An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V3) Cambridge International A Level International Mathematics (0607) paper. Not affiliated with or reproduced from Cambridge.
卷一 (Core Non-calculator)
Answer all questions. Calculators must not be used.
17 題目 · 60.01000000000001 分
題目 1 · Short Answer
3.53 分
A shopkeeper buys a box of 40 pens for $12. She sells each pen for 45 cents. Calculate the percentage profit she makes on the box of pens.
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解題
Selling price of 40 pens is \(40 \times \$0.45 = \$18\). The profit made is \(\$18 - \$12 = \$6\). The percentage profit is \(\frac{\$6}{\$12} \times 100\% = 50\%\).
評分準則
M1 for finding the total selling price of $18. M1 for calculating the profit of $6. A1 for 50.
題目 2 · Short Answer
3.53 分
A savings account pays simple interest at a rate of 4% per year. Karl invests $350 in this account. Calculate the total amount in the account at the end of 3 years.
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解題
The simple interest earned in 3 years is \(I = \frac{350 \times 4 \times 3}{100} = \$42\). The total amount in the account is the initial investment plus the interest: \(\$350 + \$42 = \$392\).
評分準則
M1 for a correct method to find the simple interest: \(\frac{350 \times 4 \times 3}{100}\). A1 for interest of 42. A1 for total of 392.
題目 3 · Short Answer
3.53 分
Given that \(f(x) = 7 - 4x\), find the value of \(x\) when \(f(x) = -5\).
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解題
Set the function equal to \(-5\): \(7 - 4x = -5\). Subtracting 7 from both sides gives \(-4x = -12\). Dividing by \(-4\) gives \(x = 3\).
評分準則
M1 for setting up the equation \(7 - 4x = -5\). M1 for isolating the variable term to get \(4x = 12\) or \(-4x = -12\). A1 for 3.
題目 4 · Short Answer
3.53 分
Given that \(h(x) = 2x^2 - 3\), find the value of \(h(-4)\).
M1 for substituting \(-4\) to get \(2(-4)^2 - 3\). M1 for correctly squaring \(-4\) to get 16. A1 for 29.
題目 5 · Short Answer
3.53 分
Factorise fully: \(6x^2y - 9xy^2\).
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解題
The highest common factor of \(6x^2y\) and \(9xy^2\) is \(3xy\). Factoring this out from each term gives \(3xy(2x - 3y)\).
評分準則
M1 for finding a common factor of at least \(3\), \(x\), or \(y\) (e.g. \(3(2x^2y - 3xy^2)\) or \(xy(6x - 9y)\)). A1 for \(3xy(2x-3y)\).
題目 6 · Short Answer
3.53 分
Expand and simplify: \((x - 4)(x + 7)\).
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解題
Expand the brackets to get \(x^2 + 7x - 4x - 28\). Combining the like terms gives \(x^2 + 3x - 28\).
評分準則
M1 for expanding to obtain at least 3 correct terms out of 4 (e.g. \(x^2 + 7x - 4x - 28\)). A1 for \(x^2 + 3x - 28\).
題目 7 · Short Answer
3.53 分
Find the \(n\)-th term of the sequence: \(5, 11, 17, 23, 29, \dots\)
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解題
The sequence increases by 6 each time, which means the general term is of the form \(6n + c\). Using the first term where \(n=1\), we have \(6(1) + c = 5\), which gives \(c = -1\). Therefore, the \(n\)-th term is \(6n - 1\).
評分準則
M1 for finding the common difference of 6 or writing \(6n + c\). A1 for \(6n - 1\).
題目 8 · Short Answer
3.53 分
A sequence has \(n\)-th term given by \(T_n = 3n^2 - 2\). Find the 5th term of this sequence.
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解題
Substitute \(n = 5\) into the given expression: \(T_5 = 3(5)^2 - 2 = 3(25) - 2 = 75 - 2 = 73\).
評分準則
M1 for substituting \(n=5\) into the expression to get \(3(5)^2 - 2\). M1 for evaluating \(5^2 = 25\). A1 for 73.
題目 9 · Short Answer
3.53 分
Arthur changes $300 into Euros (€) when the exchange rate is €1 = $1.20. How many Euros does Arthur receive?
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解題
To convert US dollars to Euros, divide by the exchange rate of 1.20: \( \text{Euros} = \frac{300}{1.20} \) \( \frac{300}{1.20} = \frac{3000}{12} = 250 \)
Arthur receives €250.
評分準則
M1 for setting up the division: \( 300 \div 1.20 \) (or equivalent) A1 for 250
題目 10 · Short Answer
3.53 分
A shop owner buys a box of 50 notebooks for $40. She sells each notebook for $1.00. Calculate her percentage profit.
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解題
First, calculate the total selling price of the notebooks: \( 50 \times \$1.00 = \$50 \)
M1 for finding the total revenue of $50 or writing \( 50 \times 1 \) M1 for calculating the profit of $10 or writing \( \frac{10}{40} \times 100 \) A1 for 25 (accept 25%)
題目 11 · Short Answer
3.53 分
Given that \( f(x) = 3x - 5 \), find the value of \( x \) when \( f(x) = 13 \).
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解題
We are given \( f(x) = 13 \). Substitute the expression for \( f(x) \): \( 3x - 5 = 13 \)
Add 5 to both sides: \( 3x = 18 \)
Divide by 3: \( x = 6 \)
評分準則
M1 for setting up the equation: \( 3x - 5 = 13 \) A1 for 6
題目 12 · Short Answer
3.53 分
Given that \( g(x) = \frac{12}{x+1} \), find \( g(3) \).
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解題
To find \( g(3) \), substitute \( x = 3 \) into the function definition: \( g(3) = \frac{12}{3 + 1} \) \( g(3) = \frac{12}{4} = 3 \)
評分準則
M1 for substituting \( x = 3 \) into the expression: \( \frac{12}{3+1} \) A1 for 3
題目 13 · Short Answer
3.53 分
Factorise completely: \( 6ax - 9ay \)
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解題
Identify the highest common factor of the terms \( 6ax \) and \( 9ay \). The highest common factor of 6 and 9 is 3. The highest common factor of the algebraic variables is \( a \). Therefore, the overall highest common factor is \( 3a \).
Factorise \( 3a \) out of each term: \( 6ax - 9ay = 3a(2x - 3y) \)
評分準則
M1 for a correct partial factorisation (e.g. \( 3(2ax - 3ay) \) or \( a(6x - 9y) \)) or identifying \( 3a \) as the common factor A1 for \( 3a(2x - 3y) \)
題目 14 · Short Answer
3.53 分
Expand and simplify: \( (2x - 3)(x + 5) \)
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解題
To expand the brackets, multiply each term in the first bracket by each term in the second bracket: \( (2x - 3)(x + 5) = 2x(x) + 2x(5) - 3(x) - 3(5) \) \( = 2x^2 + 10x - 3x - 15 \)
Combine the like terms: \( = 2x^2 + 7x - 15 \)
評分準則
M1 for expanding to get at least 3 terms correct of: \( 2x^2 + 10x - 3x - 15 \) A1 for \( 2x^2 + 7x - 15 \)
題目 15 · Short Answer
3.53 分
Find the \( n \)-th term of the sequence: \( 5, 11, 17, 23, 29, \dots \)
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解題
Find the difference between consecutive terms: \( 11 - 5 = 6 \) \( 17 - 11 = 6 \)
Since the difference is constant, this is an arithmetic sequence with common difference \( d = 6 \). The formula for the \( n \)-th term of a linear sequence is \( dn + c \), which gives \( 6n + c \).
Using the first term \( (n = 1) \): \( 6(1) + c = 5 \) \( c = -1 \)
So, the \( n \)-th term of the sequence is \( 6n - 1 \).
評分準則
M1 for identifying the common difference is 6 (e.g. finding \( 6n \) as part of the formula) A1 for \( 6n - 1 \)
題目 16 · Short Answer
3.53 分
The \( n \)-th term of a sequence is \( n^2 + 3 \). Find the 8th term of this sequence.
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解題
To find the 8th term of the sequence, substitute \( n = 8 \) into the formula for the \( n \)-th term: \( \text{8th term} = 8^2 + 3 \) \( = 64 + 3 \) \( = 67 \)
評分準則
M1 for substituting \( n = 8 \) into the expression: \( 8^2 + 3 \) A1 for 67
題目 17 · short_answer
3.53 分
The first four terms of a sequence are \(4, 11, 18, 25, \dots\). Find an expression, in terms of \(n\), for the \(n\)-th term of this sequence.
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解題
Find the difference between consecutive terms: \(11 - 4 = 7\), \(18 - 11 = 7\), and \(25 - 18 = 7\). The common difference is \(7\), so the term-to-term rule is to add \(7\). This means the formula involves \(7n\). For \(n = 1\), \(7(1) = 7\). To find the term for \(n = 1\), we must subtract \(3\) from \(7\) to get the first term of \(4\). Thus, the \(n\)-th term of the sequence is \(7n - 3\).
評分準則
M1 for identifying the common difference of \(7\) (e.g. writing \(7n + c\) or showing a constant difference of \(7\)). A1 for \(7n - 3\) or equivalent.
卷二 (Extended Non-calculator)
Answer all questions. Calculators must not be used.
21 題目 · 74.96999999999997 分
題目 1 · Short Answer
3.57 分
A tourist changes €500 into Dollars ($) when the exchange rate is €1 = $1.24. She spends $380 of this money while on holiday. She then changes all of her remaining dollars back into Euros when the exchange rate is €1 = $1.20. Calculate the amount of Euros she receives back.
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解題
1. Convert €500 to Dollars: \(500 \times 1.24 = 620\) dollars. 2. Subtract the amount spent: \(620 - 380 = 240\) dollars. 3. Convert $240 back to Euros: \(240 \div 1.20 = 200\) Euros.
評分準則
M1 for \(500 \times 1.24\) (or 620 seen) [1.19 marks]. M1 for subtracting 380 from their dollar amount (or 240 seen) [1.19 marks]. A1 for 200 [1.19 marks].
題目 2 · Short Answer
3.57 分
The function \(f(x)\) is defined as \(f(x) = \frac{2x + 5}{3x - 4}\) for \(x \neq \frac{4}{3}\). Find an expression for \(f^{-1}(x)\).
M1 for attempting to multiply by the denominator to clear the fraction [1.19 marks]. M1 for collecting terms in \(x\) on one side and factorizing [1.19 marks]. A1 for correct expression \(\frac{4x + 5}{3x - 2}\) [1.19 marks].
Factorize the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Factorize the denominator using the difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Simplify by dividing both numerator and denominator by the common factor \((2x + 1)\): \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}\).
評分準則
M1 for factorizing numerator into \((2x + 1)(x - 3)\) [1.19 marks]. M1 for factorizing denominator into \((2x - 1)(2x + 1)\) [1.19 marks]. A1 for \(\frac{x - 3}{2x - 1}\) [1.19 marks].
題目 4 · Short Answer
3.57 分
Find the \(n\)-th term of the sequence: \(3, 8, 15, 24, 35, \dots\)
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解題
Find the first differences: \(5, 7, 9, 11\). Find the second differences: \(2, 2, 2\). Since the second difference is constant, the sequence is quadratic of the form \(an^2 + bn + c\) with \(2a = 2\), so \(a = 1\). Subtracting \(n^2\) from each term of the sequence gives: \(3 - 1^2 = 2\), \(8 - 2^2 = 4\), \(15 - 3^2 = 6\), \(24 - 4^2 = 8\), \(35 - 5^2 = 10\). The sequence of differences is \(2, 4, 6, 8, 10\), which has the \(n\)-th term \(2n\). Thus, the overall \(n\)-th term is \(n^2 + 2n\).
評分準則
M1 for finding second differences are constant (2) [1.19 marks]. M1 for subtracting \(n^2\) and finding linear part is \(2n\) [1.19 marks]. A1 for \(n^2 + 2n\) or \(n(n+2)\) [1.19 marks].
題目 5 · Short Answer
3.57 分
A shopkeeper sells a camera for $276, making a profit of 15% on the cost price. Find the cost price of the camera.
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解題
Let the cost price be \(C\). The selling price is \(1.15 \times C = 276\). Therefore, \(C = \frac{276}{1.15} = \frac{27600}{115}\). Dividing numerator and denominator by 5 gives \(\frac{5520}{23}\). Performing the division: \(5520 \div 23 = 240\). The cost price is $240.
評分準則
M1 for setting up \(1.15C = 276\) [1.19 marks]. M1 for evaluating \(\frac{27600}{115}\) or \(\frac{5520}{23}\) [1.19 marks]. A1 for 240 [1.19 marks].
題目 6 · Short Answer
3.57 分
Solve the equation: \(3^{2x - 1} = 27^{x - 2}\)
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解題
Write 27 as a power of 3: \(27 = 3^3\). The equation becomes \(3^{2x - 1} = (3^3)^{x - 2}\), which simplifies to \(3^{2x - 1} = 3^{3x - 6}\). Equating the indices: \(2x - 1 = 3x - 6\). Solving for \(x\): \(3x - 2x = 6 - 1\) which gives \(x = 5\).
評分準則
M1 for writing 27 as \(3^3\) or equivalent base 3 representation [1.19 marks]. M1 for equating the exponents: \(2x - 1 = 3(x - 2)\) [1.19 marks]. A1 for 5 [1.19 marks].
題目 7 · Short Answer
3.57 分
Rearrange the formula to make \(t\) the subject: \(w = \frac{3t + 2}{5 - t}\)
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解題
Multiply both sides by \((5 - t)\): \(w(5 - t) = 3t + 2\). Expand the brackets: \(5w - wt = 3t + 2\). Rearrange to group all terms with \(t\) on one side: \(5w - 2 = 3t + wt\). Factor out \(t\): \(5w - 2 = t(w + 3)\). Divide by \((w + 3)\): \(t = \frac{5w - 2}{w + 3}\).
評分準則
M1 for clearing the fraction to obtain \(w(5 - t) = 3t + 2\) [1.19 marks]. M1 for isolating terms in \(t\) and factorizing [1.19 marks]. A1 for \(t = \frac{5w - 2}{w + 3}\) or equivalent [1.19 marks].
題目 8 · Short Answer
3.57 分
The \(n\)-th term of a sequence is given by \(T_n = a \cdot 2^{n-1} + b\). Given that \(T_1 = 7\) and \(T_2 = 11\), calculate the value of \(T_4\).
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解題
Using \(T_1 = 7\): \(a \cdot 2^{1-1} + b = 7 \implies a + b = 7\). Using \(T_2 = 11\): \(a \cdot 2^{2-1} + b = 11 \implies 2a + b = 11\). Subtracting the first equation from the second gives: \(a = 4\). Substituting \(a = 4\) into the first equation: \(4 + b = 7 \implies b = 3\). Therefore, the formula is \(T_n = 4 \cdot 2^{n-1} + 3\). To find \(T_4\): \(T_4 = 4 \cdot 2^{4-1} + 3 = 4 \cdot 2^3 + 3 = 4 \cdot 8 + 3 = 32 + 3 = 35\).
評分準則
M1 for setting up equations \(a + b = 7\) and \(2a + b = 11\) [1.19 marks]. M1 for finding \(a = 4\) and \(b = 3\) [1.19 marks]. A1 for 35 [1.19 marks].
題目 9 · Short Answer
3.57 分
An investment of \(\$400\) earns compound interest at a rate of \(3\%\) per year. Calculate the total interest earned at the end of 2 years.
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解題
At the end of the first year, the interest earned is \(3\%\) of \(\$400\): \text{Interest Year 1} = 400 \times 0.03 = 12\text{ dollars}. The value of the investment at the start of the second year is: 400 + 12 = 412\text{ dollars}. At the end of the second year, the interest earned is \(3\%\) of \(\$412\): \text{Interest Year 2} = 412 \times 0.03 = 12.36\text{ dollars}. The total interest earned over the 2 years is: 12 + 12.36 = 24.36\text{ dollars}.
評分準則
M1 for finding the first year interest of \(12\) or for multiplying by \(1.03\) M1 for finding the second year interest of \(12.36\) or for evaluating \(400 \times 1.03^2 - 400\) A1 for the correct answer \(24.36\)
題目 10 · Short Answer
3.57 分
A book is sold in London for \(\pounds15\) and in New York for \(\$24\). The exchange rate is \(\pounds1 = \$1.45\). Calculate the difference in price of the book in dollars. Give your answer in dollars.
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解題
First, convert the price in London from pounds to dollars: \text{Price in dollars} = 15 \times 1.45 = 21.75\text{ dollars}. Next, find the difference between this and the price in New York: \text{Difference} = 24 - 21.75 = 2.25\text{ dollars}.
評分準則
M1 for an attempt to convert \(\pounds15\) to dollars by multiplying by \(1.45\) A1 for getting the converted price of \(21.75\) A1 for the correct final difference of \(2.25\)
題目 11 · Short Answer
3.57 分
Given that \(f(x) = \frac{3x - 5}{2x + 1}\) for \(x \neq -\frac{1}{2}\), find \(f^{-1}(x)\).
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解題
Let \(y = \frac{3x - 5}{2x + 1}\). Multiply both sides by \(2x + 1\): \(y(2x + 1) = 3x - 5\) \(2xy + y = 3x - 5\) Isolate the terms with \(x\) on one side: \(2xy - 3x = -y - 5\) \(x(2y - 3) = -(y + 5)\) \(x = \frac{-(y + 5)}{2y - 3} = \frac{y + 5}{3 - 2y}\). Replacing \(y\) with \(x\), we get: \(f^{-1}(x) = \frac{x + 5}{3 - 2x}\).
評分準則
M1 for setting \(y = \frac{3x-5}{2x+1}\) and multiplying to clear the fraction M1 for rearranging to group terms with \(x\) on one side and factorising A1 for the correct inverse function \(\frac{x+5}{3-2x}\) (or equivalent form such as \(\frac{-x-5}{2x-3}\))
題目 12 · Short Answer
3.57 分
The graph of \(y = f(x)\) is transformed to the graph of \(y = g(x)\) by a translation of vector \(\begin{pmatrix} 2 \\ 0 \end{pmatrix}\) followed by a stretch parallel to the \(y\)-axis with scale factor 4. Find an expression for \(g(x)\) in terms of \(f(x)\).
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解題
A translation of vector \(\begin{pmatrix} 2 \\ 0 \end{pmatrix}\) shifts the graph to the right by 2 units, which transforms the function to \(f(x - 2)\). A stretch parallel to the \(y\)-axis with scale factor 4 multiplies the entire function by 4, which transforms it to \(4f(x - 2)\). Thus, \(g(x) = 4f(x - 2)\).
評分準則
M1 for identifying that the translation results in \(f(x - 2)\) M1 for applying the stretch factor of 4 to the function A1 for the correct expression \(4f(x - 2)\)
題目 13 · Short Answer
3.57 分
Factorise completely: \(6x^2 - 5xy - 6y^2\)
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解題
We need to find two numbers that multiply to \(6 \times (-6) = -36\) and add up to \(-5\). These numbers are \(-9\) and \(4\). Rewrite the expression by splitting the middle term: \(6x^2 - 9xy + 4xy - 6y^2\) Factorise by grouping: \(3x(2x - 3y) + 2y(2x - 3y)\) Factorise out the common bracket: \((3x + 2y)(2x - 3y)\).
評分準則
M1 for splitting the middle term correctly as \(-9xy + 4xy\) or writing a product of two linear binomials that gives two of the three terms correct on expansion A1 for the fully factorised correct answer \((3x + 2y)(2x - 3y)\) (or in reverse order)
題目 14 · Short Answer
3.57 分
Rearrange the formula to make \(x\) the subject: \(t = \sqrt{\frac{x+3}{x-2}}\)
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解題
Square both sides to eliminate the square root: \(t^2 = \frac{x+3}{x-2}\) Multiply both sides by \((x-2)\): \(t^2(x - 2) = x + 3\) \(t^2 x - 2t^2 = x + 3\) Rearrange the equation to group all terms containing \(x\) on one side: \(t^2 x - x = 2t^2 + 3\) Factorise out \(x\): \(x(t^2 - 1) = 2t^2 + 3\) Divide both sides by \(t^2 - 1\): \(x = \frac{2t^2 + 3}{t^2 - 1}\).
評分準則
M1 for squaring both sides correctly: \(t^2 = \frac{x+3}{x-2}\) M1 for multiplying out brackets and collecting terms in \(x\) on one side A1 for the correct final rearranged formula \(x = \frac{2t^2 + 3}{t^2 - 1}\)
題目 15 · Short Answer
3.57 分
Find the \(n\)-th term of the sequence: \(4, 14, 30, 52, \dots\)
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解題
Let's find the first differences between consecutive terms: \(14 - 4 = 10\) \(30 - 14 = 16\) \(52 - 30 = 22\) Now find the second differences: \(16 - 10 = 6\) \(22 - 16 = 6\) Since the second differences are constant and equal to \(6\), the sequence has a quadratic term of the form \(an^2\), where \(2a = 6 \implies a = 3\). Subtract \(3n^2\) from the original sequence to find the remaining linear sequence: - For \(n=1\): \(4 - 3(1^2) = 1\) - For \(n=2\): \(14 - 3(2^2) = 2\) - For \(n=3\): \(30 - 3(3^2) = 3\) - For \(n=4\): \(52 - 3(4^2) = 4\) The remaining sequence \(1, 2, 3, 4, \dots\) has the \(n\)-th term of \(n\). Combining both parts, the \(n\)-th term of the original sequence is \(3n^2 + n\).
評分準則
M1 for finding that the second difference is \(6\) M1 for identifying the term \(3n^2\) or showing a method to solve a system of equations for \(a\), \(b\), and \(c\) A1 for the correct \(n\)-th term expression \(3n^2 + n\)
題目 16 · Short Answer
3.57 分
Find the \(n\)-th term of the sequence: \(\frac{2}{3}, \frac{5}{9}, \frac{8}{27}, \frac{11}{81}, \dots\)
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解題
Let's find the \(n\)-th term of the numerator and the denominator separately. For the numerators: \(2, 5, 8, 11, \dots\) This is an arithmetic sequence with a first term \(a = 2\) and a common difference \(d = 3\). The \(n\)-th term is: \(2 + (n-1)3 = 3n - 1\). For the denominators: \(3, 9, 27, 81, \dots\) This is a geometric sequence with a first term of \(3\) and a common ratio of \(3\). The \(n\)-th term is: \(3^n\). Combining these gives the \(n\)-th term of the sequence of fractions: \(\frac{3n - 1}{3^n}\).
評分準則
M1 for finding the correct \(n\)-th term of the numerators is \(3n - 1\) M1 for finding the correct \(n\)-th term of the denominators is \(3^n\) A1 for combining both to get the correct general term \(\frac{3n - 1}{3^n}\)
題目 17 · Short Answer
3.57 分
A bank exchange rate is $1 = 0.85 Euros. A traveler exchanges $600 into Euros. They spend 340 Euros and exchange the remaining Euros back to dollars at a rate of 1 Euro = $1.15. Calculate the final amount in dollars they have.
M1 for exchanging to Euros to get 510. M1 for subtracting 340 from their Euros. A1 for the final answer 195.50.
題目 18 · Short Answer
3.57 分
Given that \(f(x) = \frac{2x + 3}{x - 1}\) for \(x \neq 1\), find an expression for \(f^{-1}(x)\).
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解題
Let \(y = \frac{2x+3}{x-1}\). Multiply both sides by \(x-1\): \(y(x-1) = 2x+3 \implies yx - y = 2x+3\). Rearrange to group terms with \(x\): \(yx - 2x = y+3\). Factorise \(x\): \(x(y-2) = y+3\). Divide by \(y-2\): \(x = \frac{y+3}{y-2}\). Therefore, \(f^{-1}(x) = \frac{x+3}{x-2}\).
評分準則
M1 for setting \(y = f(x)\) and multiplying by \(x-1\). M1 for isolating terms with \(x\). A1 for the correct inverse function.
題目 19 · Short Answer
3.57 分
Factorise completely: \(12a^2b - 3b^3\).
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解題
First, factorise out the common factor \(3b\): \(12a^2b - 3b^3 = 3b(4a^2 - b^2)\). Next, recognise that \(4a^2 - b^2\) is a difference of two squares: \(4a^2 - b^2 = (2a-b)(2a+b)\). Thus, the fully factorised expression is \(3b(2a-b)(2a+b)\).
評分準則
M1 for extracting the common factor \(3b(4a^2 - b^2)\). M1 for factorising the difference of two squares \((2a-b)(2a+b)\). A1 for the final correct factorised expression.
題目 20 · Short Answer
3.57 分
The \(n\)-th term of a sequence is given by \(T_n = an^2 + bn - 3\). Given that the second term \(T_2 = 13\) and the third term \(T_3 = 30\), find the value of \(a\) and the value of \(b\).
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解題
Using the formula for \(T_n\): For \(T_2 = 13\): \(a(2)^2 + b(2) - 3 = 13 \implies 4a + 2b = 16 \implies 2a + b = 8\) (Equation 1). For \(T_3 = 30\): \(a(3)^2 + b(3) - 3 = 30 \implies 9a + 3b = 33 \implies 3a + b = 11\) (Equation 2). Subtract Equation 1 from Equation 2: \((3a + b) - (2a + b) = 11 - 8 \implies a = 3\). Substitute \(a = 3\) into Equation 1: \(2(3) + b = 8 \implies 6 + b = 8 \implies b = 2\). Thus, \(a = 3\) and \(b = 2\).
評分準則
M1 for setting up both equations in \(a\) and \(b\). M1 for solving the simultaneous equations. A1 for both correct values.
題目 21 · Short Answer
3.57 分
The graph of \(y = f(x)\) is mapped onto the graph of \(y = g(x)\) by a stretch, parallel to the \(y\)-axis, with scale factor 3, followed by a translation of \(\begin{pmatrix} 0 \\ -5 \end{pmatrix}\). Write an expression for \(g(x)\) in terms of \(f(x)\).
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解題
A stretch parallel to the \(y\)-axis with scale factor 3 transforms the function from \(f(x)\) to \(3f(x)\). A translation of \(\begin{pmatrix} 0 \\ -5 \end{pmatrix}\) moves the graph vertically downwards by 5 units, which transforms \(3f(x)\) to \(3f(x) - 5\). Thus, \(g(x) = 3f(x) - 5\).
評分準則
M1 for \(3f(x)\) from the stretch. M1 for subtracting 5 from the translation. A1 for the correct expression \(3f(x) - 5\).
Paper 3 (Core Calculator)
Answer all questions. GDC should be used where appropriate.
18 題目 · 59.939999999999976 分
題目 1 · short_answer
3.33 分
Elena invests \(\$4500\) at a rate of \(3.2\%\) per year compound interest. Calculate the total value of her investment at the end of 5 years. Give your answer to the nearest cent.
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解題
To find the total value of the compound interest investment, we use the formula: \(A = P \left(1 + \frac{r}{100}\right)^n\) where \(P = 4500\), \(r = 3.2\), and \(n = 5\). Substituting these values gives: \(A = 4500 \left(1 + \frac{3.2}{100}\right)^5 = 4500 \times (1.032)^5\). Using a calculator: \(A \approx 4500 \times 1.17057288 = 5267.5783\). Rounding to the nearest cent gives \(\$5267.58\).
評分準則
M1 for setting up the compound interest formula: \(4500 \times (1.032)^5\) M1 for reaching a value of \(5267.58\) or \(5267.578...\) A1 for the final answer rounded correctly to the nearest cent: \(5267.58\)
題目 2 · short_answer
3.33 分
A shopkeeper buys a box of 50 calculators for \(\$400\). He sells 40 of them at \(\$11\) each, and the remaining 10 at \(\$6\) each. Calculate his percentage profit on the cost price.
M1 for calculating total selling price: \(40 \times 11 + 10 \times 6 = 500\) M1 for calculating profit: \(500 - 400 = 100\) or setting up percentage profit: \(\frac{500 - 400}{400} \times 100\) A1 for correct final answer: \(25\)
題目 3 · short_answer
3.33 分
Given the function \(f(x) = \frac{3x + 8}{2}\), find the value of \(x\) when \(f(x) = 13\).
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解題
Set up the equation: \(\frac{3x + 8}{2} = 13\). Multiply both sides by 2: \(3x + 8 = 26\). Subtract 8 from both sides: \(3x = 18\). Divide by 3: \(x = 6\).
評分準則
M1 for setting up the equation: \(\frac{3x + 8}{2} = 13\) M1 for multiplying by 2 and subtracting 8: \(3x = 18\) A1 for the correct value: \(6\)
題目 4 · short_answer
3.33 分
The graph of the quadratic function \(y = x^2 - 4x - 12\) crosses the \(x\)-axis at two points. Find the positive \(x\)-coordinate where the graph crosses the \(x\)-axis.
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解題
The graph crosses the \(x\)-axis where \(y = 0\). Set up the equation: \(x^2 - 4x - 12 = 0\). Factorise the quadratic expression: \((x - 6)(x + 2) = 0\). This gives two solutions: \(x = 6\) or \(x = -2\). The positive \(x\)-coordinate is \(6\).
評分準則
M1 for setting \(y = 0\): \(x^2 - 4x - 12 = 0\) M1 for factorising to \((x - 6)(x + 2)\) or using the quadratic formula correctly A1 for selecting the positive root: \(6\)
題目 5 · short_answer
3.33 分
Expand and simplify the expression: \(4(2x - 3) - 3(x - 5)\)
M1 for expanding the first bracket: \(8x - 12\) M1 for expanding the second bracket: \(-3x + 15\) A1 for the correct simplified expression: \(5x + 3\)
題目 6 · short_answer
3.33 分
Solve the simultaneous equations to find the value of \(x\): \(3x + 2y = 12\) and \(4x - y = 5\)
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解題
Multiply the second equation by 2 to get \(8x - 2y = 10\). Add this to the first equation: \((3x + 2y) + (8x - 2y) = 12 + 10\), which simplifies to \(11x = 22\). Divide by 11 to find \(x = 2\).
評分準則
M1 for attempting to equate coefficients or make a substitution M1 for eliminating one variable to get a linear equation in one variable (e.g., \(11x = 22\)) A1 for the correct value of \(x\): \(2\)
題目 7 · short_answer
3.33 分
Here are the first five terms of an arithmetic sequence: \(4, 11, 18, 25, 32\). Find an expression for the \(n\)-th term of this sequence.
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解題
Find the common difference between consecutive terms: \(11 - 4 = 7\) and \(18 - 11 = 7\), so \(d = 7\). The formula for the \(n\)-th term of an arithmetic sequence is \(T_n = a + (n - 1)d\) where the first term is \(a = 4\). Substituting the values gives: \(T_n = 4 + 7(n - 1) = 4 + 7n - 7 = 7n - 3\).
評分準則
M1 for finding the common difference of \(7\) or writing \(7n + c\) M1 for substituting \(n=1\) to find \(c\): \(7(1) + c = 4\) which gives \(c = -3\) A1 for the correct expression: \(7n - 3\)
題目 8 · short_answer
3.33 分
The \(n\)-th term of a sequence is given by \(T_n = 2n^2 - 5\). Find the 12th term of this sequence.
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解題
To find the 12th term of the sequence, we substitute \(n = 12\) into the given formula: \(T_{12} = 2(12)^2 - 5\). Calculate the square of 12: \(12^2 = 144\). Multiply by 2: \(2 \times 144 = 288\). Subtract 5: \(288 - 5 = 283\).
評分準則
M1 for substituting \(n = 12\) into the formula: \(2(12)^2 - 5\) M1 for calculating \(2 \times 144 - 5\) A1 for the correct answer: \(283\)
題目 9 · Short Answer
3.33 分
Amelia changes €550 into Japanese Yen (¥). The bank charges a 2% commission fee, which is subtracted from the €550 before the conversion. The exchange rate is €1 = ¥162.40. Calculate how many Japanese Yen Amelia receives, giving your answer to the nearest Yen.
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解題
First, calculate the amount after commission: \(550 \times (1 - 0.02) = 550 \times 0.98 = 539\) Euros. Next, convert this amount to Japanese Yen: \(539 \times 162.40 = 87533.6\) Yen. Rounding to the nearest Yen gives 87534 Yen.
評分準則
M1 for calculating the amount after commission: \(550 \times 0.98 = 539\) M1 for multiplying their amount by 162.40 A1 for 87534
題目 10 · Short Answer
3.33 分
Liam invests $4500 at a rate of 3.5% per year compound interest. Calculate the total value of his investment at the end of 4 years. Give your answer to the nearest cent.
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解題
Using the compound interest formula \(A = P(1 + r/100)^n\), we get: \(A = 4500 \times (1 + 0.035)^4 = 4500 \times (1.035)^4 \approx 5163.85354...\) Rounding to the nearest cent gives $5163.85.
評分準則
M1 for \(4500 \times 1.035^4\) (or equivalent step-by-step compound interest calculation) A1 for 5163.85
題目 11 · Short Answer
3.33 分
An arithmetic sequence has first term 7 and common difference 4. Find the 50th term of this sequence.
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解題
The \(n\)-th term of an arithmetic sequence is given by \(u_n = a + (n - 1)d\). Here, \(a = 7\) and \(d = 4\). For the 50th term: \(u_{50} = 7 + (50 - 1) \times 4 = 7 + 49 \times 4 = 7 + 196 = 203\).
評分準則
M1 for setting up the expression \(7 + (50-1) \times 4\) or finding the general term \(4n + 3\) M1 for calculating \(4 \times 50 + 3\) or \(7 + 196\) A1 for 203
題目 12 · Short Answer
3.33 分
Find the \(n\)-th term of the sequence: 5, 12, 21, 32, ...
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解題
The sequence is 5, 12, 21, 32, ... First differences: 7, 9, 11 Second differences: 2, 2 Since the second differences are constant, the sequence is quadratic with leading term \(\frac{2}{2}n^2 = n^2\). Subtracting \(n^2\) from the sequence terms: \(5 - 1^2 = 4\) \(12 - 2^2 = 8\) \(21 - 3^2 = 12\) \(32 - 4^2 = 16\) The differences form a linear sequence \(4, 8, 12, 16, \dots\), which is \(4n\). Therefore, the \(n\)-th term is \(n^2 + 4n\).
評分準則
M1 for finding second differences are 2 M1 for identifying the \(n^2\) component A1 for \(n^2 + 4n\)
題目 13 · Short Answer
3.33 分
Simplify fully: \(4(2x - 3) - 3(x - 5)\)
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解題
First, expand each set of brackets: \(4(2x - 3) = 8x - 12\) \(-3(x - 5) = -3x + 15\) Combine the like terms: \(8x - 12 - 3x + 15 = (8x - 3x) + (-12 + 15) = 5x + 3\).
評分準則
M1 for \(8x - 12\) or \(-3x + 15\) M1 for grouping like terms A1 for \(5x + 3\)
題目 14 · Short Answer
3.33 分
Factorise fully: \(12a^2b - 18ab^2\)
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解題
First, find the highest common factor (HCF) of the coefficients 12 and 18, which is 6. Next, find the highest common power of \(a\) and \(b\) in both terms, which is \(ab\). So, the common factor is \(6ab\). Divide both terms by \(6ab\) to find the terms inside the brackets: \(\frac{12a^2b}{6ab} = 2a\) \(\frac{-18ab^2}{6ab} = -3b\) Thus, the fully factorised expression is \(6ab(2a - 3b)\).
評分準則
M1 for identifying a common factor of \(ab\) or \(6a\) or \(6b\) M1 for partial factorisation like \(6(2a^2b - 3ab^2)\) or \(ab(12a - 18b)\) A1 for \(6ab(2a - 3b)\)
題目 15 · Short Answer
3.33 分
Given the function \(f(x) = 3x^2 - 5x + 2\), find the positive value of \(x\) for which \(f(x) = 10\). Give your answer as a simplified fraction.
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解題
Set the function equal to 10: \(3x^2 - 5x + 2 = 10\) Subtract 10 from both sides: \(3x^2 - 5x - 8 = 0\) Factorise the quadratic equation: \((3x - 8)(x + 1) = 0\) This gives two solutions: \(3x - 8 = 0 \implies x = \frac{8}{3}\) \(x + 1 = 0 \implies x = -1\) Since we need the positive value, the answer is \(\frac{8}{3}\).
評分準則
M1 for setting up the equation \(3x^2 - 5x - 8 = 0\) M1 for factorising or using the quadratic formula to find the roots A1 for \(8/3\)
題目 16 · Short Answer
3.33 分
Find the inverse function \(g^{-1}(x)\) of \(g(x) = \frac{4x - 3}{5}\).
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解題
Let \(y = g(x)\): \(y = \frac{4x - 3}{5}\) Rearrange the equation to make \(x\) the subject: \(5y = 4x - 3\) \(5y + 3 = 4x\) \(x = \frac{5y + 3}{4}\) Replace \(y\) with \(x\) to write the inverse function: \(g^{-1}(x) = \frac{5x + 3}{4}\).
評分準則
M1 for writing \(y = \frac{4x - 3}{5}\) and attempting to rearrange M1 for \(5y = 4x - 3\) or \(5x = 4y - 3\) A1 for \((5x+3)/4\)
題目 17 · Short/Medium Answer
3.33 分
Lia invests \(\$4500\) at a rate of \(2.3\%\) per year simple interest. Calculate the total value of her investment after 6 years.
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解題
To calculate the simple interest earned, use the simple interest formula: \(I = \frac{P \times R \times T}{100}\)
Substitute the given values into the formula: \(I = \frac{4500 \times 2.3 \times 6}{100} = 45 \times 13.8 = 621\)
The simple interest earned is \(\$621\).
To find the total value of the investment, add the interest to the principal: \(\text{Total Value} = 4500 + 621 = 5121\)
評分準則
M1 for a correct method to find the simple interest: \(\frac{4500 \times 2.3 \times 6}{100}\) (or 621) M1 for adding their interest to the principal: \(4500 + 621\) A1.33 for 5121
題目 18 · Short/Medium Answer
3.33 分
Here are the first five terms of a sequence:
\[8, \quad 13, \quad 18, \quad 23, \quad 28\]
Find an expression for the \(n\)-th term of this sequence.
Since the difference is constant, this is an arithmetic sequence with common difference \(d = 5\).
The \(n\)-th term of an arithmetic sequence can be written in the form \(dn + c\), where \(d\) is the common difference. So, the \(n\)-th term is \(5n + c\).
To find \(c\), use the first term where \(n = 1\): \(5(1) + c = 8\) \(5 + c = 8\) \(c = 3\)
Therefore, the expression for the \(n\)-th term is \(5n + 3\).
評分準則
M1 for finding the common difference of 5, or writing \(5n\) or \(5n + c\) M1 for setting up an equation to find the constant term: e.g., \(5(1) + c = 8\) or finding \(c = 3\) A1.33 for \(5n + 3\) (or equivalent, e.g., \(3 + 5n\))
Paper 4 (Extended Calculator)
Answer all questions. GDC should be used where appropriate.
18 題目 · 75.06000000000002 分
題目 1 · Medium/Long Answer
4.17 分
Aravind invests $5000 USD in a savings account with an interest rate of 3.2% per year, compounded quarterly. After 5 years, he withdraws the entire amount and converts it to Euros (EUR). The exchange rate is 1 USD = 0.92 EUR, and the bank charges a 1.5% commission on the converted amount in EUR. Calculate the final amount of Euros Aravind receives. Give your answer to 2 decimal places.
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解題
First, calculate the value in USD after 5 years of quarterly compounding: \(A = P \left(1 + \frac{r}{n}\right)^{nt}\) \(A = 5000 \left(1 + \frac{0.032}{4}\right)^{4 \times 5}\) \(A = 5000 (1.008)^{20} \approx 5864.4396\) USD.
Next, convert this amount to Euros (EUR) before commission: \(5864.4396 \times 0.92 \approx 5395.2845\) EUR.
Apply the 1.5% commission, meaning he receives 98.5% of the converted amount: \(5395.2845 \times (1 - 0.015) = 5395.2845 \times 0.985 \approx 5314.3552\) EUR.
Rounding to 2 decimal places gives \(5314.36\) EUR.
評分準則
M1: For calculating the compound interest amount \(5000 \times (1.008)^{20}\) or reaching \(5864.44\) M1: For multiplying by the exchange rate of 0.92 M1: For applying the 1.5% commission (multiplying by 0.985) A1: For the final answer of 5314.36 (accept 5314.35 due to intermediate rounding)
題目 2 · Medium/Long Answer
4.17 分
The functions \(f(x)\) and \(g(x)\) are defined as follows: \(f(x) = \frac{2x - 3}{x + 4}\) for \(x \neq -4\) \(g(x) = 3x + 1\)
Find the value of \(x\) for which \(f^{-1}(x) = g(2)\).
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解題
First, find the value of \(g(2)\): \(g(2) = 3(2) + 1 = 7\)
We need to find \(x\) such that \(f^{-1}(x) = 7\). By the definition of an inverse function, this is equivalent to solving: \(f(7) = x\)
M1: For finding \(g(2) = 7\) M1: For recognizing that \(f^{-1}(x) = 7\) is equivalent to \(f(7) = x\) (or for finding \(f^{-1}(x) = \frac{4x+3}{2-x}\)) M1: For evaluating \(\frac{2(7)-3}{7+4}\) (or setting up \(\frac{4x+3}{2-x} = 7\) and solving) A1: For \(x = 1\)
To solve \(\frac{3}{2x - 1} + \frac{4}{x + 3} = 1\), we first express the left-hand side over a common denominator: \(\frac{3(x + 3) + 4(2x - 1)}{(2x - 1)(x + 3)} = 1\)
This yields the solutions: \(x = 4\) or \(x = -1\)
評分準則
M1: For combining into a single fraction with a correct numerator \(11x + 5\) M1: For expanding the denominator correctly to \(2x^2 + 5x - 3\) M1: For rearranging into the quadratic form \(2x^2 - 6x - 8 = 0\) (or equivalent) A1: For obtaining both correct solutions: \(x = 4\) and \(x = -1\)
題目 4 · Medium/Long Answer
4.17 分
The first five terms of a sequence are: 5, 14, 29, 50, 77, ...
Find an expression, in terms of \(n\), for the \(n\)-th term of this sequence.
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解題
Let's find the first differences between consecutive terms: 14 - 5 = 9 29 - 14 = 15 50 - 29 = 21 77 - 50 = 27
Since the second difference is constant (6), the sequence is quadratic and has the general form \(an^2 + bn + c\), where: \(2a = 6 \implies a = 3\).
Subtract \(3n^2\) from the terms of the original sequence to find the linear part: For \(n = 1\): \(5 - 3(1)^2 = 2\) For \(n = 2\): \(14 - 3(2)^2 = 14 - 12 = 2\) For \(n = 3\): \(29 - 3(3)^2 = 29 - 27 = 2\) For \(n = 4\): \(50 - 3(4)^2 = 50 - 48 = 2\)
Since the remainder is a constant value of 2, the \(n\)-th term of the sequence is: \(3n^2 + 2\).
評分準則
M1: For calculating second differences of 6 M1: For identifying that \(a = 3\), hence finding the term \(3n^2\) M1: For a systematic subtraction of \(3n^2\) from terms to find the constant term 2 A1: For the correct expression \(3n^2 + 2\)
題目 5 · Medium/Long Answer
4.17 分
A machine is purchased for $12,000 and depreciates at a rate of 8% per year compounded annually. At the same time, $8,000 is invested in a savings account earning 4.5% simple interest per year. Calculate the positive difference between the value of the savings account and the value of the machine after 6 years. Give your answer to 2 decimal places.
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解題
1. Value of the machine after 6 years under annual compounding depreciation: \(V_{\text{machine}} = P(1 - r)^t = 12000(1 - 0.08)^6\) \(V_{\text{machine}} = 12000(0.92)^6 \approx 7276.2597\) USD.
2. Value of the savings account after 6 years under simple interest: Interest earned: \(I = P \times R \times T = 8000 \times 0.045 \times 6 = 2160\) USD. Total value of savings account: \(V_{\text{savings}} = 8000 + 2160 = 10160\) USD.
3. Positive difference between the two values: \(\text{Difference} = V_{\text{savings}} - V_{\text{machine}} = 10160 - 7276.2597 \approx 2883.7403\) USD.
Rounding to 2 decimal places, the difference is $2883.74.
評分準則
M1: For the correct depreciation formula and values: \(12000 \times 0.92^6\) A1: For the correct machine value of \(7276.26\) (or \(7276.259...\)) M1: For the correct simple interest calculation: \(8000 + (8000 \times 0.045 \times 6) = 10160\) A1: For the final positive difference of \(2883.74\) (accept \(2883.74\) or \(2883.75\) depending on intermediate rounding)
題目 6 · Medium/Long Answer
4.17 分
A rational function is defined by \(f(x) = \frac{a}{x - 2} + b\). The graph of \(y = f(x)\) passes through the coordinates \((3, 7)\) and \((5, 4)\). Find the value of \(a\) and the value of \(b\).
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解題
Substitute the point \((3, 7)\) into the function: \(7 = \frac{a}{3 - 2} + b \implies 7 = a + b \quad \text{--- (Equation 1)}\)
Substitute the point \((5, 4)\) into the function: \(4 = \frac{a}{5 - 2} + b \implies 4 = \frac{a}{3} + b \quad \text{--- (Equation 2)}\)
Subtract Equation 2 from Equation 1: \((a + b) - \left(\frac{a}{3} + b\right) = 7 - 4\) \(\frac{2}{3}a = 3\) \(2a = 9 \implies a = 4.5\)
Substitute \(a = 4.5\) back into Equation 1: \(4.5 + b = 7 \implies b = 2.5\)
Therefore, \(a = 4.5\) and \(b = 2.5\).
評分準則
M1: For substituting \((3, 7)\) to write \(a + b = 7\) M1: For substituting \((5, 4)\) to write \(\frac{a}{3} + b = 4\) M1: For a correct method to solve the simultaneous equations (elimination or substitution) A1: For both correct values \(a = 4.5\) and \(b = 2.5\) (accept fractional equivalents)
First, let's factorise each part of the algebraic fractions: 1. Factorise the numerator of the first fraction: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)
2. Factorise the denominator of the first fraction (difference of two squares): \(4x^2 - 1 = (2x - 1)(2x + 1)\)
3. Factorise the denominator of the second fraction (difference of two squares): \(x^2 - 9 = (x - 3)(x + 3)\)
Substitute these factored expressions back into the original product: \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} \times \frac{2x - 1}{(x - 3)(x + 3)}\)
Now cancel the common terms that appear in both the numerator and the denominator: - Cancel \((2x + 1)\) - Cancel \((x - 3)\) - Cancel \((2x - 1)\)
This leaves: \(\frac{1}{x + 3}\).
評分準則
M1: For factorising \(2x^2 - 5x - 3\) as \((2x+1)(x-3)\) M1: For factorising \(4x^2 - 1\) as \((2x-1)(2x+1)\) M1: For factorising \(x^2 - 9\) as \((x-3)(x+3)\) A1: For the final simplified form \(\frac{1}{x+3}\)
題目 8 · Medium/Long Answer
4.17 分
The 1st, 3rd, and 11th terms of an arithmetic progression are the first three terms of a geometric progression. The first term of the arithmetic progression is 4. Given that the common difference of the arithmetic progression is non-zero, find the common difference, \(d\), of the arithmetic progression and the common ratio, \(r\), of the geometric progression.
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解題
Let the arithmetic progression have first term \(a = 4\) and common difference \(d\). The required terms of the arithmetic progression are: - 1st term: \(T_1 = 4\) - 3rd term: \(T_3 = 4 + 2d\) - 11th term: \(T_{11} = 4 + 10d\)
Since these three terms form a geometric progression, the common ratio is equal between consecutive terms: \(\frac{T_3}{T_1} = \frac{T_{11}}{T_3} \implies T_3^2 = T_1 \times T_{11}\)
Substitute the terms: \((4 + 2d)^2 = 4(4 + 10d)\)
Expand both sides: \(16 + 16d + 4d^2 = 16 + 40d\)
Subtract 16 and rearrange terms to form a quadratic: \(4d^2 - 24d = 0\) \(4d(d - 6) = 0\)
Since the common difference \(d\) is non-zero, we have: \(d = 6\)
Now, calculate the terms of the geometric progression: - First term: \(4\) - Second term: \(4 + 2(6) = 16\) - Third term: \(4 + 10(6) = 64\)
The common ratio \(r\) is: \(r = \frac{16}{4} = 4\)
Thus, \(d = 6\) and \(r = 4\).
評分準則
M1: For expressing terms in terms of \(d\): \(4\), \(4+2d\), and \(4+10d\) M1: For writing the geometric sequence relation: \((4+2d)^2 = 4(4+10d)\) M1: For solving the quadratic equation to obtain \(d = 6\) A1: For the final answers: \(d = 6\) and \(r = 4\)
題目 9 · Extended Calculator
4.17 分
An investment of $8000 is placed in a bank account that pays an interest rate of 4.5% per year, compounded quarterly. Find the number of years, to the nearest tenth of a year, for the value of the investment to reach $11,500.
To the nearest tenth of a year, \(t = 8.1\) years.
評分準則
M1: Set up compound interest equation: \(8000(1.01125)^{4t} = 11500\) or equivalent A1: Simplify to \(1.01125^{4t} = 1.4375\) M1: Use logarithms or GDC to solve for \(t\) A1: Correct final answer to the nearest tenth of a year: 8.1
題目 10 · Extended Calculator
4.17 分
Consider the functions \(f(x) = \frac{3x - 5}{2x + 1}\) for \(x \neq -0.5\) and \(g(x) = x^2 - 3\). Find the values of \(x\) for which \(f^{-1}(x) = g(2)\). Give your answer as a fraction in its simplest form.
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解題
First, calculate the value of \(g(2)\):
\(g(2) = 2^2 - 3 = 1\)
We need to solve \(f^{-1}(x) = 1\).
By the definition of an inverse function, this is equivalent to:
\(x = f(1)\)
Substitute \(x = 1\) into the original function \(f(x)\):
M1: Calculate \(g(2) = 1\) M1: Use the property \(f^{-1}(x) = 1 \iff x = f(1)\) OR correctly find the inverse function \(f^{-1}(x) = \frac{x+5}{3-2x}\) M1: Set up the linear equation \(3x = -2\) or equivalent A1: Correct final simplified fraction: \(-2/3\)
題目 11 · Extended Calculator
4.17 分
Solve the equation:
\(\frac{4}{x - 3} - \frac{3}{x + 2} = 2\)
Show all your working and give your answers correct to 2 decimal places.
M1: Multiply through by the common denominator to obtain \(4(x+2) - 3(x-3) = 2(x-3)(x+2)\) or equivalent A1: Simplify to standard quadratic form \(2x^2 - 3x - 29 = 0\) M1: Apply the quadratic formula or use GDC to solve the quadratic equation A1: Correct answers to 2 decimal places: -3.13 and 4.63 (accept in any order)
題目 12 · Extended Calculator
4.17 分
The first four terms of a sequence are \(7, 18, 35, 58, \dots\). Find an expression, in terms of \(n\), for the \(n\)-th term of this sequence.
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解題
We analyze the first and second differences of the terms:
Since the second differences are constant, the sequence is quadratic and has the general form \(an^2 + bn + c\).
The coefficient \(a\) is half of the constant second difference:
\(2a = 6 \implies a = 3\)
Now, subtract \(3n^2\) from each term of the sequence to find the linear part:
- For \(n = 1\): \(7 - 3(1^2) = 4\) - For \(n = 2\): \(18 - 3(2^2) = 6\) - For \(n = 3\): \(35 - 3(3^2) = 8\) - For \(n = 4\): \(58 - 3(4^2) = 10\)
The remaining linear sequence is \(4, 6, 8, 10, \dots\), which can be represented by \(2n + 2\).
Combining these parts gives the expression for the \(n\)-th term:
\(T_n = 3n^2 + 2n + 2\)
評分準則
M1: Identify that the second differences are constant and equal to 6 A1: Deduce that the coefficient of \(n^2\) is 3 M1: Set up a linear sequence for the remaining part or use simultaneous equations to find the linear terms A1: Correct final expression: \(3n^2 + 2n + 2\) (or equivalent)
題目 13 · Extended Calculator
4.17 分
A shopkeeper reduces the price of a television by 12%. Later, she increases this reduced price by 15%. The final price of the television is $556.60. Calculate the original price of the television.
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解題
Let the original price of the television be \(P\).
After a 12% reduction, the price of the television becomes:
\(P \times (1 - 0.12) = 0.88P\)
After a subsequent 15% increase, the price of the television becomes:
Thus, the original price of the television was $550.
評分準則
M1: Express the reduced price as \(0.88P\) or equivalent M1: Express the final increased price as \(0.88P \times 1.15\) (or \(1.012P\)) M1: Form the equation \(1.012P = 556.60\) A1: Correct original price: 550
題目 14 · Extended Calculator
4.17 分
The graph of a quadratic function \(y = ax^2 + bx + c\) passes through the points \((-1, -2)\), \((1, 4)\), and \((2, 13)\). Find the equation of the quadratic function in the form \(y = ax^2 + bx + c\).
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解題
Substitute the coordinates of each point into the quadratic general form \(y = ax^2 + bx + c\):
For \((-1, -2)\): \(a(-1)^2 + b(-1) + c = -2 \implies a - b + c = -2\) --- (Equation 1)
For \((1, 4)\): \(a(1)^2 + b(1) + c = 4 \implies a + b + c = 4\) --- (Equation 2)
For \((2, 13)\): \(a(2)^2 + b(2) + c = 13 \implies 4a + 2b + c = 13\) --- (Equation 3)
Subtract Equation 1 from Equation 2: \((a + b + c) - (a - b + c) = 4 - (-2)\) \(2b = 6 \implies b = 3\)
Substitute \(b = 3\) back into Equation 1 and Equation 3: \(a - 3 + c = -2 \implies a + c = 1\) --- (Equation 4) \(4a + 6 + c = 13 \implies 4a + c = 7\) --- (Equation 5)
Subtract Equation 4 from Equation 5: \((4a + c) - (a + c) = 7 - 1\) \(3a = 6 \implies a = 2\)
Substitute \(a = 2\) back into Equation 4: \(2 + c = 1 \implies c = -1\)
Therefore, the quadratic function is: \(y = 2x^2 + 3x - 1\)
評分準則
M1: Set up a system of three linear equations using the given coordinates A1: Solve for one coefficient (e.g., \(b = 3\)) M1: Eliminate a second variable and solve for \(a\) and \(c\) A1: Correct final equation: \(y = 2x^2 + 3x - 1\) (or correct values for \(a=2, b=3, c=-1\))
題目 15 · Extended Calculator
4.17 分
Write as a single fraction in its simplest form:
\(\frac{2x - 3}{x^2 - 4} - \frac{3}{x + 2}\)
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解題
First, factorize the denominator of the first term using the difference of two squares:
\(x^2 - 4 = (x - 2)(x + 2)\)
Now, rewrite the subtraction with a common denominator of \((x - 2)(x + 2)\):
M1: Factorize \(x^2 - 4\) to \((x - 2)(x + 2)\) M1: Put both fractions over the common denominator \((x - 2)(x + 2)\) M1: Correct expansion of numerator \(2x - 3 - 3(x - 2)\) A1: Correct final simplified fraction: \(\frac{3-x}{x^2-4}\) or \(\frac{3-x}{(x-2)(x+2)}\)
題目 16 · Extended Calculator
4.17 分
The \(n\)-th term of a geometric sequence is given by \(T_n = p \cdot q^{n-1}\). Given that the second term of this sequence is 6 and the fifth term is 162, calculate the sum of the first 6 terms of this sequence.
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解題
We are given: - \(T_2 = p \cdot q = 6\) - \(T_5 = p \cdot q^4 = 162\)
Divide the equation for \(T_5\) by the equation for \(T_2\):
\(\frac{p \cdot q^4}{p \cdot q} = \frac{162}{6}\)
\(q^3 = 27\)
Since \(q\) is a real number, we have:
\(q = 3\)
Substitute \(q = 3\) back into the equation for \(T_2\):
\(p \cdot 3 = 6 \implies p = 2\)
Thus, the first term \(a = 2\) and the common ratio \(r = 3\).
The sum of the first \(6\) terms is given by the formula:
M1: Set up system of equations \(pq = 6\) and \(pq^4 = 162\) A1: Correctly solve for \(q = 3\) and \(p = 2\) M1: Apply the geometric sum formula for \(n = 6\) or manually add the first six terms A1: Correct sum: 728
題目 17 · Medium/Long Answer
4.17 分
An amount of $6500 is invested in an account that pays compound interest at a rate of \(r\%\) per year. After 8 years, the total value of the investment is $8200. Find the value of \(r\), giving your answer correct to 2 decimal places.
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解題
We use the compound interest formula: \(A = P \left(1 + \frac{r}{100}\right)^n\). Here, \(P = 6500\), \(A = 8200\), and \(n = 8\). Substituting these values into the formula gives: \(6500 \left(1 + \frac{r}{100}\right)^8 = 8200\). Divide both sides by 6500: \(\left(1 + \frac{r}{100}\right)^8 = \frac{8200}{6500} = \frac{82}{65}\). Take the 8th root of both sides: \(1 + \frac{r}{100} = \left(\frac{82}{65}\right)^{\frac{1}{8}} \approx 1.029461\). Subtract 1 from both sides: \(\frac{r}{100} \approx 0.029461\). Multiply by 100: \(r \approx 2.9461\). Rounding to 2 decimal places, we get \(r = 2.95\).
評分準則
M1 for setting up the correct compound interest equation: \(6500 \left(1 + \frac{r}{100}\right)^8 = 8200\). M1 for isolating the power term: \(\left(1 + \frac{r}{100}\right)^8 = \frac{82}{65}\). M1 for taking the 8th root correctly: \(1 + \frac{r}{100} \approx 1.02946\). A1.17 for the final correct answer of 2.95.
題目 18 · Medium/Long Answer
4.17 分
The \(n\)-th term of a sequence is given by \(T_n = an^2 + bn + 7\). Given that the 3rd term of the sequence is 16 and the 5th term of the sequence is 42, find the value of the 12th term.
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解題
We are given the general term \(T_n = an^2 + bn + 7\). Using the given terms, we set up a system of linear equations. For the 3rd term: \(T_3 = a(3)^2 + b(3) + 7 = 16\), which simplifies to \(9a + 3b + 7 = 16 \implies 9a + 3b = 9 \implies 3a + b = 3\) (Equation 1). For the 5th term: \(T_5 = a(5)^2 + b(5) + 7 = 42\), which simplifies to \(25a + 5b + 7 = 42 \implies 25a + 5b = 35 \implies 5a + b = 7\) (Equation 2). Subtract Equation 1 from Equation 2: \((5a + b) - (3a + b) = 7 - 3 \implies 2a = 4 \implies a = 2\). Substitute \(a = 2\) into Equation 1: \(3(2) + b = 3 \implies 6 + b = 3 \implies b = -3\). Thus, the formula for the \(n\)-th term is \(T_n = 2n^2 - 3n + 7\). To find the 12th term, substitute \(n = 12\): \(T_{12} = 2(12)^2 - 3(12) + 7 = 2(144) - 36 + 7 = 288 - 36 + 7 = 259\).
評分準則
M1 for setting up the two correct equations in terms of \(a\) and \(b\): \(9a + 3b = 9\) and \(25a + 5b = 35\). M1 for correctly solving the simultaneous equations to find \(a = 2\) and \(b = -3\). M1 for substituting \(a\) and \(b\) back into the general formula to get \(T_n = 2n^2 - 3n + 7\). A1.17 for calculating the 12th term as 259.
Paper 5 (Core Investigation)
Investigate mathematical patterns. Provide clear reasoning and working.
7 題目 · 39.97 分
題目 1 · Structured Investigation
5.71 分
A triangular pyramid of coins is built with \( n \) rows, where Row 1 is at the top. Row 1 has 1 coin, Row 2 has 2 coins, Row 3 has 3 coins, and in general, Row \( r \) has \( r \) coins. The value of each coin in Row \( r \) is \( 5r \) cents. Let \( T(r) \) be the total value, in cents, of all the coins in Row \( r \). Find the value of \( T(4) \), the total value of the coins in Row 4.
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解題
Row 4 contains \( r = 4 \) coins. The value of each coin in Row 4 is \( 5 \times 4 = 20 \) cents. The total value of the coins in Row 4 is \( T(4) = 4 \times 20 = 80 \) cents.
評分準則
M1 for identifying the number of coins in Row 4 as 4 and the coin value as 20 cents (or writing \( 4 \times 20 \)). A1 for 80.
題目 2 · Structured Investigation
5.71 分
Using the context from Question 1, write down an expression in terms of \( r \) for the total value \( T(r) \), in cents, of the coins in Row \( r \).
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解題
Row \( r \) has \( r \) coins. Each coin has a value of \( 5r \) cents. The total value is \( T(r) = r \times 5r = 5r^2 \) cents.
評分準則
M1 for the product of number of coins and value per coin: \( r \times 5r \). A1 for \( 5r^2 \).
題目 3 · Structured Investigation
5.71 分
Let \( S(n) \) be the total value, in cents, of all the coins in a pyramid of height \( n \). Find \( S(4) \), the total value of a pyramid with 4 rows.
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解題
The total value \( S(4) \) is the sum of the values of Row 1, Row 2, Row 3, and Row 4. \( T(1) = 5 \times 1^2 = 5 \), \( T(2) = 5 \times 2^2 = 20 \), \( T(3) = 5 \times 3^2 = 45 \), and \( T(4) = 5 \times 4^2 = 80 \). Therefore, \( S(4) = 5 + 20 + 45 + 80 = 150 \) cents.
評分準則
M1 for summing the values of the first 4 rows: \( T(1) + T(2) + T(3) + T(4) \). A1 for 150.
題目 4 · Structured Investigation
5.71 分
The sum of the first \( n \) square numbers is given by the formula: \( 1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6} \). Using this and your result from Question 2, find a formula for \( S(n) \), the total value in cents of a pyramid of height \( n \), in terms of \( n \).
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解題
Since \( T(r) = 5r^2 \), the total sum is \( S(n) = 5(1^2 + 2^2 + \dots + n^2) \). Substituting the given sum of squares formula, we get: \( S(n) = \frac{5n(n+1)(2n+1)}{6} \).
評分準則
M1 for recognizing that \( S(n) = 5 \sum r^2 \). A1 for \( \frac{5n(n+1)(2n+1)}{6} \) or equivalent.
題目 5 · Structured Investigation
5.71 分
Use the formula from Question 4 to find the total value, in dollars ($), of a pyramid of height 10.
M1 for substituting \( n = 10 \) into the formula. A1 for 1925 cents. A1 for converting to dollars to get 19.25.
題目 6 · Structured Investigation
5.71 分
Suppose instead, a different pyramid is built where Row \( r \) has \( r \) coins, and each coin in Row \( r \) has a value of \( 2r - 1 \) cents. Let \( U(r) \) be the total value, in cents, of the coins in Row \( r \). Write down an expression for \( U(r) \) in terms of \( r \), and expand the brackets.
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解題
Row \( r \) has \( r \) coins with each coin having a value of \( 2r - 1 \) cents. The total value is \( U(r) = r(2r - 1) = 2r^2 - r \).
評分準則
M1 for writing \( r(2r-1) \). A1 for expanding to \( 2r^2-r \).
題目 7 · Structured Investigation
5.71 分
Let \( V(n) \) be the total value, in cents, of this new pyramid of height \( n \). Using the formula for the sum of the first \( n \) integers, \( 1 + 2 + \dots + n = \frac{n(n+1)}{2} \), and the sum of squares formula, the total value \( V(n) \) can be shown to simplify to: \( V(n) = \frac{n(n+1)(4n-1)}{a} \). Find the integer value of \( a \).
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解題
We sum the terms: \( V(n) = \sum_{r=1}^n (2r^2 - r) = 2 \sum_{r=1}^n r^2 - \sum_{r=1}^n r \). Substituting the formulas gives: \( V(n) = 2 \left( \frac{n(n+1)(2n+1)}{6} \right) - \frac{n(n+1)}{2} = \frac{2n(n+1)(2n+1) - 3n(n+1)}{6} = \frac{n(n+1)[2(2n+1) - 3]}{6} = \frac{n(n+1)(4n-1)}{6} \). Comparing this with the given expression, \( a = 6 \).
評分準則
M1 for substituting both summation formulas. M1 for factorising out \( n(n+1) \) and combining over a common denominator. A1 for \( a = 6 \).
Paper 6 (Extended Investigation and Modelling)
Complete both sections: Investigation and Modelling. GDC allowed.
5 題目 · 50 分
題目 1 · Structured
10 分
An investigation of dot patterns arranged in concentric equilateral triangles. Let \(n\) represent the side length of the outermost triangle. The total number of dots in the pattern is denoted by \(S_n\). The number of dots on the outer boundary is denoted by \(B_n\). The number of dots inside the boundary is denoted by \(I_n\). For \(n = 1\): \(S_1 = 3, B_1 = 3, I_1 = 0\). For \(n = 2\): \(S_2 = 6, B_2 = 6, I_2 = 0\). For \(n = 3\): \(S_3 = 10, B_3 = 9, I_3 = 1\). (a) State the values of \(S_4\), \(B_4\), and \(I_4\). (b) Find an expression for \(B_n\) in terms of \(n\). (c) The formula for the total number of dots is \(S_n = \frac{(n+1)(n+2)}{2}\). Use this and your answer to part (b) to show that \(I_n = \frac{n^2 - 3n + 2}{2}\). (d) Find the value of \(n\) for which the number of inside dots is exactly 45.
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解題
(a) For n=4, the total number of dots is S_4 = (4+1)(4+2)/2 = 15. The boundary dots form a triangle of side 4, giving B_4 = 3 * 4 = 12 dots. The inside dots are I_4 = S_4 - B_4 = 15 - 12 = 3. (b) For any n, there are 3 vertices and (n-1) dots along each of the 3 sides. This gives B_n = 3 + 3(n-1) = 3n. (c) Inside dots I_n = S_n - B_n = \frac{(n+1)(n+2)}{2} - 3n = \frac{n^2+3n+2}{2} - \frac{6n}{2} = \frac{n^2-3n+2}{2}. (d) Set I_n = 45: \frac{n^2-3n+2}{2} = 45 \implies n^2-3n+2 = 90 \implies n^2-3n-88 = 0. Factoring gives (n-11)(n+8) = 0. Since n must be positive, n = 11.
評分準則
(a) [3 marks] 1 mark for S_4 = 15, 1 mark for B_4 = 12, 1 mark for I_4 = 3. (b) [2 marks] 1 mark for recognizing a linear progression with common difference 3, 1 mark for the correct formula B_n = 3n. (c) [3 marks] 1 mark for the relationship I_n = S_n - B_n, 1 mark for expansion of S_n, 1 mark for completing the algebraic proof. (d) [2 marks] 1 mark for setting up the quadratic equation, 1 mark for solving to find n = 11 (rejecting n = -8).
題目 2 · Structured
10 分
This investigation looks at the behavior of infinite nested square roots of the form \(x_{k+1} = \sqrt{c + x_k}\). Let \(c = 6\) and the initial term \(x_1 = \sqrt{6}\). (a) Calculate the values of \(x_1\) and \(x_2\) correct to 3 decimal places. (b) Assuming the sequence of terms approaches a positive limit \(L\) as \(k\) increases, then \(L = \sqrt{6 + L}\). (i) Show that \(L^2 - L - 6 = 0\). (ii) Solve this equation to find the value of \(L\), explaining why only one solution is valid. (c) For a general constant \(c > 0\), the limit \(L\) satisfies \(L = \sqrt{c + L}\). (i) Write down the quadratic equation in \(L\) for this general case. (ii) Show that the positive solution to this equation is \(L = \frac{1 + \sqrt{1 + 4c}}{2}\). (d) Find the positive integer value of \(c\) for which the limit \(L = 7\).
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解題
(a) x_1 = \sqrt{6} \approx 2.449. x_2 = \sqrt{6 + 2.44949} = \sqrt{8.44949} \approx 2.907. (b)(i) Squaring both sides of L = \sqrt{6+L} gives L^2 = 6+L, which rearranges to L^2 - L - 6 = 0. (b)(ii) Factoring L^2 - L - 6 = 0 gives (L-3)(L+2) = 0, so L = 3 or L = -2. Since the terms in the sequence are all positive square roots, the limit must be positive, hence L = 3. (c)(i) Squaring both sides of L = \sqrt{c+L} gives L^2 = c+L, which rearranges to L^2 - L - c = 0. (c)(ii) Using the quadratic formula: L = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-c)}}{2(1)} = \frac{1 \pm \sqrt{1+4c}}{2}. Since c > 0, the term \sqrt{1+4c} > 1, so the negative sign would give a negative limit. Thus, the positive solution is L = \frac{1 + \sqrt{1 + 4c}}{2}. (d) Setting L = 7: 7 = \sqrt{c+7} \implies 49 = c+7 \implies c = 42.
評分準則
(a) [2 marks] 1 mark for x_1 = 2.449, 1 mark for x_2 = 2.907. (b)(i) [1 mark] Correct squaring and rearrangement. (b)(ii) [2 marks] 1 mark for finding both roots (3 and -2), 1 mark for selecting L = 3 with valid justification. (c)(i) [1 mark] Correct quadratic equation L^2 - L - c = 0. (c)(ii) [2 marks] 1 mark for correct application of the quadratic formula, 1 mark for justifying the positive sign. (d) [2 marks] 1 mark for setting up the equation, 1 mark for finding c = 42.
題目 3 · Structured
10 分
A logistics company operates delivery drones. The total average monthly cost of maintaining and operating a drone battery over \(t\) months is modelled by: \(T(t) = 15 + 3.2t + \frac{320}{t}\), for \(t \ge 1\). Here, \(15 + 3.2t\) represents the monthly operating cost, and \(\frac{320}{t}\) represents the average monthly purchase cost of the battery. (a) Complete the following table of values for \(T(t)\): At \(t = 5\), \(T(t) = \dots\); At \(t = 16\), \(T(t) = \dots\). (b) Use your graphic display calculator (GDC) to find the coordinates of the local minimum of \(T(t)\) for \(t \ge 1\). (c) State the optimal number of months to use a battery before replacing it, and the corresponding minimum average monthly cost. (d) If the purchase price of a new battery rises to $500, write down the new model for the total average monthly cost \(T_{\text{new}}(t)\), and find the new optimal replacement interval in months.
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解題
(a) For t = 5: T(5) = 15 + 3.2(5) + 320/5 = 15 + 16 + 64 = 95. For t = 16: T(16) = 15 + 3.2(16) + 320/16 = 15 + 51.2 + 20 = 86.2. (b) Plotting T(t) on the GDC for t \ge 1 shows a curve with a minimum point. Using the minimum finder function, we get the coordinates (10, 79). (c) The x-coordinate represents the time in months, so the optimal replacement interval is 10 months. The y-coordinate is the minimum total average monthly cost, which is $79. (d) The new cost model is T_{\text{new}}(t) = 15 + 3.2t + \frac{500}{t}. Using the GDC or setting the derivative to 0: 3.2 - 500/t^2 = 0 \implies t^2 = 156.25 \implies t = 12.5 months.
評分準則
(a) [2 marks] 1 mark for T(5) = 95, 1 mark for T(16) = 86.2. (b) [3 marks] 1 mark for sketching/identifying curve shape, 2 marks for coordinates (10, 79) (1 mark for each coordinate). (c) [2 marks] 1 mark for 10 months, 1 mark for $79. (d) [3 marks] 1 mark for the new equation, 1 mark for equating derivative to 0 or using GDC, 1 mark for 12.5 months.
題目 4 · Structured
10 分
The expansion of a steel bridge span under varying temperatures is modeled by the quadratic function: \(L(T) = a T^2 + b T + c\) where \(L(T)\) is the length of the span in meters at temperature \(T\) in degrees Celsius (\(^\circ\text{C}\)). The following measurements were recorded: At \(T = 0^\circ\text{C}\), \(L(0) = 150\) meters. At \(T = 10^\circ\text{C}\), \(L(10) = 150.018\) meters. At \(T = 30^\circ\text{C}\), \(L(30) = 150.066\) meters. (a) State the value of \(c\). (b) Form a system of simultaneous equations and show that \(b = 0.0016\) and \(a = 0.00002\). (c) Calculate the length of the bridge span when the temperature is \(50^\circ\text{C}\). (d) Find the temperature, to 1 decimal place, at which the bridge span reaches a length of \(150.25\) meters.
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解題
(a) When T = 0, L(0) = a(0)^2 + b(0) + c = 150 \implies c = 150. (b) Using the other two data points: For T = 10: a(100) + b(10) + 150 = 150.018 \implies 10b + 100a = 0.018. For T = 30: a(900) + b(30) + 150 = 150.066 \implies 30b + 900a = 0.066. Divide the first equation by 10: b + 10a = 0.0018. Divide the second equation by 30: b + 30a = 0.0022. Subtracting the first from the second: 20a = 0.0004 \implies a = 0.00002. Then b = 0.0018 - 10(0.00002) = 0.0016. (c) At T = 50: L(50) = 0.00002(50^2) + 0.0016(50) + 150 = 0.00002(2500) + 0.08 + 150 = 0.05 + 0.08 + 150 = 150.13 m. (d) Set L(T) = 150.25: 0.00002 T^2 + 0.0016 T + 150 = 150.25 \implies 0.00002 T^2 + 0.0016 T - 0.25 = 0. Using the quadratic formula or a GDC: T = \frac{-0.0016 \pm \sqrt{0.0016^2 - 4(0.00002)(-0.25)}}{2(0.00002)} \approx 78.7^\circ\text{C} (rejecting negative temperature).
評分準則
(a) [1 mark] c = 150. (b) [4 marks] 1 mark for each equation, 1 mark for method of elimination/substitution, 1 mark for correct values of both a and b. (c) [2 marks] 1 mark for substitution of T=50, 1 mark for L(50) = 150.13. (d) [3 marks] 1 mark for setting up the quadratic equation, 1 mark for solving method (GDC or formula), 1 mark for correct positive solution T = 78.7.
題目 5 · Structured
10 分
In this investigation, we look at the difference between the squares of consecutive terms in an arithmetic progression. Let the terms of an arithmetic progression be \(a_n\). The sequence of differences is defined as \(D_n = a_{n+1}^2 - a_n^2\). First, consider the sequence of consecutive odd numbers: \(1, 3, 5, 7, 9, \dots\) (a) (i) Write down the 6th term in this sequence and its square. (ii) The first three differences are \(D_1 = 3^2 - 1^2 = 8\), \(D_2 = 5^2 - 3^2 = 16\), \(D_3 = 7^2 - 5^2 = 24\). State the values of \(D_4\) and \(D_5\). (iii) Find an expression for \(D_n\) in terms of \(n\). Next, consider the arithmetic progression \(a_n = 3n - 1\). (b) (i) Write down the first four terms of this progression and their squares. (ii) Find the first three terms of its difference sequence \(D_n = a_{n+1}^2 - a_n^2\). Finally, consider a general arithmetic progression \(a_n = d(n - 1) + a\), where \(a\) is the first term and \(d\) is the common difference. (c) Show algebraically that the difference of squares is given by \(D_n = d(2a + d(2n - 1))\).
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解題
(a)(i) The 6th term is 11, and its square is 11^2 = 121. (a)(ii) D_4 = 9^2 - 7^2 = 81 - 49 = 32. D_5 = 11^2 - 9^2 = 121 - 81 = 40. (a)(iii) The sequence of differences is 8, 16, 24, 32, 40, which is an arithmetic progression with first term 8 and common difference 8. Thus, D_n = 8n. (b)(i) For a_n = 3n - 1: For n = 1, 2, 3, 4, the terms are 2, 5, 8, 11. Their squares are 4, 25, 64, 121. (b)(ii) D_1 = 25 - 4 = 21. D_2 = 64 - 25 = 39. D_3 = 121 - 64 = 57. (c) Using the difference of squares identity: D_n = a_{n+1}^2 - a_n^2 = (a_{n+1} - a_n)(a_{n+1} + a_n). Since a_n is an arithmetic progression with common difference d, we have a_{n+1} - a_n = d. Also, a_{n+1} + a_n = (a + dn) + (a + d(n-1)) = a + dn + a + dn - d = 2a + d(2n - 1). Substituting these in, we obtain D_n = d(2a + d(2n - 1)).
評分準則
(a)(i) [1 mark] 11 and 121. (a)(ii) [1 mark] D_4 = 32, D_5 = 40. (a)(iii) [2 marks] 1 mark for finding linear pattern, 1 mark for D_n = 8n. (b)(i) [2 marks] 1 mark for terms, 1 mark for squares. (b)(ii) [2 marks] 1 mark for first difference, 1 mark for second and third differences. (c) [2 marks] 1 mark for using the difference of squares or expanding directly, 1 mark for substituting a_n and a_{n+1} and simplifying.
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