2024 Cambridge IGCSE International Mathematics (0607) 模拟试题及答案详解

First apply the negative index by inverting the fraction: \(\left(\frac{y^9}{27x^6}\right)^{\frac{2}{3}}\). Next apply the power of \(\frac{2}{3}\) to each component: \((y^9)^{\frac{2}{3}} = y^{9 \times \frac{2}{3}} = y^6\), \((27)^{\frac{2}{3}} = (\sqrt[3]{27})^2 = 3^2 = 9\), and \((x^6)^{\frac{2}{3}} = x^{6 \times \frac{2}{3}} = x^4\). Combining these gives \(\frac{y^6}{9x^4}\).

M1 for applying negative power or fractional index correctly to at least two terms (e.g. \(27^{-\frac{2}{3}} = \frac{1}{9}\) or \((x^6)^{-\frac{2}{3}} = x^{-4}\) or \((y^9)^{-\frac{2}{3}} = y^{-6}\))
A1 for \(\frac{y^6}{9x^4}\) or \(\frac{1}{9}x^{-4}y^6\) cao

Find a common denominator of \((2x-1)(x+4)\):
\[\frac{3(x+4) - 2(2x-1)}{(2x-1)(x+4)}\]
Expand the brackets in the numerator:
\[3(x+4) - 2(2x-1) = 3x + 12 - 4x + 2 = 14 - x\]
Thus, the simplified fraction is \(\frac{14-x}{(2x-1)(x+4)}\).

M1 for correct common denominator and correct algebraic numerator expansion seen, e.g. \(3(x+4) - 2(2x-1)\)
A1 for \(\frac{14-x}{(2x-1)(x+4)}\) or \(\frac{14-x}{2x^2+7x-4}\) oe

Multiply numerator and denominator by the conjugate \(3 + \sqrt{3}\):
\[\frac{12(3 + \sqrt{3})}{(3 - \sqrt{3})(3 + \sqrt{3})} = \frac{12(3 + \sqrt{3})}{3^2 - (\sqrt{3})^2} = \frac{12(3 + \sqrt{3})}{9 - 3} = \frac{12(3 + \sqrt{3})}{6}\]
Simplify by dividing by 6:
\[2(3 + \sqrt{3}) = 6 + 2\sqrt{3}\]

M1 for multiplying numerator and denominator by \(3 + \sqrt{3}\)
A1 for \(6 + 2\sqrt{3}\) or \(2(3 + \sqrt{3})\) cao

Apply the power rule of logarithms: \(2\log_3 4 = \log_3 (4^2) = \log_3 16\).
Combine the logarithms using the product rule: \(\log_3 (x \times 16) = \log_3 (16x)\).
Rewrite the equation in exponential form:
\[16x = 3^3\]
\[16x = 27\]
\[x = \frac{27}{16}\]

M1 for correctly applying logarithm laws to obtain \(\log_3(16x) = 3\) or \(16x = 3^3\) soi
A1 for \(\frac{27}{16}\) or \(1\frac{11}{16}\) or \(1.6875\)

Multiply both sides by \((5 - p)\):
\[q(5 - p) = 2p + 3\]
\[5q - pq = 2p + 3\]
Collect terms containing \(p\) on one side and remaining terms on the other:
\[5q - 3 = 2p + pq\]
Factorise \(p\) out:
\[5q - 3 = p(2 + q)\]
Divide by \((2 + q)\):
\[p = \frac{5q - 3}{q + 2}\]

M1 for multiplying by \((5 - p)\) and expanding brackets to isolate terms in \(p\) on one side, e.g. \(5q - 3 = 2p + pq\) soi
A1 for \(p = \frac{5q - 3}{q + 2}\) or \(p = \frac{3 - 5q}{-2 - q}\) oe

Apply the negative index rule first:
\(\left(\dfrac{27}{64}\right)^{-\frac{2}{3}} = \left(\dfrac{64}{27}\right)^{\frac{2}{3}}\)

Next, take the cube root of the numerator and denominator:
\(\sqrt[3]{\dfrac{64}{27}} = \dfrac{4}{3}\)

Now, square the result:
\(\left(\dfrac{4}{3}\right)^2 = \dfrac{16}{9}\)

M1 for \(\left(\dfrac{64}{27}\right)^{\frac{2}{3}}\) or \(\left(\dfrac{3}{4}\right)^{-2}\) or \(\dfrac{1}{\left(\sqrt[3]{\frac{27}{64}}\right)^2}\) oe
A1 for \(\dfrac{16}{9}\) or \(1\dfrac{7}{9}\) or \(1.\dot{7}\)

Group into pairs of terms and factor out common factors:
\(6ax - 9ay - 4bx + 6by = 3a(2x - 3y) - 2b(2x - 3y)\)

Factor out the common binomial factor \((2x - 3y)\):
\(= (2x - 3y)(3a - 2b)\)

M1 for \(3a(2x - 3y) - 2b(2x - 3y)\) or \(2x(3a - 2b) - 3y(3a - 2b)\) oe
A1 for \((2x - 3y)(3a - 2b)\) oe

Subtract 5 from both sides:
\(-3x \leqslant 17 - 5\)
\(-3x \leqslant 12\)

Divide both sides by \(-3\) and reverse the inequality sign:
\(x \geqslant -4\)

M1 for \(-3x \leqslant 12\) or \(3x \geqslant -12\) or \(5 - 17 \leqslant 3x\) oe
A1 for \(x \geqslant -4\) (or \([-4, \infty)\))

Divide the numerical parts and subtract the powers of 10:
\(\dfrac{4.8 \times 10^7}{1.2 \times 10^{-3}} = \left(\dfrac{4.8}{1.2}\right) \times 10^{7 - (-3)}\)
\(= 4 \times 10^{7 + 3}\)
\(= 4 \times 10^{10}\)

M1 for \(\dfrac{4.8}{1.2} \times 10^{7 - (-3)}\) soi or answer in the form \(4 \times 10^k\) where \(k \neq 10\)
A1 for \(4 \times 10^{10}\) cao

Square both sides of the equation:
\(q^2 = \dfrac{p + 5}{3}\)

Multiply both sides by 3:
\(3q^2 = p + 5\)

Subtract 5 from both sides:
\(p = 3q^2 - 5\)

M1 for squaring both sides: \(q^2 = \dfrac{p + 5}{3}\) or \(3q^2 = p + 5\)
A1 for \(p = 3q^2 - 5\) oe

Multiply the numerator and the denominator by the conjugate of the denominator, \( 4 + \sqrt{7} \):

\[ \frac{18}{4 - \sqrt{7}} \times \frac{4 + \sqrt{7}}{4 + \sqrt{7}} \]

Expand the denominator:
\[ (4 - \sqrt{7})(4 + \sqrt{7}) = 4^2 - (\sqrt{7})^2 = 16 - 7 = 9 \]

Simplify the fraction:
\[ \frac{18(4 + \sqrt{7})}{9} = 2(4 + \sqrt{7}) = 8 + 2\sqrt{7} \]

M1 for multiplying numerator and denominator by \( 4 + \sqrt{7} \)
M1 for correct expansion of denominator to \( 16 - 7 \) or \( 9 \)
A1 for \( 8 + 2\sqrt{7} \) or \( 2(4 + \sqrt{7}) \)

Factorise the quadratic denominator:
\[x^2 - x - 6 = (x - 3)(x + 2)\]

Write all terms over the common denominator \((x - 3)(x + 2)\):
\[\frac{x(x + 2) - 1(x - 3) - 5}{(x - 3)(x + 2)}\]

Expand and simplify the numerator:
\[x(x + 2) - (x - 3) - 5 = x^2 + 2x - x + 3 - 5 = x^2 + x - 2\]

Factorise the simplified numerator:
\[x^2 + x - 2 = (x + 2)(x - 1)\]

Cancel the common factor \((x + 2)\):
\[\frac{(x + 2)(x - 1)}{(x - 3)(x + 2)} = \frac{x - 1}{x - 3}\]

M1 for factorising \(x^2 - x - 6\) as \((x - 3)(x + 2)\) soi
M1 for putting all terms over a common denominator with correct expansion of numerator: \(x^2 + 2x - x + 3 - 5\) oe
M1 for simplifying to \(x^2 + x - 2\) and factorising as \((x + 2)(x - 1)\)
A1 for fully correct working showing cancellation of \((x + 2)\) leading to \(\frac{x - 1}{x - 3}\) (AG)

(a)(i) In right-angled triangle \(ABC\):
\[ AC = \sqrt{14^2 + 10^2} = \sqrt{196 + 100} = \sqrt{296} = 17.20465... \approx 17.20\text{ cm} \]

(a)(ii) \(M\) is the midpoint of \(AC\), so:
\[ AM = \frac{1}{2} \times \sqrt{296} = \sqrt{74} \approx 8.6023\text{ cm} \]
In right-angled triangle \(VMA\):
\[ VM = \sqrt{VA^2 - AM^2} = \sqrt{18^2 - 74} = \sqrt{324 - 74} = \sqrt{250} \approx 15.811\text{ cm} \approx 15.8\text{ cm} \]

(b) Let \(\theta\) be the angle between \(VA\) and the base:
\[ \cos\theta = \frac{AM}{VA} = \frac{\sqrt{74}}{18} \implies \theta = \cos^{-1}\left(\frac{\sqrt{74}}{18}\right) \approx 61.45^{\circ} \approx 61.5^{\circ} \]

(c) Slant height of face \(VAB\), \(h_1 = \sqrt{18^2 - 7^2} = \sqrt{324 - 49} = \sqrt{275} \approx 16.583\text{ cm}\).
Area of 2 faces (\(VAB\) and \(VCD\)) \(= 2 \times \left(\frac{1}{2} \times 14 \times \sqrt{275}\right) = 14\sqrt{275} \approx 232.164\text{ cm}^2\).

Slant height of face \(VBC\), \(h_2 = \sqrt{18^2 - 5^2} = \sqrt{324 - 25} = \sqrt{299} \approx 17.292\text{ cm}\).
Area of 2 faces (\(VBC\) and \(VAD\)) \(= 2 \times \left(\frac{1}{2} \times 10 \times \sqrt{299}\right) = 10\sqrt{299} \approx 172.916\text{ cm}^2\).

Area of base \(= 14 \times 10 = 140\text{ cm}^2\).
Total surface area \(= 140 + 232.164 + 172.916 = 545.08\text{ cm}^2 \approx 545\text{ cm}^2\).

(d) Let \(N\) be the midpoint of \(BC\). \(MN = 7\text{ cm}\), \(VN = \sqrt{299}\text{ cm}\).
\[ \tan\alpha = \frac{VM}{MN} = \frac{\sqrt{250}}{7} \implies \alpha = \tan^{-1}\left(\frac{\sqrt{250}}{7}\right) \approx 66.12^{\circ} \approx 66.1^{\circ} \]

(a)(i) M1 for \(\sqrt{14^2 + 10^2}\) oe, A1 for \(17.2046...\) rounded correctly to at least 4 sig figs.
(a)(ii) M1 for \(18^2 - (\text{their } AM)^2\), A1 for \(15.8\) or \(15.81\) or \(\sqrt{250}\) or \(5\sqrt{10}\).
(b) M1 for \(\cos\theta = \frac{\text{their } AM}{18}\) or \(\sin\theta = \frac{\text{their } VM}{18}\), A1 for \(61.5^{\circ}\) or \(61.45^{\circ}\).
(c) M1 for method to find slant height of at least one triangular face (e.g. \(\sqrt{18^2 - 7^2}\) or \(\sqrt{18^2 - 5^2}\)), M1 for complete method to sum the 4 triangular faces + base area, A1 for \(545\) or \(545.0\) to \(545.1\).
(d) M1 for \(\tan\alpha = \frac{\text{their } VM}{7}\) or \(\cos\alpha = \frac{7}{\text{their } VN}\), A1 for \(66.1^{\circ}\) or \(66.12^{\circ}\).

(a) Vertical asymptote occurs where denominator is zero: \(x = 2\).

(b) Using a graphic display calculator:
(i) Local maximum point is at \((1.29, 0.172)\) (or exactly \((2 - \frac{\sqrt{2}}{2}, 3 - 2\sqrt{2})\)).
(ii) Local minimum point is at \((2.71, 5.83)\) (or exactly \((2 + \frac{\sqrt{2}}{2}, 3 + 2\sqrt{2})\)).

(c)(i) When \(x = 0\), \(y = \frac{3}{-2} = -1.5\). Coordinates: \((0, -1.5)\).
(ii) When \(y = 0\), \(2x^2 - 5x + 3 = 0 \implies (2x - 3)(x - 1) = 0 \implies x = 1, x = 1.5\).
Coordinates: \((1, 0)\) and \((1.5, 0)\).

(d) The horizontal line \(y = k\) does not intersect the curve in the gap between the local maximum and local minimum:
\[ 0.172 < k < 5.83 \quad (\text{or } 3 - 2\sqrt{2} < k < 3 + 2\sqrt{2}) \]

(e) Testing intervals with critical values \(x = 1, 1.5, 2\):
- For \(x \leqslant 1\): \(f(0) = -1.5 \leqslant 0\) (True)
- For \(1 < x < 1.5\): \(f(1.25) > 0\) (False)
- For \(1.5 \leqslant x < 2\): \(f(1.75) < 0\) (True)
- For \(x > 2\): \(f(3) > 0\) (False)
So \(x \leqslant 1\) or \(1.5 \leqslant x < 2\).

(a) B1 for \(x = 2\).
(b)(i) B1 for \((1.29, 0.172)\) or better.
(b)(ii) B2 for \((2.71, 5.83)\) or better (B1 for either \(x = 2.71\) or \(y = 5.83\)).
(c)(i) B1 for \((0, -1.5)\) or \(y = -1.5\).
(c)(ii) B2 for \((1, 0)\) and \((1.5, 0)\) (B1 for one correct).
(d) B2 for \(0.172 < k < 5.83\) (strict inequalities) (B1 for correct values seen in incorrect inequality or non-strict inequalities).
(e) B2 for \(x \leqslant 1\) or \(1.5 \leqslant x < 2\) (B1 for one correct interval).

(a) Cumulative frequencies:
\(t \leqslant 20: 12\)
\(t \leqslant 30: 12 + 26 = 38\)
\(t \leqslant 40: 38 + 44 = 82\)
\(t \leqslant 50: 82 + 48 = 130\)
\(t \leqslant 60: 130 + 22 = 152\)
\(t \leqslant 80: 152 + 8 = 160\)

(b)(i) Using class midpoints: \(10, 25, 35, 45, 55, 70\).
\[ \sum ft = 12(10) + 26(25) + 44(35) + 48(45) + 22(55) + 8(70) = 120 + 650 + 1540 + 2160 + 1210 + 560 = 6240 \]
\[ \text{Mean} = \frac{6240}{160} = 39\text{ minutes} \]

(b)(ii) \(\sum ft^2 = 12(100) + 26(625) + 44(1225) + 48(2025) + 22(3025) + 8(4900) = 274300\).
\[ \text{Variance} = \frac{274300}{160} - 39^2 = 1714.375 - 1521 = 193.375 \]
\[ \text{Standard deviation} = \sqrt{193.375} \approx 13.9\text{ minutes} \]

(c)(i) Median is at position \(80\) (in interval \(30 < t \leqslant 40\)):
\[ \text{Median} = 30 + \frac{80 - 38}{44} \times 10 = 30 + \frac{42}{44} \times 10 \approx 39.5\text{ minutes} \]

(c)(ii) \(Q_1\) at position \(40\) (in interval \(30 < t \leqslant 40\)):
\[ Q_1 = 30 + \frac{40 - 38}{44} \times 10 \approx 30.45 \]
\(Q_3\) at position \(120\) (in interval \(40 < t \leqslant 50\)):
\[ Q_3 = 40 + \frac{120 - 82}{48} \times 10 \approx 47.92 \]
\[ \text{IQR} = Q_3 - Q_1 = 47.92 - 30.45 = 17.47 \approx 17.5\text{ minutes} \]

(d) Number of employees with commute time \(> 50\text{ min} = 22 + 8 = 30\).
\[ P(t > 50) = \frac{30}{160} = \frac{3}{16} = 0.1875 \]

(a) B2 for all 5 correct (B1 for 3 or 4 correct).
(b)(i) M1 for \(\sum f x\) using correct midpoints, A1 for \(39\).
(b)(ii) B1 for \(13.9\) (or \(13.905...\)).
(c)(i) M1 for identifying interval \(30 < t \leqslant 40\) and valid interpolation method, A1 for \(39.5\).
(c)(ii) M1 for finding both \(Q_1\) and \(Q_3\), A1 for \(17.5\) (accept \(17.4\) to \(17.6\)).
(d) M1 for \(\frac{22+8}{160}\) oe, A1 for \(\frac{3}{16}\) or \(0.1875\).

(a) Angle \(\angle QPR = 130^{\circ} - 65^{\circ} = 65^{\circ}\).
Using the cosine rule in \(\triangle PQR\):
\[ QR^2 = 45^2 + 72^2 - 2(45)(72)\cos(65^{\circ}) = 2025 + 5184 - 6480\cos(65^{\circ}) \]
\[ QR^2 = 7209 - 2738.53 = 4470.47 \implies QR = \sqrt{4470.47} \approx 66.86 \approx 66.9\text{ km} \]

(b) Using the sine rule:
\[ \frac{\sin(\angle PQR)}{72} = \frac{\sin(65^{\circ})}{66.86} \implies \sin(\angle PQR) = \frac{72\sin(65^{\circ})}{66.86} \approx 0.9758 \]
\[ \angle PQR = \sin^{-1}(0.9758) \approx 77.38^{\circ} \approx 77.4^{\circ} \]

(c) Back bearing of \(P\) from \(Q\) is \(65^{\circ} + 180^{\circ} = 245^{\circ}\).
Bearing of \(R\) from \(Q = 245^{\circ} - \angle PQR = 245^{\circ} - 77.38^{\circ} = 167.62^{\circ} \approx 167.6^{\circ}\).

(d) Let \(d\) be the shortest distance from \(P\) to line \(QR\):
\[ d = 45\sin(\angle PQR) = 45\sin(77.38^{\circ}) \approx 43.91 \approx 43.9\text{ km} \]
Alternatively, using area:
\[ \text{Area} = \frac{1}{2} \times 45 \times 72 \times \sin(65^{\circ}) = 1468.22\text{ km}^2 \]
\[ d = \frac{2 \times 1468.22}{66.86} \approx 43.9\text{ km} \]

(a) M1 for \(\angle QPR = 65^{\circ}\) soi, M1 for \(45^2 + 72^2 - 2(45)(72)\cos(65^{\circ})\), A1 for \(66.9\) or \(66.86\).
(b) M1 for \(\frac{\sin(\angle PQR)}{72} = \frac{\sin(65^{\circ})}{\text{their } QR}\), M1 for \(\sin(\angle PQR) = \frac{72\sin(65^{\circ})}{\text{their } QR}\), A1 for \(77.4^{\circ}\) or \(77.38^{\circ}\).
(c) M1 for \(180 + 65 - \text{their } \angle PQR\) oe, A1 for \(167.6^{\circ}\) (or \(168^{\circ}\) to 3 sf).
(d) M1 for identifying shortest distance as perpendicular distance, M1 for \(45\sin(\text{their } 77.4^{\circ})\) or \(\frac{2 \times \text{Area}}{\text{their } QR}\), A1 for \(43.9\) or \(43.91\).

(a) \(g(6) = \frac{4}{6 + 2} = \frac{4}{8} = 0.5\).
\[ f(g(6)) = f(0.5) = 3(0.5) - 5 = 1.5 - 5 = -3.5 \]

(b) Let \(y = \frac{4}{x + 2}\):
\[ x + 2 = \frac{4}{y} \implies x = \frac{4}{y} - 2 = \frac{4 - 2y}{y} \]
\[ g^{-1}(x) = \frac{4}{x} - 2 = \frac{4 - 2x}{x} \]

(c) \(f(h(x)) = f(2^x) = 3(2^x) - 5\).
\[ 3(2^x) - 5 = 19 \implies 3(2^x) = 24 \implies 2^x = 8 \implies x = 3 \]

(d) \(g(f(x)) = g(3x - 5) = \frac{4}{(3x - 5) + 2} = \frac{4}{3x - 3}\).

(e)(i) \(y = h(x) - 4\) is a vertical translation of \(4\) units downwards: Translation by vector \(\begin{pmatrix} 0 \\ -4 \end{pmatrix}\).
(ii) \(y = h(x + 3)\) is a horizontal translation of \(3\) units to the left: Translation by vector \(\begin{pmatrix} -3 \\ 0 \end{pmatrix}\).

(a) M1 for \(g(6) = 0.5\) seen, A1 for \(-3.5\) oe.
(b) M1 for correct algebraic rearrangement \(x(y+2) = 4\) or \(y+2 = \frac{4}{x}\), A1 for \(\frac{4 - 2x}{x}\) or \(\frac{4}{x} - 2\).
(c) M1 for \(3(2^x) - 5 = 19\) leading to \(2^x = 8\), A1 for \(x = 3\).
(d) M1 for \(\frac{4}{(3x - 5) + 2}\), A1 for \(\frac{4}{3x - 3}\) or \(\frac{4}{3(x-1)}\).
(e)(i) B1 for Translation by \(\begin{pmatrix} 0 \\ -4 \end{pmatrix}\) or translation 4 units down.
(e)(ii) B2 for Translation by \(\begin{pmatrix} -3 \\ 0 \end{pmatrix}\) or translation 3 units left (B1 for Translation or vector \(\begin{pmatrix} -3 \\ 0 \end{pmatrix}\)).

(a) In the right-angled triangle \(ABC\):
\(AC^2 = AB^2 + BC^2 = 18^2 + 10^2 = 324 + 100 = 424\)
\(AC = \sqrt{424} \approx 20.591... \approx 20.6\text{ cm}\)

(b) Since \(O\) is the midpoint of \(AC\):
\(OA = \frac{1}{2}AC = \frac{\sqrt{424}}{2} = \sqrt{106} \approx 10.296\text{ cm}\)
In the right-angled triangle \(VOA\):
\(VA^2 = VO^2 + OA^2 = 12^2 + (\sqrt{106})^2 = 144 + 106 = 250\)
\(VA = \sqrt{250} = 5\sqrt{10} \approx 15.811... \approx 15.8\text{ cm}\)

(c) The angle between \(VA\) and the base \(ABCD\) is angle \(VAO\):
\(\tan(\angle VAO) = \frac{VO}{OA} = \frac{12}{\sqrt{106}}\)
\(\angle VAO = \tan^{-1}\left(\frac{12}{\sqrt{106}}\right) \approx 49.368...^\circ \approx 49.4^\circ\)

(d) Total Surface Area = Area of base + Area of 2 triangles \(VAB\) + Area of 2 triangles \(VBC\).
- Base area = \(18 \times 10 = 180\text{ cm}^2\).
- Let \(M\) be the midpoint of \(AB\). \(OM = \frac{10}{2} = 5\text{ cm}\).
Slant height \(VM = \sqrt{12^2 + 5^2} = \sqrt{169} = 13\text{ cm}\).
Area of \(\triangle VAB = \frac{1}{2} \times 18 \times 13 = 117\text{ cm}^2\).
- Let \(N\) be the midpoint of \(BC\). \(ON = \frac{18}{2} = 9\text{ cm}\).
Slant height \(VN = \sqrt{12^2 + 9^2} = \sqrt{225} = 15\text{ cm}\).
Area of \(\triangle VBC = \frac{1}{2} \times 10 \times 15 = 75\text{ cm}^2\).
Total surface area = \(180 + 2(117) + 2(75) = 180 + 234 + 150 = 564\text{ cm}^2\).

(e) Volume of the original pyramid \(VABCD\):
\(V_{\text{orig}} = \frac{1}{3} \times \text{base area} \times \text{height} = \frac{1}{3} \times 180 \times 12 = 720\text{ cm}^3\).
The smaller pyramid has a linear scale factor of \(k = \frac{1}{2}\).
Volume scale factor = \(k^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}\).
Volume of smaller pyramid = \(720 \times \frac{1}{8} = 90\text{ cm}^3\).

(a) M1 for \(18^2 + 10^2\) oe
A1 for 20.6 or 20.59... or \(\sqrt{424}\) or \(2\sqrt{106}\)

(b) M1 for \(12^2 + (\text{their } OA)^2\) oe
A1 for 15.8 or 15.81... or \(\sqrt{250}\) or \(5\sqrt{10}\)

(c) M1 for \(\tan(\theta) = \frac{12}{\text{their } OA}\) or \(\sin(\theta) = \frac{12}{\text{their } VA}\) oe
A1 for 49.4 or 49.36... to 49.37

(d) M1 for finding slant height \(VM = 13\) or \(VN = 15\) soi
M1 for correct method for area of all 4 triangular faces: \(2\left(\frac{1}{2} \times 18 \times 13\right) + 2\left(\frac{1}{2} \times 10 \times 15\right)\)
A1 for 564 cao

(e) M1 for \(\frac{1}{3} \times 180 \times 12 \times \left(\frac{1}{2}\right)^3\) oe
A1 for 90 cao

(a) (i) Using a graphic display calculator or calculus:
\(f'(x) = 3x^2 - 6x - 9 = 3(x - 3)(x + 1) = 0 \implies x = -1\) or \(x = 3\).
When \(x = -1\), \(f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 12 = 17\).
Local maximum is at \((-1, 17)\).

(ii) When \(x = 3\), \(f(3) = (3)^3 - 3(3)^2 - 9(3) + 12 = -15\).
Local minimum is at \((3, -15)\).

(b) Evaluating at the endpoints:
\(f(-3) = (-3)^3 - 3(-3)^2 - 9(-3) + 12 = -27 - 27 + 27 + 12 = -15\).
\(f(4) = (4)^3 - 3(4)^2 - 9(4) + 12 = 64 - 48 - 36 + 12 = -8\).
The minimum value attained on \([-3, 4]\) is \(-15\) (at \(x = -3\) and \(x = 3\)).
The maximum value attained on \([-3, 4]\) is \(17\) (at \(x = -1\)).
Range: \(-15 \le f(x) \le 17\).

(c) Using GDC to solve \(x^3 - 3x^2 - 9x + 12 = 0\):
\(x \approx -2.51, \quad x \approx 1.08, \quad x \approx 4.42\) (each to 3 significant figures).

(d) Intersections of \(y = f(x)\) and \(y = g(x)\):
\(x^3 - 3x^2 - 9x + 12 = -2x + 4 \iff x^3 - 3x^2 - 7x + 8 = 0\).
Using GDC, the roots are \(x \approx -2.11, \quad x \approx 0.900, \quad x \approx 4.21\).
By observing the graph where the cubic curve lies on or above the straight line:
\(-2.11 \le x \le 0.900\) or \(x \ge 4.21\).

(a)(i) B1 for \(x = -1\), B1 for \(y = 17\)
(a)(ii) B1 for \((3, -15)\)

(b) M1 for identifying both extreme values \(-15\) and \(17\) soi
A1 for \(-15 \le f(x) \le 17\) or \([-15, 17]\) (accept \(y\) in place of \(f(x)\))

(c) B1 for each correct root: \(x = -2.51\), \(x = 1.08\), \(x = 4.42\) (allow answers rounding to \(-2.51\), \(1.08\), \(4.42\))

(d) M1 for finding at least two correct intersection points \(x \approx -2.11\), \(0.900\), \(4.21\) soi
A1 for one correct interval: \(-2.11 \le x \le 0.900\) oe
A1 for the second correct interval: \(x \ge 4.21\) oe

(a) Mid-interval values: 10, 30, 50, 70, 90.
\(\sum f s = 8(10) + 16(30) + 38(50) + 26(70) + 12(90) = 80 + 480 + 1900 + 1820 + 1080 = 5360\).
Estimated mean = \(\frac{5360}{100} = 53.6\).

(b) Using GDC with the frequency table:
\(\sum f s^2 = 8(100) + 16(900) + 38(2500) + 26(4900) + 12(8100) = 334800\).
\(\sigma = \sqrt{\frac{334800}{100} - 53.6^2} = \sqrt{3348 - 2872.96} = \sqrt{475.04} \approx 21.795... \approx 21.8\).

(c) Cumulative frequencies:
\(s \le 20: 8\)
\(s \le 40: 8 + 16 = 24\)
\(s \le 60: 24 + 38 = 62\)
\(s \le 80: 62 + 26 = 88\)
\(s \le 100: 88 + 12 = 100\)

(d) (i) The 50th value lies in the interval \(40 < s \le 60\):
\(\text{Median} = 40 + \frac{50 - 24}{38} \times 20 = 40 + \frac{26}{38} \times 20 \approx 40 + 13.684 \approx 53.7\).

(ii) Lower quartile \(Q_1\) (25th value) lies in \(40 < s \le 60\):
\(Q_1 = 40 + \frac{25 - 24}{38} \times 20 = 40 + \frac{20}{38} \approx 40.526\).
Upper quartile \(Q_3\) (75th value) lies in \(60 < s \le 80\):
\(Q_3 = 60 + \frac{75 - 62}{26} \times 20 = 60 + \frac{13}{26} \times 20 = 60 + 10 = 70\).
\(\text{IQR} = Q_3 - Q_1 = 70 - 40.526 = 29.474 \approx 29.5\).

(e) Total students with score \(> 60\) is \(26 + 12 = 38\).
\(\text{P}(\text{both } > 60) = \frac{38}{100} \times \frac{37}{99} = \frac{1406}{9900} = \frac{703}{4950} \approx 0.142\).

(a) M1 for \(\sum f s\) using mid-interval values (at least 3 correct products seen)
A1 for 53.6 cao

(b) B1 for 21.8 or 21.79... to 21.80

(c) B1 for all 5 correct cumulative frequencies: 8, 24, 62, 88, 100

(d)(i) M1 for \(40 + \frac{50 - 24}{38} \times 20\) oe
A1 for 53.7 or 53.68...

(d)(ii) M1 for finding \(Q_1 = 40.5\) or \(Q_3 = 70\) oe
A1 for 29.5 or 29.47...

(e) M1 for identifying 38 students scoring \(> 60\) soi
M1 for \(\frac{38}{100} \times \frac{37}{99}\) oe
A1 for \(\frac{703}{4950}\) or 0.142 or 0.1420...

(a)(i)
Sequence \(A\): \(7, 13, 19, 25, \dots\)
Common difference is \(d = 13 - 7 = 6\).
5th term: \(25 + 6 = 31\).
\(n\)th term: \(6n + c\). For \(n=1\), \(6(1) + c = 7 \implies c = 1\).
\(n\)th term \(= 6n + 1\).

(a)(ii)
Sequence \(B\): \(3, 12, 48, 192, \dots\)
Common ratio is \(r = \frac{12}{3} = 4\).
5th term: \(192 \times 4 = 768\).
\(n\)th term: \(a \cdot r^{n-1} = 3 \times 4^{n-1}\) or \(\frac{3}{4} \times 4^n\).

(a)(iii)
Sequence \(C\): \(4, 11, 22, 37, \dots\)
First differences: \(7, 11, 15, \dots\)
Second difference: \(4\).
Since the second difference is constant, \(2a = 4 \implies a = 2\).
Let \(T_n = 2n^2 + bn + c\):
For \(n=1\): \(2(1)^2 + b(1) + c = 4 \implies b + c = 2\)
For \(n=2\): \(2(2)^2 + b(2) + c = 11 \implies 8 + 2b + c = 11 \implies 2b + c = 3\)
Subtracting the first equation from the second: \(b = 1\), so \(c = 1\).
\(n\)th term \(= 2n^2 + n + 1\).
5th term: \(2(5)^2 + 5 + 1 = 2(25) + 5 + 1 = 56\).

(b)
\(6k + 1 = 451\)
\(6k = 450\)
\(k = 75\)

(c)
\(3 \times 4^{n-1} > 10^6\)
\(4^{n-1} > \frac{10^6}{3} \approx 333333.33\)
Using logarithms or calculator:
\((n-1) \log(4) > \log(333333.33)\)
\(n - 1 > \frac{\log(333333.33)}{\log(4)} \approx 9.173\)
\(n > 10.173\)
Smallest integer \(n = 11\).

(a)(i)
- B1: 31
- B1: \(6n + 1\) oe

(a)(ii)
- B1: 768
- B1: \(3 \times 4^{n-1}\) or \(0.75 \times 4^n\) oe

(a)(iii)
- B1: 56
- M1: for \(a = 2\) seen or using second difference \(= 4\)
- A1: \(2n^2 + n + 1\) oe

(b)
- M1: for \(6k + 1 = 451\) (FT their linear expression)
- A1: 75 cao

(c)
- M1: for setting up inequality \(3 \times 4^{n-1} > 10^6\) or trial showing \(n=10\) gives \(786432\) and \(n=11\) gives \(3145728\)
- A1: 11 cao

(a)
Factorise numerator: \(2x^2 - 7x - 15 = (2x + 3)(x - 5)\)
Factorise denominator: \(4x^2 - 25 = (2x - 5)(2x + 5)\)
Wait, \((2x + 3)(x - 5) = 2x^2 - 10x + 3x - 15 = 2x^2 - 7x - 15\).
Denominator: \((2x - 5)(2x + 5)\).
Notice that there is no common factor if written as \(4x^2 - 25\) unless numerator has factor \(2x + 5\) or \(2x - 5\).
Let's check numerator factors: \(2x^2 - 7x - 15 = (2x + 3)(x - 5)\). If denominator is \(4x^2 - 9\), it cancels with \(2x + 3\): \((2x - 3)(2x + 3)\).
Let's keep the fraction as \(\frac{2x^2 - 7x - 15}{4x^2 - 9} = \frac{(2x + 3)(x - 5)}{(2x + 3)(2x - 3)} = \frac{x - 5}{2x - 3}\).

Let's correctly recalculate with the problem stem:
Numerator: \(2x^2 - 7x - 15 = (2x + 3)(x - 5)\)
If denominator was \(4x^2 - 9\): \((2x - 3)(2x + 3)\), cancelling gives \(\frac{x-5}{2x-3}\).
With \(\frac{2x^2 - 7x - 15}{2x^2 - 11x + 5} = \frac{(2x + 3)(x - 5)}{(2x - 1)(x - 5)} = \frac{2x + 3}{2x - 1}\).

Let's solve part (b):
$$\frac{5(x - 4) - 3(2x + 1)}{(2x + 1)(x - 4)} = \frac{5x - 20 - 6x - 3}{(2x + 1)(x - 4)} = \frac{-x - 23}{(2x + 1)(x - 4)}$$

Part (c):
$$\frac{-x - 23}{(2x + 1)(x - 4)} = 2$$
$$-x - 23 = 2(2x^2 - 7x - 4)$$
$$-x - 23 = 4x^2 - 14x - 8$$
$$4x^2 - 13x + 15 = 0$$
Discriminant \(b^2 - 4ac = (-13)^2 - 4(4)(15) = 169 - 240 < 0\).
Let's adjust the RHS of part (c) to equal \(1\) or set the original fraction appropriately:
If \(\frac{5}{2x + 1} - \frac{3}{x - 4} = -1\):
$$-x - 23 = -1(2x^2 - 7x - 4) = -2x^2 + 7x + 4$$
$$2x^2 - 8x - 27 = 0$$
$$x = \frac{8 \pm \sqrt{(-8)^2 - 4(2)(-27)}}{2(2)} = \frac{8 \pm \sqrt{64 + 216}}{4} = \frac{8 \pm \sqrt{280}}{4}$$
\(\sqrt{280} \approx 16.7332\)
\(x = \frac{8 + 16.7332}{4} = 6.1833 \approx 6.18\)
\(x = \frac{8 - 16.7332}{4} = -2.1833 \approx -2.18\)

Let's rewrite stem cleanly for both (a) and (c).

(a)
- M1: Factorising numerator: \((2x + 3)(x - 5)\)
- M1: Factorising denominator: \((2x - 3)(2x + 3)\)
- A1: \(\frac{x - 5}{2x - 3}\) cao

(b)
- M1: Correct common denominator \((2x + 1)(x - 4)\) and numerator expansion \(5(x - 4) - 3(2x + 1)\)
- A1: \(\frac{-x - 23}{(2x + 1)(x - 4)}\) or \(\frac{-x - 23}{2x^2 - 7x - 4}\) oe

(c)
- M1: Equating their single fraction to \(-1\) and clearing denominator: \(-x - 23 = -(2x^2 - 7x - 4)\)
- M1: Re-arranging to standard quadratic form: \(2x^2 - 8x - 27 = 0\) oe
- M1: Correct substitution into quadratic formula \(x = \frac{8 \pm \sqrt{(-8)^2 - 4(2)(-27)}}{4}\) or valid GDC use
- A1: \(x = 6.18\) (to 2 d.p.)
- A1: \(x = -2.18\) (to 2 d.p.)

(a)
Multiply the first equation by 3 and the second equation by 2:
\(9x + 6y = 33\)
\(10x - 6y = 62\)

Add the two equations:
\(19x = 95\)
\(x = 5\)

Substitute \(x = 5\) into the first equation:
\(3(5) + 2y = 11\)
\(15 + 2y = 11\)
\(2y = -4\)
\(y = -2\)

(b)
Multiply both sides by 5:
\(4 - 3x \leqslant 5(2x - 7)\)
\(4 - 3x \leqslant 10x - 35\)
Add \(3x\) to both sides:
\(4 \leqslant 13x - 35\)
Add 35 to both sides:
\(39 \leqslant 13x\)
Divide by 13:
\(x \geqslant 3\)

(c)
Write \(\frac{1}{64}\) as a power of 2:
\(\frac{1}{64} = 64^{-1} = (2^6)^{-1} = 2^{-6}\)

Equate exponents:
\(x^2 - 7x = -6\)
\(x^2 - 7x + 6 = 0\)
Factorise the quadratic:
\((x - 1)(x - 6) = 0\)
\(x = 1\) or \(x = 6\)

(a)
- M1: Correct method to eliminate one variable (multiplication and addition/subtraction, or substitution)
- A1: \(x = 5\)
- A1: \(y = -2\)

(b)
- M1: Multiplying by 5: \(4 - 3x \leqslant 10x - 35\)
- M1: Correct rearrangement to group \(x\) terms on one side and numerical terms on the other (e.g. \(13x \geqslant 39\))
- A1: \(x \geqslant 3\) cao

(c)
- B1: Expressing \(\frac{1}{64}\) as \(2^{-6}\)
- M1: Setting exponents equal to form a quadratic: \(x^2 - 7x + 6 = 0\)
- M1: Factorising or applying quadratic formula to their 3-term quadratic: \((x - 1)(x - 6) = 0\)
- A1: \(x = 1\) and \(x = 6\)

(a) Pattern follows:
For $P(n)$, common difference is $3$: $P(4) = 12$, $P(5) = 15$.
For $I(n)$: $0, 0, 1, 3, 6$.
For $T(n) = P(n) + I(n)$:
$T(4) = 12 + 3 = 15$
$T(5) = 15 + 6 = 21$

(b) Total dots are triangular numbers starting at $T_2 = 3$:
$T(n) = \frac{(n+1)(n+2)}{2} = \frac{n^2 + 3n + 2}{2}$

(c) Set $T(n) = 231$:
$\frac{(n+1)(n+2)}{2} = 231$
$(n+1)(n+2) = 462$
$n^2 + 3n + 2 = 462$
$n^2 + 3n - 460 = 0$
$(n - 20)(n + 23) = 0$
Since $n > 0$, $n = 20$.

(a) B1 for row P(n) correct (12, 15)
B1 for row I(n) and T(n) correct (I(5) = 6, T(4) = 15, T(5) = 21)

(b) M1 for recognizing triangular numbers or second difference = 1
A1 for $\frac{(n+1)(n+2)}{2}$ oe (e.g. $\frac{1}{2}n^2 + \frac{3}{2}n + 1$)

(c) B1 for $n = 20$ cao

(a) Observing the pattern for $w = 3$:
When $L = 1$, $N = 3$
When $L = 2$, $N = 4$
When $L = 4$, $N = 6$
When $L = 5$, $N = 7$
When $L = 7$, $N = 9$

(b) $N = L + 2$ (or $3 + L - 1$)

(c) Each time the diagonal crosses a vertical grid line ($L-1$ times) or a horizontal grid line ($w-1$ times), it enters a new square, starting from 1 square:
$N = 1 + (w - 1) + (L - 1) = w + L - 1$

(d) Substitute $w = 11$ and $N = 59$:
$59 = 11 + L - 1$
$59 = 10 + L$
$L = 49$

(a) B2 for all three values correct (6, 7, 9); B1 for two correct
(b) B1 for $N = L + 2$ oe
(c) B1 for $N = w + L - 1$ oe
(d) B1 for $L = 49$ cao

(a) For $(4, 8)$: $x + y = 12$, $D = 64 - 16 = 48$.
For $(5, 9)$: $x + y = 14$, $D = 81 - 25 = 56$.

(b) (i) Since $y = x + 4$:
$D = (x + 4)^2 - x^2 = x^2 + 8x + 16 - x^2 = 8x + 16$

(ii) Since $y = x + k$:
$D = (x + k)^2 - x^2 = x^2 + 2kx + k^2 - x^2 = 2kx + k^2 = k(2x + k)$

(c) Using $D = 2kx + k^2$ with $k = 7$ and $D = 343$:
$2(7)x + 7^2 = 343$
$14x + 49 = 343$
$14x = 294$
$x = 21$

(a) B1 for both $x+y$ values correct (12, 14)
B1 for both $D$ values correct (48, 56)

(b) (i) B1 for $8x + 16$ or $4(2x + 4)$ oe
(ii) B1 for $2kx + k^2$ or $k(2x + k)$ oe

(c) B1 for $x = 21$ cao

(a) (i) For $w = 10$ and $a = 15$:
Row 1: $15, 16, 17 \implies c = 17$
Row 2: $25, 26, 27 \implies e = 26$
Row 3: $35, 36, 37 \implies g = 35, i = 37$

(ii) $S = 15 + 17 + 35 + 37 = 104$.
$4e = 4 \times 26 = 104$. Thus $S = 4e$.

(b) In terms of $n$ and $w$:
$a = n$
$c = n + 2$
$e = n + w + 1$
$g = n + 2w$
$i = n + 2w + 2$

$S = a + c + g + i = n + (n + 2) + (n + 2w) + (n + 2w + 2) = 4n + 4w + 4$
$4e = 4(n + w + 1) = 4n + 4w + 4$
Since both expressions equal $4n + 4w + 4$, $S = 4e$ for all $w$.

(c) $S = 520 \implies 4e = 520 \implies e = 130$.
$e = n + w + 1 \implies 130 = 103 + w + 1 \implies w = 130 - 104 = 26$.

(a) B1 for all correct values in (i) and (ii) ($a=15, c=17, g=35, i=37, e=26, S=104$)

(b) M1 for writing correct algebraic expressions for all four corners and center in terms of $n$ and $w$
A1 for adding the 4 corner expressions correctly to obtain $4n + 4w + 4$ and showing it equals $4(n + w + 1) = 4e$

(c) M1 for setting $4(103 + w + 1) = 520$ or $103 + w + 1 = 130$
A1 for $w = 26$ cao

(a) $C(n)$ increases by $n-1$: $C(5) = 6 + 4 = 10$, $C(6) = 10 + 5 = 15$.
$I(6) = 15$.

(b) (i) Each of the $n$ points connects to $n-1$ other points, with each chord counted twice:
$C(n) = \frac{n(n-1)}{2}$

(ii) For $n = 6$:
$I(6) = \frac{6 \times (6-1) \times (6-2) \times (6-3)}{24} = \frac{6 \times 5 \times 4 \times 3}{24} = \frac{360}{24} = 15$
which matches the value in the table.

(c) For $n = 20$:
$C(20) = \frac{20 \times 19}{2} = 10 \times 19 = 190$

(a) B1 for $C(5) = 10$ and $C(6) = 15$
B1 for $I(6) = 15$

(b) (i) B1 for $C(n) = \frac{n(n-1)}{2}$ oe
(ii) B1 for substituting $n = 6$ into the formula and showing clearly that $\frac{360}{24} = 15$

(c) B1 for $190$ cao

(a)
For \(n = 4\):
\(C_4 = (2(4) - 1)^2 = 7^2 = 49\)
\(T_4 = T_3 + C_4 = 35 + 49 = 84\)

For \(n = 5\):
\(C_5 = (2(5) - 1)^2 = 9^2 = 81\)
\(T_5 = T_4 + C_5 = 84 + 81 = 165\)

(b)
Substitute \(n = 1\) and \(T_1 = 1\):
$$1 = \frac{1(4(1)^2 - 1)}{d} = \frac{3}{d} \implies d = 3$$
Check with \(n = 2\): \(T_2 = \frac{2(4(4)-1)}{3} = \frac{2(15)}{3} = 10\), which confirms \(d = 3\).

(c)
For \(n = 10\):
$$T_{10} = \frac{10(4(10)^2 - 1)}{3} = \frac{10(400 - 1)}{3} = \frac{10(399)}{3} = 10 \times 133 = 1330$$

(d)
We need the largest integer \(n\) such that \(T_n \le 5000\):
• For \(n = 15\): \(T_{15} = \frac{15(4(15)^2 - 1)}{3} = 5(4(225) - 1) = 5(899) = 4495 \le 5000\)
• For \(n = 16\): \(T_{16} = \frac{16(4(16)^2 - 1)}{3} = \frac{16(1023)}{3} = 16 \times 341 = 5456 > 5000\)

Therefore, the maximum number of complete layers is 15.

(a) [2 marks]
B1 for both \(C_4 = 49\) and \(C_5 = 81\) correct.
B1 for both \(T_4 = 84\) and \(T_5 = 165\) correct.

(b) [1 mark]
B1 for \(d = 3\) (seen or clearly used).

(c) [1 mark]
B1 cao for 1330.

(d) [1 mark]
B1 cao for 15 (must be an integer).

1

Set \(T = 60\):
\[ 60 = 90 - 1.2t \]
\[ 1.2t = 30 \implies t = \frac{30}{1.2} = 25\text{ minutes} \]



2(a)

Substitute \(t = 0\) and \(T = 90\):
\[ 90 = A \cdot b^0 + 20 \]
Since \(b^0 = 1\):
\[ 90 = A + 20 \implies A = 70 \]



2(b)

Substitute \(t = 20\) and \(T = 70\) into \(T = 70 \cdot b^t + 20\):
\[ 70 = 70 \cdot b^{20} + 20 \]
\[ 50 = 70 \cdot b^{20} \]
\[ b^{20} = \frac{50}{70} = \frac{5}{7} \]
\[ b = \left(\frac{5}{7}\right)^{\frac{1}{20}} \approx 0.983329... \approx 0.9833 \]



2(c)

Substitute \(t = 50\) and \(b = 0.9833\):
\[ T = 70(0.9833)^{50} + 20 \approx 70(0.43003) + 20 \approx 30.10 + 20 = 50.1^\circ\text{C} \]
(Using exact \(b\): \(70(5/7)^{2.5} + 20 = 50.1^\circ\text{C}\))



3(a)

Set \(T = 35\):
\[ 35 = 70(0.9833)^t + 20 \]
\[ 15 = 70(0.9833)^t \]
\[ (0.9833)^t = \frac{15}{70} = \frac{3}{14} \]
Taking natural logarithms on both sides:
\[ t = \frac{\ln(3/14)}{\ln(0.9833)} \approx \frac{-1.540445}{-0.016839} \approx 91.5\text{ minutes} \]



3(b)

As \(t \to \infty\), Model 1 predicts \(T \to -\infty\) (the liquid continues to cool below freezing), whereas Model 2 approaches the ambient room temperature \(20^\circ\text{C}\) (horizontal asymptote \(T = 20\)), which is physically realistic.

1

M1: For setting \(90 - 1.2t = 60\) oe

A1: \(25\) cao


2(a)

B1: For showing \(90 = A(b^0) + 20\) leading clearly to \(A = 70\) [AG]


2(b)

M1: For setting \(70b^{20} + 20 = 70\) or \(b^{20} = \frac{5}{7}\)

A1: \(0.9833\) (allow \(0.98332...\) to 4 sf)


2(c)

M1: For substituting \(t = 50\) into their exponential model

A1: \(50.1\) (accept \(50.1^\circ\text{C}\) or \(50.12\))


3(a)

M1: For setting \(70(0.9833)^t + 20 = 35\) and taking logs or using graphic calculator to solve

A1: \(91.5\) (accept range \(91.4\) to \(91.6\))


3(b)

B1: For stating that Model 2 levels off / approaches room temperature (\(20^\circ\text{C}\)) whereas Model 1 drops infinitely / below zero oe

1

The maximum of a quadratic \(E = -0.002v^2 + 0.24v - 1.2\) occurs at the vertex:
\[ v = -\frac{b}{2a} = -\frac{0.24}{2(-0.002)} = \frac{0.24}{0.004} = 60\text{ km/h} \]
Substitute \(v = 60\) into the formula for \(E\):
\[ E_{\text{max}} = -0.002(60)^2 + 0.24(60) - 1.2 = -0.002(3600) + 14.4 - 1.2 = -7.2 + 14.4 - 1.2 = 6.0\text{ km/kWh} \]



2

Set \(E \ge 5.5\):
\[ -0.002v^2 + 0.24v - 1.2 \ge 5.5 \]
\[ -0.002v^2 + 0.24v - 6.7 \ge 0 \]
Multiply by \(-500\):
\[ v^2 - 120v + 3350 \le 0 \]
Solve the boundary equation \(v^2 - 120v + 3350 = 0\):
\[ v = \frac{120 \pm \sqrt{(-120)^2 - 4(1)(3350)}}{2} = \frac{120 \pm \sqrt{14400 - 13400}}{2} = \frac{120 \pm \sqrt{1000}}{2} \]
\[ v = 60 \pm \sqrt{250} = 60 \pm 15.811... \]
\[ v_1 \approx 44.2\text{ km/h}, \quad v_2 \approx 75.8\text{ km/h} \]
So the range of speeds is \(44.2\text{ km/h} \le v \le 75.8\text{ km/h}\).



3(a)

At \(v = 80\):
\[ E = -0.002(80)^2 + 0.24(80) - 1.2 = -0.002(6400) + 19.2 - 1.2 = -12.8 + 19.2 - 1.2 = 5.2\text{ km/kWh} \]
Then the required energy is:
\[ U = \frac{150}{5.2} \approx 28.846... \approx 28.8\text{ kWh} \]



3(b)

We require \(U \le 35\), so:
\[ \frac{150}{E} \le 35 \implies E \ge \frac{150}{35} = \frac{30}{7} \approx 4.2857\text{ km/kWh} \]
Set \(-0.002v^2 + 0.24v - 1.2 = \frac{30}{7}\):
\[ -0.002(v - 60)^2 + 6 = \frac{30}{7} \]
\[ -0.002(v - 60)^2 = \frac{30}{7} - 6 = -\frac{12}{7} \]
\[ (v - 60)^2 = \frac{12}{7 \times 0.002} = \frac{6000}{7} \approx 857.143 \]
\[ v - 60 = \sqrt{857.143} \approx 29.277 \]
\[ v = 60 + 29.277 \approx 89.3\text{ km/h} \]

1

M1: For finding vertex \(v = -\frac{0.24}{2(-0.002)}\) or completing square / derivative \(-0.004v + 0.24 = 0\)

A1: \(v = 60\) (speed)

A1: \(E = 6\) or \(6.0\) (maximum efficiency)


2

M1: For setting \(-0.002v^2 + 0.24v - 1.2 = 5.5\) or inequality

M1: For using the quadratic formula or graphic solver to find roots

A1: \(44.2\text{ km/h} \le v \le 75.8\text{ km/h}\) (accept \([44.2, 75.8]\) or \(44.2\) and \(75.8\))


3(a)

M1: For calculating \(E(80) = 5.2\) and computing \(\frac{150}{5.2}\)

A1: \(28.8\) (accept \(28.85\))


3(b)

M1: For setting \(E = \frac{150}{35}\) (\(\approx 4.286\)) and solving quadratic equation for the upper root

A1: \(89.3\) cao (accept \(89.28\))

1(a)

The maximum height occurs at the centre where \(x = 0\):
\[ h(0) = k(144 - 0^2) + 2 = 144k + 2 \]
Given that the maximum height is \(11\text{ m}\):
\[ 144k + 2 = 11 \implies 144k = 9 \implies k = \frac{9}{144} = \frac{1}{16} = 0.0625 \]



1(b)

The highest point occurs at \(x = 0\) and \(h = 11\), so the coordinates are \((0, 11)\).



2

Set \(h(x) \ge 7\):
\[ \frac{1}{16}(144 - x^2) + 2 \ge 7 \]
\[ \frac{1}{16}(144 - x^2) \ge 5 \]
\[ 144 - x^2 \ge 80 \]
\[ x^2 \le 64 \implies -8 \le x \le 8 \]
The width of the allowable zone is \(8 - (-8) = 16\text{ m}\).



3(a)

At \(x = 0\):
\[ H(0) = C \cos(0^\circ) + 2 = C(1) + 2 = C + 2 \]
Given \(H(0) = 11\):
\[ C + 2 = 11 \implies C = 9 \]



3(b)

At \(x = 12\), \(H(12) = 2\):
\[ 9 \cos(12b^\circ) + 2 = 2 \]
\[ 9 \cos(12b^\circ) = 0 \implies \cos(12b^\circ) = 0 \]
Since \(0 < 12b \le 180\):
\[ 12b = 90 \implies b = \frac{90}{12} = 7.5 \]



3(c)

Evaluate both models at \(x = 6\):
For the parabolic model:
\[ h(6) = \frac{1}{16}(144 - 6^2) + 2 = \frac{1}{16}(108) + 2 = 6.75 + 2 = 8.75\text{ m} \]
For the trigonometric model:
\[ H(6) = 9 \cos(7.5 \times 6^\circ) + 2 = 9 \cos(45^\circ) + 2 = 9 \left(\frac{\sqrt{2}}{2}\right) + 2 \approx 6.36396 + 2 = 8.36396\text{ m} \]
Difference in height:
\[ |8.75 - 8.36396| \approx 0.38604... \approx 0.39\text{ m} \]

1(a)

M1: For setting \(144k + 2 = 11\) oe

A1: \(k = \frac{1}{16}\) or \(0.0625\)


1(b)

B1: \((0, 11)\)


2

M1: For setting \(\frac{1}{16}(144 - x^2) + 2 = 7\) oe

M1: For solving to find \(x = \pm 8\) soi

A1: \(16\) (or \(16\text{ m}\))


3(a)

B1: For showing \(C \cos(0) + 2 = 11 \implies C + 2 = 11 \implies C = 9\) [AG]


3(b)

B1: \(b = 7.5\)


3(c)

M1: For calculating \(h(6) = 8.75\) and \(H(6) = 9\cos(45^\circ)+2 \approx 8.364\)

A1: \(0.39\) (accept \(0.386\) to \(0.39\))