An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V3) Cambridge International A Level International Mathematics (0607) paper. Not affiliated with or reproduced from Cambridge.
Paper 13 (Core)
Answer all questions. Calculators must not be used.
24 題目 · 39.83999999999998 分
題目 1 · Short Answer
1.66 分
Factorise completely \(6x^2y - 15xy^2\).
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解題
First, identify the highest common factor of \(6x^2y\) and \(15xy^2\). The highest common factor is \(3xy\). Divide each term by this factor: \(\frac{6x^2y}{3xy} = 2x\) and \(\frac{15xy^2}{3xy} = 5y\). Therefore, the factorised expression is \(3xy(2x - 5y)\).
評分準則
M1 for finding a partial common factor, e.g., \(3x(2xy - 5y^2)\) or \(y(6x^2 - 15xy)\) [1 mark]. A1 for the fully factorised correct answer \(3xy(2x - 5y)\) [0.66 marks].
題目 2 · Short Answer
1.66 分
A cuboid has a length of \(8\text{ cm}\), a width of \(5\text{ cm}\) and a height of \(4.5\text{ cm}\). Work out the volume of the cuboid.
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解題
The volume \(V\) of a cuboid is calculated using the formula: \(V = \text{length} \times \text{width} \times \text{height}\). Substituting the given values: \(V = 8 \times 5 \times 4.5 = 40 \times 4.5 = 180\text{ cm}^3\).
評分準則
M1 for showing the correct multiplication method: \(8 \times 5 \times 4.5\) [1 mark]. A1 for \(180\) [0.66 marks].
題目 3 · Short Answer
1.66 分
Solve the equation: \(4(2x - 3) = 3x + 13\).
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解題
First, expand the brackets: \(8x - 12 = 3x + 13\). Next, rearrange the equation to group the \(x\) terms on one side and the constant terms on the other: \(8x - 3x = 13 + 12\), which simplifies to \(5x = 25\). Divide both sides by \(5\) to find \(x\): \(x = 5\).
評分準則
M1 for correctly expanding the bracket to get \(8x - 12\) or for isolating the variable terms on one side, e.g., \(5x = 25\) [1 mark]. A1 for the final correct answer of \(5\) [0.66 marks].
題目 4 · Short Answer
1.66 分
Here are the first four terms of an arithmetic sequence: 3, 10, 17, 24, ... Find an expression, in terms of \(n\), for the \(n\)-th term of this sequence.
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解題
Find the common difference between consecutive terms: \(10 - 3 = 7\). Since the difference is constant, the sequence contains the term \(7n\). To find the adjustment term, substitute \(n = 1\) into \(7n\) to get \(7(1) = 7\). We need the first term to be \(3\), so we subtract \(4\) from \(7n\). Thus, the \(n\)-th term is \(7n - 4\).
評分準則
M1 for finding the common difference is \(7\) or writing an expression of the form \(7n + c\) [1 mark]. A1 for the correct expression \(7n - 4\) [0.66 marks].
題目 5 · Short Answer
1.66 分
Find the coordinates of the point where the line \(y = 4x - 18\) crosses the \(x\)-axis.
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解題
The line crosses the \(x\)-axis when \(y = 0\). Substitute \(y = 0\) into the equation of the line: \(0 = 4x - 18\). Rearranging gives: \(4x = 18\), which simplifies to \(x = \frac{18}{4} = 4.5\). The coordinates of the point are \((4.5, 0)\).
評分準則
M1 for setting \(y = 0\) to form the equation \(4x - 18 = 0\) [1 mark]. A1 for \((4.5, 0)\) or \((\frac{9}{2}, 0)\) [0.66 marks].
題目 6 · Short Answer
1.66 分
Work out the value of \(2^4 \times 5^{-2} \times 10^2\).
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解題
Calculate the individual terms: \(2^4 = 16\), \(5^{-2} = \frac{1}{5^2} = \frac{1}{25}\), and \(10^2 = 100\). Now multiply these values together: \(16 \times \frac{1}{25} \times 100 = 16 \times 4 = 64\).
評分準則
M1 for converting \(5^{-2}\) to \(\frac{1}{25}\) or evaluating \(10^2 \times 5^{-2} = 4\) [1 mark]. A1 for the correct final answer of \(64\) [0.66 marks].
題目 7 · Short Answer
1.66 分
Simplify completely: \(\frac{18p^6q^2}{3p^2q^5}\)
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解題
Simplify the coefficients and each variable separately: 1. Coefficients: \(\frac{18}{3} = 6\), 2. \(p\) terms: \(\frac{p^6}{p^2} = p^{6-2} = p^4\), 3. \(q\) terms: \(\frac{q^2}{q^5} = q^{2-5} = q^{-3} = \frac{1}{q^3}\). Combining these results gives \(\frac{6p^4}{q^3}\) or \(6p^4q^{-3}\).
評分準則
M1 for any two of \(6\), \(p^4\), or \(q^{-3}\) (or \(\frac{1}{q^3}\)) in a simplified term [1 mark]. A1 for \(\frac{6p^4}{q^3}\) or \(6p^4q^{-3}\) [0.66 marks].
題目 8 · Short Answer
1.66 分
A right-angled triangle has a hypotenuse of length \(10\text{ cm}\) and one side of length \(8\text{ cm}\). Work out the length of the third side.
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解題
Let the length of the third side be \(x\). By Pythagoras' theorem, \(x^2 + 8^2 = 10^2\). Simplifying this gives: \(x^2 + 64 = 100\). Subtracting \(64\) from both sides gives: \(x^2 = 36\). Taking the square root gives: \(x = 6\text{ cm}\).
評分準則
M1 for correctly applying Pythagoras' theorem: \(10^2 - 8^2\) or \(x^2 + 8^2 = 10^2\) [1 mark]. A1 for \(6\) [0.66 marks].
題目 9 · Short Answer
1.66 分
Factorise completely: \(12ax - 18bx\)
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解題
Find the highest common factor of the terms \(12ax\) and \(18bx\). The highest common factor of the coefficients \(12\) and \(18\) is \(6\). The common variable is \(x\). So, the common factor is \(6x\). Divide each term by \(6x\): \(12ax \div 6x = 2a\) \(-18bx \div 6x = -3b\) Thus, the factorised expression is: \(6x(2a - 3b)\)
評分準則
M1 for taking out a common factor of \(6\), \(x\), or a multiple such as \(3x\), e.g., \(6(2ax - 3bx)\) or \(x(12a - 18b)\). A1 for completely correct factorisation: \(6x(2a - 3b)\).
題目 10 · Short Answer
1.66 分
A cuboid has a length of \(5\text{ cm}\), a width of \(4\text{ cm}\), and a height of \(h\text{ cm}\). Given that the volume of the cuboid is \(120\text{ cm}^3\), find the value of \(h\).
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解題
The volume of a cuboid is given by the formula: \(V = \text{length} \times \text{width} \times \text{height}\) Substitute the given values into the formula: \(120 = 5 \times 4 \times h\) \(120 = 20h\) Divide both sides by \(20\): \(h = 6\)
評分準則
M1 for setting up the equation \(5 \times 4 \times h = 120\) or \(20h = 120\). A1 for the correct answer of \(6\).
題目 11 · Short Answer
1.66 分
Solve the equation: \(5(x - 3) = 2x + 9\)
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解題
Expand the bracket on the left side of the equation: \(5x - 15 = 2x + 9\) Subtract \(2x\) from both sides: \(3x - 15 = 9\) Add \(15\) to both sides: \(3x = 24\) Divide by \(3\): \(x = 8\)
評分準則
M1 for correctly expanding the bracket: \(5x - 15\). M1 for correctly rearranging terms to isolate \(x\), e.g., \(3x = 24\). A1 for correct solution \(8\).
題目 12 · Short Answer
1.66 分
Find the \(n\)-th term of the sequence: \(3, 7, 11, 15, 19, \dots\)
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解題
This is an arithmetic sequence. Find the common difference by subtracting consecutive terms: \(7 - 3 = 4\) \(11 - 7 = 4\) Since the common difference is \(4\), the \(n\)-th term will involve \(4n\). Find the correction term by looking at the first term where \(n = 1\): \(4(1) + c = 3\) \(4 + c = 3 \implies c = -1\) Therefore, the \(n\)-th term of the sequence is \(4n - 1\).
評分準則
M1 for recognizing a common difference of \(4\) and writing an expression of the form \(4n + c\). A1 for the correct formula: \(4n - 1\).
題目 13 · Short Answer
1.66 分
Find the coordinates of the point where the line \(y = -3x + 12\) crosses the \(x\)-axis.
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解題
The line crosses the \(x\)-axis where \(y = 0\). Substitute \(y = 0\) into the equation of the line: \(0 = -3x + 12\) Add \(3x\) to both sides: \(3x = 12\) Divide by \(3\): \(x = 4\) Therefore, the coordinates of the point are \((4, 0)\).
評分準則
M1 for substituting \(y = 0\) to get \(0 = -3x + 12\). A1 for the correct coordinate \((4, 0)\) or \(x = 4, y = 0\).
題目 14 · Short Answer
1.66 分
Simplify as a single fraction: \(\frac{3}{x} + \frac{2}{x + 1}\)
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解題
Find a common denominator, which is \(x(x + 1)\). Express each fraction with the common denominator: \(\frac{3(x + 1)}{x(x + 1)} + \frac{2x}{x(x + 1)}\) Combine the numerators: \(\frac{3(x + 1) + 2x}{x(x + 1)}\) Expand and simplify the numerator: \(\frac{3x + 3 + 2x}{x(x + 1)} = \frac{5x + 3}{x(x + 1)}\)
評分準則
M1 for expressing fractions with a common denominator \(x(x + 1)\), having at least one numerator correct. M1 for expanding and combining terms in the numerator: \(3x + 3 + 2x\). A1 for the fully simplified correct fraction: \(\frac{5x + 3}{x(x + 1)}\) or \(\frac{5x + 3}{x^2 + x}\).
題目 15 · Short Answer
1.66 分
Write \(0.000045\) in standard form.
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解題
To write a number in standard form, it must be in the form \(a \times 10^n\), where \(1 \le a < 10\) and \(n\) is an integer. Move the decimal point \(5\) places to the right to obtain \(4.5\). Since we moved the decimal point to the right, the exponent \(n\) is negative. Thus, \(0.000045 = 4.5 \times 10^{-5}\).
評分準則
M1 for writing \(4.5 \times 10^k\) where \(k\) is a non-zero integer. A1 for the correct standard form expression: \(4.5 \times 10^{-5}\).
題目 16 · Short Answer
1.66 分
Calculate \(\frac{3}{7}\) of \(42\).
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解題
To find \(\frac{3}{7}\) of \(42\), perform the calculation: \(\frac{3}{7} \times 42\) Divide \(42\) by \(7\) first: \(42 \div 7 = 6\) Now, multiply by \(3\): \(3 \times 6 = 18\)
評分準則
M1 for completing the division \(42 \div 7\) or the multiplication \(3 \times 42\). A1 for the correct final answer: \(18\).
題目 17 · Short Answer
1.66 分
Expand and simplify: \((x - 4)(x + 7)\)
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解題
Multiply each term in the first bracket by each term in the second bracket: \((x)(x) + (x)(7) + (-4)(x) + (-4)(7) = x^2 + 7x - 4x - 28\). Combine like terms to get the final simplified expression: \(x^2 + 3x - 28\).
評分準則
M1 for showing at least 3 correct terms of the expanded expression (e.g. \(x^2\), \(7x\), \(-4x\), \(-28\)). A1 for \(x^2 + 3x - 28\).
題目 18 · Short Answer
1.66 分
Solve the equation: \(\frac{2x - 3}{5} = 3\)
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解題
Multiply both sides of the equation by 5: \(2x - 3 = 15\). Add 3 to both sides: \(2x = 18\). Divide both sides by 2: \(x = 9\).
評分準則
M1 for multiplying both sides by 5 to obtain \(2x - 3 = 15\) (or equivalent first step). A1 for \(x = 9\).
題目 19 · Short Answer
1.66 分
A rectangular block of wood has a length of \(8\text{ cm}\), a width of \(5\text{ cm}\), and a height of \(4\text{ cm}\). Calculate the volume of this block.
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解題
The volume \(V\) of a rectangular block is given by the formula: \(V = \text{length} \times \text{width} \times \text{height}\). Substituting the given values: \(V = 8 \times 5 \times 4 = 160\text{ cm}^3\).
評分準則
M1 for demonstrating the product of the three dimensions: \(8 \times 5 \times 4\). A1 for 160 (units not required).
題目 20 · Short Answer
1.66 分
Find the \(n\)-th term of the sequence: \(5, 11, 17, 23, 29, \dots\)
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解題
First, find the common difference between consecutive terms: \(11 - 5 = 6\), \(17 - 11 = 6\), which is constant. This indicates the term has the form \(6n + c\). Substitute \(n = 1\) to find \(c\): \(6(1) + c = 5 \implies c = -1\). Therefore, the \(n\)-th term is \(6n - 1\).
評分準則
M1 for identifying a common difference of 6 or writing an expression of the form \(6n + c\). A1 for the correct expression \(6n - 1\).
題目 21 · Short Answer
1.66 分
A straight line has the equation \(y = -3x + 8\). Find the coordinates of the point where this line crosses the \(y\)-axis.
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解題
A line crosses the \(y\)-axis where the \(x\)-coordinate is 0. Substituting \(x = 0\) into the equation \(y = -3x + 8\) gives \(y = -3(0) + 8 = 8\). Thus, the coordinates are \((0, 8)\).
評分準則
M1 for identifying that the \(y\)-intercept of the line is 8, or setting \(x = 0\) to solve for \(y\). A1 for the correct coordinates \((0, 8)\).
題目 22 · Short Answer
1.66 分
Factorise completely: \(6x^2 - 9x\)
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解題
Find the highest common factor of \(6x^2\) and \(-9x\). The greatest common factor of the coefficients 6 and 9 is 3, and the greatest common factor of \(x^2\) and \(x\) is \(x\). Factoring out \(3x\) gives: \(3x(2x - 3)\).
評分準則
M1 for a correct partial factorisation (such as factoring out only a 3 or only an \(x\), e.g., \(3(2x^2 - 3x)\) or \(x(6x - 9)\)). A1 for the fully factorised correct answer \(3x(2x - 3)\).
題目 23 · Short Answer
1.66 分
A closed cuboid has dimensions \(3\text{ cm}\) by \(4\text{ cm}\) by \(5\text{ cm}\). Calculate the total surface area of this cuboid.
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解題
The total surface area \(A\) of a cuboid is given by \(A = 2(lw + lh + wh)\). For the given dimensions: \(lw = 3 \times 4 = 12\), \(lh = 3 \times 5 = 15\), and \(wh = 4 \times 5 = 20\). Adding these yields \(12 + 15 + 20 = 47\). Multiplying by 2 gives \(A = 2 \times 47 = 94\text{ cm}^2\).
評分準則
M1 for an expression showing the sum of the areas of the 6 faces, e.g., \(2(3\times 4 + 3\times 5 + 4\times 5)\) or seeing at least two correct face areas of 12, 15, or 20. A1 for 94.
題目 24 · Short Answer
1.66 分
Solve the simultaneous equations: \(3x + y = 11\) and \(2x - y = 4\)
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解題
Add the two equations together to eliminate \(y\): \((3x + y) + (2x - y) = 11 + 4\), which simplifies to \(5x = 15\). Dividing by 5 gives \(x = 3\). Substitute \(x = 3\) into the first equation: \(3(3) + y = 11 \implies 9 + y = 11 \implies y = 2\). Thus, \(x = 3\) and \(y = 2\).
評分準則
M1 for an attempt to eliminate one variable (e.g., adding equations to get \(5x = 15\) or substituting \(y = 11 - 3x\)). A1 for both values correct: \(x = 3\), \(y = 2\).
Paper 23 (Extended)
Answer all questions. Calculators must not be used.
15 題目 · 39.89999999999999 分
題目 1 · Algebra & Number
2.66 分
Factorise completely \(18x^2 - 50y^2\).
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解題
Factor out the common factor of 2 first: \(18x^2 - 50y^2 = 2(9x^2 - 25y^2)\)
Now, factorise the difference of two squares inside the parentheses: \(9x^2 - 25y^2 = (3x)^2 - (5y)^2 = (3x - 5y)(3x + 5y)\)
Combining these gives the fully factorised expression: \(2(3x - 5y)(3x + 5y)\)
評分準則
M1 for factorising out 2 to get \(2(9x^2 - 25y^2)\) A1 for final answer \(2(3x - 5y)(3x + 5y)\) (or equivalent order of brackets)
題目 2 · Algebra & Number
2.66 分
Find the value of \(x\) when \(3^{2x+1} \times 9^{x-2} = 27^{x+1}\).
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解題
First, express all terms with a common base of 3: \(9^{x-2} = (3^2)^{x-2} = 3^{2x-4}\) \(27^{x+1} = (3^3)^{x+1} = 3^{3x+3}\)
Substitute these back into the equation: \(3^{2x+1} \times 3^{2x-4} = 3^{3x+3}\)
Use the multiplication law of indices to simplify the left-hand side: \(3^{(2x+1) + (2x-4)} = 3^{4x-3}\)
This gives: \(3^{4x-3} = 3^{3x+3}\)
Since the bases are equal, we equate the exponents: \(4x - 3 = 3x + 3\)
Solve for \(x\): \(x = 6\)
評分準則
M1 for expressing 9 and 27 as powers of 3 M1 for correctly simplifying indices to obtain the linear equation \(4x - 3 = 3x + 3\) A1 for the correct value \(6\)
This gives the final fraction: \(\frac{3x+26}{(x-3)(x+4)}\)
評分準則
M1 for placing both terms over the common denominator \((x-3)(x+4)\) M1 for correct expansion of the numerator with signs correct: \(5x + 20 - 2x + 6\) A1 for final answer \(\frac{3x+26}{(x-3)(x+4)}\) (accept \(\frac{3x+26}{x^2+x-12}\))
題目 5 · Algebra & Number
2.66 分
Find all the integer values of \(x\) that satisfy the inequality: \(-2 \le \frac{3x - 1}{4} < 2\)
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解題
Multiply all parts of the inequality by 4: \(-8 \le 3x - 1 < 8\)
Add 1 to all parts: \(-7 \le 3x < 9\)
Divide all parts by 3: \(-\frac{7}{3} \le x < 3\)
Since \(-\frac{7}{3} \approx -2.33\), the integers in this range are: \(x = -2, -1, 0, 1, 2\)
評分準則
M1 for multiplying by 4 to get \(-8 \le 3x - 1 < 8\) M1 for correctly isolating \(x\) to obtain \(-\frac{7}{3} \le x < 3\) A1 for list of integers: \(-2, -1, 0, 1, 2\)
題目 6 · Algebra & Number
2.66 分
Express \(\frac{8}{3 - \sqrt{5}}\) in the form \(a + b\sqrt{5}\), where \(a\) and \(b\) are integers.
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解題
To rationalise the denominator, multiply the numerator and the denominator by the conjugate \(3 + \sqrt{5}\): \(\frac{8}{3 - \sqrt{5}} = \frac{8(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})}\)
First, simplify the denominator of the fraction: \(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\). Thus, the denominator becomes \(2\sqrt{3} - \sqrt{3} = \sqrt{3}\). Now, rationalise the fraction: \(\frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}\). Next, simplify the second term: \(\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}\). Finally, add the two simplified terms: \(2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}\).
評分準則
M1 for simplifying \(\sqrt{12} - \sqrt{3}\) to \(\sqrt{3}\). M1 for simplifying \(\sqrt{27}\) to \(3\sqrt{3}\). A1 for final correct answer.
題目 10 · short answer
2.66 分
Solve the equation \(9^{2x-1} \times 27^{x+3} = 3^{10}\).
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解題
Express each base as a power of 3: \(9^{2x-1} = (3^2)^{2x-1} = 3^{4x-2}\) and \(27^{x+3} = (3^3)^{x+3} = 3^{3x+9}\). Substitute these back into the equation: \(3^{4x-2} \times 3^{3x+9} = 3^{10}\). Combine the terms on the left side: \(3^{4x-2 + 3x+9} = 3^{10} \implies 3^{7x+7} = 3^{10}\). Equating the exponents gives \(7x + 7 = 10 \implies 7x = 3 \implies x = \frac{3}{7}\).
評分準則
M1 for expressing 9 and 27 as powers of 3. M1 for combining exponents to get \(7x+7=10\). A1 for correct answer.
題目 11 · short answer
2.66 分
Write as a single fraction in its simplest form: \(\frac{3}{x-2} - \frac{2x+1}{x^2-4}\).
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解題
Factorise the denominator of the second fraction: \(x^2 - 4 = (x-2)(x+2)\). Find a common denominator, which is \((x-2)(x+2)\): \(\frac{3(x+2)}{(x-2)(x+2)} - \frac{2x+1}{(x-2)(x+2)} = \frac{3x+6 - (2x+1)}{(x-2)(x+2)} = \frac{x+5}{(x-2)(x+2)} = \frac{x+5}{x^2-4}\).
評分準則
M1 for factorising the denominator to \((x-2)(x+2)\). M1 for setting up the common denominator and subtracting numerators correctly. A1 for final simplified fraction.
題目 12 · short answer
2.66 分
Solve the equation \(\frac{12}{x} - \frac{12}{x+1} = 1\).
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解題
Multiply the entire equation by the common denominator \(x(x+1)\) to clear the fractions: \(12(x+1) - 12x = x(x+1)\). Expand both sides: \(12x + 12 - 12x = x^2 + x\). Simplify: \(12 = x^2 + x\). Rearrange into standard quadratic form: \(x^2 + x - 12 = 0\). Factorise the quadratic expression: \((x+4)(x-3) = 0\). Solve for \(x\): \(x = 3\) or \(x = -4\).
評分準則
M1 for expanding and converting to standard quadratic form \(x^2 + x - 12 = 0\). M1 for factorising to \((x+4)(x-3) = 0\). A1 for correct solutions.
題目 13 · short answer
2.66 分
Rearrange the formula to make \(p\) the subject: \(q = \frac{2p + 3}{5 - p}\).
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解題
Multiply both sides by \((5-p)\) to clear the fraction: \(q(5-p) = 2p + 3\). Expand the left side: \(5q - pq = 2p + 3\). Group all terms containing \(p\) on one side and the rest on the other side: \(5q - 3 = pq + 2p\). Factorise \(p\) on the right side: \(5q - 3 = p(q + 2)\). Divide by \(q+2\) to isolate \(p\): \(p = \frac{5q - 3}{q + 2}\).
評分準則
M1 for multiplying by \(5-p\) and expanding. M1 for collecting terms in \(p\) on one side and factorising. A1 for correct final formula.
題目 14 · short answer
2.66 分
Solve the inequality \(2x^2 - x - 6 < 0\).
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解題
First, factorise the quadratic expression: \(2x^2 - x - 6 = (2x+3)(x-2)\). Solve the equation \((2x+3)(x-2) = 0\) to find the boundary points: \(x = -\frac{3}{2}\) and \(x = 2\). Since we want the expression to be less than zero, the solution lies between these boundary points: \(-\frac{3}{2} < x < 2\).
評分準則
M1 for finding the critical values \(-1.5\) and \(2\). M1 for identifying the range between these roots. A1 for the correct inequality notation.
題目 15 · short answer
2.66 分
The first four terms of a sequence are 3, 9, 19, 33. Find an expression for the \(n\)-th term of this sequence.
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解題
Compute the first differences: \(9-3=6\), \(19-9=10\), \(33-19=14\). Compute the second differences: \(10-6=4\) and \(14-10=4\). Since the second difference is constant and equal to 4, the coefficient of the \(n^2\) term is \(\frac{4}{2} = 2\). Subtract \(2n^2\) from the terms: for \(n=1\), \(3 - 2(1)^2 = 1\); for \(n=2\), \(9 - 2(2)^2 = 1\); for \(n=3\), \(19 - 2(3)^2 = 1\). Thus, the sequence is of the form \(2n^2 + 1\).
評分準則
M1 for finding the second difference of 4. M1 for identifying the term \(2n^2\). A1 for correct final expression.
Hence, find the numerical value of this expression when \(a = 3\), \(b = -2\), \(x = 5\), and \(y = 4\).
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解題
To factorise \(12ax - 18bx - 8ay + 12by\), we group the terms:
1. Group into two pairs: \((12ax - 18bx) - (8ay - 12by)\). 2. Factorise the common factors from each pair: \(6x(2a - 3b) - 4y(2a - 3b)\). 3. Factor out the common binomial factor \((2a - 3b)\): \((6x - 4y)(2a - 3b)\). 4. Fully factorise by taking out the common factor of 2 from \((6x - 4y)\): \(2(3x - 2y)(2a - 3b)\).
To find the value when \(a = 3\), \(b = -2\), \(x = 5\), and \(y = 4\): Substitute these values into the factorised form: \(2(3(5) - 2(4))(2(3) - 3(-2))\) \(= 2(15 - 8)(6 + 6)\) \(= 2(7)(12) = 168\).
評分準則
M1 for grouping terms, e.g., \(6x(2a-3b) - 4y(2a-3b)\) or equivalent. A1 for factorising to \((6x-4y)(2a-3b)\). A1 for completely factorised form \(2(3x-2y)(2a-3b)\). M1 for substituting the given values into either the original or factorised expression. A1 for the final value of 168.
題目 2 · Structured GDC
8.7 分
A solid metal cylinder has a radius of \(4\text{ cm}\) and a height of \(12\text{ cm}\). It is melted down and recast into a solid sphere.
Calculate the radius of this sphere, giving your answer correct to 3 significant figures.
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解題
1. Calculate the volume of the cylinder using the formula \(V = \pi r^2 h\): \(V = \pi \times 4^2 \times 12 = 192\pi \approx 603.186\text{ cm}^3\).
2. The volume of the sphere is equal to the volume of the cylinder. Set up the equation for the sphere's volume: \(\frac{4}{3}\pi R^3 = 192\pi\).
3. Simplify the equation by dividing both sides by \(\pi\): \(\frac{4}{3}R^3 = 192\).
5. Find the cube root of \(144\): \(R = \sqrt[3]{144} \approx 5.24148\text{ cm}\).
Rounding to 3 significant figures gives \(5.24\text{ cm}\).
評分準則
M1 for finding the volume of the cylinder as \(192\pi\) or \(603.2\). M1 for setting up the equation \(\frac{4}{3}\pi R^3 = 192\pi\) (or their cylinder volume). M1 for isolating \(R^3 = 144\). A1 for \(R = 5.24\) or \(5.241...\) (Accept 5.24).
To solve the simultaneous equations, we can use the elimination method:
1. Multiply the first equation by 2: \(8x - 6y = 50\)
2. Multiply the second equation by 3: \(9x + 6y = 18\)
3. Add the two equations together to eliminate \(y\): \((8x + 9x) + (-6y + 6y) = 50 + 18\) \(17x = 68\)
4. Solve for \(x\): \(x = 4\)
5. Substitute \(x = 4\) back into the second equation to find \(y\): \(3(4) + 2y = 6\) \(12 + 2y = 6\) \(2y = -6\) \(y = -3\)
So, the solution is \(x = 4\) and \(y = -3\).
評分準則
M1 for an attempt to equate coefficients of one variable (e.g., multiplying equations correctly). M1 for adding or subtracting equations to eliminate one variable. A1 for finding \(x = 4\). M1 for substituting their value of \(x\) back to find \(y\). A1 for finding \(y = -3\).
題目 4 · Structured GDC
8.7 分
The first five terms of a sequence are \(5, 11, 21, 35, 53, \dots\).
(a) Find an expression, in terms of \(n\), for the \(n\)-th term of this sequence. (b) Use your expression to find the value of the 20th term.
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解題
(a) Let's find the differences between consecutive terms: First differences: \(11-5=6\), \(21-11=10\), \(35-21=14\), \(53-35=18\). Second differences: \(10-6=4\), \(14-10=4\), \(18-14=4\). Since the second differences are constant, the sequence is quadratic, with the form \(an^2 + bn + c\). The coefficient \(a = \frac{\text{second difference}}{2} = \frac{4}{2} = 2\). Subtract \(2n^2\) from each term: For \(n=1\): \(5 - 2(1)^2 = 3\) For \(n=2\): \(11 - 2(2)^2 = 11 - 8 = 3\) For \(n=3\): \(21 - 2(3)^2 = 21 - 18 = 3\) Thus, the formula is \(2n^2 + 3\).
(b) To find the 20th term, substitute \(n = 20\) into the formula: \(2(20)^2 + 3 = 2(400) + 3 = 803\).
評分準則
M1 for finding first differences (6, 10, 14, 18) and second differences (4). M1 for identifying \(a = 2\) in \(an^2 + bn + c\). A1 for the correct expression \(2n^2 + 3\). M1 for substituting \(n = 20\) into their formula. A1 for the 20th term of 803.
題目 5 · Structured GDC
8.7 分
The equation of a curve is given by \(y = x^3 - 3x^2 - 9x + 5\).
Find the coordinates of the local maximum and local minimum points of this curve.
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解題
To find the local stationary points, we first find the first derivative of \(y\) with respect to \(x\): \(\frac{dy}{dx} = 3x^2 - 6x - 9\) Set \(\frac{dy}{dx} = 0\) to find the critical points: \(3x^2 - 6x - 9 = 0\) Divide by 3: \(x^2 - 2x - 3 = 0\) Factorise the quadratic expression: \((x - 3)(x + 1) = 0\) This gives \(x = 3\) and \(x = -1\).
Now, calculate the corresponding \(y\)-coordinates: For \(x = -1\): \(y = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10\) So one stationary point is at \((-1, 10)\).
For \(x = 3\): \(y = (3)^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22\) So the other stationary point is at \((3, -22)\).
By testing intervals or using the second derivative \(\frac{d^2y}{dx^2} = 6x - 6\): - At \(x = -1\), \(\frac{d^2y}{dx^2} = -12 < 0\), so \((-1, 10)\) is a local maximum. - At \(x = 3\), \(\frac{d^2y}{dx^2} = 12 > 0\), so \((3, -22)\) is a local minimum.
評分準則
M1 for differentiating to find \(\frac{dy}{dx} = 3x^2 - 6x - 9\). M1 for setting their derivative to 0. A1 for finding the critical values \(x = 3\) and \(x = -1\). M1 for substituting both \(x\)-values back into the original curve equation. A1 for the maximum point \((-1, 10)\). A1 for the minimum point \((3, -22)\).
題目 6 · Structured GDC
8.7 分
Simplify the following algebraic fraction completely: \(\frac{2x^2 - 5x - 3}{4x^2 - 1}\)
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解題
1. Factorise the numerator \(2x^2 - 5x - 3\): We look for two numbers that multiply to \(2 \times (-3) = -6\) and add to \(-5\). These numbers are \(-6\) and \(1\). \(2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\).
2. Factorise the denominator \(4x^2 - 1\): This is a difference of two squares: \(4x^2 - 1 = (2x)^2 - 1^2 = (2x - 1)(2x + 1)\).
3. Write down the fraction with the factorised expressions and simplify: \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)}\) Cancel the common factor \((2x + 1)\) from the numerator and denominator: \(\frac{x - 3}{2x - 1}\)
評分準則
M1 for attempting to factorise the numerator \(2x^2 - 5x - 3\). A1 for the correct factorised numerator \((2x + 1)(x - 3)\). M1 for factorising the denominator as a difference of squares. A1 for the correct factorised denominator \((2x - 1)(2x + 1)\). A1 for cancelling the common factor to obtain \(\frac{x - 3}{2x - 1}\).
題目 7 · Structured GDC
8.7 分
A cone has a base radius of \(6\text{ cm}\) and a slant height of \(10\text{ cm}\).
(a) Work out the perpendicular height of the cone. (b) Find the total surface area of the cone, including its base. Give your answer as an exact multiple of \(\pi\).
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解題
(a) The radius \(r\), the perpendicular height \(h\), and the slant height \(l\) form a right-angled triangle. By Pythagoras' theorem: \(h^2 + r^2 = l^2\) \(h^2 + 6^2 = 10^2\) \(h^2 + 36 = 100\) \(h^2 = 64\) \(h = 8\text{ cm}\).
(b) The formula for the total surface area of a cone is: \(A = \pi r^2 + \pi r l\) Substitute \(r = 6\) and \(l = 10\): \(A = \pi (6^2) + \pi (6)(10)\) \(A = 36\pi + 60\pi = 96\pi\text{ cm}^2\).
評分準則
M1 for applying Pythagoras' theorem \(h^2 + 6^2 = 10^2\). A1 for finding the perpendicular height \(h = 8\text{ cm}\). M1 for using the formula for the curved surface area \(\pi r l\). M1 for using the formula for the base area \(\pi r^2\). A1 for adding the terms to find \(96\pi\text{ cm}^2\).
To solve \(3x^2 - 10x + 4 = 0\), we use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 3\), \(b = -10\), and \(c = 4\).
1. Substitute the values into the formula: \(x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(3)(4)}}{2(3)}\) \(x = \frac{10 \pm \sqrt{100 - 48}}{6}\) \(x = \frac{10 \pm \sqrt{52}}{6}\)
2. Calculate the two potential values: \(\sqrt{52} \approx 7.2111\) - First value: \(x = \frac{10 + 7.2111}{6} = \frac{17.2111}{6} \approx 2.8685 \approx 2.87\) - Second value: \(x = \frac{10 - 7.2111}{6} = \frac{2.7889}{6} \approx 0.4648 \approx 0.46\)
So, the solutions are \(x = 2.87\) or \(x = 0.46\).
評分準則
M1 for correct substitution into the quadratic formula, e.g., \(\frac{10 \pm \sqrt{(-10)^2 - 4(3)(4)}}{2(3)}\). A1 for evaluating the discriminant correctly to \(\sqrt{52}\) or equivalent numerical value. M1 for calculating individual values before rounding. A1 for \(x = 2.87\) (2 d.p.). A1 for \(x = 0.46\) (2 d.p.).
題目 9 · Structured GDC
8.7 分
A closed metal container is in the shape of a cylinder of radius \(r\) cm and height \(12\) cm, with a hemisphere of radius \(r\) cm attached to the top. The total volume of the container is \(300\text{ cm}^3\).
(i) Calculate the radius \(r\) of the container. Give your answer correct to 3 significant figures.
(ii) Calculate the total outer surface area of the container (including its flat circular base). Give your answer correct to 3 significant figures.
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解題
Volume of cylinder: \(V_{\text{cyl}} = \pi r^2 h = 12\pi r^2\) Volume of hemisphere: \(V_{\text{hemi}} = \frac{2}{3}\pi r^3\)
Total Volume: \(12\pi r^2 + \frac{2}{3}\pi r^3 = 300\) Divide by \(\pi\): \(12r^2 + \frac{2}{3}r^3 = \frac{300}{\pi} \approx 95.493\) Using a GDC to solve this equation for \(r > 0\) gives \(r \approx 2.6346\text{ cm}\), which is \(2.63\text{ cm}\) to 3 significant figures.
Total surface area \(A\) is the sum of the flat circular base area, the curved cylinder area, and the curved hemisphere area: \(A = \pi r^2 + 2\pi r h + 2\pi r^2 = 3\pi r^2 + 2\pi r h\)
To 3 significant figures, the total outer surface area is \(264\text{ cm}^2\).
評分準則
M1 for setting up the volume equation: \(12\pi r^2 + \frac{2}{3}\pi r^3 = 300\) A1 for \(r = 2.63\) (accept 2.63 to 2.64) M1 for total surface area formula \(A = 3\pi r^2 + 2\pi r h\) (or separate sum of three parts: base, curved cylinder, and hemisphere) M1 for substituting their value of \(r\) and \(h = 12\) into the surface area formula A1 for \(264\) (accept 263.5 to 264.5)
題目 10 · Structured GDC
8.7 分
A sequence of patterns is made using square tiles and triangular tiles. Pattern 1 has 3 square tiles and 4 triangular tiles. Pattern 2 has 7 square tiles and 6 triangular tiles. Pattern 3 has 11 square tiles and 8 triangular tiles. Pattern 4 has 15 square tiles and 10 triangular tiles.
(i) Find an expression, in terms of \(n\), for the number of square tiles in Pattern \(n\).
(ii) Find an expression, in terms of \(n\), for the number of triangular tiles in Pattern \(n\).
(iii) The total number of tiles (squares plus triangles) in Pattern \(k\) is 127. Find the value of \(k\).
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解題
(i) The sequence of square tiles is 3, 7, 11, 15... This is an arithmetic sequence with first term \(a = 3\) and common difference \(d = 4\). Formula: \(U_n = 3 + 4(n-1) = 4n - 1\).
(ii) The sequence of triangular tiles is 4, 6, 8, 10... This is an arithmetic sequence with first term \(a = 4\) and common difference \(d = 2\). Formula: \(T_n = 4 + 2(n-1) = 2n + 2\).
(iii) The total number of tiles in Pattern \(k\) is given by the sum of the two expressions: \((4k - 1) + (2k + 2) = 6k + 1\). Set this expression equal to 127: \(6k + 1 = 127\) \(6k = 126\) \(k = 21\).
評分準則
M1 for recognizing common difference is 4 for squares A1 for \(4n - 1\) (or equivalent linear expression) M1 for recognizing common difference is 2 for triangles A1 for \(2n + 2\) (or equivalent linear expression) M1 for setting up the equation \((4k-1) + (2k+2) = 127\) or \(6k+1 = 127\) A1 for \(k = 21\)
題目 11 · Structured GDC
8.7 分
The graph of the cubic function \(y = x^3 - 3x^2 - 2x + 5\) and the straight line \(y = 2x - 1\) intersect at three points.
(i) Use your graphic display calculator (GDC) to find the integer coordinates of one of the intersection points.
(ii) Find the coordinates of the other two points of intersection. Give your answers correct to 3 significant figures.
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解題
Using a GDC, plot both equations on the same axes: \(y_1 = x^3 - 3x^2 - 2x + 5\) \(y_2 = 2x - 1\)
(i) By tracing or calculating intersections, we find the point \((1, 1)\). Verify algebraically: \(1^3 - 3(1)^2 - 2(1) + 5 = 1 - 3 - 2 + 5 = 1\) and \(2(1) - 1 = 1\). So \((1, 1)\) is correct.
(ii) Use the GDC intersection tool to locate the remaining two intersection points: Point 2: \(x \approx -1.6457...\), \(y \approx -4.2915...\) To 3 significant figures, this is \((-1.65, -4.29)\).
Point 3: \(x \approx 3.6457...\), \(y \approx 6.2915...\) To 3 significant figures, this is \((3.65, 6.29)\).
評分準則
M1 for plotting both functions on the GDC A1 for \((1, 1)\) M1 for finding the second intersection point using the GDC tool A1 for \((-1.65, -4.29)\) (or separate coordinate components to 3 s.f.) M1 for finding the third intersection point using the GDC tool A1 for \((3.65, 6.29)\) (or separate coordinate components to 3 s.f.)
Factorise each quadratic expression completely: \(2x^2 - 13x + 15 = (2x - 3)(x - 5)\), \(x^2 - 25 = (x - 5)(x + 5)\), \(x^2 + 10x + 25 = (x + 5)^2\), \(2x^2 + 7x - 15 = (2x - 3)(x + 5)\), \(2x^2 - 3x = x(2x - 3)\), and \(x^2 - 5x = x(x - 5)\). Substitute these back into the expression: \(\frac{(2x-3)(x-5)}{(x-5)(x+5)} \times \frac{(x+5)^2}{(2x-3)(x+5)} \times \frac{x(x-5)}{x(2x-3)}\). Cancel out common terms: the first two parts simplify to 1, leaving \(\frac{x(x-5)}{x(2x-3)} = \frac{x-5}{2x-3}\).
評分準則
M2 for factorising \(2x^2 - 13x + 15\) and \(2x^2 + 7x - 15\). M2 for factoring other terms correctly. M3 for showing product simplifies to 1. M2 for correct division and cancellation of remaining terms. A1 for final correct simplified expression.
題目 2 · Advanced Extended
10 分
A solid toy consists of a hemisphere of radius \(r\) cm, a cylinder of radius \(r\) cm and height \(h\) cm, and a cone of radius \(r\) cm and height \(2r\) cm. The cylinder height is \(h = 3r\). (a) Given that the total volume of the toy is \(936\pi\) cm\(^3\), show that \(r = 6\). (b) Find the exact total surface area of the toy in the form \((a + b\sqrt{5})\pi\) cm\(^2\), where \(a\) and \(b\) are integers.
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解題
(a) Total volume is \(V = V_{\text{hemisphere}} + V_{\text{cylinder}} + V_{\text{cone}} = \frac{2}{3}\pi r^3 + \pi r^2(3r) + \frac{1}{3}\pi r^2(2r) = \frac{2}{3}\pi r^3 + 3\pi r^3 + \frac{2}{3}\pi r^3 = \frac{13}{3}\pi r^3\). Setting \(\frac{13}{3}\pi r^3 = 936\pi\) gives \(r^3 = \frac{936 \times 3}{13} = 216\), hence \(r = 6\). (b) For \(r = 6\), the hemisphere curved area is \(2\pi r^2 = 72\pi\). The cylinder curved area is \(2\pi r (3r) = 6\pi r^2 = 216\pi\). For the cone, height is \(12\), radius is \(6\), slant height \(l = \sqrt{6^2 + 12^2} = 6\sqrt{5}\). Curved surface area of the cone is \(\pi r l = 36\sqrt{5}\pi\). Total surface area is \(72\pi + 216\pi + 36\sqrt{5}\pi = (288 + 36\sqrt{5})\pi\).
評分準則
(a) M1 for sum of volumes formula. M2 for simplifying to \(\frac{13}{3}\pi r^3 = 936\pi\). A1 for showing \(r=6\). (b) M1 for hemisphere curved area. M1 for cylinder curved area. M2 for finding cone slant height \(l = 6\sqrt{5}\). M1 for cone curved area. A1 for correct exact total surface area.
題目 3 · Advanced Extended
10 分
Solve the simultaneous equations: \(8x^2 + 5xy + y^2 = 128\) and \(3x + y = 12\). Show all your working.
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解題
From the second equation, \(y = 12 - 3x\). Substitute this into the first equation: \(8x^2 + 5x(12 - 3x) + (12 - 3x)^2 = 128\). Expanding the terms: \(8x^2 + 60x - 15x^2 + 144 - 72x + 9x^2 = 128\). Grouping like terms: \(2x^2 - 12x + 144 = 128\) which simplifies to \(2x^2 - 12x + 16 = 0\), or \(x^2 - 6x + 8 = 0\). Factoring gives \((x-2)(x-4) = 0\), so \(x=2\) or \(x=4\). If \(x=2\), \(y = 12 - 3(2) = 6\). If \(x=4\), \(y = 12 - 3(4) = 0\).
評分準則
M2 for rearranging second equation to make \(y\) or \(x\) the subject. M3 for substituting into the first equation and expanding correctly. M2 for simplifying to a quadratic equation in one variable. M2 for solving the quadratic equation to find two values of \(x\) (or \(y\)). A1 for matching correct pairs of \(x\) and \(y\).
題目 4 · Advanced Extended
10 分
Two sequences have the following terms: Sequence A: \(5, 12, 23, 38, 57, \dots\) Sequence B: \(2, 6, 18, 54, 162, \dots\) (a) Find an expression for the \(n\)-th term of Sequence A. (b) Find an expression for the \(n\)-th term of Sequence B. (c) A third sequence, Sequence C, has its \(n\)-th term defined by \(u_n = \text{the } n\text{-th term of Sequence A} + \text{the } n\text{-th term of Sequence B}\). Find the 10th term of Sequence C.
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解題
(a) For Sequence A, the first differences are \(7, 11, 15, 19, \dots\) and the second differences are constant at \(4\). This is a quadratic sequence with \(an^2 + bn + c\) where \(2a = 4 \implies a = 2\). Subtracting \(2n^2\) from the terms yields the linear sequence \(3, 4, 5, 6, \dots\), which has \(n\)-th term \(n+2\). Thus, the \(n\)-th term of Sequence A is \(2n^2 + n + 2\). (b) Sequence B is geometric with first term \(a = 2\) and common ratio \(r = 3\). The \(n\)-th term is \(2 \times 3^{n-1}\). (c) The 10th term of Sequence C is \(u_{10} = (2(10)^2 + 10 + 2) + 2 \times 3^{10-1} = 212 + 2 \times 19683 = 212 + 39366 = 39578\).
評分準則
(a) M2 for finding second differences and deducing \(2n^2\). M1 for finding the remaining linear part. A1 for \(2n^2 + n + 2\). (b) M1 for identifying geometric sequence. M1 for identifying ratio \(3\). A1 for \(2 \times 3^{n-1}\). (c) M2 for substituting \(n=10\) into both expressions and adding. A1 for \(39578\).
題目 5 · Advanced Extended
10 分
Let \(f(x) = x^3 - 3x^2 - 9x + 12\). (a) Find the coordinates of the local maximum and local minimum points on the curve \(y = f(x)\). (b) Find the range of values of the constant \(k\) for which the equation \(f(x) = k\) has exactly three distinct real solutions. (c) Find the gradient of the tangent to the curve \(y = f(x)\) at its \(y\)-intercept.
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解題
(a) Differentiating gives \(f'(x) = 3x^2 - 6x - 9\). Setting \(f'(x) = 0 \implies 3(x^2 - 2x - 3) = 0 \implies 3(x-3)(x+1) = 0\), so turning points are at \(x = 3\) and \(x = -1\). Substituting back: \(f(-1) = 17\) (local maximum) and \(f(3) = -15\) (local minimum). (b) The equation \(f(x) = k\) has three distinct real roots when the horizontal line \(y = k\) intersects the curve thrice, which occurs between the local minimum and local maximum: \(-15 < k < 17\). (c) The \(y\)-intercept is at \(x = 0\). The gradient of the tangent at \(x=0\) is given by \(f'(0) = 3(0)^2 - 6(0) - 9 = -9\).
評分準則
(a) M1 for finding \(f'(x)\). M1 for setting to 0 and solving to get \(x = -1, 3\). A1 for \((-1, 17)\) and A1 for \((3, -15)\). (b) M2 for recognizing boundaries are the y-values of the turning points. A1 for correct inequality \(-15 < k < 17\). (c) M1 for identifying \(y\)-intercept at \(x=0\). A1 for substituting \(x=0\) into derivative to get \(-9\).
題目 6 · Advanced Extended
10 分
(a) Rearrange the formula \(T = 2\pi \sqrt{\frac{m + 3M}{3k}}\) to make \(M\) the subject. (b) Simplify fully: \(\frac{x^4 - 16}{2x^2 - 8} \times \frac{6x}{x^2 + 4}\).
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解題
(a) Square both sides: \(T^2 = 4\pi^2 \left(\frac{m + 3M}{3k}\right)\). Multiply by \(3k\): \(3kT^2 = 4\pi^2(m + 3M)\). Expand or divide: \(\frac{3kT^2}{4\pi^2} = m + 3M \implies 3M = \frac{3kT^2}{4\pi^2} - m \implies M = \frac{kT^2}{4\pi^2} - \frac{m}{3}\). This can also be written as \(M = \frac{3kT^2 - 4\pi^2 m}{12\pi^2}\). (b) Factorise \(x^4 - 16 = (x^2 - 4)(x^2 + 4)\) and \(2x^2 - 8 = 2(x^2 - 4)\). Substituting gives: \(\frac{(x^2 - 4)(x^2 + 4)}{2(x^2 - 4)} \times \frac{6x}{x^2 + 4} = \frac{x^2 + 4}{2} \times \frac{6x}{x^2 + 4} = \frac{6x}{2} = 3x\).
評分準則
(a) M1 for squaring both sides. M1 for removing fraction by multiplying by \(3k\). M1 for isolating terms with \(M\). M1 for dividing to solve for \(M\). A1 for final correct formula. (b) M2 for factoring \(x^4-16\) and \(2x^2-8\). M2 for canceling out common factors. A1 for \(3x\).
題目 7 · Advanced Extended
10 分
A right pyramid has a rectangular base \(ABCD\) with dimensions \(AB = 12\) cm and \(BC = 9\) cm. The vertex \(V\) is directly above the centre of the base \(O\). The volume of the pyramid is \(360\) cm\(^3\). (a) Find the height \(VO\) of the pyramid. (b) Find the length of the slant edge \(VA\). (c) Find the total surface area of the pyramid, giving your answer correct to 3 significant figures.
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解題
(a) Volume of pyramid is \(V = \frac{1}{3} \times \text{Base Area} \times H\). Base area is \(12 \times 9 = 108\) cm\(^2\). Thus, \(360 = \frac{1}{3} \times 108 \times VO \implies 360 = 36 VO \implies VO = 10\) cm. (b) Diagonal \(AC = \sqrt{12^2 + 9^2} = \sqrt{225} = 15\) cm. Thus, \(AO = 7.5\) cm. Slant edge \(VA = \sqrt{VO^2 + AO^2} = \sqrt{10^2 + 7.5^2} = \sqrt{156.25} = 12.5\) cm. (c) Slant height of face \(VAB\) is \(h_{AB} = \sqrt{12.5^2 - 6^2} = \sqrt{120.25}\) cm. Area of \(VAB = \frac{1}{2} \times 12 \times \sqrt{120.25} = 6\sqrt{120.25} \approx 65.795\) cm\(^2\). Slant height of face \(VBC\) is \(h_{BC} = \sqrt{12.5^2 - 4.5^2} = \sqrt{136}\) cm. Area of \(VBC = \frac{1}{2} \times 9 \times \sqrt{136} = 4.5\sqrt{136} \approx 52.479\) cm\(^2\). Total Surface Area = \(108 + 2(65.795) + 2(52.479) \approx 344.55\) cm\(^2\). Rounding to 3 s.f. gives \(345\) cm\(^2\).
評分準則
(a) M2 for volume formula with numbers substituted. A1 for \(10\). (b) M1 for finding base diagonal \(15\). M1 for applying Pythagoras. A1 for \(12.5\). (c) M2 for finding both slant heights (or utilizing correct triangular area formula). M1 for total area sum. A1 for \(345\).
題目 8 · Advanced Extended
10 分
An express train travels \(240\) km at an average speed of \(x\) km/h. A slow train travels the same distance at an average speed that is \(20\) km/h slower than the express train. The slow train's journey takes \(1\) hour and \(36\) minutes longer. (a) Write down an expression in terms of \(x\) for the time taken by the express train. (b) Write down an expression in terms of \(x\) for the time taken by the slow train. (c) Show that the information leads to the equation \(x^2 - 20x - 3000 = 0\). (d) Solve the equation to find the time taken by the slow train for this journey.
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解題
(a) Time taken by express train: \(\frac{240}{x}\) hours. (b) Time taken by slow train: \(\frac{240}{x-20}\) hours. (c) Note that \(1\) hour and \(36\) minutes is \(1.6\) hours. So, \(\frac{240}{x-20} - \frac{240}{x} = 1.6\). Multiply by \(x(x-20)\): \(240x - 240(x-20) = 1.6x(x-20) \implies 4800 = 1.6(x^2 - 20x)\). Dividing by \(1.6\) yields \(3000 = x^2 - 20x\), which simplifies to \(x^2 - 20x - 3000 = 0\). (d) Factoring: \((x-60)(x+50) = 0\). Since speed is positive, \(x = 60\) km/h. The slow train speed is \(40\) km/h, so its time is \(\frac{240}{40} = 6\) hours.
評分準則
(a) B1 for \(240/x\). (b) B1 for \(240/(x-20)\). (c) M1 for converting time to \(1.6\) or \(8/5\). M2 for algebraic steps showing fractional elimination. A1 for reaching the given quadratic. (d) M2 for factoring or using quadratic formula to get \(x=60\). M1 for calculating the slow train's time. A1 for \(6\) hours.
題目 9 · Advanced Extended
10 分
A rectangular garden has a length of \((3x + 1)\text{ m}\) and a width of \((2x - 3)\text{ m}\). An ornamental square pond inside the garden has a side length of \((x + 2)\text{ m}\). The remaining area of the garden is \(38\text{ m}^2\).
(a) Show that \(5x^2 - 11x - 45 = 0\). [3]
(b) Solve the equation \(5x^2 - 11x - 45 = 0\), showing your working. Give your answers correct to 2 decimal places. [3]
(c) Find the perimeter of the garden, using your positive value of \(x\) from part (b). Give your answer correct to 1 decimal place. [2]
(d) Express \(\frac{3}{3x+1} - \frac{2}{2x-3}\) as a single fraction in its simplest form. [2]
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解題
(a) Area of the rectangular garden \(= (3x+1)(2x-3) = 6x^2 - 9x + 2x - 3 = 6x^2 - 7x - 3\). Area of the square pond \(= (x+2)^2 = x^2 + 4x + 4\). Remaining area \(= \text{Garden Area} - \text{Pond Area}\) \(= (6x^2 - 7x - 3) - (x^2 + 4x + 4) = 5x^2 - 11x - 7\). Since the remaining area is \(38\text{ m}^2\): \(5x^2 - 11x - 7 = 38 \implies 5x^2 - 11x - 45 = 0\). (Shown)
(b) Using the quadratic formula for \(5x^2 - 11x - 45 = 0\): \(x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(5)(-45)}}{2(5)}\) \(x = \frac{11 \pm \sqrt{121 + 900}}{10} = \frac{11 \pm \sqrt{1021}}{10}\) \(x = \frac{11 + 31.9531}{10} \approx 4.30\) \(x = \frac{11 - 31.9531}{10} \approx -2.10\) So, \(x = 4.30\) or \(x = -2.10\).
(c) Perimeter of the garden \(= 2\left(\text{length} + \text{width}\right) = 2((3x+1) + (2x-3)) = 2(5x-2) = 10x - 4\). Using the positive value of \(x = 4.2953...\): Perimeter \(= 10(4.2953) - 4 = 38.953 \approx 39.0\text{ m}\).
(a) M1 for finding the area of the garden: \(6x^2 - 7x - 3\). M1 for finding the area of the pond: \(x^2 + 4x + 4\). A1 for setting up the subtraction and successfully reducing to the given equation: \(5x^2 - 11x - 45 = 0\).
(b) M1 for correct substitution into the quadratic formula. A1 for \(\sqrt{1021}\) or equivalent. A1 for both correct decimal answers: \(4.30\) and \(-2.10\).
(c) M1 for writing down the perimeter expression \(10x-4\) or equivalent. A1 for the correct value \(39.0\text{ m}\) (accept \(39\)).
(d) M1 for a common denominator of \((3x+1)(2x-3)\). A1 for the correct simplified fraction: \(\frac{-11}{(3x+1)(2x-3)}\).
題目 10 · Advanced Extended
10 分
A solid metal toy is made from three parts: a cylinder of radius \(r\) and height \(2r\), a hemisphere of radius \(r\) attached to one circular base, and a cone of radius \(r\) and height \(h\) attached to the other circular base. The total height of the toy is \(18\text{ cm}\) and its radius is \(r = 4\text{ cm}\).
(a) Show that the height of the cone, \(h\), is \(6\text{ cm}\). [2]
(b) Calculate the total volume of the toy. Give your answer correct to 3 significant figures. [3]
(c) The toy is melted down and recast into a single large solid sphere. Calculate the radius of this sphere, correct to 2 decimal places. [3]
(d) Calculate the total surface area of the original toy (excluding the internal joined circular faces). Give your answer correct to 3 significant figures. [2]
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解題
(a) The total height of the toy is the sum of the radius of the hemisphere, the height of the cylinder, and the height of the cone: \(\text{Total height} = r + 2r + h = 3r + h\). Given \(r = 4\text{ cm}\) and total height \(= 18\text{ cm}\): \(3(4) + h = 18 \implies 12 + h = 18 \implies h = 6\text{ cm}\). (Shown)
(b) Total volume \(V\) of the toy is: \(V = V_{\text{hemisphere}} + V_{\text{cylinder}} + V_{\text{cone}}\) \(V = \frac{2}{3}\pi r^3 + \pi r^2 (2r) + \frac{1}{3}\pi r^2 h\) \(V = \frac{2}{3}\pi (4)^3 + \pi (4)^2 (8) + \frac{1}{3}\pi (4)^2 (6)\) \(V = \frac{128}{3}\pi + 128\pi + 32\pi = \frac{608}{3}\pi \approx 636.696\text{ cm}^3\). Correct to 3 significant figures, \(V = 637\text{ cm}^3\).
(c) Let \(R\) be the radius of the sphere. \(V_{\text{sphere}} = \frac{4}{3}\pi R^3\) \(\frac{4}{3}\pi R^3 = \frac{608}{3}\pi\) \(4R^3 = 608 \implies R^3 = 152\) \(R = \sqrt[3]{152} \approx 5.3368\text{ cm}\). Correct to 2 decimal places, \(R = 5.34\text{ cm}\).
(d) The total surface area consists of the curved surface of the hemisphere, the curved surface of the cylinder, and the curved surface of the cone. - Curved surface of hemisphere: \(A_1 = 2\pi r^2 = 2\pi (4)^2 = 32\pi\). - Curved surface of cylinder: \(A_2 = 2\pi r (2r) = 4\pi r^2 = 64\pi\). - Curved surface of cone: \(A_3 = \pi r l\), where \(l = \sqrt{r^2 + h^2} = \sqrt{4^2 + 6^2} = \sqrt{52}\). \(A_3 = \pi (4)\sqrt{52} = 4\sqrt{52}\pi \approx 28.844\pi\). Total surface area \(= 32\pi + 64\pi + 4\sqrt{52}\pi = (96 + 4\sqrt{52})\pi \approx 392.21\text{ cm}^2\). Correct to 3 significant figures, the surface area is \(392\text{ cm}^2\).
評分準則
(a) M1 for equating the total height to \(3r + h\). A1 for showing \(12 + h = 18 \implies h = 6\).
(b) M1 for finding the sum of the three volumes with correct formulas. M1 for substituting \(r=4\) and \(h=6\) correctly into at least two of the volume formulas. A1 for \(637\) (accept \(636.5\) to \(637.5\)).
(c) M1 for equating their volume from part (b) to \(\frac{4}{3}\pi R^3\). M1 for rearranging to find \(R^3 = 152\) or equivalent. A1 for \(5.34\).
(d) M1 for finding the slant height \(l = \sqrt{52}\) (or \(7.21\)). A1 for \(392\) (accept \(391.5\) to \(392.5\)).
題目 11 · Advanced Extended
10 分
Three sequences, Sequence A, Sequence B, and Sequence C, have the terms shown in the table below:
(a) Find the next term (the 5th term) of: (i) Sequence A, [1] (ii) Sequence B, [1] (iii) Sequence C. [1]
(b) Find an expression, in terms of \(n\), for the \(n\)-th term of: (i) Sequence A, [3] (ii) Sequence B. [2]
(c) Write down an expression, in terms of \(n\), for the \(n\)-th term of Sequence C. [1]
(d) Find the value of \(n\) for which the \(n\)-th term of Sequence B is \(12288\). [1]
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解題
(a) (i) Sequence A has differences of 6, 10, 14. The next difference is 18, so the 5th term of Sequence A is \(35 + 18 = 53\). (ii) Sequence B has terms multiplying by 2. The 5th term of Sequence B is \(24 \times 2 = 48\). (iii) The terms of Sequence C are found by multiplying corresponding terms of Sequence A and Sequence B: \(15 = 5 \times 3\), \(66 = 11 \times 6\), etc. So the 5th term of Sequence C is \(53 \times 48 = 2544\).
(b) (i) For Sequence A: \(5, 11, 21, 35\). First differences: \(6, 10, 14\). Second differences: \(4, 4\). Since the second difference is constant, the sequence is quadratic of the form \(u_n = p n^2 + q n + r\). \(2p = 4 \implies p = 2\). Subtract \(2n^2\) from each term: For \(n=1\): \(5 - 2(1)^2 = 3\) For \(n=2\): \(11 - 2(2)^2 = 3\) For \(n=3\): \(21 - 2(3)^2 = 3\) Thus, the \(n\)-th term of Sequence A is \(2n^2 + 3\). (ii) For Sequence B: \(3, 6, 12, 24\). This is a geometric sequence with first term \(a = 3\) and common ratio \(r = 2\). The \(n\)-th term of Sequence B is \(3 \times 2^{n-1}\).
(c) The \(n\)-th term of Sequence C is the product of the \(n\)-th terms of Sequence A and Sequence B: \(n\)-th term \(= 3 \times 2^{n-1}(2n^2 + 3)\).
(d) Set the \(n\)-th term of Sequence B to 12288: \(3 \times 2^{n-1} = 12288\) \(2^{n-1} = 4096\) Since \(2^{12} = 4096\): \(n - 1 = 12 \implies n = 13\).
評分準則
(a) (i) B1 for 53. (ii) B1 for 48. (iii) B1 for 2544.
(b) (i) M1 for recognizing a quadratic sequence and establishing second difference is 4. M1 for establishing \(2n^2\) is the lead term. A1 for \(2n^2 + 3\). (ii) M1 for recognizing geometric sequence with common ratio 2. A1 for \(3 \times 2^{n-1}\).
(c) B1 for \(3 \times 2^{n-1}(2n^2 + 3)\) or equivalent product of their (b)(i) and (b)(ii).
(d) B1 for \(n = 13\).
題目 12 · Advanced Extended
10 分
Consider the function \(f(x) = \frac{x^2 - 4x + 5}{x - 2}\) for \(x \neq 2\).
(a) Write down the equation of the vertical asymptote of the graph of \(y = f(x)\). [1]
(b) Find the coordinates of the local maximum and local minimum of \(y = f(x)\). [3]
(c) Show algebraically that \(f(x)\) can be written as \(x - 2 + \frac{1}{x - 2}\). [2]
(d) Find the coordinates of any points where the graph of \(y = f(x)\) crosses the coordinate axes. [2]
(e) Solve the inequality \(f(x) \le 2x - 3\) for \(x \neq 2\). Give your answers correct to 3 significant figures. [2]
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解題
(a) The vertical asymptote occurs when the denominator is zero: \(x - 2 = 0 \implies x = 2\).
(b) Using the GDC or algebraic differentiation: Using the quotient rule or the rewritten form from (c): \(f'(x) = 1 - \frac{1}{(x-2)^2}\). Setting \(f'(x) = 0\): \(1 - \frac{1}{(x-2)^2} = 0 \implies (x-2)^2 = 1 \implies x - 2 = \pm 1\). Thus \(x = 3\) or \(x = 1\). When \(x = 3\), \(f(3) = \frac{3^2 - 4(3) + 5}{3 - 2} = 2\). (Local minimum: \((3, 2)\)) When \(x = 1\), \(f(1) = \frac{1^2 - 4(1) + 5}{1 - 2} = -2\). (Local maximum: \((1, -2)\))
(d) - y-intercept: Set \(x = 0\): \(y = \frac{0^2 - 4(0) + 5}{0 - 2} = -2.5\). So the y-intercept is \((0, -2.5)\). - x-intercept: Set \(y = 0 \implies x^2 - 4x + 5 = 0\). The discriminant is \((-4)^2 - 4(1)(5) = 16 - 20 = -4 < 0\), so there are no real roots. Thus the graph does not cross the x-axis. Coordinates of the axes crossings: \((0, -2.5)\) only.
(e) Find the points of intersection of \(y = f(x)\) and \(y = 2x - 3\): \(\frac{x^2 - 4x + 5}{x - 2} = 2x - 3 \implies x^2 - 4x + 5 = (2x - 3)(x - 2)\) \(x^2 - 4x + 5 = 2x^2 - 7x + 6 \implies x^2 - 3x + 1 = 0\). Solving \(x^2 - 3x + 1 = 0\) using the quadratic formula: \(x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(1)}}{2} = \frac{3 \pm \sqrt{5}}{2}\). \(x_1 = \frac{3 - \sqrt{5}}{2} \approx 0.382\), and \(x_2 = \frac{3 + \sqrt{5}}{2} \approx 2.62\). By sketching both graphs on a GDC, we find that \(f(x) \le 2x - 3\) when \(0.382 \le x < 2\) or \(x \ge 2.62\).
評分準則
(a) B1 for \(x = 2\).
(b) M1 for finding or stating turning points occur when \(x=1\) and \(x=3\) (algebraically or using GDC). A1 for Local maximum: \((1, -2)\). A1 for Local minimum: \((3, 2)\).
(c) M1 for combining the terms over a common denominator. A1 for algebraic completion to \(f(x)\).
(d) B1 for y-intercept \((0, -2.5)\) or \(y = -2.5\). B1 for indicating no x-intercepts.
(e) M1 for finding critical values \(0.382\) and \(2.62\). A1 for both intervals correctly written: \(0.382 \le x < 2\) or \(x \ge 2.62\) (accept equivalent notation; must not include \(x=2\)).
Paper 53 (Core Investigation)
Answer all questions. Show working and explain patterns.
3 題目 · 36 分
題目 1 · Pattern Investigation
12 分
This investigation is about patterns of regular hexagonal tiles.
* **Ring 1** is a single hexagon. It has 6 outer edges (perimeter) and 0 internal edges. * **Ring 2** is formed by surrounding the single tile with a ring of hexagons. This forms a larger hexagon shape of side length 2. * **Ring 3** is formed by surrounding Ring 2 with another ring of hexagons, forming a hexagon shape of side length 3.
Let \(n\) be the ring number. \(H(n)\) is the total number of hexagons. \(P(n)\) is the perimeter (number of outer edges). \(I(n)\) is the number of internal edges.
(b) Find an expression, in terms of \(n\), for the perimeter \(P(n)\).
(c) The total number of hexagons is given by the quadratic formula \(H(n) = an^2 + bn + c\). Find the values of \(a\), \(b\), and \(c\).
(d) The relationship between the variables is \(2I(n) + P(n) = 6H(n)\). Use this formula and your answers to part (b) and part (c) to find an expression, in terms of \(n\), for the number of internal edges \(I(n)\). Show your working.
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解題
**(a)** From the patterns: * For \(n = 3\), the perimeter increases by 6: \(P(3) = 12 + 6 = 18\). * For \(n = 4\), the number of hexagons increases by 18 (since the sequence of differences of \(H(n)\) is 6, 12, 18, so \(H(4) = 19 + 18 = 37\)). * Using the formula \(2I(n) + P(n) = 6H(n)\): * For \(n = 3\): \(2I(3) + 18 = 6(19) \implies 2I(3) = 114 - 18 = 96 \implies I(3) = 48\). * For \(n = 4\): \(2I(4) + 24 = 6(37) \implies 2I(4) = 222 - 24 = 198 \implies I(4) = 99\).
**(b)** The perimeter sequence is 6, 12, 18, 24, ... This is an arithmetic sequence with a common difference of 6. So, \(P(n) = 6n\).
**(c)** We have \(H(n) = an^2 + bn + c\). Using values for \(n=1, 2, 3\): 1) \(a(1)^2 + b(1) + c = 1 \implies a + b + c = 1\) 2) \(a(2)^2 + b(2) + c = 7 \implies 4a + 2b + c = 7\) 3) \(a(3)^2 + b(3) + c = 19 \implies 9a + 3b + c = 19\)
Subtracting equation (1) from (2) gives: \(3a + b = 6\) (equation 4)
Subtracting equation (2) from (3) gives: \(5a + b = 12\) (equation 5)
Subtracting equation (4) from (5) gives: \(2a = 6 \implies a = 3\)
Substitute \(a=3\) into (4): \(3(3) + b = 6 \implies b = -3\)
Substitute \(a=3, b=-3\) into (1): \(3 - 3 + c = 1 \implies c = 1\)
Thus, \(H(n) = 3n^2 - 3n + 1\), which means \(a = 3\), \(b = -3\), and \(c = 1\).
**(d)** We are given \(2I(n) + P(n) = 6H(n)\). Rearranging for \(I(n)\): \(2I(n) = 6H(n) - P(n)\) \(I(n) = 3H(n) - \frac{1}{2}P(n)\)
**(a)** [4 marks] * Award 1 mark for each correct value: * i) 18 * ii) 48 * iii) 37 * iv) 99
**(b)** [2 marks] * M1 for recognizing common difference of 6 or writing \(6n + k\). * A1 for \(6n\).
**(c)** [3 marks] * M1 for setting up at least two simultaneous equations using \(H(n) = an^2 + bn + c\). * A1 for finding \(a = 3\) and \(b = -3\). * A1 for finding \(c = 1\).
**(d)** [3 marks] * M1 for substituting \(H(n)\) and \(P(n)\) expressions into the formula. * M1 for expanding and simplifying, e.g., \(2I(n) = 6(3n^2 - 3n + 1) - 6n\). * A1 for final simplified expression: \(I(n) = 9n^2 - 12n + 3\) (or fully correct factored equivalent).
題目 2 · Pattern Investigation
12 分
This investigation is about grids of squares made from toothpicks. Each individual \(1 \times 1\) square also has two diagonal toothpicks.
* A \(1 \times 1\) grid contains 4 small triangles, has 5 vertices (intersection points), and uses 6 toothpicks in total. * A \(2 \times 2\) grid contains 16 small triangles, has 13 vertices, and uses 20 toothpicks in total.
Let \(n\) be the grid size (\(n \times n\)). \(A(n)\) is the number of small triangles. \(V(n)\) is the number of vertices. \(T(n)\) is the total number of toothpicks.
(b) Find an expression, in terms of \(n\), for: (i) the number of triangles, \(A(n)\). (ii) the total number of toothpicks, \(T(n)\). (iii) the number of vertices, \(V(n)\).
(c) Show algebraically that \(T(n) + V(n) - A(n) = 2n^2 + 4n + 1\).
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解題
**(a)** * For \(n=3\): * Vertices \(V(3)\): Grid intersections are \(4 \times 4 = 16\). Adding the center of each of the 9 squares: \(16 + 9 = 25\). * Total toothpicks \(T(3)\): Grid lines = \(2 \times 3 \times 4 = 24\). Diagonals = \(2 \times 9 = 18\). Total = \(24 + 18 = 42\). * For \(n=4\): * Triangles \(A(4)\): Each of the 16 squares contains 4 triangles. Total = \(16 \times 4 = 64\). * Total toothpicks \(T(4)\): Grid lines = \(2 \times 4 \times 5 = 40\). Diagonals = \(2 \times 16 = 32\). Total = \(40 + 32 = 72\).
**(b)** * (i) Since each of the \(n^2\) squares is split into 4 triangles, \(A(n) = 4n^2\). * (ii) Grid lines use \(2n(n+1)\) toothpicks. Diagonals use \(2n^2\) toothpicks. Total \(T(n) = 2n(n+1) + 2n^2 = 2n^2 + 2n + 2n^2 = 4n^2 + 2n\). * (iii) Grid intersections are \((n+1)^2\). Centers of squares are \(n^2\). Total \(V(n) = (n+1)^2 + n^2 = n^2 + 2n + 1 + n^2 = 2n^2 + 2n + 1\).
**(c)** Substituting the expressions from part (b): \(T(n) + V(n) - A(n) = (4n^2 + 2n) + (2n^2 + 2n + 1) - 4n^2\) Combining like terms: \(= (4n^2 + 2n^2 - 4n^2) + (2n + 2n) + 1\) \(= 2n^2 + 4n + 1\). This matches the required expression.
評分準則
**(a)** [4 marks] * Award 1 mark for each correct value: * i) 25 * ii) 42 * iii) 64 * iv) 72
**(b)** [5 marks] * (i) [1 mark] A1 for \(4n^2\) (or equivalent). * (ii) [2 marks] M1 for identifying the component for grid lines \(2n(n+1)\) or diagonals \(2n^2\), or for identifying second differences are 8. A1 for \(4n^2 + 2n\). * (iii) [2 marks] M1 for identifying the component \((n+1)^2\) or \(n^2\), or for identifying second differences are 4. A1 for \(2n^2 + 2n + 1\).
**(c)** [3 marks] * M1 for substituting their expressions for \(T(n)\), \(V(n)\), and \(A(n)\) into \(T(n) + V(n) - A(n)\). * M1 for expanding terms correctly. * A1 for full algebraic proof showing simplification to \(2n^2 + 4n + 1\).
題目 3 · Pattern Investigation
12 分
This investigation is about overlapping circles.
A **Linear Chain** of circles consists of \(n\) circles in a line, where each circle after the first overlaps only with the previous circle. * A chain of 1 circle has 0 intersection points and 1 region inside. * A chain of 2 circles has 2 intersection points and 3 regions inside.
A **Closed Loop** of circles consists of \(n\) circles (where \(n \ge 3\)) arranged in a ring, where each circle overlaps with exactly two neighboring circles.
Let \(n\) be the number of circles.
(a) Complete the table below for a **Linear Chain**:
(b) Find an expression, in terms of \(n\), for: (i) the number of intersection points in a Linear Chain, \(P_L(n)\). (ii) the number of regions in a Linear Chain, \(R_L(n)\).
(c) Complete the table below for a **Closed Loop** (\(n \ge 3\)):
(d) Write down an expression, in terms of \(n\), for: (i) the number of intersection points in a Closed Loop, \(P_C(n)\). (ii) the number of regions in a Closed Loop, \(R_C(n)\).
(e) A Closed Loop of circles has 132 regions. Find the number of circles in this loop.
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解題
**(a)** * For \(n=3\): * The intersection points increase by 2 each time: \(P_L(3) = 4\). * The regions increase by 2 each time: \(R_L(3) = 5\). * For \(n=5\): * \(P_L(5) = 8\). * \(R_L(5) = 9\).
**(b)** * (i) The sequence for \(P_L(n)\) is 0, 2, 4, 6, 8, ... which is \(2n - 2\). * (ii) The sequence for \(R_L(n)\) is 1, 3, 5, 7, 9, ... which is \(2n - 1\).
**(c)** For a Closed Loop, both the number of intersection points and regions are twice the number of circles because each circle has 2 intersections with each of its 2 neighbors. * For \(n=5\): \(P_C(5) = 10\) and \(R_C(5) = 10\). * For \(n=6\): \(P_C(6) = 12\) and \(R_C(6) = 12\).
**(e)** Using \(R_C(n) = 2n\): \(2n = 132 \implies n = 66\) circles.
評分準則
**(a)** [3 marks] * Award 1 mark for each group of correct values: * i) 4 and ii) 5 (1 mark) * iii) 8 (1 mark) * iv) 9 (1 mark)
**(b)** [2 marks] * (i) A1 for \(2n - 2\) (or equivalent, e.g., \(2(n-1)\)). * (ii) A1 for \(2n - 1\).
**(c)** [3 marks] * Award 1 mark for each group of correct values: * i) 10 and ii) 10 (1 mark) * iii) 12 (1 mark) * iv) 12 (1 mark)
**(d)** [2 marks] * (i) A1 for \(2n\). * (ii) A1 for \(2n\).
**(e)** [2 marks] * M1 for setting \(2n = 132\) (or using their expression from d(ii)). * A1 for 66.
Paper 63 (Extended Investigation & Modelling)
Answer both Part A and Part B.
7 題目 · 59.5 分
題目 1 · Investigation & Modelling
8.5 分
Part A: Investigation. An equilateral triangle of side length \(n\) is built from smaller equilateral triangles of side length 1. Let \(T_n\) be the total number of unit triangles in a triangle of side \(n\). Let \(P_n\) be the total number of vertices (points where unit triangles meet). Let \(L_n\) be the total number of unit line segments of length 1 in the pattern.
For \(n = 1\), \(T_1 = 1\). For \(n = 2\), \(T_2 = 4\). For \(n = 3\), \(T_3 = 9\).
Find the formula for \(T_n\) in terms of \(n\).
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解題
The sequence of unit triangles \(T_n\) for \(n = 1, 2, 3, \dots\) is \(1, 4, 9, \dots\). This sequence corresponds to the squares of the integers: \(T_n = n^2\).
評分準則
M1 for writing down first few values (1, 4, 9, 16) M1 for identifying a quadratic progression A1.5 for the correct final formula \(T_n = n^2\)
題目 2 · Investigation & Modelling
8.5 分
Part A: Investigation (continued).
Let \(P_n\) be the total number of vertices (points where unit triangles meet) in a triangle of side length \(n\). For \(n = 1\), \(P_1 = 3\). For \(n = 2\), \(P_2 = 6\). For \(n = 3\), \(P_3 = 10\).
Find an expression for \(P_n\) in terms of \(n\) in the form \(a n^2 + b n + c\) where \(a, b, c\) are constants.
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解題
The vertices of the triangular grid form triangular numbers of the next order. Specifically, \(P_n = 1 + 2 + \dots + (n+1) = \frac{(n+1)(n+2)}{2}\). Expanding this expression: \(P_n = \frac{n^2 + 3n + 2}{2} = \frac{1}{2}n^2 + \frac{3}{2}n + 1\). Thus \(a = 0.5\), \(b = 1.5\), and \(c = 1\).
評分準則
M1 for recognizing triangular numbers pattern \(\frac{(n+1)(n+2)}{2}\) M1 for expanding the algebraic expression A1.5 for finding \(a = 0.5, b = 1.5, c = 1\)
題目 3 · Investigation & Modelling
8.5 分
Part A: Investigation (continued).
Let \(L_n\) be the total number of unit line segments of length 1 in the triangle pattern of side length \(n\). For \(n = 1\), \(L_1 = 3\). For \(n = 2\), \(L_2 = 9\). For \(n = 3\), \(L_3 = 18\).
Find an expression for \(L_n\) in terms of \(n\) in the form \(d n^2 + e n\) where \(d\) and \(e\) are constants.
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解題
The number of line segments is three times the triangular numbers of order \(n\). \(L_n = 3 \times \frac{n(n+1)}{2} = \frac{3}{2}n(n+1) = 1.5n^2 + 1.5n\). Thus \(d = 1.5\) and \(e = 1.5\).
評分準則
M1 for writing down the terms: 3, 9, 18, 30... M1 for determining the common second differences (3) or linking to triangular numbers A1.5 for the formula \(L_n = 1.5n^2 + 1.5n\)
題目 4 · Investigation & Modelling
8.5 分
Part A: Investigation (continued).
Find the constants \(A, B, C\) such that \(L_n = A \cdot P_n + B \cdot T_n + C\) for all positive integers \(n\).
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解題
We substitute the formulas for \(P_n\), \(T_n\), and \(L_n\) into the equation: \(1.5n^2 + 1.5n = A(0.5n^2 + 1.5n + 1) + B(n^2) + C\) \(1.5n^2 + 1.5n = (0.5A + B)n^2 + 1.5An + (A + C)\) Equating coefficients: For \(n\): \(1.5A = 1.5 \implies A = 1\). For constant term: \(A + C = 0 \implies 1 + C = 0 \implies C = -1\). For \(n^2\): \(0.5A + B = 1.5 \implies 0.5(1) + B = 1.5 \implies B = 1\). Thus, \(A=1, B=1, C=-1\).
評分準則
M1 for substituting formulas into the given equation M1 for equating coefficients of \(n^2\), \(n\), and the constant terms A1.5 for finding \(A=1\), \(B=1\), and \(C=-1\)
題目 5 · Investigation & Modelling
8.5 分
Part B: Modelling.
A scientist models the temperature \(H(t)\) of a hot cup of coffee cooling in a room. The temperature (in \(^\circ\)C) at time \(t\) minutes is modelled by the function: \(H(t) = 20 + 65 \cdot 2^{-k t}\) where \(k\) is a positive constant.
If the temperature of the coffee is \(65^\circ\)C after 10 minutes, find the exact value of \(k\).
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解題
Substitute \(t = 10\) and \(H(10) = 65\) into the model equation: \(65 = 20 + 65 \cdot 2^{-10k}\) \(45 = 65 cdot 2^{-10k}\) \(2^{-10k} = \frac{45}{65} = \frac{9}{13}\) \(2^{10k} = \frac{13}{9}\) \(10k = \log_2\left(\frac{13}{9}\right)\) \(k = \frac{1}{10}\log_2\left(\frac{13}{9}\right)\) or \(0.1\log_2\left(\frac{13}{9}\right)\).
評分準則
M1 for writing \(20 + 65 \cdot 2^{-10k} = 65\) M1 for isolating the power: \(2^{-10k} = \frac{9}{13}\) A1.5 for converting to logarithm and obtaining the exact value \(k = \frac{1}{10}\log_2\left(\frac{13}{9}\right)\)
題目 6 · Investigation & Modelling
8.5 分
Part B: Modelling (continued).
According to the model \(H(t) = 20 + 65 \cdot 2^{-k t}\), find the temperature of the room in which the coffee is cooling. State the equation of the horizontal asymptote of \(H(t)\) as \(t \to \infty\).
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解題
As time \(t \to \infty\), \(2^{-kt} \to 0\) because \(k > 0\). Make \(H(t) \to 20 + 65(0) = 20^\circ\)C. The horizontal asymptote is the line where the function values stabilize, which is \(H = 20\).
評分準則
M1 for explaining that as \(t \to \infty\), \(2^{-kt} \to 0\) A1.5 for stating room temperature is \(20^\circ\)C A1 for stating the horizontal asymptote equation \(H = 20\)
題目 7 · Investigation & Modelling
8.5 分
Part B: Modelling (continued).
Using the value \(k = 0.0531\), calculate the time, in minutes, it takes for the temperature of the coffee to fall to \(30^\circ\)C. Give your answer correct to 1 decimal place.
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解題
We set \(H(t) = 30\) and solve for \(t\) using \(k = 0.0531\): \(30 = 20 + 65 \cdot 2^{-0.0531 t}\) \(10 = 65 \cdot 2^{-0.0531 t}\) \(2^{-0.0531 t} = \frac{10}{65} = \frac{2}{13}\) Taking base-2 logarithms: \(-0.0531 t = \log_2\left(\frac{2}{13}\right) = \frac{\ln(2/13)}{\ln 2} \approx -2.70044\) \(t = \frac{-2.70044}{-0.0531} \approx 50.855\) minutes. Rounding to 1 decimal place gives \(t = 50.9\) minutes.
評分準則
M1 for setting up the equation \(20 + 65 \cdot 2^{-0.0531 t} = 30\) M1 for simplifying to \(2^{-0.0531 t} = \frac{2}{13}\) M1 for taking logarithms correctly A0.5 for rounding to \(50.9\)
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