \(\sigma^2 = \frac{\sum x_i^2}{10} - \bar{x}^2\), we have \(8^2 = \frac{\sum x_i^2}{10} - 64^2 \implies 64 = \frac{\sum x_i^2}{10} - 4096 \implies \sum x_i^2 = 41600\). (b)(i) The new mean is still 64, so \(\frac{640 + x + y}{12} = 64 \implies 640 + x + y = 768 \implies x + y = 128\). (b)(ii) Since the standard deviation of the 12 students is 8, the new variance is 64: \(\frac{41600 + x^2 + y^2}{12} - 64^2 = 8^2 \implies \frac{41600 + x^2 + y^2}{12} - 4096 = 64 \implies 41600 + x^2 + y^2 = 49920 \implies x^2 + y^2 = 8320\). Substitute \(y = 128 - x\): \(x^2 + (128 - x)^2 = 8320 \implies 2x^2 - 256x + 16384 = 8320 \implies 2x^2 - 256x + 8064 = 0 \implies x^2 - 128x + 4032 = 0 \implies (x - 56)(x - 72) = 0\). Since \(x \le y\), we have \(x = 56\) and \(y = 72\).

(a) 1A for sum = 640; 1A for sum of squares = 41600. (b)(i) 1A for x + y = 128. (b)(ii) 1M for setting up variance equation; 1M for reducing to x^2 + y^2 = 8320; 1M for quadratic equation in x; 1A for both x = 56 and y = 72.

(a) Since the vertex of \(y = f(x)\) is \(V(3, -4)\), we can write \(f(x) = a(x - 3)^2 - 4\). Since it passes through \((0, 5)\), we have \(5 = a(0 - 3)^2 - 4 \implies 9a = 9 \implies a = 1\). Thus, \(f(x) = (x - 3)^2 - 4 = x^2 - 6x + 5\). Therefore, \(a = 1\), \(b = -6\), and \(c = 5\). (b)(i) Under the horizontal translation, the vertex becomes \((3 - h, -4)\). Under the reflection along the x-axis, the vertex becomes \((3 - h, 4)\). Since the vertex lies on the y-axis, its x-coordinate is 0, so \(3 - h = 0 \implies h = 3\). (b)(ii) Since \(h = 3\), we have \(g(x) = -f(x + 3) = -[((x + 3) - 3)^2 - 4] = -(x^2 - 4) = 4 - x^2\). (b)(iii) \(g(x) = k \implies 4 - x^2 = k \implies x^2 = 4 - k \implies x = \pm \sqrt{4 - k}\). The length of \(AB\) is \(2\sqrt{4 - k} = 6 \implies \sqrt{4 - k} = 3 \implies 4 - k = 9 \implies k = -5\).

(a) 1M for setting up vertex form; 1A for finding a = 1; 1A for finding b = -6 and c = 5. (b)(i) 1A for h = 3. (b)(ii) 1A for g(x) = 4 - x^2. (b)(iii) 1M for setting up equation in k; 1A for k = -5.

(a) \(\frac{b_{n+1}}{b_n} = \frac{2^{a_{n+1}}}{2^{a_n}} = 2^{a_{n+1} - a_n} = 2^d\). Since \(2^d\) is a constant (independent of \(n\)), \(b_n\) is a geometric sequence with common ratio \(r = 2^d\). (b)(i) Since \(b_1 b_2 b_3 = 512 \implies b_1 \times b_1 r \times b_1 r^2 = 512 \implies (b_1 r)^3 = 512 \implies b_2 = 8\). Thus, \(b_1 = \frac{8}{r}\) and \(b_3 = 8r\). Substituting these into the second equation: \(\frac{8}{r} + 8 + 8r = 28 \implies 8r^2 - 20r + 8 = 0 \implies 2r^2 - 5r + 2 = 0 \implies (2r - 1)(r - 2) = 0\). So \(r = 2\) or \(r = \frac{1}{2}\). Since \(r = 2^d\), we have \(2^d = 2 \implies d = 1\), or \(2^d = \frac{1}{2} \implies d = -1\). (b)(ii) If \(d > 0\), we have \(d = 1\). Since \(a_2 = \log_2(b_2) = \log_2(8) = 3\), the first term \(a_1 = a_2 - d = 3 - 1 = 2\). The sum of the first \(n\) terms of \(a_n\) is \(S_n = \frac{n}{2}[2a_1 + (n-1)d] = \frac{n}{2}[4 + n - 1] = \frac{n(n+3)}{2}\). We want \(\frac{n(n+3)}{2} > 2000 \implies n^2 + 3n - 4000 > 0\). Solving \(n^2 + 3n - 4000 = 0\) gives \(n \approx \frac{-3 + \sqrt{9 + 16000}}{2} \approx 61.76\). Thus, the minimum integer value of \(n\) is 62.

(a) 1M for showing the ratio of consecutive terms; 1A for stating it is a GS and ratio is 2^d. (b)(i) 1M for finding b_2 = 8; 1A for finding r = 2 or r = 1/2; 1A for d = 1 or d = -1. (b)(ii) 1M for setting up inequality Sn > 2000; 1A for n = 62.

(a) Since \(2x - 3\) is a factor of \(P(x)\), by the factor theorem, we have \(P\left(\frac{3}{2}\right) = 0\). \(2\left(\frac{3}{2}\right)^3 - 5\left(\frac{3}{2}\right)^2 + k\left(\frac{3}{2}\right) - 3 = 0 \implies 2\left(\frac{27}{8}\right) - 5\left(\frac{9}{4}\right) + \frac{3k}{2} - 3 = 0 \implies \frac{27}{4} - \frac{45}{4} + \frac{3k}{2} - 3 = 0 \implies -\frac{18}{4} - 3 + \frac{3k}{2} = 0 \implies -7.5 + 1.5k = 0 \implies k = 5\). (b) Thus, \(P(x) = 2x^3 - 5x^2 + 5x - 3\). By performing algebraic division (or synthetic division) of \(P(x)\) by \((2x - 3)\), we obtain: \(P(x) = (2x - 3)(x^2 - x + 1)\). For the equation \(P(x) = 0\), either \(2x - 3 = 0\) or \(x^2 - x + 1 = 0\). For the quadratic equation \(x^2 - x + 1 = 0\), the discriminant is \(\Delta = (-1)^2 - 4(1)(1) = 1 - 4 = -3\). Since \(\Delta < 0\), the equation \(x^2 - x + 1 = 0\) has no real roots. Thus, the equation \(P(x) = 0\) has only one real root (which is \(\frac{3}{2}\)) and two non-real roots. Therefore, I disagree with the claim.

(a) 1M for substituting 3/2 into P(x) and setting to 0; 1A for k = 5. (b) 1M for division method to find quadratic factor; 1A for getting x^2 - x + 1; 1M for computing the discriminant of the quadratic factor; 1A for identifying no real roots; 1A for concluding disagreement.