An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V2) Cambridge IGCSE International Mathematics (0607) paper. Not affiliated with or reproduced from Cambridge.
Paper 22 (Extended Non-calculator)
Answer all questions. Calculators must not be used.
22 题目 · 72.27999999999999 分
题目 1 · short_answer
3 分
Find an expression, in terms of \(n\), for the \(n\)th term of the sequence:
\(5, 12, 25, 44, 69, \dots\)
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解题
First, we find the first differences between consecutive terms: \(12 - 5 = 7\) \(25 - 12 = 13\) \(44 - 25 = 19\) \(69 - 44 = 25\)
Next, we find the second differences: \(13 - 7 = 6\) \(19 - 13 = 6\) \(25 - 19 = 6\)
Since the second differences are constant and equal to 6, the sequence is quadratic and has the term \(an^2\), where: \(2a = 6 \implies a = 3\)
Subtracting \(3n^2\) from each term of the sequence: - For \(n = 1\): \(5 - 3(1)^2 = 2\) - For \(n = 2\): \(12 - 3(2)^2 = 0\) - For \(n = 3\): \(25 - 3(3)^2 = -2\) - For \(n = 4\): \(44 - 3(4)^2 = -4\)
The linear sequence of differences is \(2, 0, -2, -4, \dots\), which has a first term of 2 and a common difference of \(-2\). Its formula is \(2 + (n - 1)(-2) = 4 - 2n\).
Combining the quadratic and linear parts, the \(n\)th term of the sequence is: \(3n^2 - 2n + 4\)
评分标准
M1 for realizing second differences are constant and finding \(a = 3\) (or setting up simultaneous equations) M1 for finding the linear part \(4 - 2n\) A1 for the correct final expression \(3n^2 - 2n + 4\)
题目 2 · short_answer
3 分
The first three terms of a sequence are \(12, 6, 3\). Find the 8th term of this sequence.
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解题
This is a geometric sequence because the ratio between consecutive terms is constant: \(r = \frac{6}{12} = \frac{1}{2}\)
The first term is \(a = 12\).
The formula for the \(n\)th term of a geometric sequence is: \(u_n = a \cdot r^{n-1}\)
For the 8th term (\(n = 8\)): \(u_8 = 12 \cdot \left(\frac{1}{2}\right)^{8-1} = 12 \cdot \left(\frac{1}{2}\right)^7 = 12 \cdot \frac{1}{128} = \frac{12}{128} = \frac{3}{32}\)
评分标准
M1 for identifying the common ratio \(r = \frac{1}{2}\) oe M1 for attempting to calculate \(12 \times \left(\frac{1}{2}\right)^7\) oe A1 for the correct final fraction \(\frac{3}{32}\) (or decimal equivalent \(0.09375\))
题目 3 · short_answer
3 分
Find the quadratic function, in the form \(ax^2 + bx + c\), that has a vertex at \((1, 5)\) and passes through the point \((3, -3)\).
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解题
The vertex form of a quadratic function is: \(y = a(x - h)^2 + k\)
Given the vertex is at \((1, 5)\): \(y = a(x - 1)^2 + 5\)
Since the curve passes through \((3, -3)\), substitute \(x = 3\) and \(y = -3\): \(-3 = a(3 - 1)^2 + 5\) \(-3 = 4a + 5\) \(4a = -8 \implies a = -2\)
Now, expand the equation into the standard form \(ax^2 + bx + c\): \(y = -2(x - 1)^2 + 5\) \(y = -2(x^2 - 2x + 1) + 5\) \(y = -2x^2 + 4x - 2 + 5\) \(y = -2x^2 + 4x + 3\)
评分标准
M1 for writing the equation in vertex form: \(y = a(x - 1)^2 + 5\) M1 for substituting the coordinates \((3, -3)\) to find \(a = -2\) A1 for expanding correctly to obtain the final expression \(-2x^2 + 4x + 3\)
题目 4 · short_answer
3 分
A quadratic curve passes through the points \((-2, 0)\), \((5, 0)\), and \((0, -20)\). Find the equation of this curve in the form \(y = ax^2 + bx + c\).
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解题
Since the curve passes through the points \((-2, 0)\) and \((5, 0)\), these points are the \(x\)-intercepts. The equation can be written in factored form as: \(y = a(x + 2)(x - 5)\)
Using the given \(y\)-intercept at \((0, -20)\), we substitute \(x = 0\) and \(y = -20\): \(-20 = a(0 + 2)(0 - 5)\) \(-20 = -10a \implies a = 2\)
Now we expand the factored form to obtain standard form: \(y = 2(x + 2)(x - 5)\) \(y = 2(x^2 - 3x - 10)\) \(y = 2x^2 - 6x - 20\)
评分标准
M1 for setting up factored form: \(y = a(x + 2)(x - 5)\) M1 for substituting the point \((0, -20)\) to find \(a = 2\) A1 for standard form equation \(y = 2x^2 - 6x - 20\) oe
M1 for expanding \((2x - 3)(x + 4)\) correctly with at most one error M1 for expanding \((x - 2)^2\) correctly to get \(x^2 - 4x + 4\) (or with at most one sign error) A1 for the correct simplified expression \(x^2 + 9x - 16\)
题目 6 · short_answer
3 分
Factorise fully:
\(12x^3 - 3x\)
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解题
First, extract the greatest common factor, which is \(3x\): \(12x^3 - 3x = 3x(4x^2 - 1)\)
Next, recognise that \(4x^2 - 1\) is a difference of squares: \(4x^2 - 1 = (2x)^2 - 1^2 = (2x - 1)(2x + 1)\)
Substitute this back to get the fully factored expression: \(3x(2x - 1)(2x + 1)\)
评分标准
M1 for extracting \(3x\) to get \(3x(4x^2 - 1)\) M1 for factoring \(4x^2 - 1\) as a difference of squares A1 for the correct fully factorised expression \(3x(2x - 1)(2x + 1)\) oe
题目 7 · short_answer
3 分
Given that \(\text{f}(x) = 3 - 2x\) and \(\text{g}(x) = \frac{5}{x+1}\), find the value of \(\text{g}(\text{f}(4))\).
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解题
First, find the value of \(\text{f}(4)\): \(\text{f}(4) = 3 - 2(4) = 3 - 8 = -5\)
Now substitute this result into \(\text{g}(x)\): \(\text{g}(\text{f}(4)) = \text{g}(-5) = \frac{5}{-5 + 1} = \frac{5}{-4} = -\frac{5}{4}\)
评分标准
M1 for correctly finding \(\text{f}(4) = -5\) M1 for substituting their value of \(\text{f}(4)\) into \(\text{g}(x)\) A1 for the correct final answer \(-\frac{5}{4}\) (or \(-1.25\))
题目 8 · short_answer
3 分
Given that \(\text{f}(x) = \frac{2x + 1}{x - 3}\) where \(x \neq 3\), find \(\text{f}^{-1}(x)\).
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解题
Let \(y = \frac{2x + 1}{x - 3}\).
To find the inverse, swap \(x\) and \(y\): \(x = \frac{2y + 1}{y - 3}\)
Therefore, the inverse function is: \(\text{f}^{-1}(x) = \frac{3x + 1}{x - 2}\) where \(x \neq 2\).
评分标准
M1 for setting up the equation \(x = \frac{2y + 1}{y - 3}\) oe M1 for expanding and rearranging terms to group \(y\) on one side A1 for the correct final inverse function expression \(\frac{3x + 1}{x - 2}\) oe
题目 9 · short_answer
3.41 分
Find an expression, in terms of \(n\), for the \(n\)th term of the sequence: \(4, 7, 14, 25, 40, \dots\)
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解题
Let the general term of the quadratic sequence be \(u_n = an^2 + bn + c\). The first differences are \(3, 7, 11, 15, \dots\) and the second differences are constant at \(4\). Therefore, \(2a = 4\), which gives \(a = 2\). Subtracting \(2n^2\) from each term of the sequence yields the linear sequence: \(2, -1, -4, -7, \dots\). This is an arithmetic sequence with a common difference of \(-3\) and a first term of \(2\), which can be written as \(-3n + 5\). Combining these gives \(u_n = 2n^2 - 3n + 5\).
评分标准
M1 for finding the second difference of 4 and setting \(2a = 4\) to find \(a = 2\). M1 for subtracting \(2n^2\) and finding the linear part \(-3n + 5\). A1 for the correct final answer: \(2n^2 - 3n + 5\).
题目 10 · short_answer
3.41 分
The graph of the quadratic function \(y = ax^2 + bx + c\) has a vertex at \((2, -8)\) and passes through the point \((0, 4)\). Find the values of \(a\), \(b\) and \(c\).
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解题
Using the vertex form of a quadratic, we have \(y = a(x - h)^2 + k\), which gives \(y = a(x - 2)^2 - 8\). Substituting the point \((0, 4)\) into this equation: \(4 = a(0 - 2)^2 - 8\) simplifies to \(12 = 4a\), hence \(a = 3\). Expanding \(y = 3(x - 2)^2 - 8\) gives \(y = 3(x^2 - 4x + 4) - 8 = 3x^2 - 12x + 12 - 8 = 3x^2 - 12x + 4\). Therefore, \(a = 3\), \(b = -12\), and \(c = 4\).
评分标准
M1 for writing the vertex form \(y = a(x - 2)^2 - 8\). M1 for substituting \((0, 4)\) to find \(a = 3\). A1 for correctly expanding to standard form and identifying \(a = 3, b = -12, c = 4\).
题目 11 · short_answer
3.41 分
Factorise fully: \(2x^3 - 3x^2 - 8x + 12\)
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解题
We factorise by grouping terms: \(2x^3 - 3x^2 - 8x + 12 = x^2(2x - 3) - 4(2x - 3)\). Factoring out the common bracket \((2x - 3)\) gives \((x^2 - 4)(2x - 3)\). Since \(x^2 - 4\) is a difference of two squares, it factorises to \((x - 2)(x + 2)\). Thus, the fully factorised expression is \((x - 2)(x + 2)(2x - 3)\).
评分标准
M1 for grouping the terms, e.g., \(x^2(2x - 3) - 4(2x - 3)\). M1 for factorising into a linear and quadratic product \((x^2 - 4)(2x - 3)\). A1 for fully factorised form: \((x - 2)(x + 2)(2x - 3)\).
题目 12 · short_answer
3.41 分
The function \(f(x) = \frac{3x - 5}{2x + 4}\), where \(x \neq -2\). Find \(f^{-1}(x)\).
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解题
Set \(y = f(x)\): \(y = \frac{3x - 5}{2x + 4}\). Multiply both sides by \((2x + 4)\) to get \(y(2x + 4) = 3x - 5\). Expanding gives \(2xy + 4y = 3x - 5\). Rearranging to group \(x\) terms: \(2xy - 3x = -4y - 5\), which is \(x(2y - 3) = -(4y + 5)\). Solving for \(x\) gives \(x = \frac{4y + 5}{3 - 2y}\). Replacing \(y\) with \(x\) gives \(f^{-1}(x) = \frac{4x + 5}{3 - 2x}\).
评分标准
M1 for multiplying both sides by \(2x+4\) and expanding. M1 for rearranging to isolate \(x\) on one side of the equation. A1 for the correct inverse function: \(\frac{4x + 5}{3 - 2x}\) or equivalent.
题目 13 · short_answer
3.41 分
Find an expression, in terms of \(n\), for the \(n\)th term of the geometric sequence: \(\frac{3}{2}, -3, 6, -12, 24, \dots\)
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解题
The first term of the geometric sequence is \(a = \frac{3}{2}\). The common ratio \(r\) is found by dividing any term by its preceding term: \(r = \frac{-3}{3/2} = -2\). Using the general formula for the \(n\)th term of a geometric sequence, \(u_n = a r^{n-1}\), we substitute the values to get \(u_n = \frac{3}{2}(-2)^{n-1}\).
评分标准
M1 for identifying the common ratio \(r = -2\). M1 for using the formula \(a r^{n-1}\) with their \(a\) and \(r\). A1 for the correct expression: \(\frac{3}{2}(-2)^{n-1}\).
题目 14 · short_answer
3.41 分
A quadratic curve passes through the points \((-1, 0)\) and \((5, 0)\). The maximum value of the curve is \(18\). Find the equation of the quadratic curve in the form \(y = ax^2 + bx + c\).
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解题
Since the x-intercepts are at \(x = -1\) and \(x = 5\), the axis of symmetry is \(x = \frac{-1 + 5}{2} = 2\). The maximum value is at the vertex, so the vertex is \((2, 18)\). In factored form, the equation is \(y = a(x + 1)(x - 5)\). Substituting the vertex \((2, 18)\) gives: \(18 = a(2 + 1)(2 - 5) \Rightarrow 18 = -9a \Rightarrow a = -2\). Expanding \(y = -2(x + 1)(x - 5)\) yields \(y = -2(x^2 - 4x - 5) = -2x^2 + 8x + 10\).
评分标准
M1 for finding the vertex coordinate \((2, 18)\). M1 for substituting the vertex into either the intercept form or the vertex form to find \(a = -2\). A1 for the correct standard form equation: \(y = -2x^2 + 8x + 10\).
题目 15 · short_answer
3.41 分
Simplify fully: \(\frac{4x^2 - 9}{2x^2 - x - 3}\)
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解题
Factorise the numerator using the difference of two squares: \(4x^2 - 9 = (2x - 3)(2x + 3)\). Factorise the denominator: \(2x^2 - x - 3 = (2x - 3)(x + 1)\). Dividing both terms by the common factor \((2x - 3)\) gives the simplified fraction: \(\frac{2x + 3}{x + 1}\).
评分标准
M1 for factorising the numerator to \((2x - 3)(2x + 3)\). M1 for factorising the denominator to \((2x - 3)(x + 1)\). A1 for the fully simplified form: \(\frac{2x + 3}{x + 1}\).
题目 16 · short_answer
3.41 分
Given that \(f(x) = 2x - 3\) and \(g(x) = x^2 - 1\), solve the equation \(g(f(x)) = 8\).
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解题
First find the composite function \(g(f(x))\): \(g(f(x)) = (2x - 3)^2 - 1\). Set this expression equal to 8: \((2x - 3)^2 - 1 = 8 \Rightarrow (2x - 3)^2 = 9\). Taking the square root of both sides gives \(2x - 3 = 3\) or \(2x - 3 = -3\). Solving these linear equations: \(2x = 6 \Rightarrow x = 3\), or \(2x = 0 \Rightarrow x = 0\). The solutions are \(x = 0\) or \(x = 3\).
评分标准
M1 for obtaining the expression \((2x - 3)^2 - 1\). M1 for solving \((2x - 3)^2 = 9\) (or expanding to \(4x^2 - 12x = 0\) and factoring). A1 for both correct values: \(x = 0\) and \(x = 3\).
题目 17 · short_answer
3 分
The \(n\)th term of a sequence is given by \(u_n = an^2 + bn\). The second term is 10 and the fourth term is 36. Find the value of \(a\) and the value of \(b\).
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解题
Substitute the given terms into the formula for the \(n\)th term:
For \(n = 2\): \(u_2 = a(2^2) + b(2) = 4a + 2b = 10\) Dividing by 2 gives: \(2a + b = 5\) (Equation 1)
For \(n = 4\): \(u_4 = a(4^2) + b(4) = 16a + 4b = 36\) Dividing by 4 gives: \(4a + b = 9\) (Equation 2)
Subtract Equation 1 from Equation 2: \((4a + b) - (2a + b) = 9 - 5\) \(2a = 4\) \(a = 2\)
Substitute \(a = 2\) into Equation 1: \(2(2) + b = 5\) \(4 + b = 5\) \(b = 1\)
Therefore, \(a = 2\) and \(b = 1\).
评分标准
M1 for setting up at least one correct equation using the given terms. M1 for correctly solving the simultaneous equations to find one of the variables. A1 for both \(a = 2\) and \(b = 1\).
题目 18 · short_answer
3 分
The graph of a quadratic function \(y = ax^2 + bx + c\) has its vertex at \((3, -4)\) and passes through the point \((1, 8)\). Find the values of \(a\), \(b\) and \(c\).
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解题
Using the vertex form of a quadratic equation: \(y = a(x - h)^2 + k\) where the vertex is \((h, k) = (3, -4)\).
Now, substitute \(a = 3\) back into the vertex form and expand to the form \(y = ax^2 + bx + c\): \(y = 3(x - 3)^2 - 4\) \(y = 3(x^2 - 6x + 9) - 4\) \(y = 3x^2 - 18x + 27 - 4\) \(y = 3x^2 - 18x + 23\)
Thus, \(a = 3\), \(b = -18\), and \(c = 23\).
评分标准
M1 for writing the quadratic function in vertex form: \(y = a(x - 3)^2 - 4\). M1 for substituting the point \((1, 8)\) to solve for \(a\). A1 for expanding correctly to obtain the values \(a = 3\), \(b = -18\), and \(c = 23\).
题目 19 · short_answer
3 分
Factorise completely: \(12x^3 - 3xy^2\)
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解题
First, identify the highest common factor (HCF) of both terms, which is \(3x\): \(12x^3 - 3xy^2 = 3x(4x^2 - y^2)\)
Next, notice that the expression inside the brackets, \(4x^2 - y^2\), is a difference of two squares: \(4x^2 - y^2 = (2x)^2 - (y)^2 = (2x - y)(2x + y)\)
Combining these gives the completely factorised expression: \(3x(2x - y)(2x + y)\)
评分标准
M1 for extracting the common factor to get \(3x(4x^2 - y^2)\). M1 for factorising the difference of two squares to get \((2x - y)(2x + y)\). A1 for the complete final expression: \(3x(2x - y)(2x + y)\).
题目 20 · short_answer
4 分
Given that \(\mathrm{f}(x) = 2x - 5\) and \(\mathrm{g}(x) = \frac{3}{x+1}\), \(x \neq -1\). Find the composite function \(\mathrm{g}(\mathrm{f}(x))\) in its simplest form, and find the value of \(x\) for which \(\mathrm{g}(\mathrm{f}(x))\) is undefined.
The function is undefined when the denominator is zero: \(2x - 4 = 0\) \(2x = 4\) \(x = 2\)
评分标准
M1 for correctly substituting \(\mathrm{f}(x)\) into \(\mathrm{g}(x)\). A1 for simplifying the algebraic fraction to \[ \frac{3}{2x - 4} \]\] or \[ \frac{3}{2(x - 2)} \]. M1 for setting their denominator to 0 to find the undefined value. A1 for \(x = 2\).
题目 21 · short_answer
4 分
The first four terms of a sequence are: \(-1, 2, 7, 14, \dots\) Find the \(n\)th term of this sequence.
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解题
Analyze the differences between successive terms of the sequence: Terms: \(-1, \quad 2, \quad 7, \quad 14\) First differences: \(3, \quad 5, \quad 7\) Second differences: \(2, \quad 2\)
Since the second difference is constant and equal to 2, the sequence is quadratic, with the general formula: \(u_n = an^2 + bn + c\) where \(2a = 2 \implies a = 1\).
Subtract \(n^2\) from each term in the sequence to find the remaining linear/constant part: For \(n = 1\): \(-1 - (1)^2 = -2\) For \(n = 2\): \(2 - (2)^2 = -2\) For \(n = 3\): \(7 - (3)^2 = -2\) For \(n = 4\): \(14 - (4)^2 = -2\)
Since the remainder is a constant \(-2\), the linear term \(b = 0\) and the constant term \(c = -2\). Therefore, the \(n\)th term is \(n^2 - 2\).
评分标准
M1 for calculating first differences (3, 5, 7) and second differences (2). M1 for identifying that \(a = 1\), leading to a quadratic term of \(n^2\). M1 for a method to find the constant term (e.g., subtracting \(n^2\) from terms of the sequence). A1 for \(n^2 - 2\).
题目 22 · short_answer
4 分
A quadratic curve has \(x\)-intercepts at \(x = -2\) and \(x = 5\). It has a \(y\)-intercept at \((0, -20)\). Find the equation of the quadratic curve in the form \(y = ax^2 + bx + c\).
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解题
Using the intercept form of a quadratic equation: \(y = k(x - p)(x - q)\) where the intercepts are \(p = -2\) and \(q = 5\).
Now, substitute \(k = 2\) back into the equation and expand to the required form: \(y = 2(x + 2)(x - 5)\) \(y = 2(x^2 - 3x - 10)\) \(y = 2x^2 - 6x - 20\)
评分标准
M1 for writing the quadratic equation in factored form using intercepts: \(y = k(x + 2)(x - 5)\). M1 for substituting the point \((0, -20)\) to find \(k\). A1 for finding \(k = 2\). A1 for expanding correctly to obtain \(y = 2x^2 - 6x - 20\).
Answer all questions. Electronic graphic display calculators are permitted.
17 题目 · 114 分
题目 1 · structured_multi_part
4 分
The first four terms of a sequence are: \(11, 18, 27, 38, \dots\)
(a) Find the next term of this sequence.
(b) Find an expression, in terms of \(n\), for the \(n\)-th term of this sequence.
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解题
(a) The differences between consecutive terms are \(18 - 11 = 7\), \(27 - 18 = 9\), \(38 - 27 = 11\), which form an arithmetic progression. The next difference is \(13\). Thus, the next term of the sequence is \(38 + 13 = 51\).
(b) The terms and their differences are: Sequence: \(11, 18, 27, 38\) First differences: \(7, 9, 11\) Second differences: \(2, 2\) Since the second differences are constant and equal to 2, the formula is quadratic of the form \(a n^2 + b n + c\) with \(2a = 2 \implies a = 1\). Subtracting \(n^2\) from each term: \(11 - 1 = 10\) \(18 - 4 = 14\) \(27 - 9 = 18\) \(38 - 16 = 22\) The sequence \(10, 14, 18, 22\) is linear with first term 10 and common difference 4. Thus, \(b n + c = 4n + 6\). So the \(n\)-th term of the sequence is \(n^2 + 4n + 6\).
评分标准
(a) B1 for 51 (b) M1 for finding second difference is 2 or identifying the \(n^2\) term M1 for attempting to find the linear part \(4n + 6\) A1 for \(n^2 + 4n + 6\)
题目 2 · structured_multi_part
4 分
The first four terms of another sequence are: \(\frac{3}{5}, \frac{6}{11}, \frac{9}{17}, \frac{12}{23}, \dots\)
(a) Write down the next term of this sequence as a fraction.
(b) Find an expression, in terms of \(n\), for the \(n\)-th term of this sequence.
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解题
(a) The numerators are: \(3, 6, 9, 12, \dots\). The next numerator is 15. The denominators are: \(5, 11, 17, 23, \dots\). This is an arithmetic sequence with a common difference of 6. The next denominator is \(23 + 6 = 29\). Therefore, the next term is \(\frac{15}{29}\).
(b) The numerator for term \(n\) is \(3n\). The denominator is an arithmetic progression with first term \(a = 5\) and common difference \(d = 6\): Denominator = \(5 + 6(n - 1) = 6n - 1\). Thus, the general term is \(\frac{3n}{6n - 1}\).
评分标准
(a) B1 for \(\frac{15}{29}\) (b) M1 for finding the numerator expression \(3n\) M1 for finding the denominator expression \(6n - 1\) A1 for \(\frac{3n}{6n - 1}\)
题目 3 · structured_multi_part
4 分
A quadratic curve of the form \(y = a x^2 + b x + c\) passes through the points \((0, -5)\), \((1, -2)\), and \((3, 16)\).
(a) Show that \(c = -5\).
(b) Find the values of \(a\) and \(b\).
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解题
(a) Substituting \(x = 0\) and \(y = -5\) into the equation \(y = a x^2 + b x + c\): \(-5 = a(0)^2 + b(0) + c \implies c = -5\).
Subtracting the first equation from the second: \((3a + b) - (a + b) = 7 - 3 \implies 2a = 4 \implies a = 2\). Substitute \(a = 2\) back into the first equation: \(2 + b = 3 \implies b = 1\).
评分标准
(a) B1 for substituting \((0, -5)\) into the equation to show \(c = -5\) (b) M1 for substituting \((1, -2)\) and \((3, 16)\) to form two linear equations in \(a\) and \(b\) M1 for solving the simultaneous equations correctly A1 for \(a = 2\) and \(b = 1\)
题目 4 · structured_multi_part
4 分
A quadratic function \(f(x) = p x^2 + q x + r\) has a vertex at \((2, 3)\) and passes through the point \((4, -5)\).
(a) Write the quadratic function in the vertex form \(f(x) = p(x - h)^2 + k\), stating the values of \(h\) and \(k\).
(b) Find the values of \(p\), \(q\), and \(r\).
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解题
(a) The vertex form is \(f(x) = p(x - h)^2 + k\). Since the vertex is at \((2, 3)\), we have: \(h = 2\) and \(k = 3\).
(b) The equation is \(f(x) = p(x - 2)^2 + 3\). Substituting the point \((4, -5)\): \(-5 = p(4 - 2)^2 + 3\) \(-5 = 4p + 3 \implies 4p = -8 \implies p = -2\).
(a) B1 for identifying \(h = 2\) and \(k = 3\) (b) M1 for substituting the point \((4, -5)\) to solve for \(p = -2\) M1 for expanding the vertex form correctly A1 for \(p = -2, q = 8, r = -5\)
题目 5 · structured_multi_part
4 分
(a) Factorise completely: \(6 a x - 9 b x - 4 a y + 6 b y\).
(b) Factorise completely: \(18 x^2 - 50 y^2\).
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解题
(a) Grouping the terms: \(6 a x - 9 b x - 4 a y + 6 b y = 3x(2a - 3b) - 2y(2a - 3b)\) Factoring out the common binomial \((2a - 3b)\): \(= (3x - 2y)(2a - 3b)\).
(b) Factoring out the common numeric factor \(2\): \(18 x^2 - 50 y^2 = 2(9x^2 - 25y^2)\) Recognizing the difference of two squares inside the parentheses: \(= 2((3x)^2 - (5y)^2) = 2(3x - 5y)(3x + 5y)\).
评分标准
(a) M1 for a correct partial factorisation, e.g. \(3x(2a - 3b) - 2y(2a - 3b)\) A1 for \((3x - 2y)(2a - 3b)\) (b) M1 for factoring out 2 first to get \(2(9x^2 - 25y^2)\) A1 for \(2(3x - 5y)(3x + 5y)\)
(a) M1 for writing \(y = \frac{2x + 5}{3}\) and attempting to isolate \(x\) A1 for \(\frac{3x - 5}{2}\) (b) M1 for evaluating \(f(2) = 3\) A1 for 8
题目 8 · structured_multi_part
4 分
The curve \(y = x^3 - 3x^2 + 4\) has a local maximum at point \(A\) and a local minimum at point \(B\).
(a) Find the coordinates of \(A\) and \(B\).
(b) State the number of real solutions to the equation \(x^3 - 3x^2 + 4 = 1\).
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解题
(a) To find the stationary points, differentiate \(y\) with respect to \(x\): \(y' = 3x^2 - 6x = 0\) \(3x(x - 2) = 0 \implies x = 0\) or \(x = 2\). At \(x = 0\), \(y = 0^3 - 3(0)^2 + 4 = 4\). At \(x = 2\), \(y = 2^3 - 3(2)^2 + 4 = 8 - 12 + 4 = 0\). Since the lead coefficient is positive, \(A(0, 4)\) is the local maximum and \(B(2, 0)\) is the local minimum.
(b) The horizontal line \(y = 1\) lies strictly between the local minimum value \(y = 0\) and local maximum value \(y = 4\). Therefore, \(y = 1\) intersects the curve at exactly three distinct points. Hence, the equation has 3 real solutions.
评分标准
(a) M1 for finding the derivative and setting it to zero to solve for \(x = 0\) and \(x = 2\) A1 for both coordinates \(A(0, 4)\) and \(B(2, 0)\) (b) M1 for recognizing that \(y = 1\) lies between \(y = 0\) and \(y = 4\) A1 for 3
题目 9 · structured_multi_part
10 分
The first five terms of a sequence, A, are 2, 8, 16, 26, 38, ... (a) Write down the next two terms of sequence A. (b) Find an expression, in terms of n, for the n-th term of sequence A. (c) The first five terms of a different sequence, B, are 8, 10, 14, 22, 38, ... (i) Write down the next term of sequence B. (ii) The n-th term of sequence B is 2^n + c. Find the value of c. (d) Find the value of n for which the n-th term of sequence A is equal to the n-th term of sequence B.
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解题
(a) The differences between consecutive terms of sequence A are 6, 8, 10, 12, ... The next differences are 14 and 16. So the next terms are 38 + 14 = 52 and 52 + 16 = 68. (b) The second differences are constant and equal to 2, so the sequence is quadratic, of the form an^2 + bn + c. Here, 2a = 2 so a = 1. Using u_n = n^2 + bn + c: for n = 1, 1 + b + c = 2; for n = 2, 4 + 2b + c = 8. Solving these gives b = 3 and c = -2. The expression is n^2 + 3n - 2. (c)(i) The differences of B are powers of 2: 2, 4, 8, 16, ... The next difference is 32, so the next term is 38 + 32 = 70. (c)(ii) For n = 1, 2^1 + c = 8, which yields c = 6. (d) Setting the two expressions equal: n^2 + 3n - 2 = 2^n + 6. By trial and error or using a calculator, we find that for n = 5, both sides equal 38.
评分标准
(a) B1 for 52, B1 for 68 (b) M1 for quadratic form, M1 for finding coefficients, A1 for n^2 + 3n - 2 (c)(i) B1 for 70 (c)(ii) M1 for 2^1 + c = 8, A1 for c = 6 (d) M1 for equating both expressions, A1 for n = 5
题目 10 · structured_multi_part
10 分
Decorative fences of design n are built using vertical posts and diagonal bars. The data for the first four fence designs is given as: Design number (n): 1, 2, 3, 4 | Number of vertical posts: 2, 3, 4, 5 | Number of diagonal bars: 2, 6, 12, 20 | Total number of components: 4, 9, 16, 25. (a) Complete the data values for Design 5. (b) Find an expression, in terms of n, for: (i) the number of vertical posts in Design n, (ii) the number of diagonal bars in Design n, (iii) the total number of components in Design n. (c) A fence has 240 diagonal bars. Find the design number n of this fence.
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解题
(a) Following the pattern: posts increase by 1 each time, so Design 5 has 6 posts. Diagonal bars are n(n+1), so Design 5 has 5 * 6 = 30 diagonal bars. Total components = 6 + 30 = 36. (b)(i) The posts are 2, 3, 4, 5, so the expression is n + 1. (b)(ii) The diagonal bars are 2, 6, 12, 20, which is n(n + 1) = n^2 + n. (b)(iii) The totals are 4, 9, 16, 25, which are perfect squares: (n + 1)^2. (c) Setting the diagonal bars expression to 240: n(n + 1) = 240, which expands to n^2 + n - 240 = 0. Factoring gives (n - 15)(n + 16) = 0. Since n must be positive, n = 15.
评分标准
(a) B1 for posts = 6 and diagonal bars = 30, B1 for total = 36 (b)(i) B1 for n + 1 (b)(ii) M1 for attempting to find quadratic form, A1 for n^2 + n (b)(iii) M1 for adding parts (b)(i) and (b)(ii), A1 for (n + 1)^2 (c) M1 for n(n + 1) = 240, M1 for solving quadratic equation, A1 for n = 15
题目 11 · structured_multi_part
8 分
The first four terms of a sequence are 1, 15, 53, 127, ... (a) Write down the next term of the sequence. (b) The n-th term of this sequence is of the form a*n^3 + b. Find the value of a and the value of b. (c) Find the 10th term of this sequence. (d) Show that 2001 is not a term of this sequence.
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解题
(a) The sequence follows the formula 2n^3 - 1. For n = 5, the term is 2(5^3) - 1 = 2(125) - 1 = 249. (b) Using the formula: for n = 1, a(1)^3 + b = 1 => a + b = 1. For n = 2, a(2)^3 + b = 15 => 8a + b = 15. Subtracting the two equations: 7a = 14 => a = 2. Then, 2 + b = 1 => b = -1. (c) For n = 10, the term is 2(10^3) - 1 = 2000 - 1 = 1999. (d) Setting 2n^3 - 1 = 2001 => 2n^3 = 2002 => n^3 = 1001. Since 10^3 = 1000 and 11^3 = 1331, 1001 is not a perfect cube. Thus, n is not an integer, so 2001 is not a term of this sequence.
评分标准
(a) B1 for 249 (b) M1 for setting up equations, M1 for solving equations, A1 for a = 2, b = -1 (c) B1 for 1999 (d) M1 for 2n^3 - 1 = 2001, A1 for n^3 = 1001, A1 for explaining that 1001 is not a perfect cube
题目 12 · structured_multi_part
12 分
A quadratic function f(x) = a*x^2 + b*x + c has a vertex at (2, -5) and passes through the point (4, 3). (a) Show that a = 2. (b) Find the values of b and c. (c) Find the x-intercepts of the graph of f(x), giving your answers correct to 2 decimal places. (d) The line y = 2x + k is a tangent to the curve y = f(x). Find the value of k.
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解题
(a) Using the vertex form: f(x) = a(x - 2)^2 - 5. Substituting the point (4, 3): 3 = a(4 - 2)^2 - 5 => 3 = 4a - 5 => 4a = 8 => a = 2. (b) Expanding the vertex form f(x) = 2(x - 2)^2 - 5 = 2(x^2 - 4x + 4) - 5 = 2x^2 - 8x + 3. Hence b = -8 and c = 3. (c) Solving 2x^2 - 8x + 3 = 0 using quadratic formula: x = (8 +- sqrt(64 - 24))/4 = (8 +- sqrt(40))/4 = 2 +- sqrt(10)/2. This gives x = 0.42 and x = 3.58. (d) Equating the tangent and the curve: 2x^2 - 8x + 3 = 2x + k => 2x^2 - 10x + (3 - k) = 0. For a tangent, the discriminant must be zero: (-10)^2 - 4(2)(3 - k) = 0 => 100 - 8(3 - k) = 0 => 100 - 24 + 8k = 0 => 8k = -76 => k = -9.5.
评分标准
(a) M1 for f(x) = a(x - 2)^2 - 5, M1 for substituting (4, 3), A1 for showing a = 2 (b) M1 for expanding 2(x - 2)^2 - 5, A1 for b = -8, A1 for c = 3 (c) M1 for setting f(x) = 0, A1 for 0.42, A1 for 3.58 (d) M1 for equating 2x^2 - 8x + 3 = 2x + k, M1 for setting discriminant equal to zero, A1 for k = -9.5
题目 13 · structured_multi_part
13 分
The graph of a quadratic function y = g(x), where g(x) = a*x^2 + b*x + c, passes through the points (1, 4), (2, 9), and (-1, 6). (a) Show that b = -1. (b) Find the value of a and the value of c. (c) Write down the coordinates of the y-intercept of the graph of g(x). (d) Find the coordinates of the vertex of this quadratic function. (e) Solve the equation g(x) = 24.
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解题
(a) Setting up equations: g(1) = a + b + c = 4 (Eq 1). g(-1) = a - b + c = 6 (Eq 2). Subtracting Eq 2 from Eq 1: (a + b + c) - (a - b + c) = 4 - 6 => 2b = -2 => b = -1. (b) Substituting b = -1 into Eq 1 and g(2) = 9: a - 1 + c = 4 => a + c = 5 (Eq 3). 4a - 2 + c = 9 => 4a + c = 11 (Eq 4). Subtracting Eq 3 from Eq 4: 3a = 6 => a = 2. Then 2 + c = 5 => c = 3. (c) At x = 0, y = c = 3, so coordinates are (0, 3). (d) The x-coordinate of the vertex is x = -b/(2a) = 1/(2*2) = 0.25. The y-coordinate is g(0.25) = 2(0.25)^2 - 0.25 + 3 = 2.875. Coordinates of vertex are (0.25, 2.875). (e) Solving 2x^2 - x + 3 = 24 => 2x^2 - x - 21 = 0 => (2x - 7)(x + 3) = 0 => x = 3.5 or x = -3.
评分标准
(a) M1 for establishing equations at x=1 and x=-1, M1 for subtracting equations, A1 for b = -1 (b) M1 for substituting b=-1, M1 for solving the linear system, A1 for a = 2, c = 3 (c) B1 for (0, 3) (d) M1 for x = -b/(2a), M1 for substituting x = 0.25 into function, A1 for (0.25, 2.875) (e) M1 for 2x^2 - x - 21 = 0, M1 for factoring or using quadratic formula, A1 for -3 and 3.5
(a)(i) M1 for splitting middle term, A1 for (3x + 2)(2x - 3) (a)(ii) B1 for (2a - 3b)(2a + 3b) (b) M1 for factorising numerator, M1 for factorising denominator, A1 for (2x + 1)(x - 4), A1 for (x + 4)/(2x + 1)
题目 15 · structured_multi_part
6 分
The formula for a ratio is given by R = sqrt((k + w)/(k - w)). (a) Express w in terms of R and k. (b) Given that k = 12 and R = 2, calculate the value of w.
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解题
(a) Squaring both sides: R^2 = (k + w)/(k - w) => R^2(k - w) = k + w => k*R^2 - w*R^2 = k + w => k*R^2 - k = w*R^2 + w => k(R^2 - 1) = w(R^2 + 1) => w = k(R^2 - 1)/(R^2 + 1). (b) Substituting k = 12 and R = 2: w = 12(2^2 - 1)/(2^2 + 1) = 12(3)/5 = 36/5 = 7.2.
评分标准
(a) M1 for squaring both sides, M1 for clearing denominator, M1 for grouping terms in w, A1 for w = k(R^2 - 1)/(R^2 + 1) (b) M1 for substituting k = 12 and R = 2, A1 for 7.2
题目 16 · structured_multi_part
12 分
Let f(x) = (3x + 1)/(x - 2) for x != 2, and g(x) = x^2 - 3. (a) Find: (i) f(5), (ii) g(f(3)). (b) Find f^-1(x). (c) Find the values of x for which g(x) = 13. (d) Solve the equation f(x) = g(2).
(a)(i) B1 for 16/3 (a)(ii) M1 for finding f(3) = 10, A1 for 97 (b) M1 for setting y = f(x), M1 for rearranging to express x in terms of y, A1 for (2x + 1)/(x - 3) (c) M1 for x^2 - 3 = 13, A1 for -4 and 4 (d) M1 for g(2) = 1, M1 for (3x+1)/(x-2) = 1, M1 for solving equation, A1 for -1.5
题目 17 · structured_multi_part
4 分
A quadratic curve has the equation \(y = ax^2 + bx + c\). The curve has a \(y\)-intercept at \((0, 2)\) and passes through the points \((1, 5)\) and \((2, 12)\).
(a) Show that \(a = 2\) and \(b = 1\).
(b) Find the \(x\)-coordinate of the vertex (minimum point) of this curve.
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解题
(a) Since the \(y\)-intercept is at \((0, 2)\), substituting \(x = 0, y = 2\) into the equation gives \(c = 2\).
Substituting \((1, 5)\) into \(y = ax^2 + bx + 2\) gives: \(5 = a(1)^2 + b(1) + 2 \implies a + b = 3\).
(a) B1 for identifying \(c = 2\) (or substituting \(x = 0, y = 2\) to get \(c = 2\)) M1 for writing down two equations from the given coordinates, e.g. \(a + b = 3\) and \(4a + 2b = 10\) (or equivalent) A1 for fully correct working to show \(a = 2\) and \(b = 1\)
(b) B1 for \(-0.25\) or \(-\frac{1}{4}\)
Paper 62 (Extended Investigation & Modelling)
Answer all questions in Section A (Investigation) and Section B (Modelling). Graphic display calculators are permitted.
5 题目 · 50 分
题目 1 · investigation_modelling
10 分
### Section A: Investigation — Triangular Grid of Matchsticks
This investigation is about the total number of matchsticks, \(M\), required to build triangular grids of different sizes. - Pattern 1 has 1 small triangle and is made from 3 matchsticks. - Pattern 2 has a side length of 2 small triangles and is made from 9 matchsticks. - Pattern 3 has a side length of 3 small triangles and is made from 18 matchsticks.
(a) Write down the number of matchsticks in Pattern 4.
(c) The formula for the number of matchsticks in Pattern \(n\) can be written in the form \(M = an^2 + bn\). Find the values of \(a\) and \(b\).
(d) Calculate the number of matchsticks in Pattern 15.
(e) Find the pattern number which requires exactly 1890 matchsticks.
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解题
(a) Pattern 4 contains \(30\) matchsticks.
(b) At \(n = 4\), \(M = 30\). At \(n = 5\), \(M = 45\).
(c) Using \(M = an^2 + bn\): For \(n=1\): \(a + b = 3\) For \(n=2\): \(4a + 2b = 9\) Solving simultaneously: Multiply the first equation by 2: \(2a + 2b = 6\) Subtracting this from the second equation: \(2a = 3 \implies a = 1.5\). Then, \(1.5 + b = 3 \implies b = 1.5\). Thus, \(a = 1.5\) and \(b = 1.5\).
(a) B1 for 30 (b) B1 for 30 (or FT from part a), B1 for 45 (c) M1 for substituting values to obtain at least one correct equation, M1 for a valid method to solve simultaneous equations, A1 for \(a = 1.5\) and \(b = 1.5\) (accept equivalent fractions) (d) B1 for 360 (or FT their formula) (e) M1 for setting their quadratic formula equal to 1890, M1 for attempting to factorise or solve the quadratic equation, A1 for 35 (ignore \(-36\))
题目 2 · investigation_modelling
10 分
### Section B: Modelling — Drone Flight Path
The height, \(h\) metres, of a flight-testing drone above the ground, \(t\) seconds after launch, is modelled by the quadratic function: \[h = at^2 + bt + c\]
At launch (\(t = 0\)), the drone is on a launchpad at a height of 20 metres. The drone reaches its maximum height of 36 metres after 4 seconds.
(a) Write down the value of \(c\).
(b) Show that \(4a + b = 4\).
(c) Given that the maximum height occurs at \(t = -\frac{b}{2a}\), find the values of \(a\) and \(b\).
(d) Find the time, in seconds, when the drone lands on the ground.
(e) Write down a suitable domain for \(t\) for which this model is valid.
(c) Since \(-\frac{b}{2a} = 4 \implies b = -8a\). Substituting \(b = -8a\) into \(4a + b = 4\): \[4a - 8a = 4 \implies -4a = 4 \implies a = -1\] Then, \(b = -8(-1) = 8\). So \(a = -1\) and \(b = 8\).
(d) The model is \(h = -t^2 + 8t + 20\). When the drone lands, \(h = 0\): \[-t^2 + 8t + 20 = 0 \implies t^2 - 8t - 20 = 0\] \[(t - 10)(t + 2) = 0\] Since \(t \ge 0\), \(t = 10\) seconds.
(e) Since height cannot be negative and the drone lands at \(t = 10\), the model is valid for \(0 \le t \le 10\).
评分标准
(a) B1 for 20 (b) M1 for substituting \(t=4\), \(h=36\), and their \(c\) into the model, A1 for completing the algebraic simplification to show \(4a + b = 4\) (c) M1 for writing \(b = -8a\) or equivalent, M1 for substituting to find \(a\) or \(b\), A1 for \(a = -1\) and \(b = 8\) (d) M1 for setting \(h = 0\) with their quadratic model and attempting to factorise or solve, A1 for 10 (e) B1 for \(0 \le t \le 10\) (accept interval notation \([0, 10]\) or equivalent verbal explanation that time goes from 0 to 10 seconds)
题目 3 · investigation_modelling
10 分
### Section A: Investigation — Number Grid Diagonals
This investigation is about the difference between diagonal products of square blocks of numbers in grids of various widths.
The numbers are arranged in a grid with a column width of \(W\): Row 1: \(1, 2, \dots, W\) Row 2: \(W+1, W+2, \dots, 2W\) and so on.
For any \(2 \times 2\) block of numbers: \[\begin{pmatrix} A & B \\ C & D \end{pmatrix}\] Let \(P_1 = A \times D\) and \(P_2 = B \times C\).
(a) If \(W = 7\) (a calendar grid), and the block of numbers is: \[\begin{pmatrix} 8 & 9 \\ 15 & 16 \end{pmatrix}\] (i) Calculate \(P_1\) and \(P_2\). (ii) Work out the difference \(P_2 - P_1\).
(b) Let the top-left number in the \(2 \times 2\) block be \(x\). The block in a grid of column width \(W\) is: \[\begin{pmatrix} x & x+1 \\ x+W & x+W+1 \end{pmatrix}\] Show algebraically that the difference \(P_2 - P_1\) is always equal to \(W\).
(c) Now consider a \(3 \times 3\) block of numbers in the same grid: \[\begin{pmatrix} x & \dots & x+2 \\ \dots & \dots & \dots \\ x+2W & \dots & x+2W+2 \end{pmatrix}\] Let \(P_1\) be the product of the top-left and bottom-right corners: \(P_1 = x(x+2W+2)\). Let \(P_2\) be the product of the top-right and bottom-left corners: \(P_2 = (x+2)(x+2W)\). (i) Find and simplify an expression for the difference \(P_2 - P_1\) in terms of \(W\). (ii) If this difference is 36, find the column width \(W\) of the grid.
(a) (i) B1 for \(P_1 = 128\), B1 for \(P_2 = 135\) (ii) B1 for 7 (b) M1 for expanding \(P_1\) to \(x^2 + Wx + x\), M1 for expanding \(P_2\) to \(x^2 + Wx + x + W\), A1 for subtracting and showing the final result is \(W\) clearly (c) (i) M1 for expanding \(P_1 = x^2 + 2Wx + 2x\), M1 for expanding \(P_2 = x^2 + 2Wx + 2x + 4W\), A1 for final simplified expression \(4W\) (ii) B1 for 9
题目 4 · investigation_modelling
10 分
### Section B: Modelling — Fish Population
The population \(P\) of a species of fish introduced into a lake is modelled as a function of time \(t\) (in years) by: \[P(t) = \frac{1200}{1 + 5 \times 2^{-0.2t}}\]
(a) Find the initial population of fish introduced into the lake.
(b) Calculate the population of fish after 10 years. Give your answer to the nearest integer.
(c) On a set of axes, sketch the graph of \(P(t)\) for \(0 \le t \le 50\).
(d) Use algebra or your graphic display calculator to find the number of years it takes for the population to reach 1000. Give your answer to 3 significant figures.
(e) Write down the equation of the horizontal asymptote of the graph of \(P(t)\) as \(t\) becomes very large, and explain what this represents in the context of the model.
(c) Sketch a logistic curve starting at \((0, 200)\), passing through \((10, 533)\), with an S-shape, approaching a horizontal asymptote at \(P = 1200\).
(e) As \(t \to \infty\), \(2^{-0.2t} \to 0\), so \(P(t) \to 1200\). The asymptote is \(P = 1200\) (or \(y = 1200\)). This represents the carrying capacity of the lake (the maximum population of fish the lake can support).
评分标准
(a) B1 for 200 (b) M1 for substituting \(t = 10\) into the model correctly, A1 for 533 (accept 533.3) (c) B1 for correct general shape starting at positive y-intercept, B1 for demonstrating asymptotic behavior toward a line below the top of the axis (d) M1 for setting \(P(t) = 1000\), M1 for correct algebraic steps or GDC method to isolate/solve the exponential term, A1 for 23.2 (accept 23.2 to 23.3) (e) B1 for \(P = 1200\) (accept \(y = 1200\)), B1 for explaining it is the carrying capacity or maximum sustainable population of fish in the lake
题目 5 · investigation_modelling
10 分
### Section A: Investigation — Branching Fractal Tree
This investigation is about a mathematical model of a branching fractal tree. - At Stage 0, the tree consists of a single trunk of length 1 unit. - At Stage 1, the trunk splits into 2 branches, each of length \(\frac{1}{2}\) unit. - At Stage 2, each of the Stage 1 branches splits into 2 new branches, each of length \(\frac{1}{4}\) unit. - This pattern of branching continues, where at each Stage \(n\), every new branch from the previous stage splits into 2 branches of length \(\left(\frac{1}{2}\right)^n\) unit.
(a) Complete the table for Stage 3 and Stage 4: | Stage (\(n\)) | 0 | 1 | 2 | 3 | 4 | | :--- | :---: | :---: | :---: | :---: | :---: | | Number of new branches added (\(B_n\)) | 1 | 2 | 4 | | | | Total number of branches (\(T_n\)) | 1 | 3 | 7 | | | | Total length of all branches (\(L_n\)) | 1 | 2 | 3 | | |
(b) Write down an expression, in terms of \(n\), for: (i) \(B_n\) (ii) \(T_n\) (iii) \(L_n\)
(c) Find the Stage \(n\) at which the total number of branches \(T_n\) is first greater than 1000.
(d) Calculate the total length of all branches at Stage 50.
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解题
(a) Completed table: - For \(B_n\) at \(n=3\), it is 8. At \(n=4\), it is 16. - For \(T_n\) at \(n=3\), it is 15. At \(n=4\), it is 31. - For \(L_n\) at \(n=3\), it is 4. At \(n=4\), it is 5.
(b) (i) \(B_n = 2^n\) (ii) \(T_n\) is the sum of the geometric series: \(T_n = 1 + 2 + 4 + \dots + 2^n = 2^{n+1} - 1\) (iii) \(L_n\) increases by 1 at each stage, so \(L_n = n + 1\)
(c) Solve \(2^{n+1} - 1 > 1000 \implies 2^{n+1} > 1001\). Since \(2^9 = 512\) and \(2^{10} = 1024\): \(n + 1 = 10 \implies n = 9\).
(d) At Stage 50, the total length is \(L_{50} = 50 + 1 = 51\) units.
评分标准
(a) B1 for correct \(B_n\) row (8, 16), B1 for correct \(T_n\) row (15, 31), B1 for correct \(L_n\) row (4, 5) (b) (i) B1 for \(2^n\) (ii) M1 for noticing the pattern or using geometric sum formula \(\frac{1(2^{n+1}-1)}{2-1}\), A1 for \(2^{n+1} - 1\) (iii) B1 for \(n+1\) (c) M1 for setting \(2^{n+1} - 1 > 1000\) or identifying powers of 2 near 1000, A1 for 9 (d) B1 for 51 (or FT from their formula in b(iii))
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