2023 HKDSE Mathematics Answers & Marking Scheme

By Factor Theorem, since \( p(x) \) is divisible by \( 3x - 4 \), we have \( p\left(\frac{4}{3}\right) = 0 \). Substituting \( x = \frac{4}{3} \) into \( p(x) \), we get \( 3\left(\frac{4}{3}\right)^3 - h\left(\frac{4}{3}\right)^2 - 5\left(\frac{4}{3}\right) + 12 = 0 \). This simplifies to \( \frac{64}{9} - \frac{16h}{9} - \frac{20}{3} + 12 = 0 \). Multiplying both sides by 9, we have \( 64 - 16h - 60 + 108 = 0 \), which gives \( 112 - 16h = 0 \). Thus, \( h = 7 \).

For setting \( p\left(\frac{4}{3}\right) = 0 \) (or equivalent): 1M; For simplifying to a linear equation in \( h \), e.g., \( 112 - 16h = 0 \): 1M; For obtaining \( h = 7 \): 1A.

Let \( y = k_1 + k_2 x^2 \), where \( k_1 \) and \( k_2 \) are non-zero constants. When \( x = 2 \), \( y = 18 \), so \( 18 = k_1 + 4k_2 \) (Eq. 1). When \( x = 3 \), \( y = 33 \), so \( 33 = k_1 + 9k_2 \) (Eq. 2). Subtracting Eq. 1 from Eq. 2, we get \( 5k_2 = 15 \), which gives \( k_2 = 3 \). Substituting \( k_2 = 3 \) into Eq. 1, we get \( 18 = k_1 + 12 \), which gives \( k_1 = 6 \). Therefore, the relation is \( y = 6 + 3x^2 \). When \( x = 4 \), \( y = 6 + 3(4^2) = 6 + 48 = 54 \).

For writing \( y = k_1 + k_2 x^2 \): 1M; For finding \( k_1 = 6 \) and \( k_2 = 3 \): 1M; For obtaining \( y = 54 \): 1A.

Let the original 7 numbers be \( x_1, x_2, \dots, x_7 \). Since the original mean is 12 and the added number is 12, the new mean \( \mu' \) remains 12. For the original set, the variance is \( 4^2 = 16 \). Using the variance formula, the sum of squared deviations is \( \sum_{i=1}^7 (x_i - 12)^2 = 16 \times 7 = 112 \). When the number 12 is added, the new sum of squared deviations is \( \sum_{i=1}^7 (x_i - 12)^2 + (12 - 12)^2 = 112 + 0 = 112 \). The new variance is the new sum of squared deviations divided by the new total number of data (8): \( \sigma'^2 = \frac{112}{8} = 14 \). Thus, the new standard deviation is \( \sqrt{14} \).

For finding the original sum of squared deviations to be 112 (or equivalent): 1M; For finding the new variance to be 14 (or equivalent): 1M; For obtaining the new standard deviation of \( \sqrt{14} \): 1A.

(a) Since \(2x - 1\) is a factor of \(f(x)\), by the factor theorem, we have:
\(f\left(\frac{1}{2}\right) = 0\)
\(2\left(\frac{1}{2}\right)^3 - k\left(\frac{1}{2}\right)^2 - 13\left(\frac{1}{2}\right) + 6 = 0\)
\ \frac{1}{4} - \frac{k}{4} - \frac{13}{2} + 6 = 0\)
\(1 - k - 26 + 24 = 0\)
\(k = -1\)

(b) Since \(k = -1\), \(f(x) = 2x^3 + x^2 - 13x + 6\).
By dividing \(f(x)\) by \(2x - 1\), we have:
\(f(x) = (2x - 1)(x^2 + x - 6)\)
\(f(x) = (2x - 1)(x + 3)(x - 2)\)

(a)
- For \(f\left(\frac{1}{2}\right) = 0\) (or substituting \(x = \frac{1}{2}\) to solve) [1M]
- \(k = -1\) [1A]
(b)
- For \(f(x) = (2x - 1)(x^2 + x - 6)\) or finding the quadratic factor [1M]
- For \((2x - 1)(x + 3)(x - 2)\) [1A]

(a) Let \(z = a + \frac{b}{y^2}\), where \(a\) and \(b\) are non-zero constants.
Substituting \(y = 2, z = 11\), we have:
\(11 = a + \frac{b}{4} \Rightarrow 4a + b = 44\) --- (1)
Substituting \(y = 4, z = 5\), we have:
\(5 = a + \frac{b}{16} \Rightarrow 16a + b = 80\) --- (2)
Subtracting (1) from (2), we get:
\(12a = 36 \Rightarrow a = 3\)
Substituting \(a = 3\) into (1), we get:
\(4(3) + b = 44 \Rightarrow b = 32\)
Therefore, \(z = 3 + \frac{32}{y^2}\).

(b) When \(y = \frac{1}{2}\):
\(z = 3 + \frac{32}{(1/2)^2} = 3 + \frac{32}{1/4} = 3 + 128 = 131\).

(a)
- For writing \(z = a + \frac{b}{y^2}\) [1M]
- For setting up a system of linear equations [1M]
- \(z = 3 + \frac{32}{y^2}\) [1A]
(b)
- \(z = 131\) [1A]

(a) The coordinates of the center \(M\) of \(C\) are:
\(M = \left(\frac{-2 + 6}{2}, \frac{1 + 7}{2}\right) = (2, 4)\).
The radius \(r\) of \(C\) is:
\(r = \sqrt{(6 - 2)^2 + (7 - 4)^2} = \sqrt{16 + 9} = 5\).
Thus, the equation of \(C\) is:
\((x - 2)^2 + (y - 4)^2 = 5^2\)
\((x - 2)^2 + (y - 4)^2 = 25\) (or \(x^2 + y^2 - 4x - 8y - 5 = 0\))

(b) The slope of \(MB\) is:
\(m = \frac{7 - 4}{6 - 2} = \frac{3}{4}\).
Since the tangent at \(B\) is perpendicular to \(MB\), the slope of the tangent is:
\(m' = -\frac{1}{m} = -\frac{4}{3}\).
The equation of the tangent at \(B(6, 7)\) is:
\(y - 7 = -\frac{4}{3}(x - 6)\)
\(3(y - 7) = -4(x - 6)\)
\(3y - 21 = -4x + 24\)
\(4x + 3y - 45 = 0\)

(a)
- For finding the center \((2, 4)\) or radius \(5\) [1M]
- For \((x - 2)^2 + (y - 4)^2 = 25\) (or equivalent form) [1A]
(b)
- For finding the slope of the tangent \(= -\frac{4}{3}\) [1M]
- For \(4x + 3y - 45 = 0\) (or equivalent form) [1A]

(a) From the stem-and-leaf diagram, the weights sorted in ascending order are:
\(42, 45, 45, 48, 51, 51, 53, 54, 56, 57, 59, 60, 62, 65, 68\)
Median: Since there are 15 data, the 8th data is the median.
Median = \(54\) kg
Range = \(68 - 42 = 26\) kg
Lower quartile \(Q_1 = 48\) kg (the 4th data)
Upper quartile \(Q_3 = 60\) kg (the 12th data)
Interquartile range = \(Q_3 - Q_1 = 60 - 48 = 12\) kg

(b) If the student of weight \(53\) kg leaves, the remaining 14 weights in ascending order are:
\(42, 45, 45, 48, 51, 51, 54, 56, 57, 59, 60, 62, 65, 68\)
New median = \(\frac{54 + 56}{2} = 55\) kg
The change in the median is \(55 - 54 = 1\) kg (or the median increases by \(1\) kg).

(a)
- For median = \(54\) kg [1A]
- For range = \(26\) kg [1A]
- For interquartile range = \(12\) kg [1A]
(b)
- For the change = \(1\) kg (or increased by \(1\) kg) [1A]

(a) Since \( x - 2 \) is a factor of \( f(x) \), by the factor theorem, we have:
\( f(2) = 0 \)
\( 2(2)^3 + k(2)^2 - 13(2) - 6 = 0 \)
\( 16 + 4k - 26 - 6 = 0 \)
\( 4k - 16 = 0 \)
\( k = 4 \)
(b) With \( k = 4 \), \( f(x) = 2x^3 + 4x^2 - 13x - 6 \).
By performing polynomial division, we find:
\( f(x) = (x - 2)(2x^2 + 8x + 3) \)
To solve \( f(x) = 0 \):
\( (x - 2)(2x^2 + 8x + 3) = 0 \)
\( x - 2 = 0 \) or \( 2x^2 + 8x + 3 = 0 \)
For \( 2x^2 + 8x + 3 = 0 \), by using the quadratic formula:
\( x = \frac{-8 \pm \sqrt{8^2 - 4(2)(3)}}{2(2)} = \frac{-8 \pm \sqrt{40}}{4} = \frac{-8 \pm 2\sqrt{10}}{4} = \frac{-4 \pm \sqrt{10}}{2} \)
Thus, the roots are \( x = 2 \), \( x = \frac{-4 + \sqrt{10}}{2} \), and \( x = \frac{-4 - \sqrt{10}}{2} \).

(a) 1M for setting \( f(2) = 0 \); 1A for obtaining \( k = 4 \).
(b) 1M for factorizing \( f(x) \) to \( (x - 2)(2x^2 + 8x + 3) \); 1M for attempting to solve the quadratic equation; 1A for all three correct roots (accept equivalent exact surd forms).

(a) Let \( y = a x^2 + b \), where \( a \) and \( b \) are non-zero constants.
Substituting \( x = 2, y = 13 \):
\( 13 = a(2)^2 + b \implies 4a + b = 13 \) --- (1)
Substituting \( x = 4, y = 37 \):
\( 37 = a(4)^2 + b \implies 16a + b = 37 \) --- (2)
Subtracting (1) from (2):
\( 12a = 24 \implies a = 2 \)
Substituting \( a = 2 \) into (1):
\( 4(2) + b = 13 \implies 8 + b = 13 \implies b = 5 \)
Therefore, the formula is \( y = 2x^2 + 5 \).
(b) Substituting \( y = 55 \) into the formula:
\( 55 = 2x^2 + 5 \)
\( 2x^2 = 50 \)
\( x^2 = 25 \)
\( x = 5 \) or \( x = -5 \)

(a) 1M for setting up the equation \( y = a x^2 + b \); 1M for solving the system of equations for \( a \) and \( b \); 1A for the correct formula \( y = 2x^2 + 5 \).
(b) 1M for substituting \( y = 55 \) into the formula; 1A for obtaining both correct values of \( x \) (i.e. \( x = \pm 5 \), deduct 1 mark if \( x = -5 \) is missed).

(a) Let \(C = k_1 + \frac{k_2 h^2}{t}\), where \(k_1\) and \(k_2\) are non-zero constants. Substitute \(h=10\), \(t=4\), and \(C=350\): \(350 = k_1 + \frac{k_2 (10^2)}{4} \Rightarrow 350 = k_1 + 25k_2\). Substitute \(h=6\), \(t=3\), and \(C=220\): \(220 = k_1 + \frac{k_2 (6^2)}{3} \Rightarrow 220 = k_1 + 12k_2\). Subtracting the two equations: \(13k_2 = 130 \Rightarrow k_2 = 10\). Substitute \(k_2 = 10\) into the first equation: \(350 = k_1 + 25(10) \Rightarrow k_1 = 100\). Thus, \(C = 100 + \frac{10h^2}{t}\). When \(h = 12\) and \(t = 5\): \(C = 100 + \frac{10(12^2)}{5} = 100 + 288 = 388\). Therefore, the manufacturing cost is \(\$388\). (b) When \(C = 600\) and \(t = 8\): \(600 = 100 + \frac{10h^2}{8} \Rightarrow 500 = \frac{5}{4}h^2 \Rightarrow h^2 = 400\). Since \(h > 0\), we have \(h = 20\). Therefore, the height of the trophy is \(20\text{ cm}\).

(a) 1M for setting up \(C = k_1 + \frac{k_2 h^2}{t}\); 1M for substituting both pairs of values to obtain a system of linear equations; 1A for finding both \(k_1 = 100\) and \(k_2 = 10\); 1A for finding \(C = 388\). (b) 1M for substituting \(C = 600\) and \(t = 8\) to solve for \(h^2\); 1A for finding \(h = 20\) (rejecting \(h = -20\)).

For (a), by the remainder theorem, we have \(f(2) = 0\), which gives \(2(2)^3 + a(2)^2 + b(2) + 6 = 0\), so \(2a + b = -11\). We also have \(f(-1) = 9\), which gives \(2(-1)^3 + a(-1)^2 + b(-1) + 6 = 9\), so \(a - b = 5\). Solving the system of linear equations, we get \(a = -2\) and \(b = -7\). For (b), the equation is \(2x^3 - 2x^2 - 7x + 6 = 0\). Since \(x - 2\) is a factor, we can factorize \(f(x)\) as \((x - 2)(2x^2 + 2x - 3) = 0\). The other roots are given by \(2x^2 + 2x - 3 = 0\), which yields \(x = \frac{-1 \pm \sqrt{7}}{2}\). Since these roots are irrational, not all real roots of the equation are rational numbers. Thus, the claim is disagreed.

For (a): \(f(2) = 0\) or \(f(-1) = 9\) (1M); \(2a + b = -11\) or \(a - b = 5\) (1M); \(a = -2\) and \(b = -7\) (2A, 1A for each). For (b): \(2x^2 + 2x - 3 = 0\) (1M); Finding the roots \(x = \frac{-1 \pm \sqrt{7}}{2}\) or showing discriminant is 28 (1M); Explaining that these roots are irrational and concluding disagreement (1F).

For (a), the equation of circle \(C\) is \(x^2 + y^2 - 8x - 6y = 0\). The center is \(\left(-\frac{-8}{2}, -\frac{-6}{2}\right) = (4, 3)\). The radius is \(\sqrt{4^2 + 3^2 - 0} = 5\). For (b), let \(G(4, 3)\) be the center of \(C\). Let \(M\) be the midpoint of \(AB\). Then \(GM \perp AB\) and \(AM = \frac{1}{2}AB = 4\). In right-angled triangle \(GMA\), \(GM = \sqrt{GA^2 - AM^2} = \sqrt{5^2 - 4^2} = 3\). The perpendicular distance from \(G(4, 3)\) to the line \(L: 3x - 4y + c = 0\) is given by \(\frac{|3(4) - 4(3) + c|}{\sqrt{3^2 + (-4)^2}} = \frac{|c|}{5}\). Setting this distance to \(3\), we have \(\frac{|c|}{5} = 3\), which gives \(|c| = 15\). Thus, \(c = 15\) or \(c = -15\).

For (a): Center \((4, 3)\) (1A); Radius \(5\) (1A). For (b): \(AM = 4\) (1M); \(GM = \sqrt{5^2 - 4^2} = 3\) (1M for finding perpendicular distance); Using distance formula \(\frac{|3(4) - 4(3) + c|}{\sqrt{3^2 + (-4)^2}}\) (1M); Setting distance equal to 3 (1M); \(c = 15\) or \(c = -15\) (1A, both correct).

For (a), let \(V = ah^2 + bh^3\), where \(a\) and \(b\) are non-zero constants. When \(h = 2\), \(V = 36\), which gives \(4a + 8b = 36\), so \(a + 2b = 9\). When \(h = 3\), \(V = 99\), which gives \(9a + 27b = 99\), so \(a + 3b = 11\). Solving these equations, we get \(a = 5\) and \(b = 2\). So, the formula is \(V = 5h^2 + 2h^3\). When \(h = 4\), \(V = 5(4)^2 + 2(4)^3 = 80 + 128 = 208\). For (b), when \(h = 2\), the initial volume is \(36\). When \(h = 1.5\), the new volume is \(5(1.5)^2 + 2(1.5)^3 = 11.25 + 6.75 = 18\). The percentage decrease in volume is \(\frac{36 - 18}{36} \times 100\% = 50\%\).

For (a): \(V = ah^2 + bh^3\) (1M); Substituting both sets of values (1M); \(a = 5, b = 2\) (1A); \(V = 208\) (1A). For (b): Substituting \(h = 1.5\) to find new volume (1M); New volume is \(18\) (1A); Percentage decrease is \(50\%\) (1A).

(a) Method 1: Let the center of \(C\) be \(G(h, h)\) since the perpendicular bisector of \(PQ\) is \(y = x\). Since \(G\) lies on \(2x + 3y - 15 = 0\), we have \(2h + 3h - 15 = 0\), which gives \(5h = 15 \Rightarrow h = 3\). So the center is \(G(3, 3)\). The radius of \(C\) is \(r = \sqrt{(3-2)^2 + (3-8)^2} = \sqrt{26}\). Thus, the equation of \(C\) is \((x - 3)^2 + (y - 3)^2 = 26\) (or \(x^2 + y^2 - 6x - 6y - 8 = 0\)). Method 2: Let the equation of \(C\) be \(x^2 + y^2 + Dx + Ey + F = 0\). Since \(P(2, 8)\) and \(Q(8, 2)\) lie on \(C\), we have \(2D + 8E + F = -68\) and \(8D + 2E + F = -68\). Subtracting the equations gives \(D = E\). The center of \(C\) is \((-\frac{D}{2}, -\frac{D}{2})\). Since it lies on \(2x + 3y - 15 = 0\), we have \(2(-\frac{D}{2}) + 3(-\frac{D}{2}) - 15 = 0\), which gives \(D = -6\). Thus, \(E = -6\) and \(F = -8\). The equation of \(C\) is \(x^2 + y^2 - 6x - 6y - 8 = 0\). (b) The center of \(C\) is \(G(3, 3)\). The slope of \(GP\) is \(\frac{8-3}{2-3} = -5\). Since \(PR\) is tangent to \(C\) at \(P\), \(GP \perp PR\). So, the slope of \(PR\) is \(\frac{-1}{-5} = \frac{1}{5}\). Since \(R\) is \((k, k+10)\), we have \(\frac{(k+10)-8}{k-2} = \frac{1}{5} \Rightarrow \frac{k+2}{k-2} = \frac{1}{5} \Rightarrow 5k + 10 = k - 2 \Rightarrow 4k = -12 \Rightarrow k = -3\).

Part (a): For finding the midpoint of PQ (5, 5) or setting up equations of D, E, F (1M); For finding the perpendicular bisector equation y = x or getting D = E (1M); For finding the coordinates of the center (3, 3) or getting D = -6 and E = -6 (1M); For the correct equation of C (1A). Part (b): For finding the slope of GP as -5 (1M); For using GP perp PR to find slope of PR as 1/5 (1M); For setting up the equation of slope of PR with k (1M); For obtaining k = -3 (1A).

(a) The complement event is that all 3 selected balls have different colors (i.e., 1 red, 1 green, and 1 blue).
Number of ways to choose 3 balls of different colors = \(C^5_1 \times C^4_1 \times C^3_1 = 5 \times 4 \times 3 = 60\).
Total number of ways to choose 3 balls from 12 = \(C^{12}_3 = 220\).
The probability of choosing 3 balls of different colors = \(\frac{60}{220} = \frac{3}{11}\).
Therefore, the required probability = \(1 - \frac{3}{11} = \frac{8}{11}\).

(b) Let \(R\), \(G\), and \(B\) denote Red, Green, and Blue respectively.
The total number of possible outcomes when drawing 4 balls with replacement is \(12^4 = 20736\).
For the color of the selected balls to change exactly once, the sequence of colors must be of the form \(XXXY\), \(XXYY\), or \(XYYY\) for two distinct colors \(X, Y \in \{R, G, B\}\).
For each pair of distinct colors, the number of successful outcomes is:
- For \(\{R, G\}\):
\(RGGG\): \(5 \times 4^3 = 320\)
\(RRGG\): \(5^2 \times 4^2 = 400\)
\(RRRG\): \(5^3 \times 4 = 500\)
\(GRRR\): \(4 \times 5^3 = 500\)
\(GGRR\): \(4^2 \times 5^2 = 400\)
\(GGGR\): \(4^3 \times 5 = 320\)
Total for \(\{R, G\}\) = 2440.
- For \(\{R, B\}\):
\(RBBB\): \(5 \times 3^3 = 135\)
\(RRBB\): \(5^2 \times 3^2 = 225\)
\(RRRB\): \(5^3 \times 3 = 375\)
\(BRRR\): \(3 \times 5^3 = 375\)
\(BBRR\): \(3^2 \times 5^2 = 225\)
\(BBBR\): \(3^3 \times 5 = 135\)
Total for \(\{R, B\}\) = 1470.
- For \(\{G, B\}\):
\(GBBB\): \(4 \times 3^3 = 108\)
\(GGBB\): \(4^2 \times 3^2 = 144\)
\(GGGB\): \(4^3 \times 3 = 192\)
\(BGGG\): \(3 \times 4^3 = 192\)
\(BBGG\): \(3^2 \times 4^2 = 144\)
\(BBBG\): \(3^3 \times 4 = 108\)
Total for \(\{G, B\}\) = 888.

Total number of successful outcomes = \(2440 + 1470 + 888 = 4798\).
The required probability = \(\frac{4798}{20736} = \frac{2399}{10368}\).

(a) 1M for \(1 - \frac{C^5_1 C^4_1 C^3_1}{C^{12}_3}\) or equivalent; 1A for \(\frac{8}{11}\) (or equivalent, e.g., 0.727).
(b) 1M for identifying the pattern of exactly one change (e.g., listing cases or writing down probability expressions with 3 cases); 1A for \(\frac{2399}{10368}\) (or equivalent, e.g., 0.231).

(a) Let the center of \(C\) be \(G(h, k)\).
Since \(G\) lies on \(L: x - y - 1 = 0\), we have:
\(h - k - 1 = 0 \implies h = k + 1\) --- (1)

Since \(A(2, 8)\) and \(B(6, 0)\) lie on \(C\), the distances from \(G\) to \(A\) and \(B\) are equal:
\(GA^2 = GB^2\)
\((h - 2)^2 + (k - 8)^2 = (h - 6)^2 + k^2\) --- (2)

Substitute (1) into (2):
\((k + 1 - 2)^2 + (k - 8)^2 = (k + 1 - 6)^2 + k^2\)
\((k - 1)^2 + (k - 8)^2 = (k - 5)^2 + k^2\)
\(k^2 - 2k + 1 + k^2 - 16k + 64 = k^2 - 10k + 25 + k^2\)
\(2k^2 - 18k + 65 = 2k^2 - 10k + 25\)
\(-18k + 65 = -10k + 25\)
\(8k = 40\)
\(k = 5\)

Substitute \(k = 5\) into (1):
\(h = 5 + 1 = 6\)
So, the center of \(C\) is \(G(6, 5)\).

The radius \(r\) is:
\(r = \sqrt{(6 - 6)^2 + (5 - 0)^2} = 5\)

Thus, the equation of \(C\) is:
\((x - 6)^2 + (y - 5)^2 = 25\) (or \(x^2 + y^2 - 12x - 10y + 36 = 0\))

(b) Since the tangent lines are parallel to \(L: x - y - 1 = 0\), let their equations be \(x - y + c = 0\).
The perpendicular distance from the center \(G(6, 5)\) to the tangent lines is equal to the radius \(r = 5\):

\(\frac{|6 - 5 + c|}{\sqrt{1^2 + (-1)^2}} = 5\)
\(\frac{|1 + c|}{\sqrt{2}} = 5\)
\(|1 + c| = 5\sqrt{2}\)
\(1 + c = 5\sqrt{2}\) or \(1 + c = -5\sqrt{2}\)
\(c = 5\sqrt{2} - 1\) or \(c = -5\sqrt{2} - 1\)

Thus, the equations of the tangent lines are:
\(x - y + 5\sqrt{2} - 1 = 0\) and \(x - y - 5\sqrt{2} - 1 = 0\).

(a)
- 1M: For setting up distance equation \(GA = GB\) (or substituting coordinates into the general equation and setting up equations for \(D, E, F\)).
- 1M: For finding the center of the circle \(G(6, 5)\) (or finding \(D = -12\), \(E = -10\), and \(F = 36\)).
- 1A: For correct equation of \(C\): \((x - 6)^2 + (y - 5)^2 = 25\) (or equivalent).

(b)
- 1M: For using distance from center to line equals radius, i.e., \(\frac{|1+c|}{\sqrt{2}} = 5\), or substituting \(y = x + c\) into the circle equation and setting discriminant \(\Delta = 0\).
- 1A: For correct equations: \(x - y + 5\sqrt{2} - 1 = 0\) and \(x - y - 5\sqrt{2} - 1 = 0\) (or equivalent, e.g., \(x - y - 1 \pm 5\sqrt{2} = 0\)).

**Part (a)**

**Method 1:**
Let the center of \(C\) be \(G(h, k)\).
Since \(G(h, k)\) lies on \(L\), we have \(h - k - 1 = 0\), so \(h = k + 1\).
Since \(C\) passes through \(A(0, 8)\) and \(B(6, 0)\), the distance from \(G\) to \(A\) is equal to the distance from \(G\) to \(B\) (\(GA = GB\)):
\(GA^2 = GB^2\)
\((h - 0)^2 + (k - 8)^2 = (h - 6)^2 + (k - 0)^2\)
Substitute \(h = k + 1\):
\((k + 1)^2 + (k - 8)^2 = (k - 5)^2 + k^2\)
\(k^2 + 2k + 1 + k^2 - 16k + 64 = k^2 - 10k + 25 + k^2\)
\(2k^2 - 14k + 65 = 2k^2 - 10k + 25\)
\(-14k + 65 = -10k + 25\)
\(4k = 40 \implies k = 10\)
Then \(h = 10 + 1 = 11\).
So the center of \(C\) is \(G(11, 10)\).
The radius squared is \(R^2 = (11 - 0)^2 + (10 - 8)^2 = 121 + 4 = 125\).
The equation of \(C\) is:
\((x - 11)^2 + (y - 10)^2 = 125\) (or \(x^2 + y^2 - 22x - 20y + 96 = 0\)).

**Method 2:**
The midpoint of \(AB\) is \(M\left(\frac{0+6}{2}, \frac{8+0}{2}\right) = (3, 4)\).
The slope of \(AB\) is \(m_{AB} = \frac{0 - 8}{6 - 0} = -\frac{4}{3}\).
The slope of the perpendicular bisector of \(AB\) is \(m = -\frac{1}{m_{AB}} = \frac{3}{4}\).
The equation of the perpendicular bisector of \(AB\) is:
\(y - 4 = \frac{3}{4}(x - 3) \implies 3x - 4y + 7 = 0\).
The center of \(C\) lies on the intersection of the perpendicular bisector of \(AB\) and \(L: x - y - 1 = 0\):
\(\begin{cases} 3x - 4y + 7 = 0 \\ x - y - 1 = 0 \end{cases}\)
From the second equation, \(x = y + 1\). Substituting into the first equation:
\(3(y + 1) - 4y + 7 = 0 \implies -y + 10 = 0 \implies y = 10\).
Then \(x = 11\).
So the center of \(C\) is \(G(11, 10)\).
The radius of \(C\) is \(R = \sqrt{(11 - 0)^2 + (10 - 8)^2} = \sqrt{125}\).
The equation of \(C\) is \((x - 11)^2 + (y - 10)^2 = 125\).

---

**Part (b)**

**Method 1:**
Since \(L_1\) is parallel to \(L: x - y - 1 = 0\), we can write the equation of \(L_1\) as \(x - y + c = 0\), where the \(y\)-intercept is \(c\).
Since \(L_1\) is tangent to \(C\), the perpendicular distance from the center \(G(11, 10)\) to \(L_1\) is equal to the radius \(R = \sqrt{125}\).
\(\frac{|11 - 10 + c|}{\sqrt{1^2 + (-1)^2}} = \sqrt{125}\)
\(\frac{|1 + c|}{\sqrt{2}} = 5\sqrt{5}\)
\(|1 + c| = 5\sqrt{10}\)
\(1 + c = \pm 5\sqrt{10}\)
\(c = -1 \pm 5\sqrt{10}\)
So the \(y\)-intercepts of \(L_1\) are \(-1 + 5\sqrt{10}\) and \(-1 - 5\sqrt{10}\) (or written as \(-1 \pm 5\sqrt{10}\)).

**Method 2:**
Let the equation of \(L_1\) be \(y = x + c\).
Substitute \(y = x + c\) into the equation of \(C\):
\(x^2 + (x+c)^2 - 22x - 20(x+c) + 96 = 0\)
\(2x^2 + (2c - 42)x + (c^2 - 20c + 96) = 0\)
Since \(L_1\) is tangent to \(C\), the discriminant \(\Delta = 0\):
\((2c - 42)^2 - 4(2)(c^2 - 20c + 96) = 0\)
\(4(c - 21)^2 - 8(c^2 - 20c + 96) = 0\)
\(c^2 - 42c + 441 - 2(c^2 - 20c + 96) = 0\)
\(-c^2 - 2c + 249 = 0\)
\(c^2 + 2c - 249 = 0\)
\(c = \frac{-2 \pm \sqrt{2^2 - 4(1)(-249)}}{2} = \frac{-2 \pm \sqrt{1000}}{2} = -1 \pm 5\sqrt{10}\).
So the \(y\)-intercepts are \(-1 \pm 5\sqrt{10}\).

**(a)**
- Let center be \( (k+1, k) \) OR finding midpoint \( (3, 4) \) and slope of perpendicular bisector \( \frac{3}{4} \) [1M]
- Setting up distance equality equation OR setting up equation for perpendicular bisector and forming system of equations [1M]
- Finding the center \( (11, 10) \) [1A]
- Finding the equation of \( C \): \( (x - 11)^2 + (y - 10)^2 = 125 \) or \( x^2 + y^2 - 22x - 20y + 96 = 0 \) (Accept equivalent forms) [1A]

**(b)**
- Setting up tangent distance formula \( \frac{|11 - 10 + c|}{\sqrt{1^2 + (-1)^2}} = \sqrt{125} \) OR substituting \( y = x + c \) into the circle equation and setting \( \Delta = 0 \) [1M]
- Finding the correct \( y \)-intercepts: \( -1 \pm 5\sqrt{10} \) (Accept \( -1 + 5\sqrt{10} \) and \( -1 - 5\sqrt{10} \)) [1A]

(a) Let the center of \(C\) be \(G(h, 2h-7)\).
Since \(C\) passes through \(A(6,0)\) and \(B(-1,-7)\), we have \(GA = GB\).
\(GA^2 = GB^2\)
\((h-6)^2 + (2h-7-0)^2 = (h-(-1))^2 + (2h-7-(-7))^2\)
\((h-6)^2 + (2h-7)^2 = (h+1)^2 + (4h^2)\)
\(h^2 - 12h + 36 + 4h^2 - 28h + 49 = h^2 + 2h + 1 + 4h^2\)
\(5h^2 - 40h + 85 = 5h^2 + 2h + 1\)
\(-42h = -84\)
\(h = 2\)
So, the center of \(C\) is \(G(2, 2(2)-7) = (2, -3)\).
The radius of \(C\) is \(R = \sqrt{(2-6)^2 + (-3-0)^2} = \sqrt{16+9} = 5\).
Thus, the equation of \(C\) is:
\((x-2)^2 + (y+3)^2 = 25\) (or \(x^2 + y^2 - 4x + 6y - 12 = 0\))

(b) Since \(L_2: 3x - 4y + k = 0\) is tangent to \(C\), the perpendicular distance from the center \(G(2, -3)\) to \(L_2\) is equal to the radius \(5\).
\(\frac{|3(2) - 4(-3) + k|}{\sqrt{3^2 + (-4)^2}} = 5\)
\(\frac{|6 + 12 + k|}{5} = 5\)
\(|18 + k| = 25\)
\(18 + k = 25\) or \(18 + k = -25\)
\(k = 7\) or \(k = -43\)

(a)
- 1M: For setting up distance equation \(GA^2 = GB^2\) using \((h, 2h-7)\)
- 1M: For expanding and simplifying the quadratic equation to linear form
- 1A: For finding the center \((2, -3)\) or equivalent coordinates
- 1A: For the correct equation of circle (either standard or general form)

(b)
- 1M: For using the perpendicular distance formula from center to line
- 1M: For equating the distance to the radius of the circle
- 1M: For setting up two linear equations with absolute value
- 1A: For both correct values of \(k = 7\) and \(k = -43\)

(a) Rewrite the equation of \(C\):
\(x^2 - 12x + 36 + y^2 - 16y + 64 = -64 + 36 + 64\)
\((x - 6)^2 + (y - 8)^2 = 6^2\)
So, the center of \(C\) is \(G(6, 8)\) and the radius is \(6\).

(b)(i) Distance \(GP = \sqrt{(14 - 6)^2 + (14 - 8)^2} = \sqrt{8^2 + 6^2} = 10\).
Since the distance \(GP = 10\) is greater than the radius \(6\), \(P\) lies outside \(C\).

(b)(ii) Let the equation of the tangent passing through \(P(14, 14)\) with slope \(m\) be:
\(y - 14 = m(x - 14) \implies mx - y + 14 - 14m = 0\).
The perpendicular distance from \(G(6, 8)\) to the tangent is equal to the radius \(6\):
\(\frac{|6m - 8 + 14 - 14m|}{\sqrt{m^2 + 1}} = 6\)
\(|6 - 8m| = 6\sqrt{m^2 + 1}\)
\(3 - 4m| = 3\sqrt{m^2 + 1}\)
\((3 - 4m)^2 = 9(m^2 + 1)\)
\(9 - 24m + 16m^2 = 9m^2 + 9\)
\(7m^2 - 24m = 0 \implies m(7m - 24) = 0\)
Hence, \(m = 0\) or \(m = \frac{24}{7}\).
The equations of the tangents are:
For \(m = 0\): \(y = 14\)
For \(m = \frac{24}{7}\): \(y - 14 = \frac{24}{7}(x - 14) \implies 24x - 7y - 238 = 0\).

(c)(i) Since \(GQ_1 \perp PQ_1\), \(\triangle GQ_1P\) is a right-angled triangle.
\(GQ_1 = 6\) (radius) and \(GP = 10\).
\(PQ_1 = \sqrt{GP^2 - GQ_1^2} = \sqrt{10^2 - 6^2} = 8\).
The area of \(\triangle GQ_1P = \frac{1}{2} \times 6 \times 8 = 24\).
Since \(\triangle GQ_1P \cong \triangle GQ_2P\), the area of the quadrilateral \(GQ_1PQ_2 = 2 \times 24 = 48\).

(c)(ii) Since \(\angle GQ_1P = \angle GQ_2P = 90^\circ\), the quadrilateral \(GQ_1PQ_2\) is cyclic with \(GP\) as the diameter of its circumcircle.
Thus, the circumcircle \(C'\) of \(\triangle GQ_1Q_2\) is also the circle with \(GP\) as its diameter.
Center of \(C'\) is the midpoint of \(GP\):
\(\left(\frac{6 + 14}{2}, \frac{8 + 14}{2}\right) = (10, 11)\).
Radius of \(C'\) is \(\frac{GP}{2} = 5\).
Thus, the equation of \(C'\) is:
\((x - 10)^2 + (y - 11)^2 = 25\) (or \(x^2 + y^2 - 20x - 22y + 196 = 0\)).

(a) Center = (6, 8) [1M]
Radius = 6 [1A]

(b)(i) Correct calculation of GP = 10 and comparing with radius 6 to draw conclusion [1M/A]

(b)(ii) For setting up the linear equation of the tangent with slope m [1M]
For using the distance formula from G to the tangent = radius [1M]
For solving m = 0 or m = 24/7 [1A]
For both correct tangent equations: y = 14 and 24x - 7y - 238 = 0 [1A]

(c)(i) For finding tangent length PQ1 = 8 [1M]
For area of quadrilateral = 48 [1A]

(c)(ii) For identifying GP is the diameter of C' [1M]
For finding center = (10, 11) and radius = 5 [1M]
For the equation of C': (x-10)^2 + (y-11)^2 = 25 (or general form) [1A]

Let the original data be \(x_i\). The new data are \(y_i = -3x_i + 5\). Since adding a constant does not change measures of dispersion (range, IQR, variance), and multiplying by a factor \(c\) multiplies the range and IQR by \(|c|\) and variance by \(c^2\), we have: New range = \(|-3|R = 3R\), New interquartile range = \(|-3|I = 3I\), New variance = \((-3)^2 V = 9V\).

1 mark for correct identification of the scaling factors for range, IQR, and variance under the linear transformation.

For Statement I: In Group B, since \(Q_3 = 70\), about 25% of students (which is 10 students) scored 70 or above. In Group A, \(Q_3 = 75\), which means at least 10 students scored 75 or above, so the number of students scoring 70 or above in Group A could be greater than or equal to 10. Thus, I is not necessarily true. For Statement II: Range of A = \(95 - 20 = 75\), Range of B = \(90 - 30 = 60\). Since \(75 > 60\), II must be true. For Statement III: IQR of A = \(75 - 45 = 30\), IQR of B = \(70 - 50 = 20\). Since \(30 > 20\), III must be true.

1 mark for analyzing all three statements correctly.

Let the original 10 numbers be \(x_1, \dots, x_{10}\). Since the mean is 15, \(\sum x_i = 150\). The new mean after adding 11 and 19 is \(\frac{150 + 11 + 19}{12} = 15\), which is unchanged. The original variance is \(4^2 = 16\), so \(\frac{\sum x_i^2}{10} - 15^2 = 16 \implies \sum x_i^2 = 10 \times (225 + 16) = 2410\). The new sum of squares is \(2410 + 11^2 + 19^2 = 2410 + 121 + 361 = 2892\). The new variance is \(\frac{2892}{12} - 15^2 = 241 - 225 = 16\). Thus, the new standard deviation is \(\sqrt{16} = 4\).

1 mark for finding the correct new standard deviation.

The center of the circle \(C: x^2 + y^2 - 4x + 6y - 12 = 0\) is \((2, -3)\). The slope of the line \(3x - 4y + 5 = 0\) is \(\frac{3}{4}\). Since \(L\) is perpendicular to this line, the slope of \(L\) is \(-\frac{4}{3}\). Using the point-slope form, the equation of \(L\) is \(y - (-3) = -\frac{4}{3}(x - 2) \implies 3(y+3) = -4(x-2) \implies 4x + 3y + 1 = 0\).

1 mark for obtaining the correct equation of the straight line.

The circle is \(x^2 + y^2 - 2x - 2y - 7 = 0\). Its center is \((1, 1)\) and radius is \(r = \sqrt{1^2 + 1^2 - (-7)} = 3\). Since the line \(3x - 4y + k = 0\) is tangent to the circle, the perpendicular distance from the center \((1, 1)\) to the line is equal to the radius \(3\). Thus, \(\frac{|3(1) - 4(1) + k|}{\sqrt{3^2 + (-4)^2}} = 3 \implies \frac{|k - 1|}{5} = 3 \implies |k - 1| = 15 \implies k - 1 = 15 \text{ or } k - 1 = -15\). This gives \(k = 16\) or \(k = -14\).

1 mark for finding the correct values of \(k\).

Since \(AB\) is a diameter of \(C\), the center is the midpoint of \(AB\), which is \((3, -2)\). The radius squared is \(r^2 = (3-1)^2 + (-2-2)^2 = 20\). The equation of \(C\) is \((x-3)^2 + (y+2)^2 = 20 \implies x^2 + y^2 - 6x + 4y - 7 = 0\). Thus, I is true. For II: The distance from the origin \((0,0)\) to the center \((3, -2)\) squared is \(3^2 + (-2)^2 = 13 < r^2 = 20\), so the origin lies inside \(C\). Thus, II is true. For III: Substitute the center \((3, -2)\) into \(y = 2x - 8\): LHS = \(-2\), RHS = \(2(3) - 8 = -2\). Since LHS = RHS, the line passes through the center. Thus, III is true.

1 mark for proving all three statements are true.

From \(P(1) = -8\), we have \(1 + a + b - 6 = -8 \implies a + b = -3\). Since \(x+1\) is a factor, \(P(-1) = 0 \implies -1 + a - b - 6 = 0 \implies a - b = 7\). Solving the system of equations gives \(a = 2\) and \(b = -5\). Thus, \(P(x) = x^3 + 2x^2 - 5x - 6\). The remainder when \(P(x)\) is divided by \(x-3\) is \(P(3) = 3^3 + 2(3^2) - 5(3) - 6 = 27 + 18 - 15 - 6 = 24\).

1 mark for finding the correct remainder.

To find the LCM: For the numerical coefficients 12, 18, and 8, the LCM is 72. For the variables, we take the highest power of each base appearing in any of the expressions: the highest power of \(x\) is \(x^3\), the highest power of \(y\) is \(y^3\), the highest power of \(z\) is \(z\), and the highest power of \((z-1)\) is \((z-1)^2\). Thus, the LCM is \(72x^3 y^3 z (z - 1)^2\).

1 mark for identifying the correct LCM.

Let \(z = \frac{k x^2}{\sqrt{y}}\). When \(x\) is increased by 20% and \(y\) is decreased by 36%, the new values are \(x' = 1.2x\) and \(y' = 0.64y\). The new value of \(z\) is \(z' = \frac{k (1.2x)^2}{\sqrt{0.64y}} = \frac{1.44 k x^2}{0.8 \sqrt{y}} = 1.8 z\). Thus, \(z\) is increased by \((1.8 - 1) \times 100\% = 80\%\).

1 mark for finding the correct percentage change.

Let \(u = k_1 v + k_2 v^2\). When \(v = 2\), \(10 = 2k_1 + 4k_2 \implies k_1 + 2k_2 = 5\). When \(v = 3\), \(21 = 3k_1 + 9k_2 \implies k_1 + 3k_2 = 7\). Solving these equations, we get \(k_2 = 2\) and \(k_1 = 1\). So \(u = v + 2v^2\). When \(v = 5\), \(u = 5 + 2(5^2) = 55\).

1 mark for finding the correct value of \(u\).

Let the remainder be \(ax + b\). Since \(2x^2-3x-2 = (2x+1)(x-2)\), we can write \(f(x) = Q(x)(2x+1)(x-2) + ax + b\). By the Remainder Theorem: \(f(2) = 2a + b = 5\) and \(f(-1/2) = -a/2 + b = -5\). Solving the system of linear equations, we get \(a = 4\) and \(b = -3\). Thus, the remainder is \(4x-3\).

Method Mark: 1 mark for setting up the remainder as \(ax+b\) and using the Remainder Theorem to form equations. Accuracy Mark: 1 mark for obtaining the correct coefficients and final remainder.

Let \(z = \frac{k x^2}{\sqrt{y}}\) where \(k\) is a non-zero constant. Let \(x' = 0.9x\) and \(y' = 1.44y\). The new value \(z' = \frac{k (0.9x)^2}{\sqrt{1.44y}} = \frac{0.81 k x^2}{1.2 \sqrt{y}} = 0.675 z\). The percentage change in \(z\) is \((0.675 - 1) \times 100\% = -32.5\%\), which represents a decrease of \(32.5\%\).

Method Mark: 1 mark for expressing \(z'\) in terms of original variables and finding the factor \(0.675\). Accuracy Mark: 1 mark for identifying a decrease of \(32.5\%\).

The midpoint of \(PQ\) is \((3, 4)\). The slope of \(PQ\) is \(\frac{0-8}{6-0} = -\frac{4}{3}\). The equation of the perpendicular bisector of \(PQ\) is \(y - 4 = \frac{3}{4}(x-3)\), which simplifies to \(3x - 4y + 7 = 0\). The center of the circle is the intersection of \(3x - 4y + 7 = 0\) and \(x + y - 7 = 0\). Solving these, we get the center \(H(3, 4)\). The radius \(R = \sqrt{(3-0)^2 + (4-8)^2} = 5\). Thus, the equation of \(C\) is \((x-3)^2 + (y-4)^2 = 25\), which is \(x^2 + y^2 - 6x - 8y = 0\).

Method Mark: 1 mark for finding the perpendicular bisector of \(PQ\) and finding the center. Accuracy Mark: 1 mark for the correct final circle equation.

For the original 10 data points: \(\sum x_i = 10 \times 50 = 500\). Since \(\sigma^2 = 64\), we have \(\sum x_i^2 = 10(64 + 50^2) = 25640\). After adding 42 and 58: the new sum is \(500 + 42 + 58 = 600\), so the new mean is \(600/12 = 50\). The new sum of squares is \(25640 + 42^2 + 58^2 = 30768\). The new variance is \(\sigma'^2 = \frac{30768}{12} - 50^2 = 2564 - 2500 = 64\). Therefore, the new standard deviation is \(\sqrt{64} = 8\).

Method Mark: 1 mark for finding the new sum of squares and new mean. Accuracy Mark: 1 mark for the correct new standard deviation.

Let \(f(x) = 3x^3 + ax^2 + bx - 12\). Since \(f(x)\) is divisible by \(x^2 - x - 6 = (x-3)(x+2)\), we have \(f(3) = 0\) and \(f(-2) = 0\). From \(f(3) = 0\), we get \(3(27) + 9a + 3b - 12 = 0 \implies 3a + b = -23\). From \(f(-2) = 0\), we get \(3(-8) + 4a - 2b - 12 = 0 \implies 2a - b = 18\). Solving these equations gives \(a = -1\) and \(b = -20\). Therefore, \(a - b = -1 - (-20) = 19\).

Method Mark: 1 mark for applying the Factor Theorem and setting up the system of equations. Accuracy Mark: 1 mark for finding \(a-b = 19\).

The square of the tangent length from \(P(k, 1)\) to the circle \(x^2 + y^2 + Dx + Ey + F = 0\) is given by \(k^2 + 1^2 + Dk + E(1) + F\). Therefore, \(k^2 + 1^2 - 4k + 6(1) - 12 = 4^2\). This simplifies to \(k^2 - 4k - 5 = 16\), or \(k^2 - 4k - 21 = 0\). Factoring gives \((k-7)(k+3) = 0\), so \(k = 7\) or \(k = -3\).

Method Mark: 1 mark for using the formula for the length of tangent to set up a quadratic equation in \(k\). Accuracy Mark: 1 mark for solving \(k = -3\) or \(7\).

For I: Range = \(75 - 42 = 33\). So I is true.
For II: The total number of students \(N = 16\). The first quartile \(Q_1 = \frac{48 + 50}{2} = 49\). The third quartile \(Q_3 = \frac{64 + 64}{2} = 64\). The interquartile range is \(64 - 49 = 15\). So II is false.
For III: The value 53 appears 3 times, which is the highest frequency. So the mode is 53, and III is true. Hence, I and III only are true.

Method Mark: 1 mark for analyzing the range, quartiles, and mode from the stem-and-leaf diagram. Accuracy Mark: 1 mark for choosing I and III only.

Using the properties of variance, for any constants \(a\) and \(b\), \(\text{Var}(aX + b) = a^2 \text{Var}(X)\). Here, the transformation is \(y_i = -2x_i + 3\), so \(a = -2\). The new variance is \((-2)^2 \times 12 = 4 \times 12 = 48\).

Method Mark: 1 mark for applying the property \(\text{Var}(aX+b) = a^2 \text{Var}(X)\). Accuracy Mark: 1 mark for obtaining 48.

The center of the circle is \(H(4, -5)\) and the radius is \(R = \sqrt{(-4)^2 + 5^2 - 5} = \sqrt{36} = 6\). Since the line is tangent to the circle, the perpendicular distance from \(H\) to \(L\) equals the radius: \(\frac{|3(4) - 4(-5) + k|}{\sqrt{3^2 + (-4)^2}} = 6\), which simplifies to \(\frac{|32 + k|}{5} = 6\). Thus, \(|32 + k| = 30\), leading to \(32 + k = 30 \implies k = -2\) or \(32 + k = -30 \implies k = -62\).

Method Mark: 1 mark for using the point-to-line distance formula and equating it to the radius. Accuracy Mark: 1 mark for finding both values of \(k\).

The combined mean is \(\bar{x} = \frac{20(65) + 30(75)}{50} = 71\). Let \(\sigma_1 = 8\) and \(\sigma_2 = 8\). The combined variance is \(\sigma^2 = \frac{N_1 (\sigma_1^2 + (\bar{x}_1 - \bar{x})^2) + N_2 (\sigma_2^2 + (\bar{x}_2 - \bar{x})^2)}{N_1 + N_2} = \frac{20(8^2 + (65 - 71)^2) + 30(8^2 + (75 - 71)^2)}{50} = \frac{20(64 + 36) + 30(64 + 16)}{50} = \frac{2000 + 2400}{50} = 88\). The combined standard deviation is \(\sqrt{88}\).

Method Mark: 1 mark for calculating the combined mean and setting up the combined variance expression. Accuracy Mark: 1 mark for obtaining \(\sqrt{88}\).

The relation is \(y_i = 5 - 3x_i\). The standard deviation of \(y\) is \(|-3| \times \text{SD}(x) = 3 \times 4 = 12\). The variance of \(y\) is \(12^2 = 144\).

Award 1 mark for the correct answer C. No marks for incorrect options.

Sum of the 10 known numbers is 174. The mean of 11 numbers is 18, so the total sum is \(18 \times 11 = 198\). Thus, \(x = 198 - 174 = 24\). Since \(24\) is the maximum and \(14\) is the minimum, the range is \(24 - 14 = 10\).

Award 1 mark for the correct answer B. No marks for incorrect options.

Mary's standard score in Mathematics \(= (76-64)/12 = 1.0\). Mary's standard score in English \(= (66-56)/8 = 1.25\). Since \(1.25 > 1.0\), statement I is true. John's standard score in English is equal to Mary's in Mathematics, which is \(1.0\), so John's English score is \(56 + 1.0 \times 8 = 64\), meaning statement II is true. John's standard score in English is \(1.0\) instead of \(1.25\), so statement III is false. Thus, only I and II are true.

Award 1 mark for the correct answer C. No marks for incorrect options.

The center of \(C\) is \((2, -3)\) and its radius is \(\sqrt{2^2 + (-3)^2 - k} = \sqrt{13 - k}\). Since the line \(3x - 4y + 2 = 0\) is tangent to \(C\), the distance from the center to the line equals the radius: \(R = \frac{|3(2) - 4(-3) + 2|}{\sqrt{3^2 + 4^2}} = \frac{20}{5} = 4\). Therefore, \(\sqrt{13-k} = 4 \implies 13-k = 16 \implies k = -3\).

Award 1 mark for the correct answer A. No marks for incorrect options.

The slope of the radius connecting the center \((3, 4)\) and the origin \(O(0,0)\) is \(m_1 = \frac{4-0}{3-0} = \frac{4}{3}\). Since the tangent is perpendicular to the radius, its slope is \(m_2 = -\frac{3}{4}\). Thus, the equation of the tangent passing through \((0,0)\) is \(y = -\frac{3}{4}x\), which simplifies to \(3x + 4y = 0\).

Award 1 mark for the correct answer A. No marks for incorrect options.

The center of \(C\) is \((3, 1)\), and its radius is \(R = \sqrt{3^2 + 1^2 - (-15)} = 5\). The distance from the center to the line \(L\) is \(d = \frac{|3(3) + 4(1) - 28|}{\sqrt{3^2 + 4^2}} = \frac{15}{5} = 3\). Let the chord length be \(2h\). By Pythagoras' theorem, \(h = \sqrt{R^2 - d^2} = \sqrt{5^2 - 3^2} = 4\). Thus, the chord length is \(2h = 8\).

Award 1 mark for the correct answer C. No marks for incorrect options.

By Remainder Theorem, \(P(1) = 2 + a + b - 6 = -4 \implies a + b = 0\). Also, \(P(-2) = 2(-2)^3 + a(-2)^2 + b(-2) - 6 = -10 \implies -16 + 4a - 2b - 6 = -10 \implies 2a - b = 6\). Solving the system, we get \(a = 2\) and \(b = -2\). Thus, \(P(x) = 2x^3 + 2x^2 - 2x - 6\). The remainder when \(P(x)\) is divided by \(2x-1\) is \(P(1/2) = 2(1/8) + 2(1/4) - 2(1/2) - 6 = -25/4\).

Award 1 mark for the correct answer B. No marks for incorrect options.

The LCM of the numerical coefficients \(12, 18, 8\) is \(72\). For the variables, we take the highest power of each base: \(a^3\), \(b^4\), \(c^2\), \(d^2\). Therefore, the LCM is \(72a^3b^4c^2d^2\).

Award 1 mark for the correct answer B. No marks for incorrect options.

Let \(z = k_1 x + \frac{k_2}{y}\). Since \(z=10\) when \(x=2, y=3\), we have \(10 = 2k_1 + \frac{k_2}{3} \implies 6k_1 + k_2 = 30\). Since \(z=21\) when \(x=3, y=1\), we have \(21 = 3k_1 + k_2\). Solving these, we get \(k_1=3\) and \(k_2=12\). So \(z = 3x + \frac{12}{y}\). When \(x=4, y=2\), \(z = 3(4) + \frac{12}{2} = 12 + 6 = 18\).

Award 1 mark for the correct answer C. No marks for incorrect options.

Let \(u = \frac{kv^2}{\sqrt{w}}\). The new values of \(v\) and \(w\) are \(v' = 1.2v\) and \(w' = 0.64w\). The new value of \(u\) is \(u' = \frac{k(1.2v)^2}{\sqrt{0.64w}} = \frac{1.44kv^2}{0.8\sqrt{w}} = 1.8u\). Therefore, \(u\) is increased by \((1.8 - 1) \times 100\% = 80\%\).

Award 1 mark for the correct answer A. No marks for incorrect options.

The center of the circle is \(G(2, 3)\). The slope of the line \(L: x - 2y + k = 0\) is \(\frac{1}{2}\). The line perpendicular to \(L\) passing through the center \(G(2,3)\) has slope \(-2\). Its equation is \(y - 3 = -2(x - 2)\), which simplifies to \(2x + y - 7 = 0\). The midpoint of the chord \(AB\) must lie on this perpendicular line. Since the \(x\)-coordinate of the midpoint is given as \(1\), we substitute \(x = 1\) into the perpendicular line: \(2(1) + y - 7 = 0 \implies y = 5\). Thus, the midpoint is \((1, 5)\). Since the midpoint also lies on \(L\), we substitute \((1, 5)\) into the equation of \(L\): \(1 - 2(5) + k = 0 \implies k = 9\). Verify that the distance from \(G\) to the line is \(\sqrt{5}\), which is less than the radius \(5\), so they indeed intersect at two points.

Award 1 mark for correct answer B. No partial marks are awarded for multiple-choice questions.

Since the circle passes through \((1, 8)\), which lies in the first quadrant, the center of the circle must be in the first quadrant and can be represented as \((r, r)\), where \(r\) is the radius. The equation of the circle is \((x - r)^2 + (y - r)^2 = r^2\). Substituting the point \((1, 8)\) yields: \((1 - r)^2 + (8 - r)^2 = r^2 \implies 1 - 2r + r^2 + 64 - 16r + r^2 = r^2 \implies r^2 - 18r + 65 = 0\). Factoring the equation gives \((r - 5)(r - 13) = 0\), so \(r = 5\) or \(r = 13\). The sum of the possible radii is \(5 + 13 = 18\).

Award 1 mark for correct answer C. No partial marks are awarded for multiple-choice questions.

By the Remainder Theorem, \(P(1) = 3\) and \(P(-2) = -3\). Let the remainder when \(Q(x) = (x+1)P(x)\) is divided by \(x^2 + x - 2 = (x-1)(x+2)\) be \(ax + b\). Thus, \(Q(x) = (x^2 + x - 2)q(x) + ax + b\). Substituting \(x = 1\) yields \(Q(1) = (1+1)P(1) = 2(3) = 6\), so \(a + b = 6\). Substituting \(x = -2\) yields \(Q(-2) = (-2+1)P(-2) = (-1)(-3) = 3\), so \(-2a + b = 3\). Solving the system of equations: subtracting the second equation from the first gives \(3a = 3 \implies a = 1\), and thus \(b = 5\). The remainder is \(x + 5\).

Award 1 mark for correct answer A. No partial marks are awarded for multiple-choice questions.

Factorize each expression completely: 1) \(3x^2 - 12 = 3(x-2)(x+2)\); 2) \(x^2 - 4x + 4 = (x-2)^2\); 3) \(2x^2 - 4x = 2x(x-2)\). The LCM of the numerical coefficients \(3\), \(1\), and \(2\) is \(6\). The algebraic factors are \(x\), \(x-2\), and \(x+2\). Taking the highest power of each factor, the LCM is \(6x(x-2)^2(x+2)\).

Award 1 mark for correct answer B. No partial marks are awarded for multiple-choice questions.

Let \(z = \frac{k x^2}{\sqrt{y}}\) for some non-zero constant \(k\). Let the new values be \(x' = 0.7x\) and \(y' = 1.96y\). The new value of \(z\) is \(z' = \frac{k (x')^2}{\sqrt{y'}} = \frac{k (0.7x)^2}{\sqrt{1.96y}} = \frac{0.49 k x^2}{1.4 \sqrt{y}} = 0.35 \frac{k x^2}{\sqrt{y}} = 0.35 z\). The percentage change in \(z\) is \(\frac{z' - z}{z} \times 100\% = \frac{0.35z - z}{z} \times 100\% = -65\%\). Therefore, \(z\) is decreased by \(65\%\).

Award 1 mark for correct answer A. No partial marks are awarded for multiple-choice questions.

Let \(z = ax + by^2\) where \(a\) and \(b\) are non-zero constants. From the given conditions: 1) \(2a + 9b = 22\); 2) \(3a + 4b = 14\). Multiplying equation (1) by 3 and equation (2) by 2, we get: \(6a + 27b = 66\) and \(6a + 8b = 28\). Subtracting these equations: \(19b = 38 \implies b = 2\). Substituting \(b = 2\) back into (1): \(2a + 18 = 22 \implies a = 2\). Thus, \(z = 2x + 2y^2\). When \(x = -1\) and \(y = 4\), \(z = 2(-1) + 2(4)^2 = -2 + 32 = 30\).

Award 1 mark for correct answer A. No partial marks are awarded for multiple-choice questions.

For any linear transformation \(y_i = ax_i + b\):
1. The new mean is \(a \times \text{old mean} + b = -2m + 3 = 3 - 2m\). Thus, I is true.
2. The new standard deviation is \(|a| \times \text{old SD} = |-2|s = 2s\). Thus, II is false.
3. The new variance is the square of the new standard deviation, which is \((2s)^2 = 4s^2\). Thus, III is true.
Hence, only I and III must be true.

Award 1 mark for correct answer C. No partial marks are awarded for multiple-choice questions.

First, calculate Alan's actual score: \(X = \text{Mean} + z \times \text{SD} = 64 + 1.5 \times 8 = 76\). After the adjustment, Alan's actual mark is still \(76\), the new mean is \(70\), and his new standard score is \(1.2\). Using the standard score formula: \(z' = \frac{X - \text{Mean}'}{\text{SD}'} \implies 1.2 = \frac{76 - 70}{\text{SD}'} \implies 1.2 = \frac{6}{\text{SD}'} \implies \text{SD}' = \frac{6}{1.2} = 5\). Thus, the new standard deviation is \(5\) marks.

Award 1 mark for correct answer A. No partial marks are awarded for multiple-choice questions.

We can calculate this by taking the total number of ways to form the committee without restrictions and subtracting the number of ways where both Mary and Susan are selected.
1. Total ways without restrictions: Choose 4 boys from 7 and 4 girls from 6: \(C_{4}^{7} \times C_{4}^{6} = 35 \times 15 = 525\).
2. Ways where both Mary and Susan are selected: Mary and Susan are fixed (1 way). Choose the remaining 2 girls from the other 4 girls: \(C_{2}^{4} = 6\) ways. Choose 4 boys from 7: \(C_{4}^{7} = 35\) ways. Thus, \(35 \times 6 = 210\) ways.
3. Number of committees where they are not both selected: \(525 - 210 = 315\).

Award 1 mark for correct answer C. No partial marks are awarded for multiple-choice questions.

Rewrite the equation using the identity \(\cos^2 x = 1 - \sin^2 x\):
\(4(1 - \sin^2 x) + 5 \sin x - 5 = 0 \implies 4 - 4 \sin^2 x + 5 \sin x - 5 = 0 \implies 4 \sin^2 x - 5 \sin x + 1 = 0\).
Let \(u = \sin x\). The equation becomes \(4u^2 - 5u + 1 = 0\).
Factoring gives \((4u - 1)(u - 1) = 0\), which yields \(u = \frac{1}{4}\) or \(u = 1\).
Now solve for \(x\) in the interval \(0 \le x \le 2\pi\):
1. For \(\sin x = \frac{1}{4}\): Since \(0 < \frac{1}{4} < 1\), there are exactly \(2\) solutions in the first and second quadrants.
2. For \(\sin x = 1\): There is exactly \(1\) solution, which is \(x = \frac{\pi}{2}\).
Thus, the total number of roots is \(2 + 1 = 3\).

Award 1 mark for correct answer C. No partial marks are awarded for multiple-choice questions.

Let \(\bar{x}\) and \(v\) be the mean and variance of \(\{x_1, \dots, x_n\}\). Since \(x_{n+1} = \bar{x}\), the new mean \(\bar{y}\) is: \(\bar{y} = \frac{1}{n+1} \left( \sum_{i=1}^n x_i + x_{n+1} \right) = \frac{n \bar{x} + \bar{x}}{n+1} = \bar{x}\). Thus, I is true. The new variance \(u\) is: \(u = \frac{1}{n+1} \left( \sum_{i=1}^n (x_i - \bar{y})^2 + (x_{n+1} - \bar{y})^2 \right) = \frac{1}{n+1} \left( \sum_{i=1}^n (x_i - \bar{x})^2 + 0 \right) = \frac{n}{n+1} v\). Since \(v > 0\) and \(n > 1\), \(\frac{n}{n+1} < 1\), so \(u < v\). Thus, II is true. The new standard deviation is \(\sqrt{u} = \sqrt{\frac{n}{n+1} v} = \sqrt{\frac{n}{n+1}} \sqrt{v}\), which is \(\sqrt{\frac{n}{n+1}}\) times the original standard deviation. Thus, III is true. Therefore, the correct answer is D.

1 mark for the correct option (A/B/C/D). No marks for incorrect or multiple answers.

Rewrite the circle \(C: (x-3)^2 + (y-1)^2 = 5\). The center of \(C\) is \(G(3, 1)\) and the radius is \(R = \sqrt{5}\). Let \(d\) be the perpendicular distance from \(G\) to the line \(L\). Since the chord length \(AB = 4\), we have \(d^2 + (AB/2)^2 = R^2 \implies d^2 + 2^2 = 5 \implies d^2 = 1 \implies d = 1\). The distance from \(G(3, 1)\) to the line \(kx - y = 0\) is given by: \(d = \frac{|k(3) - 1|}{\sqrt{k^2 + 1}} = 1\). Squaring both sides: \((3k - 1)^2 = k^2 + 1 \implies 9k^2 - 6k + 1 = k^2 + 1 \implies 8k^2 - 6k = 0 \implies 2k(4k - 3) = 0\). Thus, \(k = 0\) or \(k = \frac{3}{4}\).

1 mark for the correct option (A/B/C/D). No marks for incorrect or multiple answers.

Since \(2x - 3\) is a factor of \(p(x)\), by the factor theorem, \(p(3/2) = 0 \implies 2(3/2)^3 - k(3/2)^2 + h(3/2) - 6 = 0 \implies 3k - 2h = 1\) (Eq. 1). By the remainder theorem, \(p(-1) = -15 \implies 2(-1)^3 - k(-1)^2 + h(-1) - 6 = -15 \implies k + h = 7\) (Eq. 2). Solving (Eq. 1) and (Eq. 2), we get \(k = 3\) and \(h = 4\). For the cubic equation \(2x^3 - kx^2 + hx - 6 = 0\), the sum of roots is \(\alpha + \beta + \gamma = \frac{k}{2} = \frac{3}{2}\) and the product of roots is \(\alpha\beta\gamma = -\frac{-6}{2} = 3\). The required expression is \(\alpha^2 \beta \gamma + \alpha \beta^2 \gamma + \alpha \beta \gamma^2 = \alpha\beta\gamma(\alpha + \beta + \gamma) = 3 \times \frac{3}{2} = \frac{9}{2}\).

1 mark for the correct option (A/B/C/D). No marks for incorrect or multiple answers.

Let \(Y = \log_4 y\) and \(X = \log_2 x\). Since the vertical intercept of the straight line is \(3\) and the slope is \(-2\), the equation of the line is: \(Y = -2X + 3 \implies \log_4 y = -2\log_2 x + 3\). Using the change of base formula, \(\log_4 y = \frac{\log_2 y}{\log_2 4} = \frac{1}{2}\log_2 y\). Thus, \(\frac{1}{2}\log_2 y = -2\log_2 x + 3 \implies \log_2 y = -4\log_2 x + 6 \implies \log_2 y = \log_2(x^{-4}) + \log_2(64) \implies y = 64x^{-4} = \frac{64}{x^4}\).

1 mark for the correct option (A/B/C/D). No marks for incorrect or multiple answers.

Let \(a_n = 16 r^{n-1}\) for \(n = 1, 2, 3, \dots\). Given \(a_2 + a_4 = \frac{10}{3} a_3\), we have: \(16r + 16r^3 = \frac{10}{3} (16r^2)\). Since \(0 < r < 1\), dividing both sides by \(16r\) gives: \(1 + r^2 = \frac{10}{3} r \implies 3r^2 - 10r + 3 = 0\). Solving this quadratic equation: \((3r - 1)(r - 3) = 0 \implies r = \frac{1}{3}\) or \(r = 3\) (rejected as \(0 < r < 1\)). The sum to infinity \(S_{\infty}\) of the sequence is: \(S_{\infty} = \frac{a_1}{1 - r} = \frac{16}{1 - 1/3} = 24\).

1 mark for the correct option (A/B/C/D). No marks for incorrect or multiple answers.