2021 HKDSE Mathematics Answers & Marking Scheme

Multiply both sides by \( 5 + y \):
\( 3x - 2y = 4a(5 + y) \)
\( 3x - 2y = 20a + 4ay \)

Group all terms containing \( y \) on one side:
\( 3x - 20a = 4ay + 2y \)

Factorize out \( y \):
\( 3x - 20a = y(4a + 2) \)

Divide both sides by \( 4a + 2 \):
\( y = \frac{3x - 20a}{4a + 2} \)

M1: Clearing fraction \( 3x - 2y = 4a(5 + y) \)
M1: Grouping terms in \( y \) to one side \( y(4a + 2) = 3x - 20a \)
A1.89: Correct answer \( y = \frac{3x - 20a}{4a + 2} \) (or equivalent)

First, expand the numerator using power laws:
\( (u^3 v^{-2})^4 = u^{12} v^{-8} \)

Then, divide the powers with the same bases:
\( \frac{u^{12} v^{-8}}{u^{-5} v^3} = u^{12 - (-5)} v^{-8 - 3} = u^{17} v^{-11} \)

Express with positive indices:
\( \frac{u^{17}}{v^{11}} \)

M1: Simplification of numerator to \( u^{12} v^{-8} \)
M1: Applying subtraction rule of indices \( u^{12 - (-5)} \) or \( v^{-8-3} \)
A1.89: Correct final simplified positive index form \( \frac{u^{17}}{v^{11}} \)

By Remainder Theorem, \( f(1) = -6 \):
\( 2(1)^3 + a(1)^2 - 7(1) + b = -6 \)
\( 2 + a - 7 + b = -6 \)
\( a + b = -1 \) --- (1)

By Factor Theorem, \( f(-2) = 0 \):
\( 2(-2)^3 + a(-2)^2 - 7(-2) + b = 0 \)
\( -16 + 4a + 14 + b = 0 \)
\( 4a + b = 2 \) --- (2)

Subtract (1) from (2):
\( 3a = 3 \Rightarrow a = 1 \)

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

Therefore, \( a = 1 \) and \( b = -2 \).

M1: Stating \( f(1) = -6 \) to get \( a + b = -1 \) (or equivalent)
M1: Stating \( f(-2) = 0 \) to get \( 4a + b = 2 \) (or equivalent)
A0.94: Finding \( a = 1 \)
A0.95: Finding \( b = -2 \)

For the quadratic equation to have no real roots, the discriminant \( \Delta \) must be negative:
\( \Delta < 0 \)
\( (-2k)^2 - 4(1)(3k - 2) < 0 \)
\( 4k^2 - 12k + 8 < 0 \)

Divide by 4:
\( k^2 - 3k + 2 < 0 \)

Factorize the quadratic expression:
\( (k - 1)(k - 2) < 0 \)

Therefore, the range of values of \( k \) is \( 1 < k < 2 \).

M1: Using discriminant \( \Delta = (-2k)^2 - 4(1)(3k - 2) < 0 \)
M1: Simplifying to \( k^2 - 3k + 2 < 0 \) (or equivalent quadratic inequality)
A1: Factorizing to \( (k - 1)(k - 2) < 0 \) or finding critical values \( k=1, 2 \)
A0.89: Correct range \( 1 < k < 2 \)

Let \( z = a + bx^2 \), where \( a \) and \( b \) are non-zero constants.

Substitute \( x = 2, z = 14 \):
\( 14 = a + b(2)^2 \Rightarrow a + 4b = 14 \) --- (1)

Substitute \( x = 5, z = 77 \):
\( 77 = a + b(5)^2 \Rightarrow a + 25b = 77 \) --- (2)

Subtract (1) from (2):
\( 21b = 63 \Rightarrow b = 3 \)

Substitute \( b = 3 \) into (1):
\( a + 4(3) = 14 \Rightarrow a = 2 \)

So, the equation is \( z = 2 + 3x^2 \).

When \( x = -3 \):
\( z = 2 + 3(-3)^2 = 2 + 3(9) = 29 \).

M1: Writing down \( z = a + bx^2 \)
M1: Setting up simultaneous equations \( a + 4b = 14 \) and \( a + 25b = 77 \)
A1: Solving to get \( a = 2 \) and \( b = 3 \)
A0.89: Substituting \( x = -3 \) to get \( z = 29 \)

Solve the first inequality:
\( 3(x + 2) > 5x - 4 \)
\( 3x + 6 > 5x - 4 \)
\( 10 > 2x \)
\( x < 5 \)

Solve the second inequality:
\( \frac{3 - x}{2} \le x + 3 \)
\( 3 - x \le 2(x + 3) \)
\( 3 - x \le 2x + 6 \)
\( -3 \le 3x \)
\( x \ge -1 \)

Combining the two results with 'and':
The solution is \( -1 \le x < 5 \).

The integers that satisfy this compound inequality are \( -1, 0, 1, 2, 3, 4 \).
There are 6 such integers.

M1: Solving the first inequality to get \( x < 5 \)
M1: Solving the second inequality to get \( x \ge -1 \)
A0.94: Writing the compound inequality solution \( -1 \le x < 5 \)
A0.95: Stating the number of integers is 6

(a) Let \( a \) be the first term and \( d \) be the common difference.
\( T_3 = a + 2d = 11 \) --- (1)
\( T_8 = a + 7d = 31 \) --- (2)

Subtracting (1) from (2):
\( 5d = 20 \Rightarrow d = 4 \)

Substitute \( d = 4 \) into (1):
\( a + 2(4) = 11 \Rightarrow a = 3 \)

So, the first term is \( 3 \) and the common difference is \( 4 \).

(b) The sum of the first 20 terms:
\( S_{20} = \frac{20}{2} [2a + (20 - 1)d] \)
\( S_{20} = 10 [2(3) + 19(4)] \)
\( S_{20} = 10 [6 + 76] = 10 [82] = 820 \).

M1: Setting up \( a + 2d = 11 \) and \( a + 7d = 31 \)
A1: Finding \( a = 3 \) and \( d = 4 \) (both correct for 1 mark)
M1: Using sum formula \( S_{20} = \frac{20}{2}[2(3) + 19(4)] \)
A0.89: Finding \( S_{20} = 820 \)

First, find the midpoint \( M \) of \( AB \):
\( M = \left(\frac{2 + 6}{2}, \frac{5 + (-3)}{2}\right) = (4, 1) \)

Next, find the slope of \( AB \):
\( m_{AB} = \frac{-3 - 5}{6 - 2} = \frac{-8}{4} = -2 \)

Since \( L \) is perpendicular to \( AB \), the slope of \( L \) is:
\( m_L = -\frac{1}{m_{AB}} = \frac{1}{2} \)

The equation of \( L \) using point-slope form:
\( y - 1 = \frac{1}{2}(x - 4) \)
\( 2y - 2 = x - 4 \)
\( x - 2y - 2 = 0 \)

M1: Finding midpoint of \( AB \) as \( (4, 1) \)
M1: Finding slope of \( L \) as \( \frac{1}{2} \) (by perpendicular rule \( m_1 m_2 = -1 \))
M1: Using point-slope form with midpoint and perpendicular slope
A0.89: Getting correct equation \( x - 2y - 2 = 0 \) (or equivalent)

Arrange the scores in ascending order:
52, 54, 57, 61, 63, 63, 65, 68, 70, 72, 74, 76, 81, 85, 89.

Since there are 15 data points, the median is the 8th term:
Median = 68

Range = Maximum value - Minimum value
Range = 89 - 52 = 37

To find the standard deviation, use a calculator in SD mode:
Mean \( \mu = 67 \)
Standard deviation \( \sigma = \sqrt{\frac{(52-67)^2 + (54-67)^2 + \dots + (89-67)^2}{15}} = \sqrt{339} \approx 18.41 \)

A1: Median = 68
A1: Range = 37
M0.89: Applying standard deviation formula or using calculator
A1: Standard deviation \( \approx 18.41 \) (accept 18.4)

(a) Since \(x-2\) is a factor of \(p(x)\), by Factor Theorem, \(p(2) = 0\). \(3(2)^3 + a(2)^2 + b(2) + 12 = 0 \implies 24 + 4a + 2b + 12 = 0 \implies 2a + b = -18\) --- (1). Since the remainder of \(p(x)\) divided by \(x+1\) is \(15\), by Remainder Theorem, \(p(-1) = 15\). \(3(-1)^3 + a(-1)^2 + b(-1) + 12 = 15 \implies -3 + a - b + 12 = 15 \implies a - b = 6\) --- (2). Adding (1) and (2): \(3a = -12 \implies a = -4\). Substituting \(a = -4\) into (2): \(-4 - b = 6 \implies b = -10\). (b) Substituting \(a = -4\) and \(b = -10\), we get \(p(x) = 3x^3 - 4x^2 - 10x + 12\). Since \(x-2\) is a factor of \(p(x)\), by performing division, we have: \(p(x) = (x-2)(3x^2 + 2x - 6)\). For the equation \(p(x) = 0\), we have \(x-2 = 0\) or \(3x^2 + 2x - 6 = 0\). The roots of \(3x^2 + 2x - 6 = 0\) are: \(x = \frac{-2 \pm \sqrt{2^2 - 4(3)(-6)}}{2(3)} = \frac{-2 \pm \sqrt{76}}{6} = \frac{-1 \pm \sqrt{19}}{3}\). Since \(\sqrt{19}\) is an irrational number, the roots \(\frac{-1 \pm \sqrt{19}}{3}\) are irrational. Thus, not all the roots of \(p(x) = 0\) are rational numbers. The claim is incorrect. Therefore, I disagree.

(a) 1M for setting up either p(2) = 0 or p(-1) = 15; 1M for solving the simultaneous equations; 1A for both a = -4 and b = -10. (b) 1M for factorizing p(x) to get (x-2)(3x^2 + 2x - 6); 1M for attempting to solve 3x^2 + 2x - 6 = 0; 1A for identifying that the roots are irrational; 1A for correct conclusion with explanation.

(a) Let \(C = k_1 + k_2 r^2 h\), where \(k_1\) and \(k_2\) are non-zero constants. Substitute \(r=3, h=5, C=130\): \(k_1 + k_2 (3)^2 (5) = 130 \implies k_1 + 45k_2 = 130\) --- (1). Substitute \(r=4, h=10, C=360\): \(k_1 + k_2 (4)^2 (10) = 360 \implies k_1 + 160k_2 = 360\) --- (2). Subtracting (1) from (2): \(115k_2 = 230 \implies k_2 = 2\). Substitute \(k_2 = 2\) into (1): \(k_1 + 45(2) = 130 \implies k_1 = 40\). Thus, \(C = 40 + 2r^2 h\). When \(r=5\) and \(h=8\): \(C = 40 + 2(5)^2 (8) = 40 + 400 = 440\). The cost of manufacturing the cylinder is \(\$440\). (b) Let the varying part of the cost be \(V = k_2 r^2 h = 2r^2 h\). Let the new base radius be \(r' = 0.5r\) and the new height be \(h' = 2h\). The new varying part \(V' = 2(r')^2 (h') = 2(0.5r)^2 (2h) = 2(0.25 r^2)(2h) = 0.5(2r^2 h) = 0.5V\). The percentage change in the varying part of the cost is: \(\frac{V' - V}{V} \times 100\% = \frac{0.5V - V}{V} \times 100\% = -50\%\). Thus, the varying part of the cost decreases by \(50\%\).

(a) 1M for setting up the relation C = k_1 + k_2 r^2 h; 1M for solving to find k_1 = 40 and k_2 = 2; 1M for substituting r=5 and h=8; 1A for the cost $440 (or 440). (b) 1M for expressing the new varying part in terms of the original varying part; 1M for setting up the percentage change formula; 1A for -50% (or decreases by 50%).

(a) Comparing \(x^2 + y^2 - 12x - 4y + 15 = 0\) with the general form, we have: Center of \(C\) is \(G = \left(-\frac{-12}{2}, -\frac{-4}{2}\right) = (6, 2)\). Radius of \(C\) is \(R = \sqrt{6^2 + 2^2 - 15} = \sqrt{36 + 4 - 15} = \sqrt{25} = 5\). (b)(i) The slope of \(3x - 4y + 5 = 0\) is \(-\frac{3}{-4} = \frac{3}{4}\). Since \(L\) is parallel to the line \(3x - 4y + 5 = 0\), the slope of \(L\) is also \(\frac{3}{4}\). The equation of \(L\) is: \(y - 1 = \frac{3}{4}(x - 2) \implies 4(y - 1) = 3(x - 2) \implies 4y - 4 = 3x - 6 \implies 3x - 4y - 2 = 0\). (b)(ii) Method 1: The perpendicular distance \(d\) from the center \(G(6, 2)\) of circle \(C\) to the line \(L: 3x - 4y - 2 = 0\) is: \(d = \frac{|3(6) - 4(2) - 2|}{\sqrt{3^2 + (-4)^2}} = \frac{|18 - 8 - 2|}{5} = \frac{8}{5} = 1.6\). Since the perpendicular distance \(d = 1.6 < R = 5\), the line \(L\) intersects the circle \(C\) at two distinct points. Thus, \(L\) intersects \(C\). Method 2: From \(L\), we have \(y = \frac{3x - 2}{4}\). Substituting this into the equation of \(C\): \(x^2 + \left(\frac{3x - 2}{4}\right)^2 - 12x - 4\left(\frac{3x - 2}{4}\right) + 15 = 0 \implies 16x^2 + (9x^2 - 12x + 4) - 192x - 16(3x - 2) + 240 = 0 \implies 25x^2 - 252x + 276 = 0\). The discriminant is \(\Delta = (-252)^2 - 4(25)(276) = 63504 - 27600 = 35904 > 0\). Since \(\Delta > 0\), the equation has two distinct real roots. Thus, \(L\) intersects \(C\).

(a) 1A for Center = (6, 2); 1A for Radius = 5. (b)(i) 1M for finding the slope of L is 3/4 or setting up L: 3x - 4y + k = 0; 1A for 3x - 4y - 2 = 0. (b)(ii) 1M for attempting to find the perpendicular distance from center to L (or substituting y into circle equation to get a quadratic equation in x); 1A for finding distance d = 1.6 (or finding discriminant discriminant = 35904); 1A for comparing and concluding L intersects C.

(a) Let \(a\) be the first term and \(r\) be the common ratio of the geometric sequence. We have: \(G_2 = ar = 12\) --- (1). \(G_5 = ar^4 = 96\) --- (2). Dividing (2) by (1): \(\frac{ar^4}{ar} = \frac{96}{12} \implies r^3 = 8 \implies r = 2\). Substitute \(r = 2\) into (1): \(a(2) = 12 \implies a = 6\). So, the first term is \(6\) and the common ratio is \(2\). (b)(i) The sum of the first \(n\) terms is: \(\sum_{i=1}^n G_i = \frac{a(r^n - 1)}{r - 1} = \frac{6(2^n - 1)}{2 - 1} = 6(2^n - 1)\). We require: \(6(2^n - 1) > 10^5 \implies 2^n - 1 > \frac{100000}{6} \implies 2^n > 16666.67 + 1 \implies 2^n > 16667.67\). Taking logarithm on both sides: \(n \log 2 > \log(16667.67) \implies n > \frac{\log(16667.67)}{\log 2} \approx 14.02\). Since \(n\) must be an integer, the least value of \(n\) is \(15\). (b)(ii) The general term is \(G_n = a r^{n-1} = 6 \cdot 2^{n-1}\). Then: \(A_n = \log_2 (G_n) = \log_2 (6 \cdot 2^{n-1}) = \log_2 6 + \log_2 (2^{n-1}) = \log_2 6 + (n - 1) \log_2 2 = \log_2 6 + n - 1\). Consider the difference between consecutive terms: \(A_{n+1} - A_n = (\log_2 6 + (n + 1) - 1) - (\log_2 6 + n - 1) = (\log_2 6 + n) - (\log_2 6 + n - 1) = 1\). Since the difference \(A_{n+1} - A_n\) is a constant (which is \(1\)) independent of \(n\), \(A_1, A_2, A_3, \dots\) is an arithmetic sequence.

(a) 1M for setting up equations ar = 12 and ar^4 = 96; 1A for both a = 6 and r = 2. (b)(i) 1M for setting up 6(2^n - 1) > 10^5; 1M for using logarithms to solve the inequality; 1A for n = 15. (b)(ii) 1M for expressing A_n or A_{n+1} - A_n in algebraic terms; 1A for showing the difference is constant and concluding.

(a) The 20 scores sorted in ascending order are: 42, 45, 45, 48, 50, 53, 54, 54, 57, 58, 61, 61, 61, 65, 65, 73, 74, 75, 77, 82. - Median: Since \(N=20\), the median is the average of the 10th and 11th scores: \(\text{Median} = \frac{58 + 61}{2} = 59.5\) marks. - Range: \(\text{Range} = 82 - 42 = 40\) marks. - Interquartile range (IQR): The lower quartile \(Q_1\) is the median of the first 10 scores (5th and 6th scores): \(Q_1 = \frac{50 + 53}{2} = 51.5\) marks. The upper quartile \(Q_3\) is the median of the last 10 scores (15th and 16th scores): \(Q_3 = \frac{65 + 73}{2} = 69\) marks. \(\text{IQR} = Q_3 - Q_1 = 69 - 51.5 = 17.5\) marks. (b)(i) Sum of the 20 scores: \(42+45+45+48+50+53+54+54+57+58+61+61+61+65+65+73+74+75+77+82 = 1200\). Mean of the original 20 scores \(= \frac{1200}{20} = 60\) marks. (b)(ii) Since the mean of the 22 scores remains unchanged, the new mean is still 60. Sum of the 22 scores \(= 60 \times 22 = 1320\). Thus, \(x + y = 1320 - 1200 = 120\) --- (1). The original range is 40. Since the range increases by 6 marks, the new range is \(40 + 6 = 46\). If only one score (say \(x\)) lies outside the original range [42, 82] and the other (say \(y\)) lies within or on the boundaries: Case 1: \(x < 42\) and \(y \le 82\). New range \(= 82 - x = 46 \implies x = 36\). From (1), \(y = 120 - 36 = 84\), which contradicts \(y \le 82\). Case 2: \(x \ge 42\) and \(y > 82\). New range \(= y - 42 = 46 \implies y = 88\). From (1), \(x = 120 - 88 = 32\), which contradicts \(x \ge 42\). Therefore, both \(x\) and \(y\) must lie outside the original range. Assume \(x < 42\) and \(y > 82\). New range \(= y - x = 46\) --- (2). Adding (1) and (2): \(2y = 166 \implies y = 83\). Substituting \(y = 83\) into (1): \(x = 120 - 83 = 37\). Since \(37 < 42\) and \(83 > 82\), the values are consistent. Thus, the values of \(x\) and \(y\) are \(37\) and \(83\).

(a) 1A for Median = 59.5 and Range = 40; 1A for IQR = 17.5. (b)(i) 1A for mean = 60. (b)(ii) 1M for setting up x + y = 120; 1M for analyzing and setting up y - x = 46; 1M for solving the simultaneous equations; 1A for x = 37, y = 83 (or vice versa).

(a) Let \(a\) be the first term and \(r\) be the common ratio of \(G\).
We have \(ar = 12\) and \(\frac{a}{1-r} = 64\).
From the second equation, \(a = 64(1-r)\).
Substituting this into the first equation, we get:
\(64r(1-r) = 12\)
\(64r^2 - 64r + 12 = 0\)
\(16r^2 - 16r + 3 = 0\)
\((4r-1)(4r-3) = 0\)
Thus, \(r = \frac{1}{3}\) or \(r = \frac{3}{4}\).

(b) Since the larger common ratio is \(r = \frac{3}{4}\), we have:
\(a = \frac{12}{3/4} = 16\).
The sum of the first \(k\) terms is given by:
\(S_k = \frac{16(1 - (3/4)^k)}{1 - 3/4} = 64(1 - 0.75^k)\).
To find the least \(k\) such that \(S_k > 47.9\):
\(64(1 - 0.75^k) > 47.9\)
\(1 - 0.75^k > 0.7484375\)
\(0.75^k < 0.2515625\)
\(k > \frac{\log(0.2515625)}{\log(0.75)}\)
\(k > 4.797\)
Thus, the least integer value of \(k\) is \(5\).

- (a) Establish equations: 1M
- Solve for quadratic equation in r: 1M
- Factorize and solve: 1M
- State both answers: 1A
- (b) Find a = 16: 1M
- Set up inequality: 1M
- Correct answer k = 5: 1A

(a) Rewrite the equation of \(C\):
\((x-5)^2 - 25 + (y-12)^2 - 144 + 144 = 0\)
\((x-5)^2 + (y-12)^2 = 25\)
Thus, the center of \(C\) is \(Q(5, 12)\) and the radius is \(\sqrt{25} = 5\).

(b) Let the equation of a tangent from \(O(0,0)\) with slope \(m\) be \(y = mx\), which is \(mx - y = 0\).
Since it is tangent to \(C\), the distance from \(Q(5, 12)\) to the line is equal to the radius \(5\):
\(\frac{|5m - 12|}{\sqrt{m^2 + 1}} = 5\)
\((5m - 12)^2 = 25(m^2 + 1)\)
\(25m^2 - 120m + 144 = 25m^2 + 25\)
\(-120m = -119\)
\(m = \frac{119}{120}\)
So, one tangent is \(y = \frac{119}{120}x\), or \(119x - 120y = 0\).
Since only one value of \(m\) is obtained, the other tangent must be vertical:
\(x = 0\).
Thus, the equations of the two tangents are \(x = 0\) and \(119x - 120y = 0\).

(c) Quadrilateral \(OPQR\) consists of two congruent right-angled triangles, \(\triangle OPQ\) and \(\triangle ORQ\).
The distance from \(O(0,0)\) to \(Q(5,12)\) is \(OQ = \sqrt{5^2 + 12^2} = 13\).
Since \(PQ = 5\) (radius), in right-angled \(\triangle OPQ\):
\(OP = \sqrt{OQ^2 - PQ^2} = \sqrt{13^2 - 5^2} = 12\).
Area of \(\triangle OPQ = \frac{1}{2} \times OP \times PQ = \frac{1}{2} \times 12 \times 5 = 30\).
Therefore, the area of quadrilateral \(OPQR = 2 \times 30 = 60\).

- (a) Correct center (5, 12): 1A
- Correct radius 5: 1A
- (b) Set up distance equation: 1M
- Find slope m = 119/120: 1A
- Identify the vertical tangent x = 0: 1A
- (c) Find tangent length OP = 12: 1M
- Find total area = 60: 1A

(a) Total number of ways to choose 3 balls from 10 is \(C^{10}_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120\).
The number of ways to choose 1 red, 1 blue, and 1 yellow ball is:
\(C^4_1 \times C^3_1 \times C^3_1 = 4 \times 3 \times 3 = 36\).
The required probability is \(\frac{36}{120} = \frac{3}{10}\) (or \(0.3\)).

(b) "Otherwise" means all 3 balls are of the same color.
The number of ways to choose 3 balls of the same color is:
\(C^4_3 + C^3_3 + C^3_3 = 4 + 1 + 1 = 6\).
Thus, the probability of drawing 3 balls of the same color is \(\frac{6}{120} = \frac{1}{20}\) (or \(0.05\)).
The probability of drawing exactly 2 balls of the same color is:
\(1 - 0.3 - 0.05 = 0.65\).
The expected number of tokens won is:
\(E = 20 \times 0.3 + 5 \times 0.65 + (-10) \times 0.05\)
\(E = 6 + 3.25 - 0.5 = 8.75\).
Thus, the expected number of tokens won is \(8.75\).

- (a) Total combinations (120) or method: 1M
- Favorable outcomes (36): 1M
- Correct probability 3/10: 1A
- (b) Find probability of same color (0.05): 1A
- Find probability of exactly 2 same color (0.65): 1M
- Set up expectation formula: 1M
- Correct expected value 8.75: 1A

(a) In the right-angled triangle \(\triangle TAB\) (where \(\angle TAB = 90^\circ\)):
\(TA = AB \tan 30^\circ = 8 \times \frac{1}{\sqrt{3}} = \frac{8}{\sqrt{3}}\text{ m} \approx 4.62\text{ m}\).

(b) In \(\triangle ABC\), by the Cosine Formula:
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos \angle ABC\)
\(AC^2 = 8^2 + 7^2 - 2(8)(7)\cos 60^\circ\)
\(AC^2 = 64 + 49 - 56 = 57\)
\(AC = \sqrt{57}\text{ m} \approx 7.550\text{ m}\).
In right-angled triangle \(\triangle TAC\) (where \(\angle TAC = 90^\circ\)):
\(TC = \sqrt{TA^2 + AC^2} = \sqrt{\frac{64}{3} + 57} = \sqrt{\frac{235}{3}} \approx 8.85\text{ m}\).

(c) Let \(\theta\) be the angle of elevation of \(T\) from \(C\), which is \(\angle TCA\).
In right-angled triangle \(\triangle TAC\):
\(\tan \theta = \frac{TA}{AC} = \frac{8/\sqrt{3}}{\sqrt{57}} = \frac{8}{\sqrt{171}} \approx 0.6118\)
\(\theta \approx 31.5^\circ\) (or \(31.455^\circ\)).

- (a) Use tan 30: 1M
- Correct height TA (accept 4.62 m): 1A
- (b) Cosine formula for AC^2: 1M
- Pythagoras theorem for TC: 1M
- Correct distance TC (accept 8.85 m): 1A
- (c) Use tan ratio for elevation angle: 1M
- Correct angle (accept 31.5 degrees): 1A

(a) Let \(x_i\) be the original scores of the students for \(i = 1, 2, \dots, 20\).
Sum of the test scores: \(\sum x_i = 20 \times 65 = 1300\).
Using the formula for standard deviation:
\(\sigma^2 = \frac{\sum x_i^2}{N} - \bar{x}^2\)
\(8^2 = \frac{\sum x_i^2}{20} - 65^2\)
\(64 = \frac{\sum x_i^2}{20} - 4225\)
\(\frac{\sum x_i^2}{20} = 4289\)
\(\sum x_i^2 = 85780\).

(b) (i) Correct sum of scores:
\(\text{New Sum} = 1300 - 50 - 80 + 55 + 75 = 1300\).
Correct mean \(= \frac{1300}{20} = 65\).

(ii) Correct sum of squares of scores:
\(\text{New Sum of squares} = 85780 - 50^2 - 80^2 + 55^2 + 75^2\)
\(= 85780 - 2500 - 6400 + 3025 + 5625 = 85530\).
Correct variance:
\(\sigma_{\text{new}}^2 = \frac{85530}{20} - 65^2 = 4276.5 - 4225 = 51.5\).
Correct standard deviation:
\(\sigma_{\text{new}} = \sqrt{51.5} \approx 7.18\).

- (a) Correct sum (1300): 1A
- Correct formula set up for sum of squares: 1M
- Correct sum of squares (85780): 1A
- (b)(i) Correct mean 65: 1A
- (b)(ii) Method to find new sum of squares: 1M
- Find new variance (51.5) or standard deviation formula: 1M
- Correct standard deviation (accept 7.18): 1A

From the quadratic equation, we have:
\(\alpha + \beta = \frac{5}{2}\)
\(\alpha\beta = \frac{1}{2}\)

Using the identity:
\(\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2)\)
\(\alpha^3 + \beta^3 = (\alpha + \beta)[(\alpha + \beta)^2 - 3\alpha\beta]\)

Substitute the values:
\(\alpha^3 + \beta^3 = \frac{5}{2} \left[ \left(\frac{5}{2}\right)^2 - 3\left(\frac{1}{2}\right) \right]\)
\(\alpha^3 + \beta^3 = \frac{5}{2} \left[ \frac{25}{4} - \frac{6}{4} \right]\)
\(\alpha^3 + \beta^3 = \frac{5}{2} \left( \frac{19}{4} \right) = \frac{95}{8}\).

1 mark for the correct answer A.

By the remainder theorem, \(P(1) = -6\) and \(P(-2) = -24\).
\(2(1)^3 + a(1)^2 + b(1) - 6 = -6 \Rightarrow a + b = -2\) --- (1)
\(2(-2)^3 + a(-2)^2 + b(-2) - 6 = -24 \Rightarrow -16 + 4a - 2b - 6 = -24 \Rightarrow 4a - 2b = -2 \Rightarrow 2a - b = -1\) --- (2)

Solving (1) and (2) simultaneously:
\((a+b) + (2a-b) = -2 - 1 \Rightarrow 3a = -3 \Rightarrow a = -1\)
Substituting \(a = -1\) into (1): \(-1 + b = -2 \Rightarrow b = -1\)

So, \(P(x) = 2x^3 - x^2 - x - 6\).
When \(P(x)\) is divided by \(2x-1\), the remainder is:
\(P\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) - 6\)
\(P\left(\frac{1}{2}\right) = \frac{1}{4} - \frac{1}{4} - \frac{1}{2} - 6 = -\frac{13}{2}\).

1 mark for the correct answer A.

The equation can be rewritten as:
\(3 \cdot (3^x)^2 - 10 \cdot 3^x + 3 = 0\)

Let \(y = 3^x\). The equation becomes:
\(3y^2 - 10y + 3 = 0\)
\((3y - 1)(y - 3) = 0\)
\(y = \frac{1}{3}\) or \(y = 3\)

Since \(y = 3^x\):
\(3^x = 3^{-1} \Rightarrow x = -1\)
or
\(3^x = 3^1 \Rightarrow x = 1\)

Thus, \(x = 1\) or \(x = -1\).

1 mark for the correct answer A.

Let \(z = \frac{k x^2}{\sqrt{y}}\), where \(k\) is a non-zero constant.
Let the new values be \(x' = 1.2x\) and \(y' = 0.81y\).

Then the new value of \(z\) is:
\(z' = \frac{k (1.2x)^2}{\sqrt{0.81y}} = \frac{k (1.44 x^2)}{0.9 \sqrt{y}} = 1.6 \left( \frac{k x^2}{\sqrt{y}} \right) = 1.6z\)

Therefore, the percentage change of \(z\) is:
\(\frac{1.6z - z}{z} \times 100\% = 60\%\) (increased by \(60\%\)).

1 mark for the correct answer A.

The 10th term of the sequence \(T_{10}\) is given by:
\(T_{10} = S_{10} - S_9\)

Calculate \(S_{10}\):
\(S_{10} = 2(10)^2 + 5(10) = 2(100) + 50 = 250\)

Calculate \(S_9\):
\(S_9 = 2(9)^2 + 5(9) = 2(81) + 45 = 162 + 45 = 207\)

Subtract the values:
\(T_{10} = 250 - 207 = 43\).

1 mark for the correct answer A.

For the quadratic expression \(x^2 + kx + (k+3)\) to be strictly positive for all real values of \(x\), the coefficient of \(x^2\) must be positive (which is \(1 > 0\)) and its discriminant \(\Delta\) must be strictly negative.

\(\Delta = k^2 - 4(1)(k+3) < 0\)
\(k^2 - 4k - 12 < 0\)
\((k - 6)(k + 2) < 0\)

Therefore, we have:
\(-2 < k < 6\).

1 mark for the correct answer A.

Using trigonometric identities for allied angles:
1) \(\sin(180^\circ - \theta) = \sin\theta\)
2) \(\cos(90^\circ + \theta) = -\sin\theta\)
3) \(\tan(360^\circ - \theta) = -\tan\theta = -\frac{\sin\theta}{\cos\theta}\)

Substitute these into the expression:
\(\frac{\sin\theta \cdot (-\sin\theta)}{-\frac{\sin\theta}{\cos\theta}} = \frac{-\sin^2\theta}{-\frac{\sin\theta}{\cos\theta}} = \sin^2\theta \cdot \frac{\cos\theta}{\sin\theta} = \sin\theta\cos\theta\).

1 mark for the correct answer A.

Find the center \((h, g)\) of the circle:
\(h = -\frac{-6}{2} = 3\)
\(g = -\frac{8}{2} = -4\)
So the center is \((3, -4)\).

The radius \(r\) of the circle is:
\(r = \sqrt{3^2 + (-4)^2 - k} = \sqrt{25 - k}\)

Since the line \(3x - 4y + 5 = 0\) is tangent to \(C\), the perpendicular distance from the center \((3, -4)\) to the line is equal to the radius \(r\):
\(\frac{|3(3) - 4(-4) + 5|}{\sqrt{3^2 + (-4)^2}} = r\)
\(\frac{|9 + 16 + 5|}{\sqrt{25}} = r\)
\(\frac{30}{5} = r \Rightarrow r = 6\)

Now equate \(r\):
\(\sqrt{25 - k} = 6\)
\(25 - k = 36\)
\(k = -11\).

1 mark for the correct answer A.

We can calculate the number of different committees by considering three mutually exclusive cases:

Case 1: Exactly 2 women and 3 men are chosen.
Number of ways = \(C^4_2 \times C^6_3 = 6 \times 20 = 120\)

Case 2: Exactly 3 women and 2 men are chosen.
Number of ways = \(C^4_3 \times C^6_2 = 4 \times 15 = 60\)

Case 3: Exactly 4 women and 1 man are chosen.
Number of ways = \(C^4_4 \times C^6_1 = 1 \times 6 = 6\)

Total number of ways = \(120 + 60 + 6 = 186\).

Alternative method:
Total possible combinations of 5 members from 10 is \(C^{10}_5 = 252\).
We subtract the unwanted cases:
- No women (5 men): \(C^4_0 \times C^6_5 = 1 \times 6 = 6\)
- Exactly 1 woman (4 men): \(C^4_1 \times C^6_4 = 4 \times 15 = 60\)

Total ways = \(252 - (6 + 60) = 186\).

1 mark for the correct answer A.

Let the original set of 10 numbers be \(x_1, x_2, \dots, x_{10}\).
Original Mean \(\mu = 20\).
Original Sum \(\sum_{i=1}^{10} x_i = 10 \times 20 = 200\).

When a new data 20 is added, the new sum is \(200 + 20 = 220\), and the total number of data is 11.
New Mean \(\mu' = \frac{220}{11} = 20\).

Original Standard Deviation \(\sigma = 4 \Rightarrow \text{Variance } \sigma^2 = 16\).
Using the formula \(\sigma^2 = \frac{\sum x_i^2}{10} - \mu^2\):
\(16 = \frac{\sum x_i^2}{10} - 20^2 \Rightarrow \frac{\sum x_i^2}{10} = 416 \Rightarrow \sum_{i=1}^{10} x_i^2 = 4160\).

When the new data 20 is added, the new sum of squares is:
\(\sum_{i=1}^{11} x_i^2 = 4160 + 20^2 = 4160 + 400 = 4560\).

The new variance is:
\(\sigma'^2 = \frac{4560}{11} - \mu'^2 = \frac{4560}{11} - 20^2 = \frac{4560}{11} - 400 = \frac{4560 - 4400}{11} = \frac{160}{11}\).

Thus, the new standard deviation is:
\(\sigma' = \sqrt{\frac{160}{11}} = \sqrt{\frac{16 \times 10}{11}} = 4\sqrt{\frac{10}{11}}\).

1 mark for the correct answer A.

Since \(x-2\) is a factor of \(p(x)\), we have \(p(2) = 0\):
\(a(2)^3 + b(2)^2 - 11(2) - 6 = 0 \Rightarrow 8a + 4b = 28 \Rightarrow 2a + b = 7\) --- (1)

Since \(2x+1\) is a factor of \(p(x)\), we have \(p(-1/2) = 0\):
\(a(-1/2)^3 + b(-1/2)^2 - 11(-1/2) - 6 = 0 \Rightarrow -\frac{a}{8} + \frac{b}{4} + \frac{11}{2} - 6 = 0 \Rightarrow -a + 2b = 4\) --- (2)

From (2), we have \(a = 2b - 4\). Substituting into (1), we get:
\(2(2b-4) + b = 7 \Rightarrow 5b = 15 \Rightarrow b = 3\).
Thus, \(a = 2(3) - 4 = 2\).

Therefore, \(p(x) = 2x^3 + 3x^2 - 11x - 6\).

By the Remainder Theorem, the remainder when \(p(x)\) is divided by \(x-1\) is:
\(p(1) = 2(1)^3 + 3(1)^2 - 11(1) - 6 = 2 + 3 - 11 - 6 = -12\).

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

For the quadratic equation \(x^2 - 2(k-1)x + k^2 - 5k = 0\):
Sum of roots: \(\alpha + \beta = 2(k-1)\)
Product of roots: \(\alpha\beta = k^2 - 5k\)

We are given \(\alpha^2 + \beta^2 = 28\):
\((\alpha + \beta)^2 - 2\alpha\beta = 28\)
\([2(k-1)]^2 - 2(k^2 - 5k) = 28\)
\(4(k^2 - 2k + 1) - 2k^2 + 10k = 28\)
\(2k^2 + 2k + 4 = 28\)
\(2k^2 + 2k - 24 = 0\)
\(k^2 + k - 12 = 0\)
\((k+4)(k-3) = 0\)
Thus, \(k = 3\) or \(k = -4\).

Since \(\alpha\) and \(\beta\) are distinct real roots, the discriminant \(\Delta > 0\):
\(\Delta = [-2(k-1)]^2 - 4(1)(k^2 - 5k) = 4(k^2 - 2k + 1) - 4k^2 + 20k = 12k + 4 > 0 \Rightarrow k > -\frac{1}{3}\).

Therefore, \(k = -4\) must be rejected because it leads to complex roots. The only valid solution is \(k = 3\).

Award 1 mark for the correct answer A. Rejecting \(k = -4\) is the key step that excludes option C.

Using the change of base formula:
\(\log_9 x = \frac{\log_3 x}{\log_3 9} = \frac{\log_3 x}{2} = \log_3 \sqrt{x}\).

Substitute this back into the original equation:
\(\log_3 \sqrt{x} - \log_3 y = 1\)
\(\log_3 \left(\frac{\sqrt{x}}{y}\right) = 1\)
\(\frac{\sqrt{x}}{y} = 3^1 = 3\)
\
\sqrt{x} = 3y\)

Squaring both sides of the equation, we get:
\(x = 9y^2\).

Award 1 mark for the correct answer B.

Let \(a\) be the first term and \(r\) be the common ratio.
\(T_3 = a r^2 = 12\) --- (1)
\(T_6 = a r^5 = 96\) --- (2)

Dividing (2) by (1):
\(\frac{a r^5}{a r^2} = \frac{96}{12} \Rightarrow r^3 = 8 \Rightarrow r = 2\).

Substitute \(r = 2\) into (1):
\(a(2)^2 = 12 \Rightarrow 4a = 12 \Rightarrow a = 3\).

Using the sum formula for a geometric sequence:
\(S_{10} = \frac{a(r^{10} - 1)}{r - 1} = \frac{3(2^{10} - 1)}{2 - 1} = 3(1024 - 1) = 3069\).

Award 1 mark for the correct answer B.

Solve the first inequality:
\(\frac{3x - 5}{2} < 2x + 1 \Rightarrow 3x - 5 < 4x + 2 \Rightarrow x > -7\).

Solve the second inequality:
\(4x - 7 \le 2(x + 3) \Rightarrow 4x - 7 \le 2x + 6 \Rightarrow 2x \le 13 \Rightarrow x \le 6.5\).

Combining the two solutions, we have:
\(-7 < x \le 6.5\).

Since \(x\) must be a non-negative integer, \(x\) can take any integer values from \(\{0, 1, 2, 3, 4, 5, 6\}\).
Thus, there are exactly 7 non-negative integers satisfying the inequalities.

Award 1 mark for the correct answer B. Note that 0 must be included since it is a non-negative integer.

Given the equation of \(C: x^2 + y^2 - 8x + 6y - 11 = 0\):

I. Coordinates of the center \(H = \left(-\frac{-8}{2}, -\frac{6}{2}\right) = (4, -3)\). (Statement I is true)

II. Radius \(R = \sqrt{4^2 + (-3)^2 - (-11)} = \sqrt{16 + 9 + 11} = \sqrt{36} = 6\). (Statement II is true)

III. Distance from point \((1, 2)\) to center \((4, -3)\):
\(d = \sqrt{(1-4)^2 + (2 - (-3))^2} = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} \approx 5.83\).
Since \(d \approx 5.83 < R = 6\), the point \((1, 2)\) lies inside \(C\). (Statement III is true)

Therefore, all statements I, II, and III are true.

Award 1 mark for the correct answer D.

Rewrite the equation using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\):
\(3(1 - \cos^2 \theta) - 5 \cos \theta - 1 = 0\)
\(3 - 3 \cos^2 \theta - 5 \cos \theta - 1 = 0\)
\(3 \cos^2 \theta + 5 \cos \theta - 2 = 0\)
\((3 \cos \theta - 1)(\cos \theta + 2) = 0\)

This gives:
\(\cos \theta = \frac{1}{3}\) or \(\cos \theta = -2\).

Since \(-1 \le \cos \theta \le 1\), the equation \(\cos \theta = -2\) has no real solution.
For \(\cos \theta = \frac{1}{3}\), there are exactly two solutions in the range \(0^\circ \le \theta < 360^\circ\) (one in Quadrant I and one in Quadrant IV).

Therefore, the equation has 2 roots.

Award 1 mark for the correct answer B.

Since \(AP \perp BP\), the product of the slopes of \(AP\) and \(BP\) is \(-1\).
Let the coordinates of \(P\) be \((x, y)\).

\(\frac{y-5}{x-2} \cdot \frac{y-(-3)}{x-8} = -1\)
\(\frac{y-5}{x-2} \cdot \frac{y+3}{x-8} = -1\)
\((y-5)(y+3) = -(x-2)(x-8)\)
\(y^2 - 2y - 15 = -(x^2 - 10x + 16)\)
\(y^2 - 2y - 15 = -x^2 + 10x - 16\)
\(x^2 + y^2 - 10x - 2y + 1 = 0\).

Award 1 mark for the correct answer A.

Total number of balls = \(4 + 5 + 3 = 12\).
The total number of ways to choose 3 balls from 12 without replacement is:
\(C^{12}_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220\).

"At least 2 blue balls" means drawing either exactly 2 blue balls (and 1 non-blue ball) or exactly 3 blue balls.

Case 1: Exactly 2 blue balls and 1 non-blue ball
Number of ways = \(C^5_2 \times C^7_1 = 10 \times 7 = 70\).

Case 2: Exactly 3 blue balls
Number of ways = \(C^5_3 = 10\).

Total number of favorable outcomes = \(70 + 10 = 80\).

Therefore, the required probability is:
\(P = \frac{80}{220} = \frac{4}{11}\).

Award 1 mark for the correct answer C.

Let \(X\) represent the original dataset, with mean \(\bar{X} = 48\) and standard deviation \(\sigma_X = 8\).
The new dataset is given by \(Y = -3X + 10\).

New mean:
\(\bar{Y} = -3\bar{X} + 10 = -3(48) + 10 = -144 + 10 = -134\).

New standard deviation:
\(\sigma_Y = |-3| \times \sigma_X = 3 \times 8 = 24\).
Note that standard deviation must be non-negative and is unaffected by adding a constant.

Thus, the new mean is \(-134\) and the new standard deviation is \(24\).

Award 1 mark for the correct answer A. Note that standard deviation is always non-negative.

According to the remainder theorem, we have:
\( f(2) = 2(2)^3 + a(2)^2 + b(2) - 5 = 21 \implies 16 + 4a + 2b - 5 = 21 \implies 2a + b = 5 \) --- (1)
\( f(-1) = 2(-1)^3 + a(-1)^2 + b(-1) - 5 = -9 \implies -2 + a - b - 5 = -9 \implies a - b = -2 \) --- (2)
Adding (1) and (2):
\( 3a = 3 \implies a = 1 \).
Substituting \( a = 1 \) into (2):
\( 1 - b = -2 \implies b = 3 \).
Hence, \( f(x) = 2x^3 + x^2 + 3x - 5 \).
The remainder when \( f(x) \) is divided by \( x-1 \) is:
\( f(1) = 2(1)^3 + 1(1)^2 + 3(1) - 5 = 1 \).

1 mark is awarded for the correct option B. No mark for incorrect or blank answers.

The center of the circle \( C \) is \( (3, -4) \).
The radius of the circle is \( r = \sqrt{3^2 + (-4)^2 - k} = \sqrt{25 - k} \).
Since the line \( 3x - 4y + 5 = 0 \) is tangent to \( C \), the perpendicular distance from the center \( (3, -4) \) to the line is equal to the radius \( r \).
\( d = \frac{|3(3) - 4(-4) + 5|}{\sqrt{3^2 + (-4)^2}} = \frac{|9 + 16 + 5|}{5} = 6 \).
Hence, \( r = 6 \implies r^2 = 36 \).
\( 25 - k = 36 \implies k = -11 \).

1 mark is awarded for the correct option A. No mark for incorrect or blank answers.

We can rewrite each term using trigonometric identities:
1. \( \sin(360^\circ - \theta) = -\sin\theta \)
2. \( \cos(90^\circ - \theta) = \sin\theta \)
3. \( \sin(180^\circ + \theta) = -\sin\theta \)
4. \( \tan(180^\circ - \theta) = -\tan\theta = -\frac{\sin\theta}{\cos\theta} \)

Substituting these into the expression:
\( \frac{(-\sin\theta)(\sin\theta)}{(-\sin\theta)(-\tan\theta)} = \frac{-\sin^2\theta}{\sin\theta\tan\theta} = -\frac{\sin\theta}{\tan\theta} = -\cos\theta \).

1 mark is awarded for the correct option B. No mark for incorrect or blank answers.

Using the change-of-base formula, we have:
\( \log_{16} y = \frac{\log_4 y}{\log_4 16} = \frac{\log_4 y}{2} = \frac{1}{2}\log_4 y = \log_4 \sqrt{y} \).
Thus, the equation becomes:
\( \log_4 x - \log_4 \sqrt{y} = 1 \)
\( \log_4\left(\frac{x}{\sqrt{y}}\right) = 1 \)
\( \frac{x}{\sqrt{y}} = 4^1 = 4 \)
\( \sqrt{y} = \frac{x}{4} \)
Squaring both sides:
\( y = \frac{x^2}{16} \).

1 mark is awarded for the correct option B. No mark for incorrect or blank answers.

Let \( z = \frac{k x^2}{\sqrt{y}} \), where \( k \) is a non-zero constant.
Let the new values of \( x \) and \( y \) be \( x' \) and \( y' \) respectively.
\( x' = (1 + 20\%)x = 1.2x \)
\( y' = (1 - 36\%)y = 0.64y \)
The new value of \( z \), denoted by \( z' \), is:
\( z' =
\frac{k (x')^2}{\sqrt{y'}} =
\frac{k (1.2x)^2}{\sqrt{0.64y}} =
\frac{1.44 k x^2}{0.8 \sqrt{y}} =
1.8 \left(\frac{k x^2}{\sqrt{y}}\right) = 1.8 z \).
Thus, the percentage change in \( z \) is:
\( \frac{1.8z - z}{z} \times 100\% = 80\% \) (increase).

1 mark is awarded for the correct option A. No mark for incorrect or blank answers.

From the given equation \( 2x^2 - 6x + 3 = 0 \), we have:
\( \alpha + \beta = -\frac{-6}{2} = 3 \)
\( \alpha\beta = \frac{3}{2} \)
Now, we simplify the required expression:
\( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} \)
Substituting the values:
\( \frac{3^2 - 2\left(\frac{3}{2}\right)}{\frac{3}{2}} = \frac{9 - 3}{1.5} = \frac{6}{1.5} = 4 \).

1 mark is awarded for the correct option C. No mark for incorrect or blank answers.

Let \( a \) be the first term and \( r \) be the common ratio of the geometric sequence.
We are given:
\( T_2 = ar = 12 \) --- (1)
\( T_5 = ar^4 = 324 \) --- (2)
Dividing (2) by (1):
\( r^3 = \frac{324}{12} = 27 \implies r = 3 \).
Substituting \( r = 3 \) into (1):
\( a(3) = 12 \implies a = 4 \).
The sum of the first 6 terms \( S_6 \) is:
\( S_6 = \frac{a(r^6 - 1)}{r - 1} = \frac{4(3^6 - 1)}{3 - 1} = \frac{4(729 - 1)}{2} = 2(728) = 1456 \).

1 mark is awarded for the correct option A. No mark for incorrect or blank answers.

Let us determine the vertices of the feasible region by finding the intersection points of the boundary lines:
1. Intersection of \( x + y = 6 \) and \( 2x - y = 0 \):
Adding the two equations: \( 3x = 6 \implies x = 2 \), so \( y = 4 \). Vertex: \( (2, 4) \).
2. Intersection of \( x + y = 6 \) and \( y = 1 \):
\( x + 1 = 6 \implies x = 5 \). Vertex: \( (5, 1) \).
3. Intersection of \( 2x - y = 0 \) and \( y = 1 \):
\( 2x = 1 \implies x = 0.5 \). Vertex: \( (0.5, 1) \).

Now, we evaluate the objective function \( P = 3x + 2y \) at each of these three vertices:
- At \( (2, 4) \): \( P = 3(2) + 2(4) = 6 + 8 = 14 \).
- At \( (5, 1) \): \( P = 3(5) + 2(1) = 15 + 2 = 17 \).
- At \( (0.5, 1) \): \( P = 3(0.5) + 2(1) = 1.5 + 2 = 3.5 \).

Therefore, the maximum value of \( 3x + 2y \) is \( 17 \).

1 mark is awarded for the correct option C. No mark for incorrect or blank answers.

Let \( \mu_x = 40 \) and \( \sigma_x = 6 \) be the original mean and standard deviation.
Under the linear transformation \( y_i = 3 - 2x_i \):
1. The new mean is given by:
\( \mu_y = 3 - 2\mu_x = 3 - 2(40) = 3 - 80 = -77 \).
2. The new standard deviation is given by:
\( \sigma_y = |-2|\sigma_x = 2(6) = 12 \).
Therefore, the new mean is \( -77 \) and the new standard deviation is \( 12 \).

1 mark is awarded for the correct option A. No mark for incorrect or blank answers.

Since the committee of 5 must contain at least 3 teachers, we have three cases:
Case 1: 3 teachers and 2 students
Number of ways = \( C^6_3 \times C^5_2 = 20 \times 10 = 200 \).
Case 2: 4 teachers and 1 student
Number of ways = \( C^6_4 \times C^5_1 = 15 \times 5 = 75 \).
Case 3: 5 teachers and 0 students
Number of ways = \( C^6_5 \times C^5_0 = 6 \times 1 = 6 \).

Total number of ways = \( 200 + 75 + 6 = 281 \).

1 mark is awarded for the correct option A. No mark for incorrect or blank answers.

Let \(Y = \log_5 y\) and \(X = \log_5 x\). The straight line passes through the points \((3, 0)\) and \((0, -2)\).

The slope of the line is \(m = \frac{0 - (-2)}{3 - 0} = \frac{2}{3}\).

The equation of the line is:
\(Y = \frac{2}{3}X - 2\)

Substitute \(Y\) and \(X\) back:
\(\log_5 y = \frac{2}{3}\log_5 x - 2\)

Multiply the entire equation by 3:
\(3\log_5 y = 2\log_5 x - 6\)
\(\log_5 y^3 = \log_5 x^2 - \log_5 5^6\)
\(\log_5 y^3 = \log_5 \left(\frac{x^2}{15625}\right)\)
\(y^3 = \frac{x^2}{15625}\)
\(x^2 = 15625 y^3\)

1 mark for the correct option (A).

Let \(a\) be the first term and \(r\) be the common ratio.
We are given:
1) \(a + ar = 8 \implies a(1+r) = 8\)
2) \(\frac{a}{1-r} = 9 \implies a = 9(1-r)\)

Substitute (2) into (1):
\(9(1-r)(1+r) = 8\)
\(9(1-r^2) = 8\)
\(1-r^2 = \frac{8}{9}\)
\(r^2 = \frac{1}{9}\)
\(r = \pm \frac{1}{3}\)

Case 1: If \(r = \frac{1}{3}\), then \(a = 9\left(1-\frac{1}{3}\right) = 6\).
Case 2: If \(r = -\frac{1}{3}\), then \(a = 9\left(1 - \left(-\frac{1}{3}\right)\right) = 12\).

The sum of all possible values of the first term is \(6 + 12 = 18\).

1 mark for the correct option (C).

Since \(ABCD\) is a regular tetrahedron with side length 6, \(\triangle ABC\) and \(\triangle DBC\) are equilateral triangles of side length 6.
\(AN\) is the median of \(\triangle ABC\) from \(A\) to \(BC\), so:
\(AN = 6 \sin 60^\circ = 3\sqrt{3}\)
Similarly, \(DN\) is the median of \(\triangle DBC\), so:
\(DN = 3\sqrt{3}\)

Consider \(\triangle AND\). Since \(AN = DN = 3\sqrt{3}\), \(\triangle AND\) is an isosceles triangle with \(AN = DN\).
Since \(M\) is the midpoint of \(AD\), \(NM\) is the altitude of \(\triangle AND\) from vertex \(N\) to the base \(AD\), meaning \(MN \perp AD\).
In right-angled triangle \(\triangle AMN\):
\(AM = \frac{AD}{2} = 3\)
By Pythagoras' theorem:
\(MN^2 = AN^2 - AM^2\)
\(MN^2 = (3\sqrt{3})^2 - 3^2 = 27 - 9 = 18\)
\(MN = \sqrt{18} = 3\sqrt{2}\)

1 mark for the correct option (B).

To select a team of 5 representatives with at least 2 boys and 2 girls, there are two mutually exclusive cases:
Case 1: 2 boys and 3 girls
Number of ways = \(C^6_2 \times C^5_3 = 15 \times 10 = 150\)

Case 2: 3 boys and 2 girls
Number of ways = \(C^6_3 \times C^5_2 = 20 \times 10 = 200\)

Total number of different teams = \(150 + 200 = 350\).

1 mark for the correct option (C).

We rewrite the equation of the circle \(C\) in standard form:
\((x-4)^2 + (y-4)^2 = 4^2 + 4^2 - 24\)
\((x-4)^2 + (y-4)^2 = 8\)

The center of the circle is \(K(4, 4)\) and the radius is \(R = \sqrt{8}\).

The distance between \(O(0,0)\) and \(K(4,4)\) is:
\(OK = \sqrt{(4-0)^2 + (4-0)^2} = \sqrt{32}\)

Since \(OP\) is tangent to the circle at \(P\), \(\angle OPK = 90^\circ\). By Pythagoras' theorem in \(\triangle OPK\):
\(OP^2 = OK^2 - KP^2\)
\(OP^2 = (\sqrt{32})^2 - R^2 = 32 - 8 = 24\)
\(OP = \sqrt{24} = 2\sqrt{6}\)

(Alternatively, the length of the tangent from \((x_0, y_0)\) is directly given by \(\sqrt{x_0^2 + y_0^2 + Dx_0 + Ey_0 + F}\). For \((0,0)\), this is \(\sqrt{24} = 2\sqrt{6}\).)

1 mark for the correct option (C).

Let \(y_i = -3x_i + 5\) be the new set of data.

For Statement I:
The new mean \(m' = -3m + 5 = 5 - 3m\). This is correct.

For Statement II:
The variance of the data after scaling by \(k\) and shifting by \(c\) is \(k^2\) times the original variance.
Here, \(k = -3\), so the new variance is \(v' = (-3)^2 v = 9v\). This is correct.

For Statement III:
The standard deviation after scaling is \(|k|\) times the original standard deviation.
Since \(|-3| = 3\), the new standard deviation is indeed 3 times the original standard deviation. This is correct.

Therefore, all I, II, and III are correct.

1 mark for the correct option (D).

Let \(X\) be the number of draws needed to get a red ball.
He needs "at least 3 draws" if and only if the first two draws are NOT red (i.e., both the 1st and 2nd draws are blue).

Probability of drawing a blue ball on the 1st draw = \(\frac{6}{10}\).

Since it is without replacement, the probability of drawing a blue ball on the 2nd draw given the first was blue = \(\frac{5}{9}\).

Therefore, the required probability is:
\(P(X \ge 3) = P(\text{1st is blue and 2nd is blue}) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}\).

1 mark for the correct option (A).

Since \(x \ge 0\) and \(y \ge 0\), the feasible region is bounded by the axes and the two lines \(2x + y = 12\) and \(x + 3y = 11\).

Let's find the intersection points (vertices) of the feasible region:
1) Origin \((0, 0)\).
2) On the y-axis (where \(x = 0\)): \(x + 3y \le 11 \implies y \le \frac{11}{3}\). Point is \(\left(0, \frac{11}{3}\right)\).
3) On the x-axis (where \(y = 0\)): \(2x + y \le 12 \implies x \le 6\). Point is \((6, 0)\).
4) Intersection of \(2x + y = 12\) and \(x + 3y = 11\):
From \(2x + y = 12\), we get \(y = 12 - 2x\).
Substitute into \(x + 3y = 11\):
\(x + 3(12 - 2x) = 11 \implies x + 36 - 6x = 11 \implies 5x = 25 \implies x = 5\).
Thus, \(y = 12 - 2(5) = 2\). Point is \((5, 2)\).

Now we evaluate the objective function \(P = 3x + 4y\) at each vertex:
- At \((0, 0)\): \(P = 0\)
- At \((6, 0)\): \(P = 3(6) + 4(0) = 18\)
- At \(\left(0, \frac{11}{3}\right)\): \(P = 3(0) + 4\left(\frac{11}{3}\right) = \frac{44}{3} \approx 14.67\)
- At \((5, 2)\): \(P = 3(5) + 4(2) = 15 + 8 = 23\)

Comparing these values, the maximum value of \(P\) is 23.

1 mark for the correct option (B).

To find the real and imaginary parts of \(z\), we rationalize the denominator:
\(z = \frac{(a+i)(1+2i)}{(1-2i)(1+2i)}\)
\(z = \frac{a(1) + a(2i) + i(1) + 2i^2}{1^2 + 2^2}\)
\(z = \frac{a + 2ai + i - 2}{5}\)
\(z = \frac{(a-2) + (2a+1)i}{5}\)

Thus, the real part of \(z\) is \(\text{Re}(z) = \frac{a-2}{5}\), and the imaginary part is \(\text{Im}(z) = \frac{2a+1}{5}\).

Since \(\text{Re}(z) = \text{Im}(z)\):
\(\frac{a-2}{5} = \frac{2a+1}{5}\)
\(a - 2 = 2a + 1\)
\(a = -3\)

1 mark for the correct option (A).

Since \(\angle APB = 90^\circ\), by the converse of angle in a semi-circle, the locus of \(P\) (excluding the boundary points \(A\) and \(B\)) is a circle with diameter \(AB\).

The midpoint of \(AB\) is the center of the circle, \(C\):
\(C = \left(\frac{2+8}{2}, \frac{6+(-2)}{2}\right) = (5, 2)\)

The diameter is the length of \(AB\):
\(AB = \sqrt{(8-2)^2 + (-2-6)^2} = \sqrt{6^2 + (-8)^2} = 10\)
So, the radius is \(R = 5\).

The equation of the circle is:
\((x-5)^2 + (y-2)^2 = 5^2\)
\(x^2 - 10x + 25 + y^2 - 4y + 4 = 25\)
\(x^2 + y^2 - 10x - 4y + 4 = 0\)

1 mark for the correct option (A).

Let the horizontal axis be \(X = \log_3 x\) and the vertical axis be \(Y = \log_9 y\). Since the \(X\)-intercept and \(Y\)-intercept are \(4\) and \(2\) respectively, the equation of the line is given by: \(\frac{Y - 2}{X - 0} = \frac{0 - 2}{4 - 0} = -\frac{1}{2} \implies Y = -\frac{1}{2}X + 2\). Substituting \(X = \log_3 x\) and \(Y = \log_9 y\) back into the equation: \(\log_9 y = -\frac{1}{2}\log_3 x + 2\). Since \(\log_9 y = \frac{\log_3 y}{\log_3 9} = \frac{1}{2}\log_3 y\), we have: \(\frac{1}{2}\log_3 y = -\frac{1}{2}\log_3 x + 2 \implies \log_3 y = -\log_3 x + 4 \implies \log_3 (xy) = 4 \implies xy = 3^4 = 81\).

1 mark for the correct answer B.

Since the sequence is geometric, the ratio of consecutive terms must be equal: \(\frac{x}{x + 3} = \frac{x - 2}{x}\). Cross-multiplying gives: \(x^2 = (x + 3)(x - 2) \implies x^2 = x^2 + x - 6 \implies x = 6\). Thus, the first term is \(a = 6 + 3 = 9\), and the second term is \(6\). The common ratio is \(r = \frac{6}{9} = \frac{2}{3}\). Since \(|r| < 1\), the sum to infinity exists and is given by: \(S_{\infty} = \frac{a}{1 - r} = \frac{9}{1 - 2/3} = 27\).

1 mark for the correct answer B.

From the circle equation \(x^2 + y^2 - 4x + 6y - 12 = 0\), we identify the coefficients \(D = -4\), \(E = 6\), and \(F = -12\). The center of the circle is \((2, -3)\), and its radius is \(R = \sqrt{2^2 + (-3)^2 - (-12)} = \sqrt{25} = 5\). For the line to intersect the circle at two distinct points, the perpendicular distance \(d\) from the center \((2, -3)\) to the line \(3x - 4y + k = 0\) must be less than the radius: \(d = \frac{|3(2) - 4(-3) + k|}{\sqrt{3^2 + (-4)^2}} < 5 \implies \frac{|k + 18|}{5} < 5 \implies |k + 18| < 25\). This simplifies to \(-25 < k + 18 < 25 \implies -43 < k < 7\).

1 mark for the correct answer A.

The total number of ways to choose 4 members from the 11 people without any restriction is: \(^{11}C_4 = 330\). The number of ways to choose 4 members consisting of only boys is: \(^6C_4 = 15\). The number of ways to choose 4 members consisting of only girls is: \(^5C_4 = 5\). Therefore, the number of ways to form a committee with at least one boy and at least one girl is: \(330 - 15 - 5 = 310\).

1 mark for the correct answer B.

Let the denominator be \(D = 3 - \cos^2 \theta - 2\sin \theta\). Using the trigonometric identity \(\cos^2 \theta = 1 - \sin^2 \theta\), we can rewrite the denominator: \(D = 3 - (1 - \sin^2 \theta) - 2\sin \theta = \sin^2 \theta - 2\sin \theta + 2\). Let \(u = \sin \theta\). Since \(0^\circ \le \theta < 360^\circ\), we have \(-1 \le u \le 1\). Expressing \(D\) as a quadratic function of \(u\): \(D = u^2 - 2u + 2 = (u - 1)^2 + 1\). For \(-1 \le u \le 1\), the minimum value of \(D\) occurs when \(u = 1\), yielding \(D_{min} = (1 - 1)^2 + 1 = 1\). Therefore, the maximum value of the fraction \(\frac{12}{D}\) is \(\frac{12}{D_{min}} = \frac{12}{1} = 12\).

1 mark for the correct answer C.