An original Thinka practice paper modelled on the structure and difficulty of the Mar 2026 (V2) Cambridge IGCSE Mathematics (0580) paper. Not affiliated with or reproduced from Cambridge.
卷二 (Extended Non-Calculator)
Answer all questions. Calculators must not be used in this paper. Show all necessary working clearly.
14 題目 · 53 分
題目 1 · short_answer
2 分
Find the value of \( \left(\frac{27}{64}\right)^{-\frac{2}{3}} \).
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解題
Step 1: Apply the negative index rule to invert the fraction: \[ \left(\frac{27}{64}\right)^{-\frac{2}{3}} = \left(\frac{64}{27}\right)^{\frac{2}{3}} \] Step 2: Take the cube root of the numerator and the denominator: \[ \sqrt[3]{\frac{64}{27}} = \frac{4}{3} \] Step 3: Square the result: \[ \left(\frac{4}{3}\right)^2 = \frac{16}{9} \text{ (or } 1\frac{7}{9}\text{)} \]
評分準則
M1 for \( \left(\frac{64}{27}\right)^{\frac{2}{3}} \) or \( \left(\frac{3}{4}\right)^{-2} \) or \( \frac{1}{\left(\frac{3}{4}\right)^2} \) A1 for \( \frac{16}{9} \) or \( 1\frac{7}{9} \)
題目 2 · short_answer
2 分
Simplify \(\frac{18}{\sqrt{6}} + \sqrt{24}\).
Give your answer in the form \(k\sqrt{6}\), where \(k\) is an integer.
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解題
Rationalise the first term: \[\frac{18}{\sqrt{6}} = \frac{18\sqrt{6}}{6} = 3\sqrt{6}\]
Simplify the second term: \[\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}\]
Combine the terms: \[3\sqrt{6} + 2\sqrt{6} = 5\sqrt{6}\]
評分準則
M1 for rationalising \(\frac{18}{\sqrt{6}}\) to \(3\sqrt{6}\) or simplifying \(\sqrt{24}\) to \(2\sqrt{6}\) A1 for \(5\sqrt{6}\) or \(k = 5\)
(b) Solve the equation \(3^{2n-1} = \dfrac{1}{27\sqrt{3}}\).
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解題
(a) Apply the power \(\frac{3}{2}\) to each factor: \(64^{\frac{3}{2}} = (\sqrt{64})^3 = 8^3 = 512\) \((x^6)^{\frac{3}{2}} = x^{6 \times \frac{3}{2}} = x^9\) \((y^{-4})^{\frac{3}{2}} = y^{-4 \times \frac{3}{2}} = y^{-6}\) Combining these gives \(512 x^9 y^{-6}\) or \(\dfrac{512x^9}{y^6}\).
(b) Express the right-hand side as a power of 3: \(27 = 3^3\) and \(\sqrt{3} = 3^{\frac{1}{2}}\) \(27\sqrt{3} = 3^3 \times 3^{\frac{1}{2}} = 3^{\frac{7}{2}}\) \(\dfrac{1}{27\sqrt{3}} = 3^{-\frac{7}{2}}\)
(a) B1: \(512\) seen or evaluated correctly. B1: \(x^9 y^{-6}\) or \(\dfrac{x^9}{y^6}\) correct.
(b) M1: Expressing the RHS as \(3^{-\frac{7}{2}}\) or \(3^{-3.5}\) or setting up \(2n - 1 = -3 - \frac{1}{2}\). A1: \(n = -\dfrac{5}{4}\) or \(-1.25\) cao.
題目 4 · structured
4 分
(a) The point \(A(3, 2)\) is mapped onto the point \(A'\) by an enlargement with scale factor \(-2\) and centre of enlargement \((1, 4)\). Find the coordinates of \(A'\).
(b) Shape \(T_1\) is mapped onto shape \(T_2\) by a rotation of \(90^\circ\) clockwise about \((0, 0)\). Shape \(T_2\) is mapped onto shape \(T_3\) by a reflection in the line \(x = 0\). Describe fully the single transformation that maps shape \(T_1\) onto shape \(T_3\).
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解題
(a) Find the vector from the centre \((1, 4)\) to \(A(3, 2)\): \(\begin{pmatrix} 3 - 1 \\ 2 - 4 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \end{pmatrix}\). Multiply by the scale factor \(-2\): \(-2 \begin{pmatrix} 2 \\ -2 \end{pmatrix} = \begin{pmatrix} -4 \\ 4 \end{pmatrix}\). Add this vector to the centre of enlargement: \(\begin{pmatrix} 1 \\ 4 \end{pmatrix} + \begin{pmatrix} -4 \\ 4 \end{pmatrix} = \begin{pmatrix} -3 \\ 8 \end{pmatrix}\). Thus, \(A' = (-3, 8)\).
(b) Tracking a general point \((x, y)\): - Rotation \(90^\circ\) clockwise about \((0, 0)\): \((x, y) \to (y, -x)\). - Reflection in the line \(x = 0\) (the \(y\)-axis): \((y, -x) \to (-y, -x)\). The mapping \((x, y) \to (-y, -x)\) is a reflection in the line \(y = -x\).
評分準則
(a) M1: Vector from centre found as \(\begin{pmatrix} 2 \\ -2 \end{pmatrix}\) and multiplied by \(-2\) to give \(\begin{pmatrix} -4 \\ 4 \end{pmatrix}\), or correct graphical method. A1: \((-3, 8)\) cao.
(b) B1: Reflection B1: in the line \(y = -x\) (Note: award 0 marks if more than one transformation is named).
題目 5 · structured
4 分
Points \(A\), \(B\), \(C\), and \(D\) lie on the circumference of a circle, centre \(O\). \(TA\) is a tangent to the circle at \(A\). Angle \(TAB = 56^\circ\) and angle \(BDC = 38^\circ\). [NOT TO SCALE]
(a) Write down the value of angle \(BCA\), giving a geometrical reason for your answer. (b) Calculate angle \(AOB\). (c) Calculate angle \(ADC\).
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解題
(a) By the alternate segment theorem, the angle between the tangent \(TA\) and chord \(AB\) is equal to the angle subtended by the chord in the alternate segment: Angle \(BCA = 56^\circ\). Reason: Alternate segment theorem.
(b) The angle subtended by an arc at the centre is twice the angle subtended at the circumference: Angle \(AOB = 2 \times \text{Angle } BCA = 2 \times 56^\circ = 112^\circ\).
(c) Angles in the same segment subtended by chord \(AB\) are equal, so Angle \(ADB = \text{Angle } BCA = 56^\circ\). Therefore: Angle \(ADC = \text{Angle } ADB + \text{Angle } BDC = 56^\circ + 38^\circ = 94^\circ\).
Two mathematically similar vases, \(A\) and \(B\), are made of the same metal. Vase \(A\) has a height of \(12\text{ cm}\) and vase \(B\) has a height of \(18\text{ cm}\).
(a) The total surface area of vase \(A\) is \(140\text{ cm}^2\). Calculate the total surface area of vase \(B\).
(b) The mass of vase \(B\) is \(2.7\text{ kg}\). Calculate the mass of vase \(A\), giving your answer in grams.
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解題
(a) Linear scale factor \(k = \dfrac{18}{12} = \dfrac{3}{2}\). Area scale factor \(= k^2 = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}\). Surface area of \(B = 140 \times \dfrac{9}{4} = 35 \times 9 = 315\text{ cm}^2\).
(b) Volume (and mass) scale factor \(= k^3 = \left(\dfrac{3}{2}\right)^3 = \dfrac{27}{8}\). Mass of \(B = 2.7\text{ kg} = 2700\text{ g}\). Mass of \(A = \dfrac{2700}{\frac{27}{8}} = 2700 \times \dfrac{8}{27} = 100 \times 8 = 800\text{ g}\).
(b) Deal with the negative exponent by taking the reciprocal: \(\left(\dfrac{16}{81}\right)^{-\frac{3}{4}} = \left(\dfrac{81}{16}\right)^{\frac{3}{4}}\) Take the 4th root, then cube: \(\sqrt[4]{\dfrac{81}{16}} = \dfrac{3}{2}\) \(\left(\dfrac{3}{2}\right)^3 = \dfrac{27}{8}\).
評分準則
(a) M1: Expressing both sides as powers of 2, e.g. \(2^{3x+3} = 2^{7-x}\) or \(3(x+1) = 2 + 5 - x\). A1: \(x = 1\) cao.
(b) M1: Inverting fraction to \(\left(\frac{81}{16}\right)^{\frac{3}{4}}\) or evaluating 4th root to get \(\left(\frac{2}{3}\right)^{-3}\) or \(\frac{3}{2}\). A1: \(\dfrac{27}{8}\) or \(3\dfrac{3}{8}\) cao.
題目 8 · structured
3 分
The vertices of triangle \(P\) are \((1, 2)\), \((4, 2)\), and \((4, 5)\).
(a) Triangle \(P\) is mapped onto triangle \(Q\) by a translation by the vector \(\begin{pmatrix} -3 \\ 1 \end{pmatrix}\). Write down the coordinates of the vertices of triangle \(Q\).
(b) Triangle \(P\) is mapped onto triangle \(R\) with vertices \((-2, 1)\), \((-2, 4)\), and \((-5, 4)\). Describe fully the single transformation that maps triangle \(P\) onto triangle \(R\).
(b) Comparing the coordinates of corresponding vertices: \((1, 2) \to (-2, 1)\) \((4, 2) \to (-2, 4)\) \((4, 5) \to (-5, 4)\) The rule is \((x, y) \to (-y, x)\), which corresponds to a rotation of \(90^\circ\) anticlockwise (or \(270^\circ\) clockwise) about the origin \((0, 0)\).
評分準則
(a) B1: Correct coordinates for all 3 vertices: \((-2, 3)\), \((1, 3)\), \((1, 6)\).
(b) B1: Rotation of \(90^\circ\) anticlockwise (or \(270^\circ\) clockwise). B1: Centre \((0, 0)\) or origin. (Note: Award 0 for part (b) if more than one transformation is mentioned).
題目 9 · structured
4 分
(a) Each interior angle of a regular polygon is \(140^\circ\) greater than each exterior angle. Work out the number of sides of this regular polygon.
(b) The interior angle of a regular \(n\)-sided polygon is \(156^\circ\). Find the value of \(n\).
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解題
(a) Let the interior angle be \(I\) and the exterior angle be \(E\). We know that \(I + E = 180^\circ\) and \(I - E = 140^\circ\). Subtracting the two equations: \(2E = 180^\circ - 140^\circ = 40^\circ \implies E = 20^\circ\). The sum of the exterior angles of any polygon is \(360^\circ\). Number of sides \(= \dfrac{360^\circ}{E} = \dfrac{360^\circ}{20^\circ} = 18\).
Show all necessary working to simplify fully \[\frac{14}{3 - \sqrt{2}} - \sqrt{32}\] Give your answer in the form \(a + b\sqrt{2}\), where \(a\) and \(b\) are integers.
M1 for multiplying numerator and denominator by \(3 + \sqrt{2}\) A1 for obtaining \(6 + 2\sqrt{2}\) or \(\frac{42 + 14\sqrt{2}}{7}\) B1 for \(\sqrt{32} = 4\sqrt{2}\) A1 for final answer \(6 - 2\sqrt{2}\) cao (must have \(a = 6\), \(b = -2\))
題目 11 · free-response
4 分
Show that \[\frac{3}{2x - 1} - \frac{2}{2x + 1} - \frac{2x + 5}{4x^2 - 1} = 0\] for all valid values of \(x\).
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解題
Step 1: Identify that the common denominator is \((2x - 1)(2x + 1) = 4x^2 - 1\).
Step 2: Express the fractions over the common denominator: \[\frac{3(2x + 1) - 2(2x - 1) - (2x + 5)}{(2x - 1)(2x + 1)}\]
Step 3: Expand the terms in the numerator: \[3(2x + 1) = 6x + 3\] \[-2(2x - 1) = -4x + 2\] \[-(2x + 5) = -2x - 5\]
M1 for recognising \(4x^2 - 1 = (2x - 1)(2x + 1)\) and writing expressions over a common denominator \((2x - 1)(2x + 1)\) M1 for correct expansion of brackets in numerator: \(6x + 3 - 4x + 2 - (2x + 5)\) (allow 1 sign error) A1 for correct simplified numerator showing \(6x - 4x - 2x + 3 + 2 - 5 = 0\) A1 for fully correct and convincing conclusion with no steps omitted
題目 12 · free-response
5 分
The coordinates of two points are \(P(-3, 8)\) and \(Q(5, -4)\).
Find the equation of the perpendicular bisector of the line segment \(PQ\). Give your answer in the form \(ax + by + c = 0\), where \(a\), \(b\), and \(c\) are integers and \(a > 0\).
Step 2: Find the gradient of line segment \(PQ\): \[m_{PQ} = \frac{-4 - 8}{5 - (-3)} = \frac{-12}{8} = -\frac{3}{2}\]
Step 3: Find the gradient of the perpendicular bisector: \[m_{\perp} = -\frac{1}{m_{PQ}} = \frac{2}{3}\]
Step 4: Use the point-slope formula with midpoint \((1, 2)\) and gradient \(\frac{2}{3}\): \[y - 2 = \frac{2}{3}(x - 1)\] \[3(y - 2) = 2(x - 1)\] \[3y - 6 = 2x - 2\] \[2x - 3y + 4 = 0\]
評分準則
B1 for midpoint \((1, 2)\) M1 for gradient of \(PQ = \frac{-4 - 8}{5 - (-3)} = -\frac{3}{2}\) M1 for perpendicular gradient \(m = \frac{2}{3}\) (using negative reciprocal of their gradient) M1 for substituting their midpoint and their perpendicular gradient into a linear equation \(y - y_1 = m(x - x_1)\) or \(y = mx + c\) A1 for \(2x - 3y + 4 = 0\) (or any non-zero integer multiple with \(a > 0\)) cao
題目 13 · free-response
5 分
The functions \(\mathrm{f}\) and \(\mathrm{g}\) are defined by \[\mathrm{f}(x) = \frac{3x - 1}{x + 2}, \quad x \neq -2\] \[\mathrm{g}(x) = 5 - 2x\]
(a) Find \(\mathrm{f}^{-1}(x)\). (b) Solve the equation \(\mathrm{g}(\mathrm{f}(x)) = 9\).
(a) M1 for multiplying by \(x + 2\) to get \(y(x + 2) = 3x - 1\) (or \(x(y + 2) = 3y - 1\)) M1 for isolating terms in \(x\) (or \(y\)): \(x(3 - y) = 2y + 1\) A1 for \(\mathrm{f}^{-1}(x) = \frac{2x + 1}{3 - x}\) or \(\frac{-2x - 1}{x - 3}\) (b) M1 for setting up \(5 - 2\left(\frac{3x-1}{x+2}\right) = 9\) or using \(\mathrm{f}(x) = \mathrm{g}^{-1}(9)\) A1 for \(x = -\frac{3}{5}\) or \(-0.6\) cao
題目 14 · free-response
4 分
A right-angled triangle has a base of length \((\sqrt{5} - 1)\text{ cm}\) and an area of \((3\sqrt{5} + 5)\text{ cm}^2\).
Calculate the height of the triangle. Give your answer in the form \(p + q\sqrt{5}\), where \(p\) and \(q\) are integers.
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解題
Step 1: Use the formula for the area of a triangle, \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\): \[\frac{1}{2} (\sqrt{5} - 1) h = 3\sqrt{5} + 5\]
Step 2: Multiply by 2 to isolate \((\sqrt{5} - 1) h\): \[(\sqrt{5} - 1) h = 2(3\sqrt{5} + 5) = 6\sqrt{5} + 10\]
Step 3: Express \(h\) and rationalise the denominator: \[h = \frac{10 + 6\sqrt{5}}{\sqrt{5} - 1}\] Multiply numerator and denominator by \(\sqrt{5} + 1\): \[h = \frac{(10 + 6\sqrt{5})(\sqrt{5} + 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)}\]
M1 for setting up equation \(\frac{1}{2}(\sqrt{5} - 1)h = 3\sqrt{5} + 5\) or \(h = \frac{2(3\sqrt{5} + 5)}{\sqrt{5} - 1}\) M1 for multiplying numerator and denominator by \(\sqrt{5} + 1\) M1 for correct expansion of numerator to \(40 + 16\sqrt{5}\) or denominator to 4 A1 for \(10 + 4\sqrt{5}\) (accept \(p = 10, q = 4\)) cao
Answer all questions. A scientific calculator should be used where appropriate. Give non-exact numerical answers correct to 3 significant figures or 1 decimal place for angles in degrees.
23 題目 · 82 分
題目 1 · short_answer
2 分
A rectangular metal sheet has length \(l = 8.4\text{ cm}\) and width \(w = 5.2\text{ cm}\), each correct to 1 decimal place.
Calculate the upper bound for the perimeter of the metal sheet.
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解題
The upper bound for the length is \(8.4 + 0.05 = 8.45\text{ cm}\). The upper bound for the width is \(5.2 + 0.05 = 5.25\text{ cm}\).
M1 for upper bounds \(8.45\) and \(5.25\) seen A1 for 27.4 cao
題目 2 · short_answer
2 分
A sector of a circle of radius \(7.5\text{ cm}\) has a sector angle of \(112^\circ\).
Calculate the area of the sector.
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解題
The area of a sector is given by: \[ \text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 \] Substitute \(\theta = 112^\circ\) and \(r = 7.5\text{ cm}\): \[ \text{Area} = \frac{112}{360} \times \pi \times 7.5^2 = \frac{112}{360} \times \pi \times 56.25 \approx 54.97787... \] Rounding to 3 significant figures gives \(55.0\text{ cm}^2\).
評分準則
M1 for \(\frac{112}{360} \times \pi \times 7.5^2\) oe A1 for 55.0 or 54.98 or answer in range [54.97, 55.0]
題目 3 · short_answer
2 分
Factorise completely. \[ 6x^2 - 15xy \]
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解題
Find the highest common factor of \(6x^2\) and \(15xy\), which is \(3x\).
Factorise: \[ 6x^2 - 15xy = 3x(2x - 5y) \]
評分準則
B2 for \(3x(2x - 5y)\) or B1 for partial factorisation such as \(3(2x^2 - 5xy)\) or \(x(6x - 15y)\)
題目 4 · short_answer
2 分
The vector \(\mathbf{v} = \begin{pmatrix} -7 \\ 24 \end{pmatrix}\).
Calculate the magnitude of \(\mathbf{v}\), \(|\mathbf{v}|\).
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解題
The magnitude of a column vector \(\begin{pmatrix} x \\ y \end{pmatrix}\) is \(\sqrt{x^2 + y^2}\). \[ |\mathbf{v}| = \sqrt{(-7)^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \]
評分準則
M1 for \((-7)^2 + 24^2\) or \(\sqrt{(-7)^2 + 24^2}\) oe A1 for 25 cao
題目 5 · short_answer
2 分
Line \(L_1\) has equation \(y = 4x - 7\).
Line \(L_2\) is perpendicular to line \(L_1\) and passes through the point \((8, 3)\).
Find the equation of line \(L_2\).
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解題
The gradient of \(L_1\) is \(4\). The gradient of \(L_2\) is \(m = -\frac{1}{4} = -0.25\).
Using point-slope form with \((8, 3)\): \[ y - 3 = -0.25(x - 8) \] \[ y - 3 = -0.25x + 2 \implies y = -0.25x + 5 \]
評分準則
M1 for gradient of perpendicular line \(-\frac{1}{4}\) or \(-0.25\) used A1 for \(y = -0.25x + 5\) or \(y = -\frac{1}{4}x + 5\) oe
題目 6 · short_answer
2 分
A curve has equation \(y = 2x^3 - 9x^2 + 12\).
Find the gradient of the curve at the point where \(x = 4\).
M1 for derivative \(6x^2 - 18x\) (at least one term correct) A1 for 24 cao
題目 7 · short_answer
2 分
An antique vase is valued at \(\$3200\). Its value increases exponentially at a rate of \(4.5\%\) per year.
Calculate the value of the vase after \(6\) years. Give your answer correct to the nearest dollar.
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解題
Use the compound growth formula \(V = P(1 + r)^n\): \[ V = 3200 \times (1 + 0.045)^6 = 3200 \times (1.045)^6 \] \[ V = 3200 \times 1.302260... \approx 4167.23... \] Rounding to the nearest dollar gives \(\$4167\).
評分準則
M1 for \(3200 \times 1.045^6\) oe A1 for 4167 (accept 4167.23 or 4167.2 if not rounded, or rounded to nearest dollar)
題目 8 · short_answer
2 分
In a right-angled triangle \(ABC\), angle \(ABC = 90^\circ\), \(AB = 6.8\text{ cm}\) and \(BC = 9.5\text{ cm}\).
Calculate angle \(BAC\).
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解題
In triangle \(ABC\), relative to angle \(BAC\): - Opposite side = \(BC = 9.5\text{ cm}\) - Adjacent side = \(AB = 6.8\text{ cm}\)
M1 for \(\tan(BAC) = \frac{9.5}{6.8}\) or \(\tan^{-1}\left(\frac{9.5}{6.8}\right)\) oe A1 for 54.4 or 54.41...
題目 9 · structured_multi_step
4 分
A workshop produces two types of brackets, type \(P\) and type \(Q\). Each type \(P\) bracket requires \(2\text{ cm}\) of steel rod and each type \(Q\) bracket requires \(5\text{ cm}\) of steel rod. The total length of steel rod available is at most \(90\text{ cm}\). The workshop must produce at least \(12\) type \(P\) brackets. The number of type \(Q\) brackets produced must be strictly greater than half the number of type \(P\) brackets.
Let \(x\) be the number of type \(P\) brackets and \(y\) be the number of type \(Q\) brackets.
(a) Write down three inequalities in terms of \(x\) and \(y\) representing this information. (b) By considering integer values of \(x\) and \(y\), find the maximum total number of brackets, \(x + y\), that can be produced.
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解題
(a) 1. Total rod length: \(2x + 5y \le 90\) 2. Minimum type \(P\) brackets: \(x \ge 12\) 3. More than half condition: \(y > 0.5x\) (or \(2y > x\))
(b) We need to find integers \(x, y\) such that \(x \ge 12\), \(y > 0.5x\), and \(2x + 5y \le 90\) to maximise \(x + y\).
Rearranging the first inequality gives \(5y \le 90 - 2x \implies y \le \frac{90 - 2x}{5}\). Testing possible integer values of \(x\ge 12\): - If \(x = 17\): \(y \le \frac{90 - 34}{5} = 11.2\). Since \(y > 8.5\), possible integers for \(y\) are \(9, 10, 11\). Max \(x + y = 17 + 11 = 28\). - If \(x = 18\): \(y \le \frac{90 - 36}{5} = 10.8\). Since \(y > 9\), possible integer is \(y = 10\). Max \(x + y = 18 + 10 = 28\). - If \(x = 19\): \(y \le \frac{90 - 38}{5} = 10.4\). Since \(y > 9.5\), possible integer is \(y = 10\). Total \(x + y = 19 + 10 = 29\). - If \(x = 20\): \(y \le \frac{90 - 40}{5} = 10\). But \(y > 10\), so no integer \(y\) exists.
Thus, the maximum total number of brackets is \(29\) (achieved at \(x = 19, y = 10\)).
評分準則
(a) B1 for \(2x + 5y \le 90\) oe B1 for \(x \ge 12\) and \(y > 0.5x\) (or \(2y > x\))
(b) M1 for systematic testing of integer pairs \((x, y)\) satisfying all three inequalities A1 for 29 cao
題目 10 · structured_multi_step
4 分
A train accelerates uniformly from rest to a speed of \(V\text{ m/s}\) in \(30\text{ seconds}\). It then travels at this constant speed of \(V\text{ m/s}\) for \(90\text{ seconds}\). Finally, it decelerates uniformly to rest at a constant rate of \(0.5\text{ m/s}^2\).
The total distance travelled by the train during the entire journey is \(3.25\text{ km}\).
(a) Show that the time taken to decelerate is \(2V\text{ seconds}\). (b) Form an equation in \(V\) and solve it to find the value of \(V\).
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解題
(a) Deceleration is \(0.5\text{ m/s}^2\). \(\text{Time taken} = \frac{\text{change in speed}}{\text{deceleration}} = \frac{V - 0}{0.5} = 2V\text{ seconds}\).
(b) Total distance is \(3.25\text{ km} = 3250\text{ m}\). Distance in stage 1 (acceleration): \(\frac{1}{2} \times 30 \times V = 15V\) Distance in stage 2 (constant speed): \(90 \times V = 90V\) Distance in stage 3 (deceleration): \(\frac{1}{2} \times 2V \times V = V^2\)
(a) B1 for \(\frac{V}{0.5} = 2V\) with clear explanation
(b) M1 for setting up distance expression: \(15V + 90V + \frac{1}{2}(2V)(V)\) oe M1 for quadratic equation \(V^2 + 105V - 3250 = 0\) correctly simplified and factorised (or correct quadratic formula applied) A1 for 25 cao
題目 11 · structured_multi_step
4 分
The functions \(\mathrm{f}(x)\) and \(\mathrm{g}(x)\) are defined as: \[\mathrm{f}(x) = \frac{2x - 1}{x + 3}, \quad x \ne -3\] \[\mathrm{g}(x) = 4x - 5\]
(a) Find \(\mathrm{f}^{-1}(x)\). (b) Solve the equation \(\mathrm{f}\mathrm{g}(x) = 1\).
(a) M1 for multiplying by denominator and isolating terms in \(x\): \(yx - 2x = -3y - 1\) or \(2x - yx = 3y + 1\) A1 for \(\frac{3x + 1}{2 - x}\) oe
(b) M1 for setting \(\frac{8x - 11}{4x - 2} = 1\) or \(4x - 5 = \mathrm{f}^{-1}(1)\) A1 for \(2.25\) or \(\frac{9}{4}\)
題目 12 · structured_multi_step
4 分
The table shows information about the masses, \(m\text{ grams}\), of \(160\) apples picked from an orchard.
\[ \begin{array}{|c|c|} \hline \text{Mass } (m\text{ g}) & \text{Frequency} \\ \hline 100 < m \le 140 & 32 \\ 140 < m \le 160 & 48 \\ 160 < m \le 170 & 35 \\ 170 < m \le 200 & k \\ 200 < m \le 250 & 15 \\ \hline \end{array} \]
(a) Find the value of \(k\). (b) In a histogram drawn to represent this distribution, the height of the bar for the class interval \(140 < m \le 160\) is \(7.2\text{ cm}\). Calculate the height, in \(\text{cm}\), of the bar for the class interval \(170 < m \le 200\).
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解題
(a) Total frequency is \(160\). \(32 + 48 + 35 + k + 15 = 160\) \(130 + k = 160 \implies k = 30\).
(b) Frequency density \(\text{FD} = \frac{\text{Frequency}}{\text{Class width}}\). For interval \(140 < m \le 160\): \(\text{Class width} = 160 - 140 = 20\) \(\text{FD} = \frac{48}{20} = 2.4\)
The height of the bar is \(7.2\text{ cm}\), so the scale factor is: \(\text{Scale} = \frac{7.2\text{ cm}}{2.4} = 3\text{ cm per unit of FD}\).
For interval \(170 < m \le 200\): \(\text{Class width} = 200 - 170 = 30\) \(\text{Frequency} = k = 30\) \(\text{FD} = \frac{30}{30} = 1.0\)
(b) M1 for finding frequency density of \(140 < m \le 160\) as \(\frac{48}{20} = 2.4\) M1 for finding frequency density of \(170 < m \le 200\) as \(\frac{30}{30} = 1.0\) and using proportionality with \(7.2\) A1 for 3 cao
題目 13 · structured_multi_step
4 分
A curve has equation \(y = 2x^3 - 9x^2 - 24x + 15\).
(a) Find \(\frac{\mathrm{d}y}{\mathrm{d}x}\). (b) Find the coordinates of the two stationary points of the curve and determine, with working, whether each point is a maximum or a minimum.
Using the second derivative to determine nature: \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x - 18\) At \(x = -1\): \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(-1) - 18 = -30 < 0\), so \((-1, 28)\) is a local maximum. At \(x = 4\): \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(4) - 18 = 30 > 0\), so \((4, -97)\) is a local minimum.
評分準則
(a) B1 for \(6x^2 - 18x - 24\)
(b) M1 for equating their derivative to 0 and solving to find \(x = -1\) and \(x = 4\) M1 for substituting both \(x\) values into the original curve equation to find \(y = 28\) and \(y = -97\) A1 for identifying \((-1, 28)\) as maximum and \((4, -97)\) as minimum with correct justification (second derivative or gradient test)
題目 14 · structured_multi_step
4 分
A cyclist rides a distance of \(36\text{ km}\) at an average speed of \(x\text{ km/h}\). On the return journey along the same route, her average speed is \((x - 3)\text{ km/h}\). The return journey takes \(24\text{ minutes}\) longer than the outward journey.
(a) Write down an equation in terms of \(x\) and show that it simplifies to \(x^2 - 3x - 270 = 0\). (b) Solve the equation \(x^2 - 3x - 270 = 0\) to calculate the time taken for the return journey, giving your answer in hours and minutes.
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解題
(a) Time for outward journey = \(\frac{36}{x}\text{ hours}\). Time for return journey = \(\frac{36}{x - 3}\text{ hours}\). \(24\text{ minutes} = \frac{24}{60}\text{ hours} = \frac{2}{5}\text{ hours}\).
(b) \(x^2 - 3x - 270 = 0\) \((x - 18)(x + 15) = 0\) Since speed must be positive, \(x = 18\).
Return speed = \(18 - 3 = 15\text{ km/h}\). Time for return journey = \(\frac{36}{15} = 2.4\text{ hours}\). \(0.4\text{ hours} = 0.4 \times 60 = 24\text{ minutes}\). Total time = \(2\text{ hours } 24\text{ minutes}\).
評分準則
(a) M1 for setting up \(\frac{36}{x - 3} - \frac{36}{x} = \frac{24}{60}\) oe M1 for clearing fractions correctly: \(36x - 36(x - 3) = \frac{2}{5}(x^2 - 3x)\) or equivalent A1 for complete rigorous algebraic progression leading to \(x^2 - 3x - 270 = 0\) without omitted steps
(b) B1 for \(2\text{ hours } 24\text{ minutes}\) (or \(2.4\text{ hours}\) correctly converted to \(2\text{ h } 24\text{ min}\))
題目 15 · structured_multi_step
4 分
A closed region \(R\) is defined by the following set of inequalities: \[x \ge 0\] \[y \ge 2\] \[y \ge x - 1\] \[2x + 3y \le 24\]
(a) Determine the number of points with integer coordinates \((x, y)\) that lie inside or on the boundary of region \(R\). (b) Find the integer coordinate point \((x, y)\) in region \(R\) that gives the maximum value of the expression \(3x + 4y\), and state this maximum value.
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解題
(a) For integer values of \(y\ge 2\), the conditions require \(0 \le x \le y + 1\) and \(x \le \frac{24 - 3y}{2}\): - For \(y = 2\): \(x \le 3\) and \(x \le 9 \implies x \in \{0, 1, 2, 3\}\) (4 points) - For \(y = 3\): \(x \le 4\) and \(x \le 7.5 \implies x \in \{0, 1, 2, 3, 4\}\) (5 points) - For \(y = 4\): \(x \le 5\) and \(x \le 6 \implies x \in \{0, 1, 2, 3, 4, 5\}\) (6 points) - For \(y = 5\): \(x \le 6\) and \(x \le 4.5 \implies x \in \{0, 1, 2, 3, 4\}\) (5 points) - For \(y = 6\): \(x \le 7\) and \(x \le 3 \implies x \in \{0, 1, 2, 3\}\) (4 points) - For \(y = 7\): \(x \le 8\) and \(x \le 1.5 \implies x \in \{0, 1\}\) (2 points) - For \(y = 8\): \(x \le 9\) and \(x \le 0 \implies x = 0\) (1 point) - For \(y \ge 9\): \(x < 0\), no non-negative integer \(x\).
Total number of integer points = \(4 + 5 + 6 + 5 + 4 + 2 + 1 = 27\).
(a) M1 for listing or systematically counting valid integer \(x\)-values for each integer \(y\) A1 for 27 cao
(b) M1 for testing integer boundary coordinates near the top/right edges into \(3x + 4y\) A1 for point \((3, 6)\) and maximum value \(33\)
題目 16 · structured_multi_step
4 分
A survey recorded the waiting time, \(t\text{ minutes}\), of \(200\) customers at a service centre. A histogram was drawn with the following frequency densities for four of the five class intervals: - For \(0 < t \le 5\), frequency density \(= 4.8\) - For \(5 < t \le 10\), frequency density \(= 8.4\) - For \(10 < t \le 15\), frequency density \(= 9.2\) - For \(25 < t \le 40\), frequency density \(= 1.6\)
(a) Show that the number of customers who waited between \(15\) and \(25\text{ minutes}\) (\(15 < t \le 25\)) is \(64\). (b) Calculate an estimate for the number of customers who waited for more than \(18\text{ minutes}\).
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解題
(a) \(\text{Frequency} = \text{Frequency density} \times \text{Class width}\). - For \(0 < t \le 5\): \(4.8 \times 5 = 24\) - For \(5 < t \le 10\): \(8.4 \times 5 = 42\) - For \(10 < t \le 15\): \(9.2 \times 5 = 46\) - For \(25 < t \le 40\): \(1.6 \times 15 = 24\)
Sum of known frequencies = \(24 + 42 + 46 + 24 = 136\). Total number of customers is \(200\). Frequency for \(15 < t \le 25 = 200 - 136 = 64\).
(b) For \(15 < t \le 25\): Class width = \(25 - 15 = 10\). Frequency = \(64\). Customers waiting between \(18\) and \(25\text{ minutes}\): Width = \(25 - 18 = 7\). \(\text{Estimated frequency} = \frac{7}{10} \times 64 = 44.8\).
Customers waiting between \(25\) and \(40\text{ minutes} = 24\).
Total customers waiting more than \(18\text{ minutes}\) = \(44.8 + 24 = 68.8\) (or \(69\)).
評分準則
(a) M1 for calculating the frequencies of the 4 given intervals: \(24, 42, 46, 24\) A1 for complete calculation showing \(200 - (24 + 42 + 46 + 24) = 64\)
(b) M1 for \(\frac{25 - 18}{10} \times 64\) (or \(44.8\)) A1 for \(68.8\) or \(69\)
題目 17 · Structured Multi-Step
4 分
The functions \(\mathrm{f}(x)\) and \(\mathrm{g}(x)\) are defined as \[\mathrm{f}(x) = 3x - 5\] \[\mathrm{g}(x) = \frac{8}{2x + 1}, \quad x \neq -0.5\]
(a) Find \(\mathrm{f}^{-1}(x)\).
(b) Find the value of \(x\) when \(\mathrm{gf}(x) = 2\).
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解題
(a) Let \(y = 3x - 5\). Rearrange to make \(x\) the subject: \[y + 5 = 3x\] \[x = \frac{y + 5}{3}\] Therefore, \(\mathrm{f}^{-1}(x) = \frac{x + 5}{3}\).
(a) M1 for correct first step in finding inverse: \(x = 3y - 5\) or \(y + 5 = 3x\) or \(\frac{y+5}{3}\) A1 for \(\frac{x + 5}{3}\) or \(\frac{1}{3}(x + 5)\) oe
(b) M1 for setting up \(\frac{8}{2(3x - 5) + 1} = 2\) or finding \(\mathrm{g}^{-1}(2) = 1.5\) and setting \(3x - 5 = 1.5\) A1 for \(\frac{13}{6}\) or \(2\frac{1}{6}\) or \(2.17\) or \(2.166\dots\)
題目 18 · short-answer
5 分
The diagram shows a pyramid with a horizontal rectangular base \(ABCD\). \(AB = 16\text{ cm}\) and \(BC = 12\text{ cm}\). The apex \(V\) is vertically above the centre of the base, \(M\). Each sloping edge of the pyramid has a length of \(18\text{ cm}\).
(a) Calculate the height, \(VM\), of the pyramid. [2]
(b) Calculate the angle between the face \(VBC\) and the base \(ABCD\). [3]
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解題
(a) In the horizontal base \(ABCD\), calculate the diagonal \(AC\): \[AC^2 = AB^2 + BC^2 = 16^2 + 12^2 = 256 + 144 = 400\] \[AC = \sqrt{400} = 20\text{ cm}\] Since \(M\) is the centre of the base, \(AM = \frac{1}{2}AC = 10\text{ cm}\). In right-angled triangle \(VMA\): \[VM = \sqrt{VA^2 - AM^2} = \sqrt{18^2 - 10^2} = \sqrt{324 - 100} = \sqrt{224} \approx 14.9666\text{ cm}\] Rounded to 3 significant figures, \(VM = 15.0\text{ cm}\) (or \(14.97\text{ cm}\) to 4 s.f., exact \(4\sqrt{14}\)).
(b) Let \(N\) be the midpoint of \(BC\). Then \(VN \perp BC\) and \(MN \perp BC\). The distance \(MN = \frac{1}{2}AB = 8\text{ cm}\). In right-angled triangle \(VMN\), the angle between face \(VBC\) and the base is \(\angle VNM\): \[\tan(\angle VNM) = \frac{VM}{MN} = \frac{\sqrt{224}}{8} = \frac{14.9666}{8} \approx 1.8708\] \[\angle VNM = \tan^{-1}(1.8708) \approx 61.874^\circ\] Correct to 1 decimal place, the angle is \(61.9^\circ\).
評分準則
(a) M1 for \(AM = \frac{1}{2}\sqrt{16^2 + 12^2} = 10\) or \(AC = 20\) A1 for 15.0 or 14.96 to 14.97 or \(\sqrt{224}\) or \(4\sqrt{14}\)
(b) B1 for identifying/calculating \(MN = 8\) or \(VN = \sqrt{18^2 - 6^2} = \sqrt{288}\) M1 for \(\tan(\theta) = \frac{\text{their } VM}{8}\) or \(\cos(\theta) = \frac{8}{\sqrt{288}}\) or \(\sin(\theta) = \frac{\text{their } VM}{\sqrt{288}}\) A1 for 61.9 or 61.87 to 61.88
題目 19 · short-answer
5 分
The value of a rare painting was \(\$14\,000\) on 1 January 2015. The value increased exponentially at a constant rate of \(4.5\%\) per year.
(a) Calculate the value of the painting on 1 January 2023. Give your answer correct to the nearest dollar. [2]
(b) The value of a sculpture was \(\$22\,000\) on 1 January 2015 and depreciated exponentially at a constant rate of \(3\%\) per year. Find the year in which the value of the painting first exceeded the value of the sculpture on 1 January. [3]
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解題
(a) Number of years from 1 January 2015 to 1 January 2023 is \(t = 2023 - 2015 = 8\). \[\text{Value} = 14000 \times (1 + 0.045)^8 = 14000 \times 1.045^8 \approx 19909.41\] To the nearest dollar, the value is \(\$19\,909\).
(b) Let \(t\) be the number of years after 2015. Painting value: \(P(t) = 14000 \times 1.045^t\) Sculpture value: \(S(t) = 22000 \times (1 - 0.03)^t = 22000 \times 0.97^t\) We require \(P(t) > S(t)\): \[14000 \times 1.045^t > 22000 \times 0.97^t\] \[\left(\frac{1.045}{0.97}\right)^t > \frac{22000}{14000} = \frac{11}{7}\] \[(1.07732)^t > 1.57143\] Evaluating for integer values of \(t\): - For \(t = 6\): \(P(6) = 14000 \times 1.045^6 \approx 18231.67\) and \(S(6) = 22000 \times 0.97^6 \approx 18325.43\) (Painting < Sculpture) - For \(t = 7\): \(P(7) = 14000 \times 1.045^7 \approx 19052.09\) and \(S(7) = 22000 \times 0.97^7 \approx 17775.67\) (Painting > Sculpture) Therefore, the condition is first met after \(t = 7\) years. Year = \(2015 + 7 = 2022\).
評分準則
(a) M1 for \(14000 \times 1.045^8\) oe A1 for 19 909 cao
(b) M1 for setting up inequality or equation \(14000 \times 1.045^t = 22000 \times 0.97^t\) oe M1 for systematic trial of values around \(t = 6\) and \(t = 7\) or solving \(t > 6.07\) A1 for 2022 cao
題目 20 · short-answer
5 分
A solid metal cylinder has radius \(r\text{ cm}\) and height \(h\text{ cm}\). The mass of the cylinder is \(m\text{ grams}\).
\(r = 3.4\text{ cm}\), correct to 1 decimal place. \(h = 8.5\text{ cm}\), correct to 1 decimal place. \(m = 780\text{ g}\), correct to 2 significant figures.
The density, \(D\text{ g/cm}^3\), of the cylinder is given by \(D = \frac{m}{\pi r^2 h}\).
(a) Write down the lower bound of \(r\). [1]
(b) Calculate the upper bound of the density, \(D\). Show all your working and give your answer correct to 3 significant figures. [4]
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解題
(a) The radius \(r = 3.4\text{ cm}\) correct to 1 decimal place has bounds: \[3.35 \le r < 3.45\] Lower bound of \(r = 3.35\text{ cm}\).
(b) Find the bounds for each measurement: - Mass \(m = 780\text{ g}\) (correct to 2 s.f., nearest 10): Upper bound \(m_{\text{UB}} = 785\text{ g}\) - Height \(h = 8.5\text{ cm}\) (correct to 1 d.p.): Lower bound \(h_{\text{LB}} = 8.45\text{ cm}\) - Radius \(r = 3.4\text{ cm}\): Lower bound \(r_{\text{LB}} = 3.35\text{ cm}\)
To maximise \(D = \frac{m}{\pi r^2 h}\), use the maximum numerator and minimum denominator: \[D_{\text{UB}} = \frac{m_{\text{UB}}}{\pi \times (r_{\text{LB}})^2 \times h_{\text{LB}}}\] \[D_{\text{UB}} = \frac{785}{\pi \times 3.35^2 \times 8.45}\] \[\pi \times 3.35^2 \times 8.45 = \pi \times 11.2225 \times 8.45 = 94.830125 \pi \approx 297.9178\] \[D_{\text{UB}} = \frac{785}{297.9178} \approx 2.63495\text{ g/cm}^3\] Correct to 3 significant figures, \(D_{\text{UB}} = 2.63\text{ g/cm}^3\).
評分準則
(a) B1 for 3.35 cao
(b) B1 for \(m_{\text{UB}} = 785\) and \(h_{\text{LB}} = 8.45\) seen M1 for \(\frac{m_{\text{UB}}}{\pi \times (r_{\text{LB}})^2 \times h_{\text{LB}}}\) with their bounds \(780 < m_{\text{UB}} \le 785\), \(3.35 \le r_{\text{LB}} < 3.4\), \(8.45 \le h_{\text{LB}} < 8.5\) M1 for evaluation of denominator as \(297.9\dots\) or \(94.83\pi\) A1 for 2.63 or 2.634 to 2.635 nfww
題目 21 · short-answer
5 分
In a parallelogram \(OACB\), \(\vec{OA} = \mathbf{a}\) and \(\vec{OB} = \mathbf{b}\). \(P\) is the point on the side \(AC\) such that \(AP : PC = 2 : 1\). \(Q\) is the midpoint of \(BC\). The line \(OP\) is extended to the point \(X\) such that \(\vec{OX} = k\vec{OP}\), where \(k\) is a constant.
(a) Express \(\vec{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\) in its simplest form. [2]
(b) Express \(\vec{AX}\) in terms of \(k\), \(\mathbf{a}\) and \(\mathbf{b}\). [1]
(c) Given that \(A\), \(X\) and \(Q\) lie on a straight line, find the value of \(k\). [2]
(c) \(Q\) is the midpoint of \(BC\), and \(\vec{BC} = \vec{OA} = \mathbf{a}\). \[\vec{OQ} = \vec{OB} + \frac{1}{2}\vec{BC} = \mathbf{b} + \frac{1}{2}\mathbf{a}\] \[\vec{AQ} = \vec{OQ} - \vec{OA} = \mathbf{b} + \frac{1}{2}\mathbf{a} - \mathbf{a} = -\frac{1}{2}\mathbf{a} + \mathbf{b}\] Since \(A\), \(X\) and \(Q\) are collinear, \(\vec{AX} = \lambda \vec{AQ}\) for some scalar \(\lambda\): \[(k - 1)\mathbf{a} + \frac{2}{3}k\mathbf{b} = \lambda\left(-\frac{1}{2}\mathbf{a} + \mathbf{b}\right) = -\frac{1}{2}\lambda\mathbf{a} + \lambda\mathbf{b}\] Equating coefficients of \(\mathbf{b}\): \[\lambda = \frac{2}{3}k\] Equating coefficients of \(\mathbf{a}\): \[k - 1 = -\frac{1}{2}\lambda = -\frac{1}{2}\left(\frac{2}{3}k\right) = -\frac{1}{3}k\] \[k + \frac{1}{3}k = 1 \implies \frac{4}{3}k = 1 \implies k = \frac{3}{4}\]
評分準則
(a) M1 for \(\vec{AP} = \frac{2}{3}\mathbf{b}\) or valid route \(\vec{OA} + \frac{2}{3}\vec{OB}\) A1 for \(\mathbf{a} + \frac{2}{3}\mathbf{b}\) oe
(b) B1 for \((k - 1)\mathbf{a} + \frac{2}{3}k\mathbf{b}\) or \(k\mathbf{a} + \frac{2}{3}k\mathbf{b} - \mathbf{a}\)
(c) M1 for \(\vec{AQ} = -\frac{1}{2}\mathbf{a} + \mathbf{b}\) oe and setting up ratio/equation with coefficients A1 for \(k = \frac{3}{4}\) or 0.75
題目 22 · short-answer
5 分
\(A\), \(B\) and \(C\) are three points on horizontal ground. \(AB = 55\text{ m}\), \(AC = 70\text{ m}\) and angle \(BAC = 52^\circ\). A vertical pole \(BT\) of height \(24\text{ m}\) stands at \(B\).
(a) Calculate the distance \(BC\). [3]
(b) Calculate the angle of elevation of the top of the pole, \(T\), from the point \(C\). [2]
(b) Since the pole \(BT\) is vertical and \(BC\) lies on the horizontal ground, triangle \(TBC\) is right-angled at \(B\). Let \(\theta\) be the angle of elevation of \(T\) from \(C\): \[\tan(\theta) = \frac{BT}{BC} = \frac{24}{56.4305} \approx 0.42530\] \[\theta = \tan^{-1}(0.42530) \approx 23.039^\circ\] Correct to 1 decimal place, the angle of elevation is \(23.0^\circ\).
評分準則
(a) M1 for \(55^2 + 70^2 - 2(55)(70)\cos(52^\circ)\) A1 for 3184.4... seen A1 for 56.4 or 56.43... nfww
(b) M1 for \(\tan(\theta) = \frac{24}{\text{their } BC}\) A1 for 23.0 or 23.03 to 23.04
題目 23 · short-answer
5 分
The position vectors of points \(A\) and \(B\) relative to the origin \(O\) are \(\vec{OA} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}\) and \(\vec{OB} = \begin{pmatrix} 6 \\ 7 \end{pmatrix}\).
(a) Find the vector \(\vec{AB}\) as a column vector. [1]
(b) Calculate the magnitude \(|\vec{AB}|\). [2]
(c) The point \(C\) has position vector \(\vec{OC} = \begin{pmatrix} 10 \\ k \end{pmatrix}\). Given that angle \(BAC = 90^\circ\), find the value of \(k\). [2]