2022 HKDSE 數學 模擬試題連答案詳解

\(\frac{(m^{-3}n^4)^3}{m^2 n^{-5}} = \frac{m^{-9}n^{12}}{m^2 n^{-5}} = \frac{n^{12 - (-5)}}{m^{2 - (-9)}} = \frac{n^{17}}{m^{11}}\)

1M for \((m^a n^b)^c = m^{ac}n^{bc}\) applied correctly to numerator
1M for using index laws \(\frac{x^p}{x^q} = x^{p-q}\) or \(x^{-k} = \frac{1}{x^k}\)
1A for \(\frac{n^{17}}{m^{11}}\)

\(3p + 2q = 4r(5 - p)\)
\(3p + 2q = 20r - 4pr\)
\(3p + 4pr = 20r - 2q\)
\(p(3 + 4r) = 20r - 2q\)
\(p = \frac{20r - 2q}{3 + 4r}\)

1M for clearing the fraction \(3p + 2q = 4r(5 - p)\)
1M for grouping terms in \(p\) on one side
1A for \(p = \frac{20r - 2q}{3 + 4r}\) or equivalent

(a) \(4u^2 - 12uv + 9v^2 = (2u - 3v)^2\)
(b) \(4u^2 - 12uv + 9v^2 - 6u + 9v = (2u - 3v)^2 - 3(2u - 3v) = (2u - 3v)(2u - 3v - 3)\)

(a) 1A for \((2u - 3v)^2\)
(b) 1M for using the result of (a)
1A for \((2u - 3v)(2u - 3v - 3)\)

(a) Selling price \(= 480 \times (1 + 15\%) = 480 \times 1.15 = \$552\)
(b) Let \(\$M\) be the marked price.
\(M \times (1 - 20\%) = 552\)
\(0.8M = 552\)
\(M = 690\)
Thus, the marked price is \(\$690\).

(a) 1M for \(480 \times (1 + 15\%)\)
1A for \(\$552\)
(b) 1M for \(\frac{552}{1 - 20\%}\) or \(M(1 - 20\%) = 552\)
1A for \(\$690\)

(a) \(\frac{5x - 7}{3} \ge 2x - 4\)
\(5x - 7 \ge 6x - 12\)
\(-x \ge -5\)
\(x \le 5\)

(b) Solving \(3x + 14 > 0\):
\(3x > -14\)
\(x > -\frac{14}{3}\)
Combined compound inequality: \(-\frac{14}{3} < x \le 5\).
The negative integers in this range are \(-4, -3, -2, -1\).
Therefore, there are 4 negative integers satisfying both inequalities.

(a) 1M for clearing denominator and collecting terms
1A for \(x \le 5\)
(b) 1M for finding \(x > -\frac{14}{3}\) and combining inequalities
1A for 4

(a) The coordinates of \(A'\) are \((4, 6)\).
The coordinates of \(B'\) are \((-2, -2)\).

(b) Slope of \(A'B' = \frac{-2 - 6}{-2 - 4} = \frac{-8}{-6} = \frac{4}{3}\).
Equation of the line passing through \(A'\) and \(B'\):
\(y - 6 = \frac{4}{3}(x - 4)\)
\(3(y - 6) = 4(x - 4)\)
\(3y - 18 = 4x - 16\)
\(4x - 3y + 2 = 0\)

(a) 1A for \(A'(4, 6)\)
1A for \(B'(-2, -2)\)
(b) 1M for finding slope and setting up equation of straight line
1A for \(4x - 3y + 2 = 0\) (or \(y = \frac{4}{3}x + \frac{2}{3}\))

(a) Since the total number of students is 25:
\(4 + k + 8 + 5 + h = 25 \implies k + h = 8\) ... (1)

Since the mean is 4.4:
\(\frac{2(4) + 3k + 4(8) + 5(5) + 6h}{25} = 4.4\)
\(8 + 3k + 32 + 25 + 6h = 110\)
\(3k + 6h + 65 = 110\)
\(3k + 6h = 45 \implies k + 2h = 15\) ... (2)

Subtracting (1) from (2):
\(h = 7\)
Substituting \(h = 7\) into (1):
\(k = 1\)

(b) The number of students is 25. The median is the 13th datum.
Cumulative frequencies: \(2: 4\), \(3: 5\), \(4: 13\), \(5: 18\), \(6: 25\).
Thus, the median is 4 books.
The mode is the datum with the highest frequency, which is 4 books (frequency = 8).

(a) 1M for \(k + h = 8\)
1M for \(\frac{2(4) + 3k + 4(8) + 5(5) + 6h}{25} = 4.4\)
1A for \(h = 7\) and \(k = 1\)
(b) 1A for median = 4
1A for mode = 4

(a) Let the angle of the sector be \(\theta\).
Perimeter of the sector \(= 2r + \text{arc length}\)
\(2(12) + 2\pi(12)\left(\frac{\theta}{360^\circ}\right) = 24 + 5\pi\)
\(24 + \frac{24\pi \theta}{360^\circ} = 24 + 5\pi\)
\(\frac{\pi \theta}{15^\circ} = 5\pi\)
\(\theta = 75^\circ\)

(b) Area of the sector \(= \pi r^2 \left(\frac{\theta}{360^\circ}\right)\)
\(= \pi(12)^2 \left(\frac{75^\circ}{360^\circ}\right)\)
\(= 144\pi \times \frac{5}{24}\)
\(= 30\pi\text{ cm}^2\)

(a) 1M for setting up equation for the perimeter: \(2(12) + 2\pi(12)\left(\frac{\theta}{360^\circ}\right) = 24 + 5\pi\)
1A for \(\theta = 75^\circ\)
(b) 1M for \(\pi(12)^2 \left(\frac{75^\circ}{360^\circ}\right)\) or \(\frac{1}{2} r \ell = \frac{1}{2}(12)(5\pi)\)
1A for \(30\pi\text{ cm}^2\)

(a)
\(\frac{5x + 3}{2} < 3x + 4\)
\(5x + 3 < 6x + 8\)
\(-x < 5\)
\(x > -5\)

\(4 - 3x \le 16\)
\(-3x \le 12\)
\(x \ge -4\)

Since the compound inequality requires both conditions to hold simultaneously,
the solution of (*) is \(x \ge -4\).

(b)
The negative integers satisfying (*) are \(-4\), \(-3\), \(-2\), and \(-1\).
Thus, there are 4 negative integers satisfying (*).

(a)
\(x > -5\) (1M)
\(x \ge -4\) (1M)
\(x \ge -4\) (1A)

(b)
4 (1A)

(a) Let \(f(x) = ax + bx^2\), where \(a\) and \(b\) are non-zero constants.
Since \(f(2) = 14\) and \(f(5) = 65\), we have:
\(2a + 4b = 14 \implies a + 2b = 7\)
\(5a + 25b = 65 \implies a + 5b = 13\)
Solving the simultaneous equations:
\(3b = 6 \implies b = 2\)
\(a + 2(2) = 7 \implies a = 3\)
Therefore, \(f(x) = 3x + 2x^2\) (or \(f(x) = 2x^2 + 3x\)).

(b) \(f(x) = 35\)
\(2x^2 + 3x = 35\)
\(2x^2 + 3x - 35 = 0\)
\((2x - 7)(x + 5) = 0\)
\(x = \frac{7}{2}\) or \(x = -5\)

(c) \(g(x) = 2x^2 + 3x - k\).
The graph of \(y = g(x)\) does not touch or intersect the \(x\)-axis means that the equation \(2x^2 + 3x - k = 0\) has no real roots.
\(\Delta < 0\)
\(3^2 - 4(2)(-k) < 0\)
\(9 + 8k < 0\)
\(k < -\frac{9}{8}\)

(a) Let \(f(x) = ax + bx^2\) [1M]
Substituting \((2, 14)\) or \((5, 65)\) to set up linear equations [1M]
\(f(x) = 2x^2 + 3x\) [1A]

(b) Setting up \(2x^2 + 3x - 35 = 0\) [1M]
\(x = \frac{7}{2}\) or \(x = -5\) [1A]

(c) Using \(\Delta < 0\) [1M]
\(k < -\frac{9}{8}\) [1A]

(a) The number of athletes is 20.
The median is the average of the 10th and 11th data values.
From the diagram, the 10th value is \(60 + p\) and the 11th value is \(65\).
\(\frac{(60 + p) + 65}{2} = 65 \implies 60 + p = 65 \implies p = 5\).

The lower quartile \(Q_1 = \frac{58 + 60}{2} = 59\text{ kg}\).
The upper quartile \(Q_3 = \frac{72 + (70 + q)}{2} = 71 + \frac{q}{2}\text{ kg}\).
Inter-quartile range \(= Q_3 - Q_1 = 14.5\)
\(\left(71 + \frac{q}{2}\right) - 59 = 14.5\)
\(12 + \frac{q}{2} = 14.5 \implies \frac{q}{2} = 2.5 \implies q = 5\).

(b) (i) The original 20 data values have median \(65\text{ kg}\).
When 4 athletes of weights \(62\text{ kg}\), \(65\text{ kg}\), \(65\text{ kg}\), and \(68\text{ kg}\) join, the total number of athletes becomes 24.
Among the 4 new data, two are \(\le 65\) and two are \(\ge 65\).
The 12th and 13th values of the new combined data set are both \(65\text{ kg}\).
New median \(= \frac{65 + 65}{2} = 65\text{ kg}\).
Therefore, the change in the median is \(0\text{ kg}\).

(ii) Original mean \(= \frac{52+54+56+58+58+60+61+63+65+65+65+68+68+69+71+72+75+76+77+80+84}{20} = \frac{1352}{20} = 67.6\text{ kg}\).
Original standard deviation \(\sigma_1 \approx 8.7886\text{ kg}\).
For the new 24 data values, the mean is \(\frac{1352 + 62 + 65 + 65 + 68}{24} = \frac{1612}{24} \approx 67.1667\text{ kg}\).
New standard deviation \(\sigma_2 \approx 8.1275\text{ kg}\).
Since \(\sigma_2 < \sigma_1\), the standard deviation decreases.
Thus, the claim is agreed.

(a) \(\frac{60+p+65}{2} = 65 \implies p = 5\) [1A]
\(Q_1 = 59\) and setting up equation for IQR [1M]
\(q = 5\) [1A]

(b)(i) New median \(= 65\text{ kg}\) and change \(= 0\text{ kg}\) [1A]
(b)(ii) Finding original standard deviation (\(\approx 8.79\)) or calculating deviations [1M]
Finding new standard deviation (\(\approx 8.13\)) [1M]
Conclusion with correct reason [1A]

(a) Centre of \(C = \left(-\frac{-12}{2}, -\frac{6}{2}\right) = (6, -3)\).
Radius of \(C = \sqrt{6^2 + (-3)^2 - (-19)} = \sqrt{36 + 9 + 19} = \sqrt{64} = 8\).

(b) (i) Since \(L\) is tangent to \(C\), the perpendicular distance from the centre \((6, -3)\) to \(L\) is equal to the radius \(8\).
\(\frac{|4(6) - 3(-3) + k|}{\sqrt{4^2 + (-3)^2}} = 8\)
\(\frac{|24 + 9 + k|}{5} = 8\)
\(|33 + k| = 40\)
\(33 + k = 40 \implies k = 7\) or \(33 + k = -40 \implies k = -73\).

(ii) When \(k = 7 > 0\), the equation of \(L\) is \(4x - 3y + 7 = 0\).
The line passing through the centre \(G(6, -3)\) and perpendicular to \(L\) has slope \(-\frac{3}{4}\).
Equation of the normal line:
\(y - (-3) = -\frac{3}{4}(x - 6)\)
\(4(y + 3) = -3(x - 6)\)
\(3x + 4y - 6 = 0\)

Solving \(\begin{cases} 4x - 3y + 7 = 0 \\ 3x + 4y - 6 = 0 \end{cases}\):
From the second equation, \(y = \frac{6 - 3x}{4}\).
Substitute into the first equation:
\(4x - 3\left(\frac{6 - 3x}{4}\right) + 7 = 0\)
\(16x - 18 + 9x + 28 = 0\)
\(25x + 10 = 0 \implies x = -\frac{2}{5} = -0.4\).
\(y = \frac{6 - 3(-0.4)}{4} = \frac{7.2}{4} = 1.8\).
Thus, the coordinates of \(T\) are \((-0.4, 1.8)\).

(a) Centre \(= (6, -3)\) [1A]
Radius \(= 8\) [1A]

(b)(i) Using distance from centre to line \(= r\) (or substituting \(y\) into circle equation and setting \(\Delta = 0\)) [1M]
\(|33 + k| = 40\) [1M]
\(k = 7\) or \(k = -73\) [1A]

(b)(ii) Finding the equation of the normal line through centre [1M]
Solving simultaneous equations to get \(T(-0.4, 1.8)\) [1A]

(a) Since \(x - 2\) is a factor of \(f(x)\), \(f(2) = 0\).
\(3(2)^3 + h(2)^2 + k(2) + 8 = 0\)
\(24 + 4h + 2k + 8 = 0 \implies 4h + 2k = -32 \implies 2h + k = -16\) ... (1)

Since the remainder of \(f(x)\) divided by \(x + 1\) is \(9\), \(f(-1) = 9\).
\(3(-1)^3 + h(-1)^2 + k(-1) + 8 = 9\)
\(-3 + h - k + 8 = 9 \implies h - k = 4\) ... (2)

Adding (1) and (2):
\(3h = -12 \implies h = -4\).
From (2), \(-4 - k = 4 \implies k = -8\).

(b) (i) \(f(x) = 3x^3 - 4x^2 - 8x + 8\).
Dividing \(f(x)\) by \(x - 2\):
\(3x^3 - 4x^2 - 8x + 8 = (x - 2)(3x^2 + 2x - 4)\).

(ii) For the equation \(f(x) = 0\):
\((x - 2)(3x^2 + 2x - 4) = 0\)
\(x - 2 = 0\) or \(3x^2 + 2x - 4 = 0\)
For \(3x^2 + 2x - 4 = 0\):
\(x = \frac{-2 \pm \sqrt{2^2 - 4(3)(-4)}}{2(3)} = \frac{-2 \pm \sqrt{4 + 48}}{6} = \frac{-2 \pm \sqrt{52}}{6} = \frac{-1 \pm \sqrt{13}}{3}\).
Since \(\sqrt{13}\) is not an integer, \(\frac{-1 \pm \sqrt{13}}{3}\) are irrational numbers.
Therefore, not all roots of \(f(x) = 0\) are rational numbers.
The claim is disagreed.

(a) \(f(2) = 0 \implies 2h + k = -16\) [1M]
\(f(-1) = 9 \implies h - k = 4\) [1M]
\(h = -4, k = -8\) [1A]

(b)(i) \((x - 2)(3x^2 + 2x - 4)\) [1A]
(b)(ii) Solving \(3x^2 + 2x - 4 = 0\) to get \(x = \frac{-1 \pm \sqrt{13}}{3}\) [1M]
Showing that \(\frac{-1 \pm \sqrt{13}}{3}\) are irrational [1M]
Conclusion: Disagree [1A]

(a) Let \(a\) be the 1st term and \(r\) be the common ratio.
\(T_3 = ar^2 = 72\) ... (1)
\(T_6 = ar^5 = -576\) ... (2)
Dividing (2) by (1):
\(\frac{ar^5}{ar^2} = \frac{-576}{72}\)
\(r^3 = -8 \implies r = -2\).
Substitute \(r = -2\) into (1):
\(a(-2)^2 = 72 \implies 4a = 72 \implies a = 18\).

(b) (i) \(S_n = \frac{a(1 - r^n)}{1 - r} = \frac{18(1 - (-2)^n)}{1 - (-2)} = \frac{18(1 - (-2)^n)}{3} = 6(1 - (-2)^n)\).

(ii) We want \(S_n > 10^7\):
\(6(1 - (-2)^n) > 10^7\)
\(1 - (-2)^n > \frac{10^7}{6}\)
\(-(-2)^n > \frac{10^7}{6} - 1 \approx 1666665.67\)

If \(n\) is an even integer, \(-(-2)^n = -2^n < 0\), which cannot be greater than \(1666665.67\).
Thus, \(n\) must be an odd integer.
For odd \(n\), \(-(-2)^n = 2^n\).
\(2^n > \frac{10^7}{6} - 1\)
\(n \log 2 > \log\left(\frac{10^7 - 6}{6}\right)\)
\(n > \frac{\log(1666665.67)}{\log 2} \approx 20.67\)
Since \(n\) must be an odd integer, the least odd integer greater than \(20.67\) is \(n = 21\).

Check:
For \(n = 21\), \(S_{21} = 6(1 - (-2)^{21}) = 6(1 + 2097152) = 12582918 > 10^7\).
For \(n = 19\), \(S_{19} = 6(1 + 2^{19}) = 3145734 < 10^7\).
Thus, the least value of \(n\) is \(21\).

(a) Setting up \(ar^2 = 72\) and \(ar^5 = -576\) [1M]
\(r = -2\) [1A]
\(a = 18\) [1A]

(b)(i) \(S_n = 6(1 - (-2)^n)\) [1A]
(b)(ii) Setting up \(6(1 - (-2)^n) > 10^7\) [1M]
Recognizing \(n\) must be odd and solving \(2^n > 1666665.67\) (or testing odd values) [1M]
\(n = 21\) [1A]

(a) The common ratio is $\dfrac{\log_2 k^2}{\log_2 k} = \dfrac{2\log_2 k}{\log_2 k} = 2$.

(b)(i) $T(n) = T(1) \cdot r^{n-1} = (\log_2 k) \cdot 2^{n-1} = 2^{n-1}\log_2 k$.

(ii) Sum of the first $m$ terms is:
\[ \sum_{i=1}^m T(i) = \log_2 k \cdot \frac{2^m - 1}{2 - 1} = (2^m - 1)\log_2 k \]
Since $k > 1$, we have $\log_2 k > 0$.
\[ (2^m - 1)\log_2 k > 1000\log_2 k \]
\[ 2^m - 1 > 1000 \]
\[ 2^m > 1001 \]
Since $2^9 = 512$ and $2^{10} = 1024$, the least integer value of $m$ is $10$.

(a) Common ratio $= 2$ 1A

(b)(i) $T(n) = 2^{n-1}\log_2 k$ 1A

(b)(ii) $\sum_{i=1}^m T(i) = \dfrac{(\log_2 k)(2^m - 1)}{2 - 1}$ 1M
$(2^m - 1)\log_2 k > 1000\log_2 k$ 1M
$2^m > 1001$ 1M
$m \ge 10$, so the least value is $10$ 1A

(a) The total number of balls is $6 + 4 + 2 = 12$.
Number of ways to choose $3$ balls from $12$ is $C_3^{12} = 220$.
Number of ways to draw $1$ red, $1$ blue, and $1$ green ball is:
\[ C_1^6 \times C_1^4 \times C_1^2 = 6 \times 4 \times 2 = 48 \]
The required probability is:
\[ \frac{48}{220} = \frac{12}{55} \]

(b) The cases for winning a prize are drawing exactly $2$ red balls or drawing exactly $3$ red balls.
Number of non-red balls is $4 + 2 = 6$.
Number of ways to draw $2$ red and $1$ non-red ball:
\[ C_2^6 \times C_1^6 = 15 \times 6 = 90 \]
Number of ways to draw $3$ red balls:
\[ C_3^6 = 20 \]
Total favorable outcomes $= 90 + 20 = 110$.
Probability of winning a prize is:
\[ \frac{110}{220} = \frac{1}{2} = 0.5 \]
Since the probability is exactly $0.5$, it is not strictly greater than $0.5$.
Thus, the claim is disagreed.

(a) Total number of combinations $= C_3^{12} = 220$ 1M
Number of favorable outcomes $= C_1^6 \times C_1^4 \times C_1^2 = 48$ 1M
Required probability $= \dfrac{48}{220} = \dfrac{12}{55}$ 1A

(b) $\text{P(2 red and 1 non-red)} = \dfrac{C_2^6 \times C_1^6}{220} = \dfrac{90}{220}$ 1M
$\text{P(3 red)} = \dfrac{C_3^6}{220} = \dfrac{20}{220}$ 1M
$\text{P(winning)} = \dfrac{90 + 20}{220} = 0.5$ 1A
$0.5
gtr 0.5$, so the claim is disagreed 1A (f.t.)

(a) Notice that $\Delta UVW$ is symmetric about the $y$-axis because $V(-6, 0)$ and $W(6, 0)$ are reflections of each other across the $y$-axis. Thus, the centre of the circumcircle lies on the $y$-axis, so let its coordinates be $(0, c)$.
The distance from $(0, c)$ to $U(0, 8)$ equals the distance to $W(6, 0)$:
\[ (8 - c)^2 = 6^2 + (0 - c)^2 \]
\[ 64 - 16c + c^2 = 36 + c^2 \]
\[ 16c = 28 \implies c = \frac{7}{4} \]
The radius $R = 8 - \frac{7}{4} = \frac{25}{4}$.
The equation of the circumcircle is:
\[ x^2 + \left(y - \frac{7}{4}\right)^2 = \left(\frac{25}{4}\right)^2 \]
\[ x^2 + y^2 - \frac{7}{2}y - 36 = 0 \quad \text{or} \quad 2x^2 + 2y^2 - 7y - 72 = 0 \]

(b)(i) Let $K(x, y)$.
\[ KU^2 + KV^2 + KW^2 = [x^2 + (y-8)^2] + [(x+6)^2 + y^2] + [(x-6)^2 + y^2] \]
\[ = (x^2 + y^2 - 16y + 64) + (x^2 + 12x + 36 + y^2) + (x^2 - 12x + 36 + y^2) \]
\[ = 3x^2 + 3y^2 - 16y + 136 \]
Given $3x^2 + 3y^2 - 16y + 136 = 156$:
\[ 3x^2 + 3y^2 - 16y - 20 = 0 \]
\[ x^2 + y^2 - \frac{16}{3}y = \frac{20}{3} \]
\[ x^2 + \left(y - \frac{8}{3}\right)^2 = \frac{20}{3} + \frac{64}{9} = \frac{124}{9} \]
Thus, the locus of $K$ is a circle with centre $\left(0, \frac{8}{3}\right)$ and radius $\frac{\sqrt{124}}{3} \approx 3.71$.

(ii) Let $C_1$ be the circumcircle with centre $O_1\left(0, \frac{7}{4}\right)$ and radius $R_1 = \frac{25}{4} = 6.25$.
Let $C_2$ be the locus circle with centre $O_2\left(0, \frac{8}{3}\right)$ and radius $R_2 = \frac{\sqrt{124}}{3} \approx 3.71$.
Distance between centres $d = \left|\frac{8}{3} - \frac{7}{4}\right| = \frac{11}{12} \approx 0.917$.
Note that:
\[ R_1 - R_2 = 6.25 - 3.71 = 2.54 \]
Since $d = 0.917 < R_1 - R_2 = 2.54$, circle $C_2$ lies entirely inside circle $C_1$ without intersecting it.
Thus, the locus of $K$ does not intersect the circumcircle.

(a) Setting up distance equation or standard form for circumcircle 1M
Finding centre $\left(0, \dfrac{7}{4}\right)$ or radius $\dfrac{25}{4}$ 1M
Equation of circumcircle: $x^2 + y^2 - \dfrac{7}{2}y - 36 = 0$ (or equivalent) 1A

(b)(i) Expanding $KU^2 + KV^2 + KW^2 = 156$ 1M
Locus of $K$ is a circle with centre $\left(0, \dfrac{8}{3}\right)$ and radius $\dfrac{\sqrt{124}}{3}$ 1A

(b)(ii) Finding distance between centres $d = \dfrac{11}{12}$ and difference of radii $R_1 - R_2$ 1M
Conclusion: Since $d < R_1 - R_2$, they do not intersect (No) 1A (f.t.)

(a) Since $\Delta ABC$ is equilateral with side $12\text{ cm}$ and $M$ is the mid-point of $BC$:
\[ AM = 12\sin 60^\circ = 12\left(\frac{\sqrt{3}}{2}\right) = 6\sqrt{3}\text{ cm} \approx 10.3923\text{ cm} \]
In isosceles $\Delta VBC$, $VB = VC = 10\text{ cm}$ and $BM = 6\text{ cm}$. Since $VM \perp BC$:
\[ VM = \sqrt{VB^2 - BM^2} = \sqrt{10^2 - 6^2} = \sqrt{64} = 8\text{ cm} \]

(b) Let $G$ be the projection of $V$ onto the base $\Delta ABC$. Since $VABC$ is a regular pyramid, $G$ is the centroid of $\Delta ABC$.
\[ MG = \frac{1}{3}AM = \frac{1}{3}(6\sqrt{3}) = 2\sqrt{3}\text{ cm} \]
Since $VM \perp BC$ and $AM \perp BC$, the angle between face $VBC$ and base $ABC$ is $\angle VMG$.
\[ \cos \angle VMG = \frac{MG}{VM} = \frac{2\sqrt{3}}{8} = \frac{\sqrt{3}}{4} \]
\[ \angle VMG = \arccos\left(\frac{\sqrt{3}}{4}\right) \approx 64.3411^\circ \approx 64.3^\circ \]

(c) In $\Delta VAM$, $VA = 10\text{ cm}$, $VM = 8\text{ cm}$, $AM = 6\sqrt{3}\text{ cm}$.
Let the angle between $VA$ and the plane $VBC$ be considered. Note that $BC \perp VM$ and $BC \perp AM$, so $BC$ is perpendicular to the plane $VAM$. Thus the plane $VAM \perp$ plane $VBC$.
The projection of $VA$ onto plane $VBC$ lies along the line of intersection, which is $VM$ (or within plane $VBC$).
Let $\theta = \angle AVM$. By the cosine rule in $\Delta VAM$:
\[ \cos \angle AVM = \frac{VA^2 + VM^2 - AM^2}{2(VA)(VM)} = \frac{10^2 + 8^2 - (6\sqrt{3})^2}{2(10)(8)} = \frac{100 + 64 - 108}{160} = \frac{56}{160} = 0.35 \]
\[ \angle AVM = \arccos(0.35) \approx 69.5127^\circ \]
Since $BC \perp \text{plane } VAM$, the projection of $A$ on plane $VBC$ falls on the line $VM$, so the angle between $VA$ and plane $VBC$ is precisely $\angle AVM \approx 69.5^\circ$.
Since $69.5^\circ > 45^\circ$, the angle exceeds $45^\circ$.
Therefore, the claim is incorrect.

(a) $AM = 12\sin 60^\circ = 6\sqrt{3}\text{ cm}$ 1A
$VM = \sqrt{10^2 - 6^2} = 8\text{ cm}$ 1M + 1A

(b) Identifying $\angle VMG$ as the angle between the planes 1M
$\angle VMG = \arccos\left(\dfrac{2\sqrt{3}}{8}\right) \approx 64.3^\circ$ 1A

(c) $\cos \angle AVM = \dfrac{10^2 + 8^2 - (6\sqrt{3})^2}{2(10)(8)} = 0.35$ 1M
$\angle AVM \approx 69.5^\circ > 45^\circ$, hence the claim is incorrect 1A (f.t.)

(a) Completing the square for $\mathrm{f}(x)$:
\[ \mathrm{f}(x) = 2(x^2 - 4kx) + 8k^2 + 3k - 1 \]
\[ = 2(x - 2k)^2 - 2(4k^2) + 8k^2 + 3k - 1 \]
\[ = 2(x - 2k)^2 + 3k - 1 \]
Thus, the coordinates of the vertex $V_1$ are $(2k, 3k - 1)$.

(b)(i) Reflecting $y = \mathrm{f}(x)$ with respect to the $x$-axis gives $y = -\mathrm{f}(x)$. The vertex becomes $(2k, -(3k - 1)) = (2k, 1 - 3k)$.
Translating upwards by $6$ units gives the new vertex $V_2$:
\[ (2k, 1 - 3k + 6) = (2k, 7 - 3k) \]

(ii) Note that $V_1$ and $V_2$ share the same $x$-coordinate $2k$.
Therefore, the distance between $V_1$ and $V_2$ is the absolute difference between their $y$-coordinates:
\[ |(7 - 3k) - (3k - 1)| = 10 \]
\[ |8 - 6k| = 10 \]
Case 1:
\[ 8 - 6k = 10 \implies -6k = 2 \implies k = -\frac{1}{3} \]
Wait, let's recompute:
$8 - 6k = 10 \implies -6k = 2 \implies k = -\frac{1}{3}$.
Case 2:
\[ 8 - 6k = -10 \implies -6k = -18 \implies k = 3 \]
Thus, the possible values of $k$ are $3$ and $-\frac{1}{3}$.

(a) $\mathrm{f}(x) = 2(x - 2k)^2 + 3k - 1$ 1M
Vertex $V_1 = (2k, 3k - 1)$ 1A

(b)(i) Vertex after reflection is $(2k, 1 - 3k)$ 1M
$V_2 = (2k, 7 - 3k)$ 1A

(b)(ii) $|(7 - 3k) - (3k - 1)| = 10$ 1M
$|8 - 6k| = 10$ 1M
$k = 3$ or $k = -\dfrac{1}{3}$ 1A