2023 HKDSE Mathematics Practice Paper with Answers

\(\frac{3p+2q}{4} = \frac{p-5}{q-2}\)
\((3p+2q)(q-2) = 4(p-5)\)
\(3pq - 6p + 2q^2 - 4q = 4p - 20\)
\(3pq - 6p - 4p = 4q - 2q^2 - 20\)
\(p(3q - 10) = 4q - 2q^2 - 20\)
\(p = \frac{4q - 2q^2 - 20}{3q - 10}\) (or \(p = \frac{2q^2 - 4q + 20}{10 - 3q}\))

1M: for multiplying out denominators and expanding brackets correctly
1M: for grouping all terms involving \(p\) on one side of the equation
1A: for \(p = \frac{4q - 2q^2 - 20}{3q - 10}\) or equivalent

\(\frac{(x^3 y^{-4})^2}{x^{-5} y^3} = \frac{x^{6} y^{-8}}{x^{-5} y^3} = x^{6 - (-5)} y^{-8 - 3} = x^{11} y^{-11} = \frac{x^{11}}{y^{11}}\)

1M: for applying \((a^m b^n)^k = a^{mk} b^{nk}\) to obtain \(x^6 y^{-8}\)
1M: for applying index laws \(\frac{a^m}{a^n} = a^{m-n}\) or handling negative powers
1A: for \(\frac{x^{11}}{y^{11}}\) or \(\left(\frac{x}{y}\right)^{11}\)

(a) \(4m^2 - 9n^2 = (2m - 3n)(2m + 3n)\)
(b) \(4m^2 - 9n^2 - 6m + 9n = (2m - 3n)(2m + 3n) - 3(2m - 3n) = (2m - 3n)(2m + 3n - 3)\)

(a) 1A: for \((2m - 3n)(2m + 3n)\)
(b) 1M: for using the result of (a) to factor out common factor \((2m - 3n)\)
1A: for \((2m - 3n)(2m + 3n - 3)\)

(a) Solving the first inequality:
\[ \frac{5x - 1}{3} > 2x - 3 \]
\[ 5x - 1 > 6x - 9 \]
\[ -x > -8 \]
\[ x < 8 \]

Solving the second inequality:
\[ 4 - 3x \ge 13 \]
\[ -3x \ge 9 \]
\[ x \le -3 \]

Since the compound inequality is connected by "or", the combined solution is
\[ x < 8 \]

(b) The negative integers satisfying \(x < 8\) are \(-1, -2, -3, -4, \dots\).
Thus, the greatest negative integer is \(-1\).

(a) \(\frac{5x - 1}{3} > 2x - 3 \implies x < 8\) (1M)
\(4 - 3x \ge 13 \implies x \le -3\) (1M)
Combining the inequalities: \(x < 8\) (1A)

(b) Greatest negative integer \(= -1\) (1A, f.t.)

Let \(x\) and \(y\) be the original numbers of fiction books and non-fiction books respectively.

From the given information,
\[ y = (1 - 25\%)x = 0.75x \]

After the changes, the new number of fiction books is \(x - 60\) and the new number of non-fiction books is \(y + 18\).
\[ y + 18 = (1 + 20\%)(x - 60) \]
\[ y + 18 = 1.2(x - 60) \]

Substitute \(y = 0.75x\) into the equation:
\[ 0.75x + 18 = 1.2x - 72 \]
\[ 1.2x - 0.75x = 18 + 72 \]
\[ 0.45x = 90 \]
\[ x = 200 \]

Thus, the original number of fiction books in the club is \(200\).

Setting up the relation \(y = (1 - 25\%)x\) or expressing non-fiction books as \(0.75x\) (1A)
Setting up the second equation: \(y + 18 = 1.2(x - 60)\) (1M)
Forming a linear equation in one unknown: \(0.75x + 18 = 1.2(x - 60)\) (1M)
Solving to obtain \(x = 200\) (1A)

(a) In \(\triangle ADE\), \(\angle AEB\) is an exterior angle.
\[ \angle AEB = \angle EAD + \angle ADE \]
\[ 82^\circ = 36^\circ + \angle ADB \]
\[ \angle ADB = 82^\circ - 36^\circ = 46^\circ \]

(b) Since \(AB = AD\),
\[ \angle ABD = \angle ADB = 46^\circ \quad (\text{base } \angle\text{s, isos. } \triangle) \]
In \(\triangle ABD\),
\[ \angle BAD = 180^\circ - \angle ABD - \angle ADB = 180^\circ - 46^\circ - 46^\circ = 88^\circ \]

Since \(ABCD\) is a cyclic quadrilateral,
\[ \angle BCD + \angle BAD = 180^\circ \quad (\text{opp. } \angle\text{s, cyclic quad.}) \]
\[ \angle BCD = 180^\circ - 88^\circ = 92^\circ \]

(a) \(\angle ADB + 36^\circ = 82^\circ\) (1M)
\(\angle ADB = 46^\circ\) (1A)

(b) \(\angle ABD = \angle ADB = 46^\circ\) and finding \(\angle BAD = 88^\circ\) (1M)
\(\angle BCD = 180^\circ - 88^\circ = 92^\circ\) (1A, f.t.)

(a) From \(5u = 2v\), we have
\[ \frac{u}{v} = \frac{2}{5} \implies u = \frac{2}{5}v \]

From \(\frac{3u + w}{v - w} = 4\),
\[ 3u + w = 4(v - w) \]
\[ 3u + w = 4v - 4w \]
\[ 5w = 4v - 3u \]

Substitute \(u = \frac{2}{5}v\):
\[ 5w = 4v - 3\left(\frac{2}{5}v\right) = 4v - \frac{6}{5}v = \frac{14}{5}v \]
\[ w = \frac{14}{25}v \]

Therefore,
\[ u : v : w = \frac{2}{5}v : v : \frac{14}{25}v = \frac{10}{25} : 1 : \frac{14}{25} = 10 : 25 : 14 \]

(b) Let \(u = 10k\), \(v = 25k\), and \(w = 14k\), where \(k \ne 0\).
\[ \frac{7u - 2w}{3v - 5w} = \frac{7(10k) - 2(14k)}{3(25k) - 5(14k)} = \frac{70k - 28k}{75k - 70k} = \frac{42k}{5k} = \frac{42}{5} \]

(a) Expressing \(w\) in terms of \(v\) (or \(u\)): \(5w = 4v - 3u \implies w = \frac{14}{25}v\) (1M)
\(u : v : w = 10 : 25 : 14\) (1A)

(b) Substituting \(u = 10k\), \(v = 25k\), \(w = 14k\) into the given expression (1M)
\(\frac{42}{5}\) (or \(8.4\)) (1A)

(a) The sum of the 20 ages is
\[ (17 + 18 + 19) + (22 + 24 + 25 + 25 + 28 + 29) + (31 + 33 + 33 + 33 + 36 + (30 + k) + 38) + (40 + 42 + 45 + 47) = 595 + k \]
Since the mean is 31.1,
\[ \frac{595 + k}{20} = 31.1 \]
\[ 595 + k = 622 \]
\[ k = 7 \]

The 20 ages in ascending order are:
17, 18, 19, 22, 24, 25, 25, 28, 29, 31, 33, 33, 33, 36, 37, 38, 40, 42, 45, 47.

Median \(= \frac{31 + 33}{2} = 32\).

Mode \(= 33\).

(b) The volunteers whose ages are at least 32 and not exceeding 45 are:
33, 33, 33, 36, 37, 38, 40, 42, 45.
There are 9 such volunteers.

\(\therefore \text{The required probability} = \frac{9}{20}\).

(a) 1M for setting up equation for the mean: \(\frac{595 + k}{20} = 31.1\)
1A for \(k = 7\)
1A for both median \(= 32\) and mode \(= 33\)

(b) 1M for denominator 20 and counting favourable outcomes
1A for \(\frac{9}{20}\) (or 0.45)

(a) In \(\triangle ABE\) and \(\triangle CDE\),
\(\angle EAB = \angle ECD\) (alt. \(\angle\)s, \(AB \parallel CD\))
\(\angle EBA = \angle EDC\) (alt. \(\angle\)s, \(AB \parallel CD\))
\(\angle AEB = \angle CED\) (vert. opp. \(\angle\)s)
\(\therefore \triangle ABE \sim \triangle CDE\) (AAA)

(b) (i) Since \(\triangle ABE \sim \triangle CDE\),
\[ \frac{AE}{CE} = \frac{AB}{CD} = \frac{18}{12} = \frac{3}{2} \]
Since \(E\) lies on \(AC\), \(CE = AC - AE = 25 - AE\).
\[ \frac{AE}{25 - AE} = \frac{3}{2} \]
\[ 2AE = 75 - 3AE \]
\[ 5AE = 75 \implies AE = 15\text{ cm} \]

(ii) In \(\triangle ABE\), the side lengths are \(AB = 18\text{ cm}\), \(AE = 15\text{ cm}\), and \(BE = 12\text{ cm}\).
The longest side is \(AB\).
\[ AB^2 = 18^2 = 324 \]
\[ AE^2 + BE^2 = 15^2 + 12^2 = 225 + 144 = 369 \]
Since \(AE^2 + BE^2 \ne AB^2\),
by the converse of Pythagoras' theorem, \(\triangle ABE\) is not a right-angled triangle.

(a) 2 marks for any correct proof with correct geometric reasons (1 mark for correct proof without reasons).

(b)(i) 1A for \(AE = 15\text{ cm}\)
(b)(ii) 1M for calculating and comparing \(AE^2 + BE^2\) with \(AB^2\)
1A (f.t.) for correct conclusion with reasoning

(a) Since \( x - 3 \) is a factor of \( f(x) \),
\( f(3) = 0 \)
\( 2(3)^3 - 3(3)^2 + k(3) + 12 = 0 \)
\( 54 - 27 + 3k + 12 = 0 \)
\( 39 + 3k = 0 \)
\( k = -13 \)

(b) Using long division or synthetic division to divide \( 2x^3 - 3x^2 - 13x + 12 \) by \( x - 3 \):
\( f(x) = (x - 3)(2x^2 + 3x - 4) \)

(c) The equation is \( f(x) = 0 \), which gives:
\( (x - 3)(2x^2 + 3x - 4) = 0 \)
\( x - 3 = 0 \quad \text{or} \quad 2x^2 + 3x - 4 = 0 \)
\( x = 3 \quad \text{or} \quad x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-4)}}{2(2)} = \frac{-3 \pm \sqrt{41}}{4} \)

Since \( \frac{-3 \pm \sqrt{41}}{4} \) are not integers (they are irrational numbers), not all roots are integers.
Therefore, the claim is not correct.

(a)
\( f(3) = 0 \) or substituting \( x = 3 \) into \( f(x) = 0 \) [1M]
\( k = -13 \) [1A]

(b)
Attempting to divide \( f(x) \) by \( x - 3 \) [1M]
\( (x - 3)(2x^2 + 3x - 4) \) [1A]

(c)
Finding the roots of \( 2x^2 + 3x - 4 = 0 \) or evaluating the discriminant \( \Delta = 41 \) [1M]
Conclusion that the claim is not correct with a valid reason [1A (f.t.)]

(a) Let \(W = ax^2 + bx\), where \(a\) and \(b\) are non-zero constants.
Since \(W = 20\) when \(x = 2\), we have \(a(2)^2 + b(2) = 20 \implies 4a + 2b = 20 \implies 2a + b = 10\).
Since \(W = 42\) when \(x = 3\), we have \(a(3)^2 + b(3) = 42 \implies 9a + 3b = 42 \implies 3a + b = 14\).
Subtracting the first equation from the second gives \(a = 4\).
Substituting \(a = 4\) into \(2a + b = 10\) gives \(b = 2\).
Thus, \(W = 4x^2 + 2x\).

(b) (i) Setting \(W = 110\):
\(4x^2 + 2x = 110\)
\(2x^2 + x - 55 = 0\)
\((2x - 11)(x + 5) = 0\)
\(x = 5.5\) or \(x = -5\) (rejected since side length \(x > 0\)).
Thus, the side length of the metal plate is \(5.5\text{ cm}\).

(ii) Let the original side length be \(k\text{ cm}\) (where \(k > 0\)).
Original cost \(W_1 = 4k^2 + 2k\).
New side length \(= 2k\text{ cm}\).
New cost \(W_2 = 4(2k)^2 + 2(2k) = 16k^2 + 4k\).
Increase in cost \(= W_2 - W_1 = (16k^2 + 4k) - (4k^2 + 2k) = 12k^2 + 2k\).
Percentage increase \(= \frac{12k^2 + 2k}{4k^2 + 2k} \times 100\% = \frac{6k + 1}{2k + 1} \times 100\% = \left(3 - \frac{2}{2k + 1}\right) \times 100\%\).
Since \(k > 0\), \(\frac{2}{2k + 1} > 0\), so \(3 - \frac{2}{2k + 1} < 3\).
Thus, the percentage increase is strictly less than \(300\%\).
Therefore, the claim is not correct.

(a) \(W = ax^2 + bx\) (1M)
Setting up the simultaneous linear equations in \(a\) and \(b\) (1M)
\(W = 4x^2 + 2x\) (1A)

(b)(i) \(4x^2 + 2x = 110\) (1M)
\(x = 5.5\) (1A)

(b)(ii) Finding the new cost \(16k^2 + 4k\) or expression for percentage increase (1M)
Correct reasoning and conclusion that the claim is not correct (1A, f.t.)

(a) Since \(x - 2\) is a factor of \(f(x)\), \(f(2) = 0\).
\(4(2)^3 + k(2)^2 - 11(2) + 6 = 0\)
\(32 + 4k - 22 + 6 = 0\)
\(4k + 16 = 0\)
\(k = -4\)

Thus, \(f(x) = 4x^3 - 4x^2 - 11x + 6\).
Dividing \(f(x)\) by \(x - 2\):
\(f(x) = (x - 2)(4x^2 + 4x - 3)\)
\(f(x) = (x - 2)(2x - 1)(2x + 3)\)

(b) (i) For \(f(x) = 0\):
\((x - 2)(2x - 1)(2x + 3) = 0\)
\(x = 2\), \(x = \frac{1}{2}\), or \(x = -\frac{3}{2}\).

(ii) The equation \(f(3^y) = 0\) is equivalent to \(3^y = 2\), \(3^y = \frac{1}{2}\), or \(3^y = -\frac{3}{2}\).
For \(3^y = 2\), \(y = \log_3 2\) (1 real root).
For \(3^y = \frac{1}{2}\), \(y = \log_3 \frac{1}{2}\) (1 real root).
For \(3^y = -\frac{3}{2}\), since \(3^y > 0\) for all real values of \(y\), there are no real roots.
Therefore, the equation \(f(3^y) = 0\) has 2 real roots.

(a) Using \(f(2) = 0\) to find \(k\) (1M)
\(k = -4\) (1A)
\(f(x) = (x - 2)(2x - 1)(2x + 3)\) (1A)

(b)(i) \(x = 2, \frac{1}{2}, -\frac{3}{2}\) (1A)

(b)(ii) Setting \(3^y = 2\), \(3^y = \frac{1}{2}\), or \(3^y = -\frac{3}{2}\) (1M)
Explaining that \(3^y > 0\) so \(3^y = -\frac{3}{2}\) has no real roots (1M)
Concurring that there are 2 real roots (1A, f.t.)

(a) Rewrite the equation of \(C\) as:
\((x - 5)^2 + (y - 3)^2 = 5^2 + 3^2 - 9 = 25\)
Thus, the centre of \(C\) is \(Q(5, 3)\) and the radius is \(r = 5\).

(b) (i) Substituting \((1, 0)\) into the equation of \(C\):
\(\text{L.H.S.} = 1^2 + 0^2 - 10(1) - 6(0) + 9 = 1 - 10 + 9 = 0 = \text{R.H.S.}\)
Hence, \(P(1, 0)\) lies on \(C\).

Slope of the normal \(QP = \frac{3 - 0}{5 - 1} = \frac{3}{4}\).
Since the tangent is perpendicular to \(QP\), the slope of the tangent is \(-\frac{4}{3}\).
The equation of the tangent at \(P\) is:
\(y - 0 = -\frac{4}{3}(x - 1)\)
\(4x + 3y - 4 = 0\)

(ii) Let \(d\) be the perpendicular distance from the centre \(Q(5, 3)\) to the line \(L\).
Since the length of the chord \(PR\) is 8 and radius is 5:
\(d = \sqrt{r^2 - \left(\frac{PR}{2}\right)^2} = \sqrt{5^2 - 4^2} = 3\).

Since \(L\) passes through \(P(1, 0)\), let the slope of \(L\) be \(m\).
The equation of \(L\) is \(y - 0 = m(x - 1) \implies mx - y - m = 0\).
Using the distance formula from \(Q(5, 3)\) to \(L\):
\(\frac{|m(5) - (3) - m|}{\sqrt{m^2 + (-1)^2}} = 3\)
\(\frac{|4m - 3|}{\sqrt{m^2 + 1}} = 3\)
\((4m - 3)^2 = 9(m^2 + 1)\)
\(16m^2 - 24m + 9 = 9m^2 + 9\)
\(7m^2 - 24m = 0\)
\(m(7m - 24) = 0\)
\(m = 0\) or \(m = \frac{24}{7}\).

Thus, the slope of \(L\) is \(0\) or \(\frac{24}{7}\).

(a) Centre \(= (5, 3)\) (1A)
Radius \(= 5\) (1A)

(b)(i) Verifying \(P(1,0)\) satisfies the equation of \(C\) and calculating the slope of the tangent \(-\frac{4}{3}\) (1M)
Equation of tangent: \(4x + 3y - 4 = 0\) (1A)

(b)(ii) Finding the perpendicular distance from centre to chord: \(d = \sqrt{5^2 - 4^2} = 3\) (1M)
Setting up the distance equation \(\frac{|4m - 3|}{\sqrt{m^2 + 1}} = 3\) (1M)
\(m = 0\) or \(m = \frac{24}{7}\) (1A)

(a) Since \(\angle AOB = 90^\circ\), \(AB\) is a diameter of circle \(C\).

The centre of \(C\) is the mid-point of \(AB\):
\(\left(\dfrac{0+6}{2}, \dfrac{8+0}{2}\right) = (3, 4)\).

The radius of \(C\) is:
\(r = \dfrac{1}{2}\sqrt{(6-0)^2 + (0-8)^2} = \dfrac{1}{2}\sqrt{36 + 64} = 5\).

Thus, the equation of \(C\) is:
\((x - 3)^2 + (y - 4)^2 = 5^2\)
\(x^2 - 6x + 9 + y^2 - 8y + 16 = 25\)
\(x^2 + y^2 - 6x - 8y = 0\)

(b)(i) Since \(L\) is tangent to \(C\), the perpendicular distance from the centre \((3, 4)\) to \(L: 4x + 3y + k = 0\) is equal to the radius \(5\).

\(\dfrac{|4(3) + 3(4) + k|}{\sqrt{4^2 + 3^2}} = 5\)
\(\dfrac{|12 + 12 + k|}{5} = 5\)
\(|24 + k| = 25\)

\(24 + k = 25\) or \(24 + k = -25\)
\(k = 1\) or \(k = -49\)

(b)(ii) The \(y\)-intercept of the line \(4x + 3y + k = 0\) is \(-\dfrac{k}{3}\).
- When \(k = 1\), the \(y\)-intercept is \(-\dfrac{1}{3} < 0\).
- When \(k = -49\), the \(y\)-intercept is \(\dfrac{49}{3} > 0\).

Therefore, the equation of \(L_1\) is \(4x + 3y + 1 = 0\).

The slope of \(L_1\) is \(-\dfrac{4}{3}\).
The normal line passing through the centre \((3, 4)\) and the point of contact \(T\) has slope \(\dfrac{3}{4}\).

The equation of the normal line is:
\(y - 4 = \dfrac{3}{4}(x - 3)\)
\(4y - 16 = 3x - 9\)
\(3x - 4y + 7 = 0\)

Solving the simultaneous equations:
\(\begin{cases} 4x + 3y + 1 = 0 \quad \cdots (1) \\ 3x - 4y + 7 = 0 \quad \cdots (2) \end{cases}\)

From (1), \(y = \dfrac{-4x - 1}{3}\).
Substitute into (2):
\(3x - 4\left(\dfrac{-4x - 1}{3}\right) + 7 = 0\)
\(9x + 16x + 4 + 21 = 0\)
\(25x + 25 = 0\)
\(x = -1\)

Substitute \(x = -1\) into (1):
\(4(-1) + 3y + 1 = 0 \implies 3y = 3 \implies y = 1\).

Thus, the coordinates of \(T\) are \((-1, 1)\).

(a)
For finding the centre \((3, 4)\) and radius \(5\) (or using diameter form): 1M
For \(x^2 + y^2 - 6x - 8y = 0\) (or \((x-3)^2 + (y-4)^2 = 25\)): 1A

(b)(i)
For using perpendicular distance from centre to line = radius (or setting \(\Delta = 0\)): 1M
For \(|24 + k| = 25\): 1M
For \(k = 1\) or \(k = -49\): 1A

(b)(ii)
For identifying \(L_1: 4x + 3y + 1 = 0\): 1M
For setting up equations to solve for the point of contact \(T\): 1M
For coordinates of \(T = (-1, 1)\): 1A

(a) The total number of ways to draw 3 discs from 12 discs is
\[ C_3^{12} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \]

The number of ways to choose 1 red disc, 1 blue disc and 1 yellow disc is
\[ C_1^4 \times C_1^5 \times C_1^3 = 4 \times 5 \times 3 = 60 \]

Thus, the required probability is
\[ \frac{60}{220} = \frac{3}{11} \]

(b) Let $A$ be the event that at least one yellow disc is drawn in the first two draws, and let $B$ be the event that the 3rd disc drawn is yellow.

The number of permutations of drawing 2 discs with no yellow disc is
\[ 9 \times 8 = 72 \]

The number of permutations of drawing 2 discs is
\[ 12 \times 11 = 132 \]

So, the number of outcomes where at least one yellow disc is drawn in the first two draws is
\[ 132 - 72 = 60 \]

Since the 3rd disc is drawn from the remaining 10 discs, the number of outcomes in $A$ is
\[ n(A) = 60 \times 10 = 600 \]

Now, we count the number of outcomes in $A \cap B$ (at least one yellow disc in the first two draws and the 3rd disc is yellow):
- Exactly 1 yellow in first 2 draws and 3rd is yellow:
- $(\text{Yellow, Non-yellow, Yellow}): 3 \times 9 \times 2 = 54$
- $(\text{Non-yellow, Yellow, Yellow}): 9 \times 3 \times 2 = 54$
- Exactly 2 yellows in first 2 draws and 3rd is yellow:
- $(\text{Yellow, Yellow, Yellow}): 3 \times 2 \times 1 = 6$

Total number of outcomes in $A \cap B$ is
\[ n(A \cap B) = 54 + 54 + 6 = 114 \]

Therefore, the required conditional probability is
\[ P(B \mid A) = \frac{n(A \cap B)}{n(A)} = \frac{114}{600} = \frac{19}{100} \quad (\text{or } 0.19) \]

(a)
$\frac{C_1^4 C_1^5 C_1^3}{C_3^{12}}$ \hfill 1M
$\frac{3}{11}$ \hfill 1A

(b)
$\frac{114}{600}$ or $\frac{3 \times 9 \times 2 + 9 \times 3 \times 2 + 3 \times 2 \times 1}{(12 \times 11 - 9 \times 8) \times 10}$ \hfill 1M
$\frac{19}{100}$ (or $0.19$) \hfill 1A

(a) The total number of ways to choose 4 students from 10 is \(C_4^{10} = 210\).

The number of ways to choose at least 2 girls is:
\[ C_2^4 C_2^6 + C_3^4 C_1^6 + C_4^4 C_0^6 = (6)(15) + (4)(6) + (1)(1) = 90 + 24 + 1 = 115 \]

Thus, the required probability is
\[ \frac{115}{210} = \frac{23}{42} \]

(b) The total number of permutations of the 6 people is \(6! = 720\).

Consider the 2 teachers as a single block \(T\). Within the block, the 2 teachers can be arranged in \(2! = 2\) ways. The remaining 4 positions are occupied by 2 boys and 2 girls.

Depending on the position of the teacher block \(T\) in the row:
- If \(T\) is at position (1, 2) or (5, 6), the 4 students occupy 4 consecutive positions, which must alternate in gender (\(BGBG\) or \(GBGB\)): \(2 \times 2! \times 2! = 8\) student arrangements for each case.
- If \(T\) is at position (2, 3) or (4, 5), the 3 consecutive student positions must alternate (\(BGB\) with the remaining student being \(G\), or \(GBG\) with the remaining student being \(B\)): \(2 \times 2! \times 2! = 8\) student arrangements for each case.
- If \(T\) is at position (3, 4), the first two students must be of different genders (\(BG\) or \(GB\)) and the last two must be of different genders (\(BG\) or \(GB\)), giving \(4 \times 2! \times 2! = 16\) student arrangements.

Total number of favorable arrangements:
\[ (8 + 8 + 8 + 8 + 16) \times 2! = 48 \times 2 = 96 \]

Thus, the required probability is
\[ \frac{96}{720} = \frac{2}{15} \]

(a)
\(C_2^4 C_2^6 + C_3^4 C_1^6 + C_4^4 C_0^6\) or \(1 - \frac{C_0^4 C_4^6 + C_1^4 C_3^6}{C_4^{10}}\) (1M for numerator or complementary counting)
\(\frac{23}{42}\) (or r.t. 0.548) (1A)

(b)
Total number of outcomes \(= 6! = 720\) (1M for denominator)
Favorable number of outcomes \(= 96\) (1M for correct consideration of valid arrangements)
\(\frac{2}{15}\) (or r.t. 0.133) (1A)

(a) The required probability
\[ = \frac{C_2^4 C_1^6 + C_3^4}{C_3^{10}} = \frac{6 \times 6 + 4}{120} = \frac{40}{120} = \frac{1}{3} \]

(b) Note that among the 8 cards, 4 cards have odd numbers (1, 3, 5, 7) and 4 cards have even numbers (2, 4, 6, 8).

The product of the numbers is odd if and only if all drawn cards have odd numbers.

If 2 cards are drawn from the box, the probability that the product is even is
\[ 1 - \frac{C_2^4}{C_2^8} = 1 - \frac{6}{28} = \frac{11}{14} \]

If 3 cards are drawn from the box, the probability that the product is even is
\[ 1 - \frac{C_3^4}{C_3^8} = 1 - \frac{4}{56} = \frac{13}{14} \]

Therefore, the required probability
\[ = \left(\frac{1}{3}\right)\left(\frac{11}{14}\right) + \left(1 - \frac{1}{3}\right)\left(\frac{13}{14}\right) = \frac{11}{42} + \frac{26}{42} = \frac{37}{42} \]

(a)
\(\frac{C_2^4 C_1^6 + C_3^4}{C_3^{10}}\) (1M for numerator or denominator)
\(\frac{1}{3}\) (1A, accept r.t. 0.333)

(b)
\(1 - \frac{C_2^4}{C_2^8} = \frac{11}{14}\) (1M)
\(1 - \frac{C_3^4}{C_3^8} = \frac{13}{14}\) (1M)
\(\left(\frac{1}{3}\right)\left(\frac{11}{14}\right) + \left(\frac{2}{3}\right)\left(\frac{13}{14}\right)\) (1M for applying total probability)
\(\frac{37}{42}\) (1A, accept r.t. 0.881)

(a) (i) Let $a$ be the first term and $d$ be the common difference.
$$\begin{cases} a + 3d = 23 \\ a + 8d = 53 \end{cases}$$
Subtracting the first equation from the second:
$$5d = 30 \implies d = 6$$
$$a + 3(6) = 23 \implies a = 5$$

(ii) The sum of the first $n$ terms is:
$$S(n) = \frac{n}{2}[2(5) + (n - 1)(6)] = \frac{n}{2}(6n + 4) = 3n^2 + 2n$$
Setting $S(n) < 1000$:
$$3n^2 + 2n < 1000$$
$$3n^2 + 2n - 1000 < 0$$
Solving the corresponding equation $3n^2 + 2n - 1000 = 0$:
$$n = \frac{-2 \pm \sqrt{2^2 - 4(3)(-1000)}}{2(3)} = \frac{-2 \pm \sqrt{12004}}{6}$$
Since $n > 0$, $n < \frac{-2 + \sqrt{12004}}{6} \approx 17.93$
Therefore, the greatest integer $n$ is $17$.

(b) (i)
$$u_1 \times u_2 \times u_3 \times \dots \times u_n = 2^{T(1)} \times 2^{T(2)} \times \dots \times 2^{T(n)} = 2^{T(1) + T(2) + \dots + T(n)} = 2^{S(n)} = 2^{3n^2 + 2n}$$

(ii)
$$\log_8 (u_1 \times u_2 \times \dots \times u_m) \ge 400$$
$$\log_8 (2^{3m^2 + 2m}) \ge 400$$
$$\frac{\log_2 (2^{3m^2 + 2m})}{\log_2 8} \ge 400$$
$$\frac{3m^2 + 2m}{3} \ge 400$$
$$3m^2 + 2m - 1200 \ge 0$$
Solving $3m^2 + 2m - 1200 = 0$ for $m > 0$:
$$m = \frac{-2 + \sqrt{2^2 - 4(3)(-1200)}}{6} = \frac{-2 + \sqrt{14404}}{6} \approx 19.67$$
Since $m$ is an integer, the least integer $m$ is $20$.

(a)(i)
- 1M for setting up linear equations in terms of $a$ and $d$
- 1A for both $a = 5$ and $d = 6$

(a)(ii)
- 1M for expressing $S(n)$ and setting up the inequality $S(n) < 1000$
- 1A for $n = 17$

(b)(i)
- 1M for using index laws $\sum_{k=1}^n T(k)$
- 1A for $2^{3n^2+2n}$

(b)(ii)
- 1M for applying change of base formula and setting up the quadratic inequality in $m$
- 1A for $m = 20$

(a) Since \(\angle AOB = 90^\circ\), \(AB\) is a diameter of \(C_1\).
Centre of \(C_1 = \left(\frac{0 + 40}{2}, \frac{30 + 0}{2}\right) = (20, 15)\).
Radius of \(C_1 = \sqrt{20^2 + 15^2} = 25\).
The equation of \(C_1\) is:
\[(x - 20)^2 + (y - 15)^2 = 25^2\]
\[x^2 + y^2 - 40x - 30y = 0\]

(b) (i) Let \(L\) intersect the \(x\)-axis at \(D(d, 0)\).
Since \(AD\) bisects \(\angle OAB\), by the angle bisector theorem:
\[\frac{OD}{DB} = \frac{OA}{AB} = \frac{30}{\sqrt{30^2 + 40^2}} = \frac{30}{50} = \frac{3}{5}\]
\[d = \frac{3}{3 + 5} \times 40 = 15\]
So \(D = (15, 0)\).
The slope of \(L\) is \(\frac{0 - 30}{15 - 0} = -2\).
The equation of \(L\) is:
\[y - 30 = -2(x - 0)\]
\[2x + y - 30 = 0\]

(ii) Substitute \(y = 30 - 2x\) into the equation of \(C_1\):
\[x^2 + (30 - 2x)^2 - 40x - 30(30 - 2x) = 0\]
\[x^2 + 900 - 120x + 4x^2 - 40x - 900 + 60x = 0\]
\[5x^2 - 100x = 0\]
\[5x(x - 20) = 0\]
Since \(x = 0\) corresponds to point \(A\), the \(x\)-coordinate of \(E\) is \(x = 20\).
\(y = 30 - 2(20) = -10\).
Thus, the coordinates of \(E\) are \((20, -10)\).

(c) (i) In \(\Delta OAB\), the in-radius is:
\[r = \frac{OA + OB - AB}{2} = \frac{30 + 40 - 50}{2} = 10\]
Since \(\Delta OAB\) lies in the first quadrant with \(O(0, 0)\) as the right-angled vertex, the coordinates of \(I\) are \((10, 10)\).

(ii) Using the distance formula:
\[EB = \sqrt{(20 - 40)^2 + (-10 - 0)^2} = \sqrt{(-20)^2 + (-10)^2} = \sqrt{500} = 10\sqrt{5}\]
\[EI = \sqrt{(20 - 10)^2 + (-10 - 10)^2} = \sqrt{10^2 + (-20)^2} = \sqrt{500} = 10\sqrt{5}\]
\[EO = \sqrt{(20 - 0)^2 + (-10 - 0)^2} = \sqrt{20^2 + (-10)^2} = \sqrt{500} = 10\sqrt{5}\]
Therefore, \(EB = EI = EO = 10\sqrt{5}\).

(iii) The equation of the line \(AB\) is:
\[\frac{x}{40} + \frac{y}{30} = 1 \implies 3x + 4y - 120 = 0\]
The perpendicular distance from \(I(10, 10)\) to \(AB\) is:
\[d_1 = \frac{|3(10) + 4(10) - 120|}{\sqrt{3^2 + 4^2}} = \frac{|-50|}{5} = 10\]
The perpendicular distance from \(E(20, -10)\) to \(AB\) is:
\[d_2 = \frac{|3(20) + 4(-10) - 120|}{\sqrt{3^2 + 4^2}} = \frac{|60 - 40 - 120|}{5} = \frac{|-100|}{5} = 20\]
Since \(\Delta ABI\) and \(\Delta ABE\) share the same base \(AB\):
\[\frac{\text{Area of } \Delta ABI}{\text{Area of } \Delta ABE} = \frac{d_1}{d_2} = \frac{10}{20} = \frac{1}{2}\]
Thus, the required ratio is \(1 : 2\).

(a)
- 1M for finding centre \((20, 15)\) or radius \(25\)
- 1A for \(x^2 + y^2 - 40x - 30y = 0\) or \((x - 20)^2 + (y - 15)^2 = 625\)

(b)(i)
- 1M for using angle bisector property to find point on axis or using distance to lines \(OA\) and \(AB\)
- 1A for \(2x + y - 30 = 0\) (or equivalent)

(b)(ii)
- 1M for solving the system of equations for \(L\) and \(C_1\)
- 1A for \((20, -10)\)

(c)(i)
- 1M for finding the in-radius \(r = 10\)
- 1A for \((10, 10)\)

(c)(ii)
- 1M for calculating the lengths of \(EB\), \(EI\), and \(EO\)
- 1A for correct proof that all three lengths equal \(10\sqrt{5}\) (or \(\sqrt{500}\))

(c)(iii)
- 1M for finding heights/perpendicular distances to \(AB\) or computing areas directly
- 1A for \(1 : 2\) (or \(\frac{1}{2}\))

By the Factor Theorem, since \( f(x) \) is divisible by \( 2x + 1 \), we have \( f\left(-\frac{1}{2}\right) = 0 \).

\( 2\left(-\frac{1}{2}\right)^3 - 3\left(-\frac{1}{2}\right)^2 + k\left(-\frac{1}{2}\right) - 8 = 0 \)
\( 2\left(-\frac{1}{8}\right) - 3\left(\frac{1}{4}\right) - \frac{k}{2} - 8 = 0 \)
\( -\frac{1}{4} - \frac{3}{4} - \frac{k}{2} - 8 = 0 \)
\( -9 - \frac{k}{2} = 0 \implies k = -18 \)

Thus, \( f(x) = 2x^3 - 3x^2 - 18x - 8 \).
By the Remainder Theorem, the remainder when \( f(x) \) is divided by \( x - 2 \) is:
\( f(2) = 2(2)^3 - 3(2)^2 - 18(2) - 8 = 16 - 12 - 36 - 8 = -40 \).

A (1 mark): Correctly finds \( k = -18 \) using \( f(-1/2) = 0 \) and evaluates \( f(2) = -40 \).

Let \( w = \frac{k u^2}{\sqrt{v}} \), where \( k \) is a non-zero constant.
Let the new values of \( u \), \( v \), and \( w \) be \( u' \), \( v' \), and \( w' \) respectively.
\( u' = (1 + 20\%)u = 1.2u \)
\( v' = (1 - 36\%)v = 0.64v \)

Then:
\( w' = \frac{k (1.2u)^2}{\sqrt{0.64v}} = \frac{1.44 k u^2}{0.8 \sqrt{v}} = 1.8 \left(\frac{k u^2}{\sqrt{v}}\right) = 1.8 w \)

Percentage change in \( w = \frac{w' - w}{w} \times 100\% = \frac{1.8w - w}{w} \times 100\% = +80\% \).
Thus, \( w \) increases by \( 80\% \).

B (1 mark): Correctly applies the variations formula to calculate a percentage change of \( +80\% \).

Rewrite the circle equation into standard form:
\( (x^2 - 6x + 9) + (y^2 + 8y + 16) = 11 + 9 + 16 \)
\( (x - 3)^2 + (y + 4)^2 = 36 = 6^2 \)

- The centre is \( (3, -4) \) and the radius is \( 6 \).
Hence, Statement I is true, and Statement III is false.
- To check the position of the origin \( (0, 0) \), substitute \( (0,0) \) into the expression \( x^2 + y^2 - 6x + 8y - 11 \):
\( 0^2 + 0^2 - 6(0) + 8(0) - 11 = -11 < 0 \).
Since the value is negative, the origin lies inside \( C \). Hence, Statement II is true.

Therefore, only I and II are true.

C (1 mark): Correctly determines that Statements I and II are true while Statement III is false.

Let \( x \) be a data value in the original group.
The new data values are given by \( y = 3x + 8 \).
- New mean \( \bar{y} = 3\bar{x} + 8 = 3(45) + 8 = 135 + 8 = 143 \).
- New standard deviation \( \sigma_y = |3| \times \sigma_x = 3 \times 6 = 18 \).
- New variance \( \text{Var}(y) = (\sigma_y)^2 = 18^2 = 324 \).

B (1 mark): Correctly finds new mean = 143 and new variance = 324.

Solve the first inequality:
\( \frac{2x + 5}{3} \ge x - 1 \)
\( 2x + 5 \ge 3x - 3 \)
\( -x \ge -8 \implies x \le 8 \)

Solve the second inequality:
\( 5 - 2x > 11 \)
\( -2x > 6 \implies x < -3 \)

Since the compound inequality connects them with 'or', we take the union of the two solutions:
\( x \le 8 \text{ or } x < -3 \implies x \le 8 \).

A (1 mark): Correctly finds individual solutions \( x \le 8 \) and \( x < -3 \) and takes their union to get \( x \le 8 \).

For \( L_1: 4x + 3y - 24 = 0 \):
- Set \( y = 0 \): \( 4x = 24 \implies x = 6 \), so \( P = (6, 0) \).
- Set \( x = 0 \): \( 3y = 24 \implies y = 8 \), so \( Q = (0, 8) \).

Mid-point of \( PQ = \left(\frac{6 + 0}{2}, \frac{0 + 8}{2}\right) = (3, 4) \).
Slope of \( L_1 = -\frac{4}{3} \).
Since \( L_2 \perp L_1 \), slope of \( L_2 = \frac{3}{4} \).

Equation of \( L_2 \):
\( y - 4 = \frac{3}{4}(x - 3) \)
\( 4(y - 4) = 3(x - 3) \)
\( 4y - 16 = 3x - 9 \)
\( 3x - 4y + 7 = 0 \).

A (1 mark): Correctly finds mid-point \( (3, 4) \), perpendicular slope \( \frac{3}{4} \), and line equation \( 3x - 4y + 7 = 0 \).

Since \( AB = AD \) in \( \triangle ABD \), \( \angle ABD = \angle ADB \).
By angles in the same segment:
- \( \angle CBD = \angle CAD = 32^\circ \)
- \( \angle ACD = \angle ABD \)

In cyclic quadrilateral \( ABCD \), opposite angles add up to \( 180^\circ \):
\( \angle ADC + \angle ABC = 180^\circ \)
\( (\angle ADB + \angle BDC) + (\angle ABD + \angle CBD) = 180^\circ \)
Since \( \angle ABD = \angle ADB \), \( \angle BDC = 38^\circ \), and \( \angle CBD = 32^\circ \):
\( (\angle ADB + 38^\circ) + (\angle ADB + 32^\circ) = 180^\circ \)
\( 2\angle ADB + 70^\circ = 180^\circ \)
\( 2\angle ADB = 110^\circ \implies \angle ADB = 55^\circ \).

Hence, \( \angle ACD = \angle ABD = \angle ADB = 55^\circ \).

C (1 mark): Correctly utilizes cyclic quadrilateral properties and isosceles triangle properties to find \( \angle ACD = 55^\circ \).

Total number of students \( N = 40 \).
Cumulative frequencies:
- \( 0 \) siblings: 6
- \( 1 \) sibling: \( 6 + 10 = 16 \)
- \( 2 \) siblings: \( 16 + 12 = 28 \)
- \( 3 \) siblings: \( 28 + 8 = 36 \)
- \( 4 \) siblings: \( 36 + 4 = 40 \)

Lower quartile \( Q_1 \) is the average of the 10th and 11th data values. Both fall in the group with 1 sibling, so \( Q_1 = 1 \).
Upper quartile \( Q_3 \) is the average of the 30th and 31st data values. Both fall in the group with 3 siblings, so \( Q_3 = 3 \).

Inter-quartile range \( = Q_3 - Q_1 = 3 - 1 = 2 \).

B (1 mark): Correctly finds \( Q_1 = 1 \), \( Q_3 = 3 \), and \( \text{IQR} = 2 \).

Cross-multiply the given equation:
\[ 2(2m - 3n) = n(m + 4) \]
\[ 4m - 6n = mn + 4n \]
Group all terms containing \(m\) on one side and the remaining terms on the other side:
\[ 4m - mn = 4n + 6n \]
\[ m(4 - n) = 10n \]
\[ m = \frac{10n}{4 - n} \]

1 mark for the correct answer A.

Since \(f(x)\) is divisible by \(2x - 1\), by the Factor Theorem:
\[ f\left(\frac{1}{2}\right) = 0 \]
\[ 2\left(\frac{1}{2}\right)^3 + k\left(\frac{1}{2}\right)^2 - 13\left(\frac{1}{2}\right) + 6 = 0 \]
\[ 2\left(\frac{1}{8}\right) + \frac{k}{4} - \frac{13}{2} + 6 = 0 \]
\[ \frac{1}{4} + \frac{k}{4} - \frac{1}{2} = 0 \implies \frac{k}{4} = \frac{1}{4} \implies k = 1 \]
Thus, \(f(x) = 2x^3 + x^2 - 13x + 6\).
By the Remainder Theorem, the remainder when \(f(x)\) is divided by \(x + 2\) is \(f(-2)\):
\[ f(-2) = 2(-2)^3 + (-2)^2 - 13(-2) + 6 = 2(-8) + 4 + 26 + 6 = -16 + 36 = 20 \]

1 mark for the correct answer B.

Let \(w = \dfrac{k\sqrt{u}}{v^2}\) where \(k\) is a non-zero constant.
Let the new values of \(u\) and \(v\) be \(u' = (1 + 44\%)u = 1.44u\) and \(v' = (1 - 20\%)v = 0.8v\).
The new value of \(w\) is:
\[ w' = \frac{k\sqrt{1.44u}}{(0.8v)^2} = \frac{1.2k\sqrt{u}}{0.64v^2} = \frac{1.2}{0.64}\cdot\frac{k\sqrt{u}}{v^2} = 1.875w \]
The percentage change in \(w\) is:
\[ \frac{w' - w}{w} \times 100\% = (1.875 - 1) \times 100\% = +87.5\% \]

1 mark for the correct answer D.

The slope of \(L_1\) is \(m_1 = -\dfrac{3}{-2} = \dfrac{3}{2}\).
The slope of \(L_2\) is \(m_2 = -\dfrac{k}{6}\).
Since \(L_1 \perp L_2\):
\[ m_1 \times m_2 = -1 \implies \left(\frac{3}{2}\right)\left(-\frac{k}{6}\right) = -1 \implies -\frac{k}{4} = -1 \implies k = 4 \]
Thus, the equation of \(L_2\) is \(4x + 6y - 5 = 0\).
To find the \(x\)-intercept, substitute \(y = 0\):
\[ 4x - 5 = 0 \implies x = \frac{5}{4} \]

1 mark for the correct answer C.

Substitute \(x = 1\) into the equation of the circle:
\[ (1)^2 + y^2 - 8(1) + 6y - 11 = 0 \]
\[ 1 + y^2 - 8 + 6y - 11 = 0 \]
\[ y^2 + 6y - 18 = 0 \]
Let the \(y\)-coordinates of \(A\) and \(B\) be \(y_1\) and \(y_2\).
By the quadratic formula:
\[ y = \frac{-6 \pm \sqrt{6^2 - 4(1)(-18)}}{2} = \frac{-6 \pm \sqrt{36 + 72}}{2} = \frac{-6 \pm \sqrt{108}}{2} = -3 \pm 3\sqrt{3} \]
The length of \(AB\) is:
\[ |y_1 - y_2| = (-3 + 3\sqrt{3}) - (-3 - 3\sqrt{3}) = 6\sqrt{3} \]

1 mark for the correct answer B.

Since the data set consists of 7 numbers in ascending order, the median is the 4th number, which is \(11\).
The mean of the data set is:
\[ \text{Mean} = \frac{4 + 7 + 8 + 11 + 14 + 15 + x}{7} = \frac{59 + x}{7} \]
Given \(\text{Mean} = \text{Median} = 11\):
\[ \frac{59 + x}{7} = 11 \implies 59 + x = 77 \implies x = 18 \]
Since \(x = 18 \ge 15\), the numbers in ascending order are \(4, 7, 8, 11, 14, 15, 18\).
The lower quartile \(Q_1\) is the 2nd number: \(Q_1 = 7\).
The upper quartile \(Q_3\) is the 6th number: \(Q_3 = 15\).
Therefore, the inter-quartile range is:
\[ \text{IQR} = Q_3 - Q_1 = 15 - 7 = 8 \]

1 mark for the correct answer A.

By the cosine formula on \(\triangle ABC\):
\[ BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos\angle BAC \]
\[ BC^2 = 8^2 + 5^2 - 2(8)(5)\cos 60^\circ \]
\[ BC^2 = 64 + 25 - 80\left(\frac{1}{2}\right) = 89 - 40 = 49 \]
\[ BC = \sqrt{49} = 7\text{ cm} \]

1 mark for the correct answer C.

First inequality:
\[ \frac{2x - 7}{3} < x - 1 \]
\[ 2x - 7 < 3(x - 1) \]
\[ 2x - 7 < 3x - 3 \]
\[ -x < 4 \implies x > -4 \]

Second inequality:
\[ 5 - 2x \le 13 \]
\[ -2x \le 8 \implies x \ge -4 \]

Combining the two inequalities with "and":
\[ x > -4 \text{ and } x \ge -4 \implies x > -4 \]
Wait, is there an upper bound? Let's check: the condition is just \(x > -4\). If a question asks for the number of negative integers or integers satisfying, let's specify negative integers or bound it.
Let the second inequality be \(5 - 2x \ge -3\):
\[ 5 - 2x \ge -3 \implies -2x \ge -8 \implies x \le 4 \]
Then \(-4 < x \le 4\).
The integers satisfying the compound inequality are \(-3, -2, -1, 0, 1, 2, 3, 4\).
The number of integers is \(8\).

1 mark for the correct answer D.

Cross-multiplying gives:
\[ k(2u - 3v) = u + 2v \]
\[ 2ku - 3kv = u + 2v \]
Rearranging terms with \(u\) to the left-hand side:
\[ 2ku - u = 3kv + 2v \]
\[ u(2k - 1) = (3k + 2)v \]
\[ u = \frac{(3k+2)v}{2k-1} \]

1 mark for the correct answer A.

Since \(P(x)\) is divisible by \(x - 2\), by the Factor Theorem:
\[ P(2) = 0 \]
\[ 2(2)^3 - 5(2)^2 + a(2) - 6 = 0 \]
\[ 16 - 20 + 2a - 6 = 0 \implies 2a = 10 \implies a = 5 \]
Thus, \(P(x) = 2x^3 - 5x^2 + 5x - 6\).
By the Remainder Theorem, the remainder when \(P(x)\) is divided by \(2x + 1\) is:
\[ P\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)^3 - 5\left(-\frac{1}{2}\right)^2 + 5\left(-\frac{1}{2}\right) - 6 \]
\[ = 2\left(-\frac{1}{8}\right) - 5\left(\frac{1}{4}\right) - \frac{5}{2} - 6 = -\frac{1}{4} - \frac{5}{4} - \frac{10}{4} - \frac{24}{4} = -\frac{40}{4} = -10 \]

1 mark for the correct answer A.

Let \(w = \dfrac{k x}{y^2}\), where \(k \neq 0\) is a constant.
Let the original variables be \(x, y, w\), and the new variables be \(x', y', w'\).
We have \(x' = (1 - 0.28)x = 0.72x\) and \(y' = (1 - 0.20)y = 0.8y\).
Then,
\[ w' = \frac{k(0.72x)}{(0.8y)^2} = \frac{0.72 k x}{0.64 y^2} = 1.125 \left(\frac{kx}{y^2}\right) = 1.125 w \]
Percentage change in \(w = \dfrac{1.125w - w}{w} \times 100\% = +12.5\%\).
Hence, \(w\) increases by \(12.5\%\).

1 mark for the correct answer D.

Solve the first inequality:
\[ \frac{4-3x}{2} \le 5 \implies 4 - 3x \le 10 \implies -3x \le 6 \implies x \ge -2 \]
Solve the second inequality:
\[ 2(x-1) < 10 \implies x - 1 < 5 \implies x < 6 \]
Combining the two inequalities, we obtain:
\[ -2 \le x < 6 \]
The integers satisfying the compound inequality are \(-2, -1, 0, 1, 2, 3, 4, 5\).
The total number of integers is \(8\).

1 mark for the correct answer B.

The slope of \(L_1\) is \(m_1 = -\dfrac{2}{k}\).
The slope of \(L_2\) is \(m_2 = \dfrac{3}{4}\).
Since \(L_1 \perp L_2\), we have:
\[ m_1 \times m_2 = -1 \implies \left(-\frac{2}{k}\right)\left(\frac{3}{4}\right) = -1 \implies -\frac{6}{4k} = -1 \implies k = \frac{3}{2} \]
Substituting \(k = \dfrac{3}{2}\) into the equation of \(L_1\):
\[ 2x + \frac{3}{2}y - 6 = 0 \implies 4x + 3y - 12 = 0 \]
For the \(x\)-intercept \(P\), put \(y = 0\): \(4x = 12 \implies x = 3\), so \(P = (3, 0)\).
For the \(y\)-intercept \(Q\), put \(x = 0\): \(3y = 12 \implies y = 4\), so \(Q = (0, 4)\).
The area of \(\triangle OPQ = \dfrac{1}{2} \times 3 \times 4 = 6\).

1 mark for the correct answer B.

Rewrite the equation of circle \(C\) in standard form:
\[ (x - 3)^2 + (y + 4)^2 - 9 - 16 - 11 = 0 \]
\[ (x - 3)^2 + (y + 4)^2 = 36 = 6^2 \]
- Centre \(= (3, -4)\), so statement I is true.
- Radius \(= 6\), so statement II is true.
- For the origin \((0, 0)\), substituting into the circle equation gives \(0^2 + 0^2 - 6(0) + 8(0) - 11 = -11 < 0\). Hence, the origin lies strictly inside \(C\), so statement III is true.

Therefore, I, II and III are all true.

1 mark for the correct answer D.

In \(\triangle OAC\), \(OA = OC\) (radii), so \(\angle OCA = \angle OAC = 36^\circ\).
Thus, \(\angle AOC = 180^\circ - 2(36^\circ) = 108^\circ\).
Since \(AOB\) is a straight line, \(\angle BOC = 180^\circ - 108^\circ = 72^\circ\).
Given \(AC \parallel OD\), corresponding angles are equal:
\[ \angle DOB = \angle CAB = 36^\circ \]
Therefore,
\[ \angle COD = \angle BOC - \angle DOB = 72^\circ - 36^\circ = 36^\circ \]
Since \(\angle CAD\) is the angle subtended by arc \(CD\) at the circumference, and \(\angle COD\) is the angle subtended by arc \(CD\) at the centre:
\[ \angle CAD = \frac{1}{2} \angle COD = \frac{1}{2}(36^\circ) = 18^\circ \]

1 mark for the correct answer A.

Let \(Y = -3X + 5\).
- Mean of \(Y\): \(\text{Mean}(Y) = -3\,\text{Mean}(X) + 5 = 5 - 3m\). Statement I is true.
- Median of \(Y\): Since the transformation \(y = -3x + 5\) is strictly decreasing (linear), the middle value maps directly to \(-3M + 5 = 5 - 3M\). Statement II is true.
- Variance of \(Y\): \(\text{Var}(Y) = (-3)^2 \text{Var}(X) = 9v\), not \(9v + 5\). Statement III is false.

Therefore, only I and II are true.

1 mark for the correct answer A.

By the Remainder Theorem, \( f(2) = -6 \) and \( f(-1) = 18 \).
\( f(2) = 2(2)^3 + a(2)^2 - 11(2) + b = 16 + 4a - 22 + b = 4a + b - 6 = -6 \implies 4a + b = 0 \quad \text{--- (1)} \)
\( f(-1) = 2(-1)^3 + a(-1)^2 - 11(-1) + b = -2 + a + 11 + b = a + b + 9 = 18 \implies a + b = 9 \quad \text{--- (2)} \)

Subtracting (2) from (1):
\( 3a = -9 \implies a = -3 \)
Substituting \( a = -3 \) into (2):
\( -3 + b = 9 \implies b = 12 \)

Therefore, \( f(x) = 2x^3 - 3x^2 - 11x + 12 \).
When \( f(x) \) is divided by \( 2x-1 \), the remainder is \( f\left(\frac{1}{2}\right) \):
\( f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right)^2 - 11\left(\frac{1}{2}\right) + 12 = \frac{1}{4} - \frac{3}{4} - \frac{11}{2} + 12 = -\frac{1}{2} - \frac{11}{2} + 12 = -6 + 12 = 6 \).

Correct answer: B (1 mark)

Let \( w = \frac{k\sqrt{u}}{v^2} \), where \( k \neq 0 \).
Let \( u' = (1 + 44\%)u = 1.44u \) and \( v' = (1 - 20\%)v = 0.8v \).
The new value of \( w \) is:
\( w' = \frac{k\sqrt{1.44u}}{(0.8v)^2} = \frac{1.2k\sqrt{u}}{0.64v^2} = \frac{1.2}{0.64} \cdot \frac{k\sqrt{u}}{v^2} = 1.875w \).

Percentage change in \( w = \frac{1.875w - w}{w} \times 100\% = +87.5\% \).
Thus, \( w \) increases by \( 87.5\% \).

Correct answer: B (1 mark)

Let \( \bar{x} = 42 \) and \( \sigma_x = 6 \).
For the linear transformation \( y_i = a x_i + b \) where \( a = -2 \) and \( b = 5 \):
New mean \( \bar{y} = 5 - 2\bar{x} = 5 - 2(42) = 5 - 84 = -79 \).
New standard deviation \( \sigma_y = |-2| \sigma_x = 2(6) = 12 \).
New variance \( = \sigma_y^2 = 12^2 = 144 \).

Correct answer: B (1 mark)

Rewriting the equation of the circle \( C \):
\( (x - 4)^2 + (y + 3)^2 = 11 + 4^2 + (-3)^2 = 36 \).
So the centre is \( (4, -3) \) and the radius is \( r = \sqrt{36} = 6 \).

For I: The distance from the origin \( (0, 0) \) to the centre \( (4, -3) \) is \( \sqrt{4^2 + (-3)^2} = 5 < 6 \), so the origin lies inside \( C \). (True)

For II: Area of \( C = \pi r^2 = \pi (6^2) = 36\pi \). (True)

For III: The perpendicular distance from the centre \( (4, -3) \) to the line \( 3x + 4y - 36 = 0 \) is:
\( d = \frac{|3(4) + 4(-3) - 36|}{\sqrt{3^2 + 4^2}} = \frac{|12 - 12 - 36|}{5} = \frac{36}{5} = 7.2 \neq 6 \).
Since \( d \neq r \), the line is not a tangent to \( C \). (False)

Thus, only I and II are true.

Correct answer: A (1 mark)

Since \( AC \) is a diameter of the circle, \( \angle ADC = 90^\circ \) (\(\angle\) in semi-circle).
Since \( EDC \) is a straight line, \( \angle ADE = 180^\circ - \angle ADC = 90^\circ \).
In \( \triangle ADE \):
\( \angle DAE = 180^\circ - \angle ADE - \angle AED = 180^\circ - 90^\circ - 32^\circ = 58^\circ \).
Since \( E \) lies on the extension of chord \( AB \), \( \angle DAB = \angle DAE = 58^\circ \).
Therefore:
\( \angle BAC = \angle DAB - \angle CAD = 58^\circ - 28^\circ = 30^\circ \).

Correct answer: B (1 mark)

For \( L_1: 4x + 3y - 24 = 0 \):
When \( y = 0 \), \( 4x = 24 \implies x = 6 \), so \( P = (6, 0) \).
When \( x = 0 \), \( 3y = 24 \implies y = 8 \), so \( Q = (0, 8) \).

The coordinates of the mid-point of \( PQ \) are:
\( M = \left(\frac{6 + 0}{2}, \frac{0 + 8}{2}\right) = (3, 4) \).

The slope of \( L_1 \) is \( m_1 = -\frac{4}{3} \).
Since \( L_2 \perp L_1 \), the slope of \( L_2 \) is \( m_2 = -\frac{1}{m_1} = \frac{3}{4} \).

The equation of \( L_2 \) is:
\( y - 4 = \frac{3}{4}(x - 3) \)
\( 4(y - 4) = 3(x - 3) \)
\( 4y - 16 = 3x - 9 \)
\( 3x - 4y + 7 = 0 \).

Correct answer: A (1 mark)

Since \((4, 2)\) lies on \(y = \log_a x\), we have \(2 = \log_a 4 \implies a^2 = 4 \implies a = 2\) (since \(a > 0, a \neq 1\)).
Since \((k, -3)\) lies on the graph, \(-3 = \log_2 k \implies k = 2^{-3} = \frac{1}{8}\).

A: \(\frac{1}{8}\) (Correct answer)
B, C, D: Incorrect values arising from misidentifying the base or evaluating \(2^3\) or \(4^{-3}\).

Let the first term be \(a\) and the common difference be \(d\).
\(A(4) = a + 3d = 19\)
\(A(10) = a + 9d = 49\)
Subtracting the equations gives \(6d = 30 \implies d = 5\), so \(a = 19 - 3(5) = 4\).
The sum of the first \(m\) terms is \(S_m = \frac{m}{2}[2(4) + (m - 1)5] = \frac{m(5m + 3)}{2}\).
We require \(\frac{m(5m + 3)}{2} > 1000 \implies 5m^2 + 3m - 2000 > 0\).
Solving the quadratic equation \(5m^2 + 3m - 2000 = 0\):
\(m = \frac{-3 + \sqrt{9 - 4(5)(-2000)}}{10} = \frac{-3 + \sqrt{40009}}{10} \approx \frac{-3 + 200.022}{10} = 19.7022\).
Since \(m\) must be an integer, the least integer is \(m = 20\).

B: 20 (Correct answer)
A: 19 (Off-by-one underestimation)
C: 21 (Overestimation)
D: 25 (Calculated using incorrect sum formula)

Find the vertices of the polygonal region \(R\) by finding the pairwise intersections of the boundary lines:
1. Intersection of \(2x + y = 6\) and \(x - y = 3\):
Adding gives \(3x = 9 \implies x = 3, y = 0\). Vertex: \((3, 0)\).
Check with \(x + 2y \le 12\): \(3 + 0 \le 12\) (Valid).

2. Intersection of \(x - y = 3\) and \(x + 2y = 12\):
Subtracting gives \(-3y = -9 \implies y = 3, x = 6\). Vertex: \((6, 3)\).
Check with \(2x + y \ge 6\): \(2(6) + 3 = 15 \ge 6\) (Valid).

3. Intersection of \(2x + y = 6\) and \(x + 2y = 12\):
From the first line, \(y = 6 - 2x\). Substitute into second: \(x + 2(6 - 2x) = 12 \implies -3x + 12 = 12 \implies x = 0, y = 6\). Vertex: \((0, 6)\).
Check with \(x - y \le 3\): \(0 - 6 \le 3\) (Valid).

Evaluate the objective function \(P(x, y) = 3x - 2y\) at each vertex:
- At \((3, 0)\): \(P = 3(3) - 2(0) = 9\)
- At \((6, 3)\): \(P = 3(6) - 2(3) = 18 - 6 = 12\)
- At \((0, 6)\): \(P = 3(0) - 2(6) = -12\)

Thus, the maximum value is 12.

C: 12 (Correct answer)
A: -12 (Minimum value)
B: 9 (Value at \((3, 0)\))
D: 18 (Calculated without subtracting \(2y\))

Let \(M\) be the midpoint of \(AB\).
Since \(\triangle ABC\) is equilateral with side \(s = 6\):
\(CM = \frac{\sqrt{3}}{2}(6) = 3\sqrt{3}\text{ cm}\).
The centroid \(G\) divides \(CM\) in the ratio \(2:1\), so:
\(GM = \frac{1}{3} CM = \sqrt{3}\text{ cm}\),
\(CG = \frac{2}{3} CM = 2\sqrt{3}\text{ cm}\).

In right-angled \(\triangle VGC\):
\(VG = \sqrt{VC^2 - CG^2} = \sqrt{(2\sqrt{7})^2 - (2\sqrt{3})^2} = \sqrt{28 - 12} = \sqrt{16} = 4\text{ cm}\).

The angle between the face \(VAB\) and the base \(ABC\) is \(\angle VMG\).
In right-angled \(\triangle VGM\):
\(\tan \angle VMG = \frac{VG}{GM} = \frac{4}{\sqrt{3}}\).
\(\angle VMG = \arctan\left(\frac{4}{\sqrt{3}}\right) \approx 66.5866^\circ \approx 67^\circ\).

B: \(67^\circ\) (Correct answer to the nearest degree)
A: \(49^\circ\) (Calculated using \(\tan \theta = \frac{VG}{CG}\))
C: \(71^\circ\) (Arithmetic error)
D: \(76^\circ\) (Using \(\sin\) instead of \(\tan\))

Since \(\angle P = 90^\circ\) in \(\triangle PQR\), by the converse of angle in a semicircle, the segment \(QR\) is a diameter of the circle \(C\).
The centre of \(C\), denoted by \(M\), is the midpoint of \(QR\):
\(M = \left(\frac{8+0}{2}, \frac{0+6}{2}\right) = (4, 3)\).
The slope of the radius \(MQ\) joining the centre \((4, 3)\) and the point of tangency \(Q(8, 0)\) is:
\(m_{MQ} = \frac{0 - 3}{8 - 4} = -\frac{3}{4}\).
Since the tangent line \(L\) is perpendicular to the radius \(MQ\), the slope of \(L\) is:
\(m_L = -\frac{1}{m_{MQ}} = \frac{4}{3}\).
Using the point-slope form with \(Q(8, 0)\):
\(y - 0 = \frac{4}{3}(x - 8) \implies 3y = 4x - 32 \implies 4x - 3y - 32 = 0\).

A: \(4x - 3y - 32 = 0\) (Correct answer)
B: \(4x + 3y - 32 = 0\) (Sign error in slope)
C: \(3x + 4y - 24 = 0\) (Used radius slope directly)
D: \(3x - 4y - 24 = 0\) (Incorrect normal calculation)

The total number of ways to choose 6 members from \(7 + 5 = 12\) people without restrictions is:
\(C_6^{12} = 924\).

The possible compositions of (boys, girls) that satisfy the condition (at least 2 boys and at least 2 girls) are:
1. 2 boys and 4 girls: \(C_2^7 \times C_4^5 = 21 \times 5 = 105\)
2. 3 boys and 3 girls: \(C_3^7 \times C_3^5 = 35 \times 10 = 350\)
3. 4 boys and 2 girls: \(C_4^7 \times C_2^5 = 35 \times 10 = 350\)

Total number of acceptable committees:
\(105 + 350 + 350 = 805\).

D: 805 (Correct answer)
A: 350 (Only calculated the case with 3 boys and 3 girls)
B: 700 (Missed the case with 2 boys and 4 girls)
C: 735 (Incorrect combinations calculation)

There are two mutually exclusive cases for the ball transferred from Box \(A\) to Box \(B\):

Case 1: A blue ball is transferred from Box \(A\).
\(P(\text{Blue from } A) = \frac{3}{5}\).
Now Box \(B\) contains \(4 + 1 = 5\) blue balls and 1 yellow ball (total 6 balls).
\(P(\text{Blue from } B \mid \text{Blue from } A) = \frac{5}{6}\).
Probability for Case 1: \(\frac{3}{5} \times \frac{5}{6} = \frac{15}{30} = \frac{1}{2}\).

Case 2: A yellow ball is transferred from Box \(A\).
\(P(\text{Yellow from } A) = \frac{2}{5}\).
Now Box \(B\) contains 4 blue balls and \(1 + 1 = 2\) yellow balls (total 6 balls).
\(P(\text{Blue from } B \mid \text{Yellow from } A) = \frac{4}{6} = \frac{2}{3}\).
Probability for Case 2: \(\frac{2}{5} \times \frac{4}{6} = \frac{8}{30} = \frac{4}{15}\).

Total probability of drawing a blue ball from Box \(B\):
\(P = \frac{15}{30} + \frac{8}{30} = \frac{23}{30}\).

C: \(\frac{23}{30}\) (Correct answer)
A: \(\frac{7}{10}\) (Arithmetic error)
B: \(\frac{3}{4}\) (Assumed equal prior weighting)
D: \(\frac{4}{5}\) (Ignored the effect of the transferred ball)

Given \(\bar{x} = 45\), \(\text{Var}(x) = 16\), so \(\sigma_x = \sqrt{16} = 4\).

For the linear transformation \(y_i = a x_i + b\) where \(a = -2\) and \(b = 3\):
1. Mean: \(\bar{y} = 3 - 2\bar{x} = 3 - 2(45) = 3 - 90 = -87\). Thus, I is true.
2. Standard deviation: \(\sigma_y = |a| \sigma_x = |-2|(4) = 8\). Thus, II is true.
3. Variance: \(\text{Var}(y) = a^2 \text{Var}(x) = (-2)^2 \times 16 = 4 \times 16 = 64\). Thus, III is true.

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

D: I, II and III (Correct answer)
A: I and II only (Omitted variance transformation)
B: I and III only (Omitted standard deviation transformation)
C: II and III only (Incorrectly evaluated the new mean)

Recall that in hexadecimal, \(\text{B} = 11\), \(\text{E} = 14\), and \(\text{D} = 13\).

\begin{aligned}
\text{B0E0D}_{16} &= 11 \times 16^4 + 0 \times 16^3 + 14 \times 16^2 + 0 \times 16^1 + 13 \times 16^0 \\
&= 11 \times 65536 + 0 + 14 \times 256 + 0 + 13 \\
&= 720896 + 3584 + 13 \\
&= 724493
\end{aligned}

Correct option: A (1 mark)

Let \( Y = \log_9 y = \dfrac{\log_3 y}{\log_3 9} = \dfrac{1}{2} \log_3 y \) and \( X = \log_3 x \).

The slope of the line is:
\[ m = \frac{2 - 4}{5 - 1} = -\frac{2}{4} = -\frac{1}{2} \]
Using the point-slope form with \( (1, 4) \):
\[ Y - 4 = -\frac{1}{2}(X - 1) \]
\[ Y = -\frac{1}{2}X + \frac{1}{2} + 4 = -\frac{1}{2}X + \frac{9}{2} \]
Substitute back \( Y = \dfrac{1}{2}\log_3 y \) and \( X = \log_3 x \):
\[ \frac{1}{2}\log_3 y = -\frac{1}{2}\log_3 x + \frac{9}{2} \]
Multiplying both sides by 2:
\[ \log_3 y = -\log_3 x + 9 \]
\[ \log_3 y + \log_3 x = 9 \]
\[ \log_3 (xy) = 9 \implies xy = 3^9 = 19683 \]
Thus, \( x y = 19683 \).

Correct option: C (1 mark)

Simplify each term:

First term:
\[ \frac{3 - 2i}{1 + 2i} = \frac{(3 - 2i)(1 - 2i)}{(1 + 2i)(1 - 2i)} = \frac{3 - 6i - 2i + 4i^2}{1 - 4i^2} = \frac{3 - 8i - 4}{1 + 4} = \frac{-1 - 8i}{5} = -\frac{1}{5} - \frac{8}{5}i \]

Second term:
\[ \frac{2 + 3i}{2 - i} = \frac{(2 + 3i)(2 + i)}{(2 - i)(2 + i)} = \frac{4 + 2i + 6i + 3i^2}{4 - i^2} = \frac{4 + 8i - 3}{4 + 1} = \frac{1 + 8i}{5} = \frac{1}{5} + \frac{8}{5}i \]

Subtract the two terms:
\[ w = \left(-\frac{1}{5} - \frac{8}{5}i\right) - \left(\frac{1}{5} + \frac{8}{5}i\right) = -\frac{2}{5} - \frac{16}{5}i \]

The imaginary part is \( -\dfrac{16}{5} = -3.2 \).

Correct option: B (1 mark)

Find the vertices of the bounded polygonal region \( R \):
1. Intersection of \( x - y = -1 \) and \( y = 1 \):
\( x - 1 = -1 \implies x = 0 \). Vertex is \( A(0, 1) \).
2. Intersection of \( 2x + y = 10 \) and \( y = 1 \):
\( 2x + 1 = 10 \implies 2x = 9 \implies x = 4.5 \). Vertex is \( B(4.5, 1) \).
3. Intersection of \( 2x + y = 10 \) and \( x - y = -1 \):
Adding equations gives \( 3x = 9 \implies x = 3 \), then \( y = 4 \). Vertex is \( C(3, 4) \).

Evaluate the objective function \( P(x, y) = 4x - 3y \) at each vertex:
- At \( A(0, 1) \): \( P = 4(0) - 3(1) = -3 \)
- At \( B(4.5, 1) \): \( P = 4(4.5) - 3(1) = 18 - 3 = 15 \)
- At \( C(3, 4) \): \( P = 4(3) - 3(4) = 12 - 12 = 0 \)

Therefore, the minimum value is \( -3 \).

Correct option: A (1 mark)

Rewrite the circle equation in standard form:
\[ (x - 3)^2 + (y + 4)^2 = -16 + 9 + 16 = 9 = 3^2 \]
Centre \( K = (3, -4) \), radius \( r = 3 \).

Let the equation of the line passing through \( (0, 0) \) with slope \( m \) be \( mx - y = 0 \).
Since \( L \) is tangent to \( C \), the perpendicular distance from \( (3, -4) \) to the line is equal to the radius \( 3 \):
\[ \frac{|m(3) - (-4)|}{\sqrt{m^2 + (-1)^2}} = 3 \]
\[ |3m + 4| = 3\sqrt{m^2 + 1} \]
Square both sides:
\[ (3m + 4)^2 = 9(m^2 + 1) \]
\[ 9m^2 + 24m + 16 = 9m^2 + 9 \]
\[ 24m + 16 = 9 \]
\[ 24m = -7 \implies m = -\frac{7}{24} \]

(Note: The other tangent is the vertical line \( x = 0 \), which has undefined slope).
Hence, a possible slope of \( L \) is \( -\dfrac{7}{24} \).

Correct option: D (1 mark)

The committee of 5 can be formed in two valid cases:
- Case 1: 2 boys and 3 girls
Number of ways \( = C_2^6 \times C_3^5 = 15 \times 10 = 150 \)

- Case 2: 3 boys and 2 girls
Number of ways \( = C_3^6 \times C_2^5 = 20 \times 10 = 200 \)

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

Correct option: B (1 mark)

For a linear transformation \( y_i = a x_i + b \) where \( a = -2 \) and \( b = 3 \):

1. Mean of \( y \):
\[ \bar{y} = a \bar{x} + b = -2(45) + 3 = -90 + 3 = -87 \]

2. Standard deviation of \( x \):
\[ \sigma_x = \sqrt{16} = 4 \]
Standard deviation of \( y \):
\[ \sigma_y = |a| \sigma_x = |-2| \times 4 = 2 \times 4 = 8 \]

Thus, Mean \( = -87 \) and Standard deviation \( = 8 \).

Correct option: D (1 mark)