2024 HKDSE Mathematics Answers & Marking Scheme

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

(b) Substituting \( k = -1 \) into \( f(x) \), we get:
\( f(x) = 2x^3 + x^2 - 13x + 6 \)
Since \( 2x - 1 \) is a factor of \( f(x) \), by long division or synthetic division, we get:
\( f(x) = (2x - 1)(x^2 + x - 6) \)
Factoring the quadratic term:
\( f(x) = (2x - 1)(x - 2)(x + 3) \)

(a)
- For using \( f(1/2) = 0 \) (1M)
- For finding \( k = -1 \) (1A)
(b)
- For finding the quadratic factor \( x^2 + x - 6 \) (1M)
- For the correct factorization \( (2x - 1)(x - 2)(x + 3) \) (1A)

(a) The radius \( r \) of \( C \) is the perpendicular distance from the center \( (4, -3) \) to the line \( L: 3x - 4y + 1 = 0 \).
Using the point-to-line distance formula:
\( r = \frac{|3(4) - 4(-3) + 1|}{\sqrt{3^2 + (-4)^2}} = \frac{|12 + 12 + 1|}{5} = \frac{25}{5} = 5 \)
The equation of \( C \) is:
\( (x - 4)^2 + (y + 3)^2 = 5^2 \)
\( (x - 4)^2 + (y + 3)^2 = 25 \) (or \( x^2 + y^2 - 8x + 6y = 0 \))

(b) Let \( d \) be the distance between the origin \( O(0,0) \) and the center \( (4, -3) \):
\( d = \sqrt{(0 - 4)^2 + (0 - (-3))^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \)
Since the distance from \( O \) to the center is exactly equal to the radius \( r = 5 \), the origin \( O(0,0) \) lies on the circle \( C \).

(a)
- For finding the radius \( r = 5 \) (1M)
- For writing down the correct equation (1A) (standard or general form)
(b)
- For substituting \( (0,0) \) into the equation or finding distance to the center (1M)
- For concluding that \( O(0,0) \) lies on the circle with explanation (1A)

Using the trigonometric identity \( \cos^2 \theta = 1 - \sin^2 \theta \):
\( 2(1 - \sin^2 \theta) + 5\sin \theta - 4 = 0 \)
\( 2 - 2\sin^2 \theta + 5\sin \theta - 4 = 0 \)
\( -2\sin^2 \theta + 5\sin \theta - 2 = 0 \)
\( 2\sin^2 \theta - 5\sin \theta + 2 = 0 \)
\( (2\sin \theta - 1)(\sin \theta - 2) = 0 \)

Since \( -1 \le \sin \theta \le 1 \), the equation \( \sin \theta = 2 \) has no real solutions.
Therefore, we have:
\( 2\sin \theta - 1 = 0 \)
\( \sin \theta = \frac{1}{2} \)

Since \( 0^\circ \le \theta \le 360^\circ \):
\( \theta = 30^\circ \) or \( \theta = 150^\circ \)

- For using \( \cos^2 \theta = 1 - \sin^2 \theta \) to form a quadratic in \( \sin \theta \) (1M)
- For factoring to get \( (2\sin \theta - 1)(\sin \theta - 2) = 0 \) or finding \( \sin \theta = \frac{1}{2} \) (1M)
- For finding \( \theta = 30^\circ \) (1A)
- For finding \( \theta = 150^\circ \) (1A)

(a) Since the mean of the 7 data values is 7:
\( \frac{3 + 5 + 6 + 8 + 9 + x + y}{7} = 7 \)
\( 31 + x + y = 49 \)
\( x + y = 18 \)

Since the range of the data is 9, if the minimum value of the data set is 3, then the maximum value must be \( 3 + 9 = 12 \).
Thus, we can let \( y = 12 \).
Then, \( x = 18 - 12 = 6 \).
This gives the data set \( 3, 5, 6, 6, 8, 9, 12 \), whose minimum is 3 and maximum is 12, satisfying the range condition of 9.
(If the minimum value was instead \( x < 3 \), then the maximum value would be \( x + 9 \). Since \( y \le x + 9 \) and \( x + y = 18 \), we would have \( 18 - x \le x + 9 \Rightarrow 2x \ge 9 \Rightarrow x \ge 4.5 \), which contradicts \( x < 3 \). Hence, \( x = 6 \) and \( y = 12 \) is the only solution.)

(b) The data set is \( \{3, 5, 6, 6, 8, 9, 12\} \) with mean \( \mu = 7 \).
The variance \( \sigma^2 \) is:
\( \sigma^2 = \frac{(3-7)^2 + (5-7)^2 + 2(6-7)^2 + (8-7)^2 + (9-7)^2 + (12-7)^2}{7} \)
\( \sigma^2 = \frac{16 + 4 + 2(1) + 1 + 4 + 25}{7} = \frac{52}{7} \approx 7.42857 \)
The standard deviation is \( \sigma = \sqrt{\frac{52}{7}} \approx 2.7255 \approx 2.73 \).

(a)
- For establishing \( x + y = 18 \) (1M)
- For finding \( x = 6 \) and \( y = 12 \) (1A)
(b)
- For applying correct formula for standard deviation (1M)
- For obtaining \( 2.73 \) (accept \( 2.72 \le \sigma \le 2.74 \)) (1A)

(a) Since the quadratic equation has real roots, the discriminant \( \Delta \ge 0 \).
\( \Delta = [-2(k - 1)]^2 - 4(1)(2k + 1) \ge 0 \)
\( 4(k^2 - 2k + 1) - 4(2k + 1) \ge 0 \)
\( 4k^2 - 8k + 4 - 8k - 4 \ge 0 \)
\( 4k^2 - 16k \ge 0 \)
Divide by 4:
\( k^2 - 4k \ge 0 \)
\( k(k - 4) \ge 0 \)
Thus, \( k \le 0 \) or \( k \ge 4 \).

(b) The positive integers satisfying the condition are \( 4, 5, 6, \dots \) (since \( k \le 0 \) does not yield positive integers).
Thus, the smallest positive integer is \( k = 4 \).
Substituting \( k = 4 \) into the original equation:
\( x^2 - 2(4 - 1)x + (2(4) + 1) = 0 \)
\( x^2 - 6x + 9 = 0 \)
\( (x - 3)^2 = 0 \)
\( x = 3 \) (repeated root)

(a)
- For setting up \( \Delta \ge 0 \) (1M)
- For obtaining \( k \le 0 \) or \( k \ge 4 \) (1A) (must include 'or')
(b)
- For identifying \( k = 4 \) (1M)
- For solving the equation to get \( x = 3 \) (1A)

(a) By the Remainder Theorem:
\( g(1) = 6 \Rightarrow a(1)^3 + b(1)^2 - 5(1) - 2 = 6 \)
\( a + b - 7 = 6 \Rightarrow a + b = 13 \) --- (1)

\( g(-2) = -12 \Rightarrow a(-2)^3 + b(-2)^2 - 5(-2) - 2 = -12 \)
\( -8a + 4b + 10 - 2 = -12 \)
\( -8a + 4b + 8 = -12 \)
\( -8a + 4b = -20 \Rightarrow -2a + b = -5 \) --- (2)

Subtracting (2) from (1):
\( 3a = 18 \Rightarrow a = 6 \)
Substituting \( a = 6 \) into (1):
\( 6 + b = 13 \Rightarrow b = 7 \)

(b) Note that \( x^2 + x - 2 = (x - 1)(x + 2) \).
Let the remainder when \( g(x) \) is divided by \( x^2 + x - 2 \) be \( px + q \).
We can write: \( g(x) = (x^2 + x - 2)Q(x) + px + q = (x - 1)(x + 2)Q(x) + px + q \)
Substituting \( x = 1 \):
\( g(1) = p(1) + q \Rightarrow p + q = 6 \) --- (3)
Substituting \( x = -2 \):
\( g(-2) = p(-2) + q \Rightarrow -2p + q = -12 \) --- (4)

Subtracting (4) from (3):
\( 3p = 18 \Rightarrow p = 6 \)
Substituting \( p = 6 \) into (3):
\( 6 + q = 6 \Rightarrow q = 0 \)
Therefore, the remainder is \( 6x \).

(a)
- For setting up the two linear equations using Remainder Theorem (1M)
- For finding \( a = 6 \) and \( b = 7 \) (1A)
(b)
- For establishing equations for the remainder of the form \( px + q \) (1M)
- For obtaining the remainder \( 6x \) (1A)

(a) Let \( y = k_1 + \frac{k_2}{x^2} \), where \( k_1 \) and \( k_2 \) are non-zero constants.
Substituting \( x = 1 \) and \( y = 8 \):
\( 8 = k_1 + k_2 \) --- (1)
Substituting \( x = 2 \) and \( y = 5 \):
\( 5 = k_1 + \frac{k_2}{4} \) --- (2)

Subtracting (2) from (1):
\( 3 = \frac{3}{4}k_2 \Rightarrow k_2 = 4 \)
Substituting \( k_2 = 4 \) into (1):
\( 8 = k_1 + 4 \Rightarrow k_1 = 4 \)
Therefore, \( y = 4 + \frac{4}{x^2} \).

(b) Substituting \( y = 4.25 \):
\( 4.25 = 4 + \frac{4}{x^2} \)
\( 0.25 = \frac{4}{x^2} \)
\( x^2 = \frac{4}{0.25} = 16 \)
\( x = 4 \) or \( x = -4 \)

(a)
- For writing down the variation equation \( y = k_1 + \frac{k_2}{x^2} \) (1M)
- For finding \( k_1 = 4 \) and \( k_2 = 4 \) to get \( y = 4 + \frac{4}{x^2} \) (1A)
(b)
- For substituting \( y = 4.25 \) and solving for \( x^2 \) (1M)
- For finding both values \( x = 4 \) and \( x = -4 \) (1A) (deduct 1 mark if only one solution is given)

(a) Let the first term of the arithmetic sequence be \( a \) and the common difference be \( d \).
We are given:
\( T(3) = a + 2d = 14 \) --- (1)
\( T(7) = a + 6d = 30 \) --- (2)

Subtracting (1) from (2):
\( 4d = 16 \Rightarrow d = 4 \)
Substituting \( d = 4 \) into (1):
\( a + 2(4) = 14 \Rightarrow a = 6 \)
So, the first term is 6 and the common difference is 4.

(b) The sum of the first \( n \) terms is given by \( S(n) = \frac{n}{2}[2a + (n-1)d] \).
We set \( S(n) > 500 \):
\( \frac{n}{2}[2(6) + (n-1)4] > 500 \)
\( \frac{n}{2}[12 + 4n - 4] > 500 \)
\( \frac{n}{2}[4n + 8] > 500 \)
\( n(2n + 4) > 500 \)
\( 2n^2 + 4n - 500 > 0 \)
\( n^2 + 2n - 250 > 0 \)

Using the quadratic formula to find the roots of \( n^2 + 2n - 250 = 0 \):
\( n = \frac{-2 \pm \sqrt{2^2 - 4(1)(-250)}}{2} = \frac{-2 \pm \sqrt{1004}}{2} \)
For \( n > 0 \):
\( n > \frac{-2 + 31.686}{2} \approx 14.843 \)
Since \( n \) must be an integer, the least value of \( n \) is 15.

(a)
- For setting up the two linear equations in terms of \( a \) and \( d \) (1M)
- For finding \( a = 6 \) and \( d = 4 \) (1A)
(b)
- For setting up the inequality \( S(n) > 500 \) (1M)
- For finding the least value of \( n = 15 \) (1A)

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

(b) The area of \( \triangle ABC \) is:
\( \text{Area} = \frac{1}{2} \cdot AB \cdot AC \cdot \sin \angle BAC \)
\( \text{Area} = \frac{1}{2} \cdot 8 \cdot 5 \cdot \sin 60^\circ \)
\( \text{Area} = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3}\text{ cm}^2 \)

(a)
- For substituting values correctly into the cosine formula (1M)
- For finding \( BC = 7\text{ cm} \) (1A) (accept without units)
(b)
- For substituting values correctly into the triangle area formula (1M)
- For obtaining the correct area \( 10\sqrt{3}\text{ cm}^2 \) (1A) (accept without units)

(a) Since \(x-2\) is a factor of \(p(x)\), by the factor theorem:
\(p(2) = 0 \Rightarrow 3(2)^3 + a(2)^2 + b(2) - 12 = 0 \Rightarrow 4a + 2b = -12 \Rightarrow 2a + b = -6\) --- (1)
When \(p(x)\) is divided by \(x+1\), the remainder is \(-15\). By the remainder theorem:
\(p(-1) = -15 \Rightarrow 3(-1)^3 + a(-1)^2 + b(-1) - 12 = -15 \Rightarrow a - b = 0 \Rightarrow a = b\) --- (2)
Substituting (2) into (1):
\(2a + a = -6 \Rightarrow 3a = -6 \Rightarrow a = -2\)
Hence, \(a = -2\) and \(b = -2\).

(b) Substituting \(a = -2\) and \(b = -2\) into \(p(x)\):
\(p(x) = 3x^3 - 2x^2 - 2x - 12\)
Since \(x-2\) is a factor:
\(p(x) = (x-2)(3x^2 + 4x + 6)\)
Consider the equation \(p(x) = 0 \Rightarrow x-2 = 0\) or \(3x^2 + 4x + 6 = 0\).
For the quadratic equation \(3x^2 + 4x + 6 = 0\):
The discriminant \(\Delta = 4^2 - 4(3)(6) = 16 - 72 = -56 < 0\).
Since \(\Delta < 0\), the quadratic equation has no real roots.
Therefore, not all roots of \(p(x) = 0\) are real.
So, the claim is disagreed.

(a)
- For attempting to use the Factor Theorem: \(p(2) = 0\) [1M]
- For attempting to use the Remainder Theorem: \(p(-1) = -15\) [1M]
- For obtaining \(2a + b = -6\) and \(a - b = 0\) (or equivalent) [1M]
- For \(a = -2\) and \(b = -2\) [1A]
(b)
- For finding the other factor \(3x^2 + 4x + 6\) [1M]
- For checking discriminant \(\Delta = -56 < 0\) [1M]
- For concluding that not all roots are real and stating disagreement [1A]

(a) The center of circle \(C\) is \(G = \left(-\frac{-8}{2}, -\frac{2}{2}\right) = (4, -1)\).
The radius of \(C\) is \(r = \sqrt{4^2 + (-1)^2 - (-8)} = \sqrt{16 + 1 + 8} = 5\).

(b) Since \(L\) is tangent to \(C\), the perpendicular distance from the center \((4, -1)\) to \(L\) is equal to the radius \(5\).
\(\frac{|3(4) - 4(-1) + k|}{\sqrt{3^2 + (-4)^2}} = 5\)
\(\frac{|16 + k|}{5} = 5\)
\(|16 + k| = 25\)
So, \(16 + k = 25 \Rightarrow k = 9\), or \(16 + k = -25 \Rightarrow k = -41\).
The line \(L\) can be written as \(y = \frac{3}{4}x + \frac{k}{4}\), so its \(y\)-intercept is \(\frac{k}{4}\).
Since the \(y\)-intercept is positive, \(k > 0\). Thus, \(k = 9\).

(c) Substitute \(k = 9\) into \(L\): \(3x - 4y + 9 = 0 \Rightarrow y = \frac{3x+9}{4}\).
Substitute this into the equation of \(C\):
\(x^2 + \left(\frac{3x+9}{4}\right)^2 - 8x + 2\left(\frac{3x+9}{4}\right) - 8 = 0\)
\(16x^2 + (9x^2 + 54x + 81) - 128x + 24x + 72 - 128 = 0\)
\(25x^2 - 50x + 25 = 0\)
\(x^2 - 2x + 1 = 0\)
\((x-1)^2 = 0 \Rightarrow x = 1\).
When \(x = 1\), \(y = \frac{3(1)+9}{4} = 3\).
Thus, the point of contact is \((1, 3)\).

(a)
- For finding the center \((4, -1)\) [1A]
- For finding the radius \(5\) [1A]
(b)
- For using the perpendicular distance from center to line equals radius [1M]
- For solving to obtain \(k = 9\) or \(k = -41\) [1M]
- For choosing \(k = 9\) with justification [1A]
(c)
- For substituting \(y\) (or \(x\)) into the circle equation [1M]
- For obtaining the point of contact \((1, 3)\) [1A]

(a) The minimum value is 15.
The maximum value is \(40+b\).
Since the range is 32:
\((40+b) - 15 = 32 \Rightarrow b = 7\).

Since \(N = 15\), the median is the 8th term when the data is sorted.
The first 7 terms are: 15, 15, 18, 19, 20, 22, 25.
The 8th term is the median, which is 25.
Since \(5 \le a \le 8\), the term \(20+a\) must be the 8th term or larger. To have the 8th term as 25, we must have \(a = 5\) (if \(a > 5\), the 8th term would be \(20+a > 25\), making the median greater than 25).
Therefore, \(a = 5\).

(b) For \(a = 5\) and \(b = 7\), the original 15 data values in ascending order are:
15, 15, 18, 19, 20, 22, 25, 25, 28, 31, 31, 36, 40, 42, 47.
Original lower quartile \(Q_1\) = 4th term = 19.
Original upper quartile \(Q_3\) = 12th term = 36.
Original \(IQR = 36 - 19 = 17\).

With two more students (25 and 43) joining, the new 17 data values in ascending order are:
15, 15, 18, 19, 20, 22, 25, 25, 25, 28, 31, 31, 36, 40, 42, 43, 47.
New lower quartile \(Q_1'\) = median of the first 8 terms = \(\frac{19+20}{2} = 19.5\).
New upper quartile \(Q_3'\) = median of the last 8 terms = \(\frac{36+40}{2} = 38\).
New \(IQR = 38 - 19.5 = 18.5\).

The change in the interquartile range is \(18.5 - 17 = 1.5\).

(a)
- For finding \(b = 7\) [1A]
- For identifying that the 8th term is 25 [1M]
- For finding \(a = 5\) [1A]
(b)
- For finding the original \(Q_1 = 19\) and \(Q_3 = 36\) (or original \(IQR = 17\)) [1M]
- For finding the new \(Q_1 = 19.5\) [1A]
- For finding the new \(Q_3 = 38\) [1A]
- For finding the change in IQR is \(1.5\) [1A]

(a) Since \(B\) is due East of \(A\) and the bearing of \(C\) from \(A\) is \(N50^\circ E\):
\(\angle CAB = 90^\circ - 50^\circ = 40^\circ\).
Since the bearing of \(C\) from \(B\) is \(N30^\circ W\) and \(A\) is due West of \(B\):
\(\angle CBA = 90^\circ - 30^\circ = 60^\circ\).
In \(\triangle ABC\):
\(\angle ACB = 180^\circ - \angle CAB - \angle CBA = 180^\circ - 40^\circ - 60^\circ = 80^\circ\).
By the sine formula:
\(\frac{AC}{\sin 60^\circ} = \frac{120}{\sin 80^\circ}\)
\(AC = \frac{120 \sin 60^\circ}{\sin 80^\circ} \approx 105.52945 \approx 106\text{ m}\).

(b) (i) In right-angled \(\triangle ACP\):
\(\tan 35^\circ = \frac{CP}{AC} = \frac{h}{AC}\)
\(h = AC \tan 35^\circ \approx 105.52945 \times \tan 35^\circ \approx 73.8913 \approx 73.9\).

(ii) In \(\triangle ABC\), by the sine formula:
\(\frac{BC}{\sin 40^\circ} = \frac{120}{\sin 80^\circ} \Rightarrow BC = \frac{120 \sin 40^\circ}{\sin 80^\circ} \approx 78.32441\text{ m}\).
In right-angled \(\triangle BCP\), let \(\theta\) be the angle of elevation of \(P\) from \(B\):
\(\tan \theta = \frac{CP}{BC} = \frac{h}{BC}\)
\(\tan \theta \approx \frac{73.8913}{78.32441} \approx 0.943399\)
\ heta \approx 43.3297^\circ \approx 43.3^\circ\).
Therefore, the angle of elevation of \(P\) from \(B\) is \(43.3^\circ\).

(a)
- For finding \(\angle CAB = 40^\circ\), \(\angle CBA = 60^\circ\), and \(\angle ACB = 80^\circ\) [1M]
- For using the sine formula: \(\frac{AC}{\sin 60^\circ} = \frac{120}{\sin 80^\circ}\) [1M]
- For finding \(AC \approx 106\text{ m}\) (or 105.5 m) [1A]
(b)
- (i) For using \(\tan 35^\circ = \frac{h}{AC}\) to find \(h \approx 73.9\) [1A]
- (ii) For using the sine formula to find \(BC \approx 78.3\text{ m}\) [1M]
- (ii) For using \(\tan \theta = \frac{h}{BC}\) [1M]
- (ii) For finding \(\theta \approx 43.3^\circ\) [1A]

(a) Let \(d\) be the common difference of \(A(n)\) and \(r\) be the common ratio of \(G(n)\).
We are given:
\(A(1) = 4\), so \(A(2) = 4 + d\) and \(A(5) = 4 + 4d\).
\(G(1) = 4\), so \(G(2) = 4r\) and \(G(3) = 4r^2\).

Since \(A(2) = G(2)\), we have:
\(4 + d = 4r \Rightarrow d = 4r - 4\) --- (1)
Since \(A(5) = G(3)\), we have:
\(4 + 4d = 4r^2\) --- (2)

Substitute (1) into (2):
\(4 + 4(4r - 4) = 4r^2\)
\(4 + 16r - 16 = 4r^2\)
\(4r^2 - 16r + 12 = 0\)
\(r^2 - 4r + 3 = 0\)
\((r - 1)(r - 3) = 0\)
Since \(r \neq 1\), we have \(r = 3\).
Substitute \(r = 3\) into (1):
\(d = 4(3) - 4 = 8\).
Thus, the common difference is \(8\) and the common ratio is \(3\).

(b) The sum of the first \(n\) terms of \(A(n)\) is:
\(S_n = \frac{n}{2} [2(4) + (n-1)8] = \frac{n}{2} [8 + 8n - 8] = 4n^2\).
We want \(S_n > 5000\):
\(4n^2 > 5000\)
\(n^2 > 1250\)
\(n > \sqrt{1250} \approx 35.355\).
Since \(n\) must be an integer, the least value of \(n\) is \(36\).

(a)
- For setting up equations \(4 + d = 4r\) and \(4 + 4d = 4r^2\) [1M]
- For substituting to form a quadratic equation in \(r\) [1M]
- For solving to obtain \(r = 3\) (rejecting \(r = 1\)) [1A]
- For finding \(d = 8\) [1A]
(b)
- For simplifying to \(S_n = 4n^2\) [1M]
- For setting up inequality \(4n^2 > 5000\) [1M]
- For finding \(n = 36\) [1A]

(a) Since \(x-2\) is a factor of \(P(x)\), by the factor theorem:
\(P(2) = 0\)
\(2(2)^3 + a(2)^2 + b(2) - 12 = 0\)
\(16 + 4a + 2b - 12 = 0\)
\(2a + b = -2\) --- (1)

Since the remainder is \(-21\) when divided by \(x+1\), by the remainder theorem:
\(P(-1) = -21\)
\(2(-1)^3 + a(-1)^2 + b(-1) - 12 = -21\)
\(-2 + a - b - 12 = -21\)
\(a - b = -7\) --- (2)

Solving (1) and (2):
From (2), \(b = a + 7\).
Substitute into (1): \(2a + a + 7 = -2 \implies 3a = -9 \implies a = -3\).
Then, \(b = -3 + 7 = 4\).

(b) Since \(a = -3\) and \(b = 4\), \(P(x) = 2x^3 - 3x^2 + 4x - 12\).
Since \(x-2\) is a factor of \(P(x)\), we perform division to find:
\(P(x) = (x-2)(2x^2 + x + 6)\).

For the equation \(P(x) = 0\), we have:
\(x - 2 = 0\) or \(2x^2 + x + 6 = 0\).

The discriminant of the quadratic equation \(2x^2 + x + 6 = 0\) is:
\(\Delta = 1^2 - 4(2)(6) = 1 - 48 = -47 < 0\).

Since \(\Delta < 0\), the quadratic factor has no real roots.
Therefore, the equation \(P(x) = 0\) has only one real root (which is 2), and the other two roots are non-real.
The claim is disagreed.

- For using factor theorem to obtain 2a + b = -2 (1M)
- For using remainder theorem to obtain a - b = -7 (1M)
- For finding a = -3, b = 4 (1A)
- For factorising P(x) into (x-2)(2x^2 + x + 6) (1M)
- For considering the discriminant of 2x^2 + x + 6 = 0 (1M)
- For calculating discriminant = -47 and stating it is < 0 (1A)
- For correct conclusion with explanation (1f.t.)

(a) Let the elements of \(S\) be \(x_1, x_2, \dots, x_{15}\).
Mean \(\bar{x} = \frac{\sum x_i}{15} = 48 \implies \sum x_i = 15 \times 48 = 720\).

Since the standard deviation \(\sigma = 8\):
\(\sigma^2 = \frac{\sum x_i^2}{15} - \bar{x}^2\)
\(8^2 = \frac{\sum x_i^2}{15} - 48^2\)
\(64 = \frac{\sum x_i^2}{15} - 2304\)
\(\frac{\sum x_i^2}{15} = 2368 \implies \sum x_i^2 = 15 \times 2368 = 35520\).

(b) (i) The sum of elements in \(S'\) is \(720 + 36 + 60 = 816\).
The number of elements in \(S'\) is \(15 + 2 = 17\).
The mean of \(S'\) is \(\bar{y} = \frac{816}{17} = 48\).
(Alternatively: Since the average of the two added numbers 36 and 60 is \(\frac{36+60}{2} = 48\), which equals the original mean, the mean remains unchanged.)

(ii)
Method 1 (By Calculation):
The sum of squares of elements in \(S'\) is \(\sum y_i^2 = 35520 + 36^2 + 60^2 = 35520 + 1296 + 3600 = 40416\).
The new variance is \(\sigma'^2 = \frac{40416}{17} - 48^2 = 2377.412 - 2304 = 73.412\).
The new standard deviation is \(\sigma' = \sqrt{73.412} \approx 8.57\).
Since \(8.57 > 8\), the standard deviation of \(S'\) is greater than that of \(S\).

Method 2 (By Logic):
The mean of the set remains unchanged at 48.
The deviations of the two added numbers 36 and 60 from the mean 48 are \(|36 - 48| = 12\) and \(|60 - 48| = 12\) respectively.
Since both deviations (12) are greater than the original standard deviation (8), adding these two numbers increases the dispersion of the data set.
Therefore, the standard deviation of \(S'\) is greater than the standard deviation of \(S\).

- For finding the sum = 720 (1A)
- For setting up variance equation (1M)
- For finding the sum of squares = 35520 (1A)
- For finding new mean = 48 in (b)(i) (1A)
- For calculating new variance / SD OR explaining using deviations from the mean in (b)(ii) (1M)
- For obtaining new SD approx 8.57 OR stating deviations (12) are larger than 8 (1A)
- For concluding that standard deviation of S' is greater than S (1f.t.)

(a) The circle equation is \(x^2 + y^2 - 10x - 6y + 9 = 0\).
The center of \(C\) is \(G = \left(5, 3\right)\).
The radius of \(C\) is \(r = \sqrt{5^2 + 3^2 - 9} = \sqrt{25} = 5\).

For \(L\) to intersect \(C\) at two distinct points, the perpendicular distance \(d\) from \(G(5, 3)\) to \(L\) must be less than the radius \(r = 5\).
\(d = \frac{|4(5) - 3(3) + k|}{\sqrt{4^2 + (-3)^2}} < 5\)
\(\frac{|20 - 9 + k|}{5} < 5\)
\(|11 + k| < 25\)
\(-25 < 11 + k < 25\)
\(-36 < k < 14\).

(b) When \(k = -1\), the perpendicular distance from \(G(5, 3)\) to \(L: 4x - 3y - 1 = 0\) is:
\(d = \frac{|11 - 1|}{5} = \frac{10}{5} = 2\).

Let \(H\) be the projection of \(G\) onto chord \(AB\). By the properties of circles, \(H\) is the midpoint of \(AB\) and \(GH = d = 2\).
In right-angled triangle \(\triangle GAH\):
\(AH = \sqrt{GA^2 - GH^2} = \sqrt{5^2 - 2^2} = \sqrt{21}\).

The length of chord \(AB = 2 \times AH = 2\sqrt{21}\).

The area of \(\triangle GAB\) is:
\(\text{Area} = \frac{1}{2} \times AB \times GH = \frac{1}{2} \times 2\sqrt{21} \times 2 = 2\sqrt{21}\).

- For finding center G(5, 3) and radius r = 5 (1A)
- For setting up distance inequality (or discriminant > 0) (1M)
- For finding the range -36 < k < 14 (1A)
- For finding the perpendicular distance d = 2 (1M)
- For finding half chord length AH = \sqrt{21} or chord length AB = 2\sqrt{21} (1A)
- For using area formula (1M)
- For finding the area = 2\sqrt{21} (1A)

(a) In right-angled triangle \(\triangle BCD\):
\(BD = \sqrt{BC^2 + CD^2} = \sqrt{12^2 + 5^2} = 13\text{ cm}\).
Since \(AB\) is perpendicular to the horizontal plane \(BCD\), \(AB \perp BD\).
In right-angled triangle \(\triangle ABD\):
\(AD = \sqrt{AB^2 + BD^2} = \sqrt{9^2 + 13^2} = \sqrt{250} \approx 15.8\text{ cm}\).

(b) Since \(AB \perp\) plane \(BCD\), \(AB \perp CD\).
We are given \(BC \perp CD\).
Since \(CD\) is perpendicular to both \(AB\) and \(BC\), \(CD \perp\) plane \(ABC\).
Therefore, \(CD \perp AC\).
Since \(AC \perp CD\) and \(BC \perp CD\), the line of intersection is \(CD\), and the angle between plane \(ACD\) and plane \(BCD\) is \(\angle ACB\).
In right-angled triangle \(\triangle ABC\):
\(\tan \angle ACB = \frac{AB}{BC} = \frac{9}{12} = 0.75\)
\(\angle ACB = \arctan(0.75) \approx 36.8699^\circ \approx 36.9^\circ\).

(c) Since \(CD \perp\) plane \(ABC\), the projection of line \(AD\) on plane \(ABC\) is \(AC\).
Thus, the angle between the line \(AD\) and the plane \(ABC\) is \(\angle DAC\).
In right-angled triangle \(\triangle ABC\):
\(AC = \sqrt{AB^2 + BC^2} = \sqrt{9^2 + 12^2} = 15\text{ cm}\).
In right-angled triangle \(\triangle ACD\) (with \(\angle ACD = 90^\circ\)):
\(\tan \angle DAC = \frac{CD}{AC} = \frac{5}{15} = \frac{1}{3}\)
\(\angle DAC = \arctan\left(\frac{1}{3}\right) \approx 18.4349^\circ \approx 18.4^\circ\).

- For finding BD = 13 (1M)
- For finding AD = 15.8 cm (1A)
- For identifying the required angle is angle ACB (1M)
- For finding angle ACB = 36.9 degrees (1A)
- For identifying the required angle is angle DAC (or stating CD perpendicular to plane ABC) (1M)
- For finding AC = 15 cm (1M or implicit)
- For finding angle DAC = 18.4 degrees (1A)

(a) (i) Since \(2x - 1\) is a factor of \(G(x)\), by the factor theorem:
\(G(1/2) = 0\)
\(2(1/2)^2 + k(1/2) - 3 = 0\)
\(\frac{1}{2} + \frac{k}{2} - 3 = 0 \implies k = 5\).

(ii) Since \(k = 5\), \(G(x) = 2x^2 + 5x - 3 = (2x - 1)(x + 3)\).
For \(F(x) = 2x^3 - 5x^2 - 4x + 3\):
Since \(2x - 1\) is a factor of \(F(x)\), performing division:
\(F(x) = (2x - 1)(x^2 - 2x - 3) = (2x - 1)(x + 1)(x - 3)\).

Comparing \(F(x)\) and \(G(x)\):
\(F(x) = (2x - 1)(x + 1)(x - 3)\)
\(G(x) = (2x - 1)(x + 3)\)

Greatest Common Divisor (GCD) \(= 2x - 1\)
Least Common Multiple (LCM) \(= (2x - 1)(x + 1)(x - 3)(x + 3)\).

(b) For the equation \(\frac{G(x)}{F(x)} = \frac{2}{x+1}\):
The equation is defined for \(F(x) \neq 0\) and \(x + 1 \neq 0\), i.e., \(x \neq 1/2\), \(x \neq -1\), and \(x \neq 3\).

Substitute the factorised forms of \(F(x)\) and \(G(x)\):
\(\frac{(2x-1)(x+3)}{(2x-1)(x+1)(x-3)} = \frac{2}{x+1}\)

Since \(x \neq 1/2\), we can cancel \(2x - 1\):
\(\frac{x+3}{(x+1)(x-3)} = \frac{2}{x+1}\)

Since \(x \neq -1\), we can multiply both sides by \(x+1\):
\(\frac{x+3}{x-3} = 2\)
\(x + 3 = 2(x - 3)\)
\(x + 3 = 2x - 6 \implies x = 9\).

Since \(x = 9\) satisfies all constraints, the solution is \(x = 9\).

- (a)(i) For setting G(1/2) = 0 (1M)
- (a)(i) For finding k = 5 (1A)
- (a)(ii) For factorising F(x) = (2x - 1)(x + 1)(x - 3) (1M)
- (a)(ii) For GCD = 2x - 1 (1A)
- (a)(ii) For LCM = (2x - 1)(x + 1)(x - 3)(x + 3) (1A)
- (b) For simplifying the equation to (x+3)/(x-3) = 2 (1M)
- (b) For solving to find x = 9 (domain check implicit or explicit) (1A)