An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V2) Cambridge IGCSE International Mathematics (0607) paper. Not affiliated with or reproduced from Cambridge.
部分 1: Extended Non-Calculator Skills
Answer all questions. Calculators must not be used in this section. All answers should be written in simplest form.
16 题目 · 40 分
题目 1 · Short Answer
2.5 分
Rationalise the denominator and simplify:
\(\frac{8}{3 - \sqrt{5}}\)
查看答案详解收起答案详解
解题
To rationalise the denominator, multiply both the numerator and the denominator by the conjugate of the denominator, which is \(3 + \sqrt{5}\):
M1 for multiplying numerator and denominator by \(3 + \sqrt{5}\) A1 for denominator of \(4\) seen in working A0.5 for final answer of \(6 + 2\sqrt{5}\) (accept \(2(3 + \sqrt{5})\))
题目 2 · Short Answer
2.5 分
Find the \(n\)th term of the sequence:
\(3, \ 9, \ 19, \ 33, \ 51, \ \dots\)
查看答案详解收起答案详解
解题
Let the terms of the sequence be \(u_1 = 3, \ u_2 = 9, \ u_3 = 19, \ u_4 = 33, \ u_5 = 51\).
Find the second differences: \(10 - 6 = 4\) \(14 - 10 = 4\) \(18 - 14 = 4\)
Since the second difference is a constant \(4\), the sequence is quadratic of the form \(an^2 + bn + c\), where: \(a = \frac{\text{second difference}}{2} = \frac{4}{2} = 2\).
Subtracting \(2n^2\) from the terms of the sequence: For \(n=1\): \(3 - 2(1)^2 = 1\) For \(n=2\): \(9 - 2(2)^2 = 1\) For \(n=3\): \(19 - 2(3)^2 = 1\)
Since the remainder is constant and equal to \(1\), we have \(bn + c = 1\) (where \(b = 0\) and \(c = 1\)).
Thus, the \(n\)th term is \(2n^2 + 1\).
评分标准
M1 for finding first differences of \(6, 10, 14, 18\) and second difference of \(4\) (or identifying coefficient \(a = 2\)) M1 for subtracting \(2n^2\) from each term to find the linear component A0.5 for \(2n^2 + 1\) (or equivalent)
题目 3 · Short Answer
2.5 分
\(y\) is inversely proportional to the square root of \((x + 2)\). When \(x = 7\), \(y = 8\).
Find \(y\) when \(x = 14\).
查看答案详解收起答案详解
解题
Since \(y\) is inversely proportional to the square root of \((x + 2)\):
\(y = \frac{k}{\sqrt{x + 2}}\)
Substitute the given values \(x = 7\) and \(y = 8\) to find \(k\):
Now, use the point-slope form with point \((4, 3)\) and gradient \(m = \frac{1}{2}\):
\(y - 3 = \frac{1}{2}(x - 4)\)
\(y - 3 = \frac{1}{2}x - 2\)
\(y = \frac{1}{2}x + 1\)
评分标准
M1 for finding the gradient of \(L\) as \(-2\) M1 for determining the perpendicular gradient is \(\frac{1}{2}\) and attempting to write the equation of the line A0.5 for the final correct equation \(y = \frac{1}{2}x + 1\) (or \(y = 0.5x + 1\))
题目 5 · Short Answer
2.5 分
Given that \(\mathrm{f}(x) = 3x - 4\) and \(\mathrm{g}(x) = \frac{x + 2}{5\mathcal{}}\),
find \(\mathrm{g}^{-1}(\mathrm{f}(x))\) in its simplest form.
查看答案详解收起答案详解
解题
To find \(\mathrm{g}^{-1}(x)\), let \(y = \mathrm{g}(x)\):
\(y = \frac{x + 2}{5}\)
\(5y = x + 2 \implies x = 5y - 2\)
Therefore, \(\mathrm{g}^{-1}(x) = 5x - 2\).
Now, find \(\mathrm{g}^{-1}(\mathrm{f}(x))\) by substituting \(\mathrm{f}(x) = 3x - 4\) into \(\mathrm{g}^{-1}(x)\):
M1 for finding \(\mathrm{g}^{-1}(x) = 5x - 2\) (or writing the equation \(\frac{y+2}{5} = 3x - 4\)) M1 for substituting \(\mathrm{f}(x)\) to get \(5(3x - 4) - 2\) (or expanding \(y+2 = 15x - 20\)) A0.5 for final simplified expression \(15x - 22\)
题目 6 · Short Answer
2.5 分
Solve the inequality:
\(5 - 2(3x - 1) \ge 4x + 17\)
查看答案详解收起答案详解
解题
Expand the bracket on the left-hand side:
\(5 - 6x + 2 \ge 4x + 17\)
\(7 - 6x \ge 4x + 17\)
Collect the \(x\) terms on the left-hand side by subtracting \(4x\) from both sides:
\(7 - 10x \ge 17\)
Subtract \(7\) from both sides:
\(-10x \ge 10\)
Divide both sides by \(-10\) and reverse the inequality sign:
\(x \le -1\)
评分标准
M1 for expanding correctly to obtain \(7 - 6x \ge 4x + 17\) M1 for isolating \(x\) terms, e.g., \(-10x \ge 10\) (or \(10 \ge 10x\)) A0.5 for \(x \le -1\) (reversing the inequality sign is required for this mark)
题目 7 · Short Answer
2.5 分
Write as a single fraction in its simplest form:
\(\frac{3}{x - 4} - \frac{2}{x + 3}\)
查看答案详解收起答案详解
解题
Find a common denominator, which is \((x - 4)(x + 3)\):
M1 for calculating the volume of the cylinder as \(36\pi\) M1 for setting up the equation \(\frac{4}{3}\pi R^3 = 36\pi\) and solving for \(R^3 = 27\) A0.5 for final answer of \(3\)
题目 9 · Short Answer
2.5 分
These are the first five terms of a sequence: \[3, \quad 8, \quad 17, \quad 30, \quad 47\] Find an expression, in terms of \(n\), for the \(nth\) term of this sequence.
Now, the second differences: \(9 - 5 = 4\) \(13 - 9 = 4\) \(17 - 13 = 4\)
Since the second difference is constant at \(4\), the sequence has a quadratic term with coefficient \(\frac{4}{2} = 2\), which is \(2n^2\).
Subtracting \(2n^2\) from the terms of the sequence: For \(n=1\): \(3 - 2(1)^2 = 1\) For \(n=2\): \(8 - 2(2)^2 = 0\) For \(n=3\): \(17 - 2(3)^2 = -1\) For \(n=4\): \(30 - 2(4)^2 = -2\)
This forms a linear sequence: \(1, 0, -1, -2, \dots\) which has a common difference of \(-1\) and a starting term of \(1\). The formula for this linear part is \(2 - n\).
Combining both parts, the \(nth\) term of the sequence is: \(2n^2 - n + 2\)
评分标准
M1: for finding the second difference of the sequence is \(4\) (indicating a \(2n^2\) term) M1: for a method to find the linear component (e.g. generating the sequence \(1, 0, -1, \dots\) and identifying its expression as \(2 - n\)) A0.5: for the correct final expression \(2n^2 - n + 2\) (or equivalent)
题目 10 · Short Answer
2.5 分
A solid cone has a base radius of \(3\text{ cm}\) and a vertical height of \(4\text{ cm}\). Calculate the total surface area of the cone, leaving your answer in terms of \(\pi\).
查看答案详解收起答案详解
解题
First, find the slant height \(l\) of the cone using Pythagoras' theorem: \[l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ cm}\]
The total surface area \(A\) of a solid cone is the sum of the curved surface area and the base area: \[A = \pi r l + \pi r^2\]
Substitute the known values: \[A = \pi (3)(5) + \pi (3)^2\] \[A = 15\pi + 9\pi = 24\pi\text{ cm}^2\]
评分标准
M1: for finding the slant height \(l = 5\text{ cm}\) M1: for substituting the correct values into the total surface area formula, i.e., \(\pi (3)(5) + \pi (3)^2\) A0.5: for \(24\pi\)
题目 11 · Short Answer
2.5 分
Find the equation of the line perpendicular to the line \(2y - 3x = 8\) that passes through the point \((6, -1)\). Give your answer in the form \(y = mx + c\).
查看答案详解收起答案详解
解题
First, find the gradient of the given line: \[2y - 3x = 8 \implies 2y = 3x + 8 \implies y = \frac{3}{2}x + 4\]
The gradient of this line is \(m_1 = \frac{3}{2}\).
Since the perpendicular line is perpendicular to the given line, its gradient \(m_2\) must satisfy: \[m_2 = -\frac{1}{m_1} = -\frac{2}{3}\]
Now, use the point-slope form with \(m_2 = -\frac{2}{3}\) and point \((6, -1)\): \[y - y_1 = m_2(x - x_1)\] \[y - (-1) = -\frac{2}{3}(x - 6)\] \[y + 1 = -\frac{2}{3}x + 4\] \[y = -\frac{2}{3}x + 3\]
评分标准
M1: for finding the gradient of the perpendicular line, \(m = -\frac{2}{3}\) M1: for a correct substitution of point \((6, -1)\) and gradient into \(y = mx + c\) or equivalent A0.5: for \(y = -\frac{2}{3}x + 3\)
题目 12 · Short Answer
2.5 分
In triangle \(ABC\), \(AB = 5\text{ cm}\), \(AC = 8\text{ cm}\), and angle \(BAC = 60^\circ\). Calculate the length of the side \(BC\).
查看答案详解收起答案详解
解题
Using the Cosine Rule to find side \(BC\): \[BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(BAC)\]
M1: for a correct substitution into the Cosine Rule: \(BC^2 = 5^2 + 8^2 - 2(5)(8)\cos(60^\circ)\) M1: for using \\cos(60^\circ) = 0.5\ and simplifying to \(BC^2 = 49\) A0.5: for \(7\) (accept \(7\text{ cm}\))
题目 13 · Short Answer
2.5 分
Find the coordinates of the turning point of the curve with equation \(y = -2x^2 + 12x - 5\).
查看答案详解收起答案详解
解题
We can complete the square for the quadratic equation: \[y = -2(x^2 - 6x) - 5\]
This is the vertex form \(y = a(x - h)^2 + k\), where the turning point is \((h, k)\). Therefore, the coordinates of the turning point are \((3, 13)\).
评分标准
M1: for a correct method to find the x-coordinate of the turning point, e.g. usingcompleting the square to get \(x = 3\) or using \(x = -\frac{b}{2a} = -\frac{12}{2(-2)} = 3\) M1: for finding the corresponding y-coordinate, \(y = 13\) A0.5: for coordinates \((3, 13)\)
题目 14 · Short Answer
2.5 分
A bag contains 5 red balls and 3 blue balls. Two balls are selected at random from the bag without replacement.
Calculate the probability that the two selected balls are of different colours. Give your answer as a fraction in its simplest form.
查看答案详解收起答案详解
解题
Total number of balls is \(5 + 3 = 8\).
There are two possible ways to select balls of different colours: 1. Red first, then Blue (RB): \[P(\text{R followed by B}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\]
2. Blue first, then Red (BR): \[P(\text{B followed by R}) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\]
Adding these two independent possibilities: \[P(\text{different colours}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56}\]
Simplifying the fraction: \[\frac{30}{56} = \frac{15}{28}\]
评分标准
M1: for calculating the probability of a single different-colour outcome, e.g., \(\frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\) M1: for adding the two different scenarios together to get \(\frac{30}{56}\) A0.5: for simplifying to the final fraction \(\frac{15}{28}\)
题目 15 · Short Answer
2.5 分
The point \(P(3, -2)\) is rotated \(90^\circ\) anticlockwise about the origin to point \(Q\). Point \(Q\) is then reflected in the \(x\)-axis to point \(R\). Find the coordinates of point \(R\).
查看答案详解收起答案详解
解题
First, perform the rotation of \(90^\circ\) anticlockwise about the origin on point \(P(3, -2)\). The mapping rule for a \(90^\circ\) anticlockwise rotation about the origin is: \[(x, y) \to (-y, x)\]
Applying this rule to \(P(3, -2)\): \[Q = (-(-2), 3) = (2, 3)\]
Next, perform the reflection of point \(Q(2, 3)\) in the \(x\)-axis. The mapping rule for reflection in the \(x\)-axis is: \[(x, y) \to (x, -y)\]
Applying this rule to \(Q(2, 3)\): \[R = (2, -3)\]
评分标准
M1: for finding the coordinates of \(Q\) as \((2, 3)\) (or showing correct working for a \(90^\circ\) anticlockwise rotation) M1: for applying the reflection in the \(x\)-axis to their \(Q\) A0.5: for \((2, -3)\)
题目 16 · Short Answer
2.5 分
Let \(f(x) = \frac{3x + 1}{2}\) and \(g(x) = 2x - 5\). Find the value of \(f^{-1}(g(4))\). Give your answer as a fraction.
Now, we need to find \(f^{-1}(3)\). Let \(f^{-1}(3) = k\), which means \(f(k) = 3\). \[\frac{3k + 1}{2} = 3\] \[3k + 1 = 6\] \[3k = 5\] \[k = \frac{5}{3}\]
Thus, \(f^{-1}(g(4)) = \frac{5}{3}\).
评分标准
M1: for finding \(g(4) = 3\) M1: for setting up the equation \(\frac{3k + 1}{2} = 3\) or finding the inverse function \(f^{-1}(x) = \frac{2x - 1}{3}\) A0.5: for \(\frac{5}{3}\) (or \(1 \frac{2}{3}\))
Answer all questions. A graphic display calculator should be used where appropriate. Show all necessary working.
11 题目 · 121 分
题目 1 · Structured Multi-Part
11 分
Three sequences have the following properties.
**Sequence A** has first four terms: $$4, \ \ 11, \ \ 22, \ \ 37, \ \ \dots$$
(a) (i) Write down the next term of Sequence A. (ii) Find an expression for the $n$-th term of Sequence A.
**Sequence B** has first four terms: $$\frac{1}{3}, \ \ \frac{4}{5}, \ \ \frac{9}{7}, \ \ \frac{16}{9}, \ \ \dots$$
(b) (i) Write down the next term of Sequence B. (ii) Find an expression for the $n$-th term of Sequence B.
**Sequence C** has first four terms: $$3, \ \ 6, \ \ 12, \ \ 24, \ \ \dots$$
(c) (i) Write down the next term of Sequence C. (ii) Find an expression for the $n$-th term of Sequence C. (iii) Calculate the $10$-th term of Sequence C.
查看答案详解收起答案详解
解题
**(a) Sequence A:** (i) Let's find the differences between terms: $$11 - 4 = 7$$ $$22 - 11 = 11$$ $$37 - 22 = 15$$ These differences are $7, 11, 15$, which increase by $4$ each time. Thus, the next difference is $15 + 4 = 19$. The next term is $37 + 19 = 56$.
(ii) Since the second differences are constant ($4$), the sequence is quadratic: $a_n = a n^2 + b n + c$. With $2a = 4 \implies a = 2$. Let $a_n = 2n^2 + bn + c$: $$n = 1: 2(1)^2 + b(1) + c = 4 \implies b + c = 2$$ $$n = 2: 2(2)^2 + b(2) + c = 11 \implies 8 + 2b + c = 11 \implies 2b + c = 3$$ Subtracting the first equation from the second: $$b = 1 \implies c = 1$$ So, the $n$-th term is $2n^2 + n + 1$.
**(b) Sequence B:** (i) The numerators are the perfect squares: $1^2, 2^2, 3^2, 4^2$, so the next numerator is $5^2 = 25$. The denominators are consecutive odd numbers: $3, 5, 7, 9$, so the next denominator is $11$. The next term is $\frac{25}{11}$.
(ii) The $n$-th term of the numerator is $n^2$. The $n$-th term of the denominator is $2n + 1$. Thus, the $n$-th term is $\frac{n^2}{2n + 1}$.
**(c) Sequence C:** (i) This is a geometric sequence where each term is multiplied by $2$. The next term is $24 \times 2 = 48$.
(ii) The first term $a = 3$ and the common ratio $r = 2$. The $n$-th term is $3 \times 2^{n-1}$.
(iii) The $10$-th term is: $$3 \times 2^{10-1} = 3 \times 2^9 = 3 \times 512 = 1536$$
评分标准
**(a)** (i) **B1** for 56 (ii) **M1** for attempting quadratic form (e.g. finding second difference of 4) **M1** for setting up equations to find $b$ and $c$ **A1** for $2n^2 + n + 1$
**(b)** (i) **B1** for $\frac{25}{11}$ (ii) **B2** for $\frac{n^2}{2n + 1}$ (or **B1** for numerator $n^2$ or denominator $2n + 1$ seen)
**(c)** (i) **B1** for 48 (ii) **B2** for $3 \times 2^{n-1}$ (or **B1** for geometric form $a \cdot r^{n-1}$ with one correct value) (iii) **B1** for 1536
题目 2 · Structured Multi-Part
11 分
A closed storage container consists of a cylinder of radius $r$ cm surmounting a hemisphere of the same radius $r$ cm. The total height of the container is $25$ cm, and the radius $r$ is $6$ cm.
(a) Show that the height of the cylindrical part is $19$ cm. (b) Calculate the total volume of the container. Give your answer in terms of $\pi$. (c) Calculate the total external surface area of the container. Give your answer correct to $3$ significant figures. (d) A geometrically similar container has a total height of $50$ cm. Calculate the volume of this larger container, giving your answer correct to $3$ significant figures.
查看答案详解收起答案详解
解题
(a) The total height of the container is the sum of the radius of the hemisphere and the height of the cylinder: $$\text{Total Height} = r + h_{\text{cylinder}}$$ $$25 = 6 + h_{\text{cylinder}} \implies h_{\text{cylinder}} = 19\text{ cm}$$
(b) The total volume is the sum of the volume of the hemisphere and the volume of the cylinder: $$V_{\text{hemisphere}} = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (6)^3 = 144\pi \text{ cm}^3$$ $$V_{\text{cylinder}} = \pi r^2 h_{\text{cylinder}} = \pi (6)^2 (19) = 684\pi \text{ cm}^3$$ $$\text{Total Volume} = 144\pi + 684\pi = 828\pi \text{ cm}^3$$
(c) The total external surface area consists of the curved surface area of the hemisphere, the curved surface area of the cylinder, and the flat top of the cylinder: $$\text{Area}_{\text{hemisphere}} = 2\pi r^2 = 2\pi (6)^2 = 72\pi \text{ cm}^2$$ $$\text{Area}_{\text{cylinder curved}} = 2\pi r h = 2\pi (6)(19) = 228\pi \text{ cm}^2$$ $$\text{Area}_{\text{top}} = \pi r^2 = \pi (6)^2 = 36\pi \text{ cm}^2$$ $$\text{Total Area} = 72\pi + 228\pi + 36\pi = 336\pi \approx 1055.58 \approx 1060\text{ cm}^2 \text{ (3 s.f.)}$$
(d) The scale factor of enlargement from the smaller to the larger container is: $$k = \frac{50}{25} = 2$$ Therefore, the volume of the larger container is: $$V_{\text{larger}} = k^3 \times V_{\text{smaller}} = 2^3 \times 828\pi = 8 \times 828\pi = 6624\pi \approx 20810 \approx 20800\text{ cm}^3 \text{ (3 s.f.)}$$ Alternatively, $8 \times 1055.58 \times \frac{828}{336} = 20810 \approx 20800\text{ cm}^3$.
评分标准
(a) **M1** for $25 - 6$ or equation $h + 6 = 25$ **A1** for $19$ cm shown clearly
(b) **M1** for formula of volume of hemisphere $\frac{2}{3}\pi r^3$ or volume of cylinder $\pi r^2 h$ used with $r=6$ **M1** for adding both volume components **A1** for $828\pi$ (allow $2600$ if not given in terms of $\pi$)
(c) **M1** for hemisphere curved surface area $2\pi(6)^2$ or cylinder curved surface area $2\pi(6)(19)$ **M1** for including cylinder top area $\pi(6)^2$ **M1** for adding all three components together **A1** for $1060$ (allow $1055$ to $1056$)
(d) **M1** for volume scale factor $k^3 = 2^3 = 8$ used **A1** for $20800$ (accept $6624\pi$)
题目 3 · Structured Multi-Part
11 分
The coordinates of two points are $A(-3, 1)$ and $B(5, 7)$.
(a) Calculate the gradient of the line $AB$. (b) Find the equation of the line $AB$ in the form $y = mx + c$. (c) Find the coordinates of the midpoint of $AB$. (d) Find the equation of the perpendicular bisector of $AB$ in the form $ay + bx = d$, where $a$, $b$, and $d$ are integers. (e) Line $L$ is parallel to $AB$ and passes through the point $(4, -3)$. Find the equation of line $L$ in the form $y = mx + c$.
查看答案详解收起答案详解
解题
(a) The gradient $m$ of the line $AB$ is given by: $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 1}{5 - (-3)} = \frac{6}{8} = \frac{3}{4} = 0.75$$
(b) Using the point-slope formula with point $A(-3, 1)$: $$y - 1 = \frac{3}{4}(x - (-3))$$ $$y - 1 = \frac{3}{4}x + \frac{9}{4}$$ $$y = \frac{3}{4}x + \frac{13}{4} \implies y = 0.75x + 3.25$$
(c) The coordinates of the midpoint $M$ of $AB$ are: $$M = \left( \frac{-3 + 5}{2}, \ \frac{1 + 7}{2} \right) = \left( \frac{2}{2}, \ \frac{8}{2} \right) = (1, 4)$$
(d) The gradient of the perpendicular bisector is the negative reciprocal of the gradient of $AB$: $$m_{\perp} = -\frac{1}{3/4} = -\frac{4}{3}$$ Since it passes through the midpoint $M(1, 4)$: $$y - 4 = -\frac{4}{3}(x - 1)$$ Multiplying both sides by $3$: $$3(y - 4) = -4(x - 1)$$ $$3y - 12 = -4x + 4$$ $$3y + 4x = 16$$ (Thus, $a = 3$, $b = 4$, $d = 16$. Any integer multiple is acceptable, e.g., $6y + 8x = 32$.)
(e) Line $L$ is parallel to $AB$, so it has gradient $m = \frac{3}{4}$. It passes through $(4, -3)$: $$y - (-3) = \frac{3}{4}(x - 4)$$ $$y + 3 = \frac{3}{4}x - 3$$ $$y = \frac{3}{4}x - 6 \implies y = 0.75x - 6$$
评分标准
(a) **M1** for correct formula or substitution $\frac{7-1}{5-(-3)}$ **A1** for $\frac{3}{4}$ or $0.75$
(b) **M1** for substituting their gradient into $y = mx + c$ **A1** for $y = \frac{3}{4}x + \frac{13}{4}$ or $y = 0.75x + 3.25$
(c) **M1** for coordinates formula substituted correctly **A1** for $(1, 4)$
(d) **M1** for perpendicular gradient $m_{\perp} = -\frac{1}{\text{their } m} = -\frac{4}{3}$ **M1** for using their midpoint and $m_{\perp}$ to form equation **A1** for $3y + 4x = 16$ (or equivalent integer form)
(e) **M1** for using $m = \frac{3}{4}$ and $(4, -3)$ **A1** for $y = 0.75x - 6$ (or $y = \frac{3}{4}x - 6$)
题目 4 · Structured Multi-Part
11 分
In triangle $ABC$, the side lengths are $AB = 8$ cm, $BC = 11$ cm, and the angle $ABC = 65^\circ$.
(a) Calculate the length of $AC$. Give your answer correct to $3$ significant figures. (b) Calculate angle $ACB$. Give your answer correct to $1$ decimal place. (c) Calculate the area of triangle $ABC$. Give your answer correct to $3$ significant figures. (d) Calculate the shortest distance from the point $A$ to the line $BC$. Give your answer correct to $3$ significant figures.
查看答案详解收起答案详解
解题
(a) Using the Cosine Rule to find $AC$: $$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)$$ $$AC^2 = 8^2 + 11^2 - 2(8)(11)\cos(65^\circ)$$ $$AC^2 = 64 + 121 - 176(0.422618)$$ $$AC^2 = 185 - 74.3808 = 110.6192$$ $$AC = \sqrt{110.6192} \approx 10.5176\text{ cm}$$ $$AC \approx 10.5\text{ cm (3 s.f.)}$$
(b) Using the Sine Rule to find angle $ACB$: $$\frac{\sin(ACB)}{AB} = \frac{\sin(ABC)}{AC}$$ $$\sin(ACB) = \frac{8 \cdot \sin(65^\circ)}{10.5176}$$ $$\sin(ACB) = \frac{8 \cdot 0.906308}{10.5176} \approx 0.68936$$ $$\text{Angle } ACB = \arcsin(0.68936) \approx 43.58^\circ \approx 43.6^\circ\text{ (1 d.p.)}$$
(c) The area of triangle $ABC$ is: $$\text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin(ABC)$$ $$\text{Area} = \frac{1}{2} \cdot 8 \cdot 11 \cdot \sin(65^\circ)$$ $$\text{Area} = 44 \cdot 0.906308 = 39.877\text{ cm}^2 \approx 39.9\text{ cm}^2\text{ (3 s.f.)}$$
(d) The shortest distance from point $A$ to the line $BC$ is the perpendicular height $h$ from $A$ to $BC$. In the right-angled triangle formed with $AB$ as hypotenuse: $$h = AB \cdot \sin(65^\circ) = 8 \cdot \sin(65^\circ) \approx 7.2505\text{ cm} \approx 7.25\text{ cm (3 s.f.)}$$ Alternatively, using $\text{Area} = \frac{1}{2} \cdot BC \cdot h$: $$39.877 = \frac{1}{2} \cdot 11 \cdot h \implies h = \frac{79.754}{11} \approx 7.25\text{ cm (3 s.f.)}$$ These calculations are consistent.
评分标准
(a) **M1** for correct formula $8^2 + 11^2 - 2(8)(11)\cos(65^\circ)$ **M1** for $AC^2 \approx 110.6$ **A1** for $10.5$ (accept $10.51$ to $10.52$)
(b) **M1** for correct Sine Rule formula $\frac{\sin(ACB)}{8} = \frac{\sin(65^\circ)}{\text{their } AC}$ **M1** for $\sin(ACB) \approx 0.689$ **A1** for $43.6^\circ$ (accept $43.5^\circ$ to $43.6^\circ$)
(c) **M1** for $\frac{1}{2} \cdot 8 \cdot 11 \cdot \sin(65^\circ)$ **A1** for $39.9$ (accept $39.8$ to $39.9$)
(d) **M1** for $8 \cdot \sin(65^\circ)$ or setting up area equation $\frac{1}{2} \times 11 \times h = \text{their area}$ **M1** for correct rearranging **A1** for $7.25$ (accept $7.24$ to $7.26$)
题目 5 · Structured Multi-Part
11 分
Consider the function $f(x) = \frac{x^2 - 4}{x^2 - 1}$.
(a) Write down the equations of the vertical asymptotes of the graph of $y = f(x)$. (b) Write down the equation of the horizontal asymptote of the graph of $y = f(x)$. (c) Find the $x$-coordinates of the points where the graph cuts the $x$-axis. (d) Find the $y$-coordinate of the point where the graph cuts the $y$-axis. (e) Sketch the graph of $y = f(x)$, clearly showing the asymptotes and intercepts. (f) Solve the inequality $f(x) > 2$.
查看答案详解收起答案详解
解题
(a) Vertical asymptotes occur where the denominator is zero: $$x^2 - 1 = 0 \implies (x - 1)(x + 1) = 0 \implies x = 1 \text{ and } x = -1$$
(b) The horizontal asymptote is found by considering the limit as $x \to \pm\infty$: $$y = \lim_{x \to \pm\infty} \frac{x^2 - 4}{x^2 - 1} = 1 \implies y = 1$$
(c) The graph cuts the $x$-axis where $y = 0 \implies x^2 - 4 = 0 \implies x^2 = 4 \implies x = 2$ and $x = -2$.
(d) The graph cuts the $y$-axis where $x = 0 \implies y = \frac{0^2 - 4}{0^2 - 1} = \frac{-4}{-1} = 4$.
(e) Sketch outline: - Three branches divided by vertical asymptotes $x = -1$ and $x = 1$. - The middle branch is U-shaped, with a minimum at $(0, 4)$ and goes to $+\infty$ as $x \to \pm 1$. - The outer branches approach the horizontal asymptote $y = 1$ from below as $x \to \pm\infty$ and cross the $x$-axis at $(-2, 0)$ and $(2, 0)$, tending to $-\infty$ as $x$ approaches $-1$ from the left and $1$ from the right.
(f) To solve $f(x) > 2$: $$\frac{x^2 - 4}{x^2 - 1} > 2 \implies \frac{x^2 - 4}{x^2 - 1} - 2 > 0 \implies \frac{x^2 - 4 - 2(x^2 - 1)}{x^2 - 1} > 0$$ $$\frac{-x^2 - 2}{x^2 - 1} > 0 \implies \frac{x^2 + 2}{x^2 - 1} < 0$$ Since the numerator $x^2 + 2 > 0$ for all real $x$, the fraction is negative if and only if the denominator is negative: $$x^2 - 1 < 0 \implies -1 < x < 1$$
评分标准
(a) **B2** for $x = 1$ and $x = -1$ (or **B1** for either vertical asymptote)
(b) **B1** for $y = 1$
(c) **B2** for $x = 2$ and $x = -2$ (or **B1** for either)
(d) **B1** for $y = 4$ or coordinate $(0, 4)$
(e) **B3** for a fully correct sketch containing: - Three correct branches with vertical asymptotes clearly indicated (**B1**) - Mid-branch turning point at $(0,4)$ and asymptotes correctly approached (**B1**) - Outer branches crossing $x$-axis at $\pm 2$ (**B1**)
(f) **M1** for setting up inequality $\frac{x^2-4}{x^2-1} > 2$ and attempting to simplify **A1** for $-1 < x < 1$
题目 6 · Structured Multi-Part
11 分
A bag contains $5$ red marbles and $7$ blue marbles. Two marbles are chosen at random from the bag, one after the other.
(a) Show the possible outcomes and their probabilities by drawing a tree diagram: (i) if the selection is done with replacement. (ii) if the selection is done without replacement.
(b) Calculate the probability of selecting at least one red marble: (i) if the selection is done with replacement. (ii) if the selection is done without replacement.
(c) If the selection is done without replacement, state with a reason whether selecting a red marble first and selecting a blue marble second are independent events.
查看答案详解收起答案详解
解题
(a) (i) **With replacement:** First pick: Red $P(R) = \frac{5}{12}$, Blue $P(B) = \frac{7}{12}$ Second pick: - After Red: Red $P(R|R) = \frac{5}{12}$, Blue $P(B|R) = \frac{7}{12}$ - After Blue: Red $P(R|B) = \frac{5}{12}$, Blue $P(B|B) = \frac{7}{12}$
(ii) **Without replacement:** First pick: Red $P(R) = \frac{5}{12}$, Blue $P(B) = \frac{7}{12}$ Second pick: - After Red: Red $P(R|R) = \frac{4}{11}$, Blue $P(B|R) = \frac{7}{11}$ - After Blue: Red $P(R|B) = \frac{5}{11}$, Blue $P(B|B) = \frac{6}{11}$
(c) The events are **not independent** because the probability of the second marble being blue is affected by the color of the first marble. For example, $P(B_2 | R_1) = \frac{7}{11} \approx 0.636$, whereas $P(B_2 | B_1) = \frac{6}{11} \approx 0.545$.
评分标准
(a) (i) **B2** for fully correct tree diagram with correct probabilities for each branch with replacement (ii) **B2** for fully correct tree diagram with correct probabilities for each branch without replacement
(b) (i) **M1** for $1 - P(BB)$ or $\left(\frac{5}{12} \times \frac{5}{12}\right) + \left(\frac{5}{12} \times \frac{7}{12}\right) + \left(\frac{7}{12} \times \frac{5}{12}\right)$ **M1** for correct substitution $\frac{25 + 35 + 35}{144}$ or $1 - \frac{49}{144}$ **A1** for $\frac{95}{144}$ (accept $0.66$ or $0.660$) (ii) **M1** for $1 - P(BB)$ or $\left(\frac{5}{12} \times \frac{4}{11}\right) + \left(\frac{5}{12} \times \frac{7}{11}\right) + \left(\frac{7}{12} \times \frac{5}{11}\right)$ **M1** for correct substitution $\frac{20 + 35 + 35}{132}$ or $1 - \frac{42}{132}$ **A1** for $\frac{15}{22}$ (accept $0.682$ or $68.2\%$)
(c) **B1** for stating 'not independent' and providing a valid reason based on changed probabilities
题目 7 · Structured Multi-Part
11 分
Let $f(x) = 3x - 5$ and $g(x) = \frac{2}{x + 1}$, where $x eq -1$.
(a) Find $f(4)$. (b) Find $g(f(2))$. (c) Find $f^{-1}(x)$. (d) Find $g^{-1}(x)$. (e) Find and simplify the expression for $f(g(x))$, writing it as a single algebraic fraction.
(c) Let $y = 3x - 5$. Swap $x$ and $y$ to find the inverse: $$x = 3y - 5 \implies 3y = x + 5 \implies y = \frac{x + 5}{3}$$ $$f^{-1}(x) = \frac{x + 5}{3}$$
(d) Let $y = \frac{2}{x + 1}$. Swap $x$ and $y$: $$x = \frac{2}{y + 1}$$ $$x(y + 1) = 2 \implies xy + x = 2 \implies xy = 2 - x \implies y = \frac{2 - x}{x}$$ $$g^{-1}(x) = \frac{2 - x}{x} = \frac{2}{x} - 1$$
(e) Substitute $g(x)$ into $f(x)$: $$f(g(x)) = 3\left(\frac{2}{x + 1}\right) - 5 = \frac{6}{x + 1} - 5$$ To write it as a single fraction: $$f(g(x)) = \frac{6 - 5(x + 1)}{x + 1} = \frac{6 - 5x - 5}{x + 1} = \frac{1 - 5x}{x + 1}$$
评分标准
(a) **B1** for $7$
(b) **M1** for $f(2) = 1$ seen **A1** for $1$
(c) **M1** for correct method of rearrangement (e.g., $y + 5 = 3x$) **A1** for $\frac{x + 5}{3}$
(d) **M1** for switching $x$ and $y$ and multiplying: $x(y+1) = 2$ **M1** for expanding and isolating the $y$ term: $xy = 2 - x$ **A1** for $\frac{2-x}{x}$ (or $\frac{2}{x}-1$)
(e) **M1** for substituting $g(x)$ correctly into $f(x)$: $3\left(\frac{2}{x+1}\right) - 5$ **M1** for common denominator step: $\frac{6 - 5(x+1)}{x+1}$ **A1** for $\frac{1-5x}{x+1}$
题目 8 · Structured Multi-Part
11 分
The population of a species of insects in a forest is modeled by $P(t) = P_0 \times k^t$, where $t$ is the time in weeks and $P(t)$ is the population. Initially, when $t = 0$, the population is $1500$. After $3$ weeks, the population is $2592$.
(a) Show that $k = 1.2$. (b) Find the population after $6$ weeks, correct to the nearest integer. (c) Calculate the percentage increase in the population each week. (d) Find the number of complete weeks it takes for the population to exceed $10,000$.
(b) After $6$ weeks ($t = 6$): $$P(6) = 1500 \times (1.2)^6$$ $$P(6) = 1500 \times 2.985984 = 4478.976$$ Rounding to the nearest integer gives $4480$.
(c) The weekly multiplier is $k = 1.2$, which can be written as $1 + 0.20$. The percentage increase each week is: $$\text{Percentage Increase} = (1.2 - 1) \times 100\% = 20\%$$
(d) We want to find the smallest integer $t$ such that $P(t) > 10000$: $$1500 \times (1.2)^t > 10000$$ $$(1.2)^t > \frac{10000}{1500} = \frac{20}{3}$$ $$(1.2)^t > 6.6667$$ Taking the natural logarithm (or logarithm base 10) of both sides: $$t \ln(1.2) > \ln(6.6667)$$ $$t > \frac{\ln(6.6667)}{\ln(1.2)} \approx \frac{1.89712}{0.18232} \approx 10.41$$ Since $t$ must be a number of complete weeks, we round $10.41$ up to the next integer: $$t = 11\text{ weeks}$$
评分标准
(a) **M1** for $P_0 = 1500$ **M1** for setting up equation $1500 \times k^3 = 2592 \implies k^3 = 1.728$ **A1** for $k = \sqrt[3]{1.728} = 1.2$ shown clearly
(b) **M1** for evaluating $1500 \times (1.2)^6$ **A1** for $4480$ (accept $4479$)
(c) **M1** for $(1.2 - 1) \times 100\%$ or similar method **A1** for $20\%$
(d) **M1** for setting up inequality $1500 \times (1.2)^t > 10000$ **M1** for using logs to solve: $t \log(1.2) > \log(6.667)$ **A1** for $t > 10.4$ **A1** for $11$ complete weeks
题目 9 · Structured Multi-Part
11 分
A surveyor is mapping a triangular park, \(PQR\).
The distance \(PQ\) is 120 metres and the bearing of \(Q\) from \(P\) is \(055^\circ\). The distance \(PR\) is 150 metres and the bearing of \(R\) from \(P\) is \(135^\circ\).
(a) Show that the angle \(QPR = 80^\circ\). (b) Calculate the distance \(QR\), giving your answer to the nearest metre. (c) Calculate the angle \(PQR\), correct to 1 decimal place. (d) Calculate the bearing of \(R\) from \(Q\), correct to 1 decimal place. (e) Calculate the area of the triangular park, giving your answer correct to 3 significant figures.
(c) Using the Sine Rule: \(\frac{\sin(PQR)}{PR} = \frac{\sin(QPR)}{QR}\) \(\frac{\sin(PQR)}{150} = \frac{\sin(80^\circ)}{175.07}\) \(\sin(PQR) = \frac{150 \times \sin(80^\circ)}{175.07} \approx 0.84378\) \(\text{Angle } PQR = \arcsin(0.84378) \approx 57.54^\circ \approx 57.5^\circ\).
(d) The bearing of \(P\) from \(Q\) is \(55^\circ + 180^\circ = 235^\circ\). To find the bearing of \(R\) from \(Q\), we subtract the interior angle \(PQR\): \(\text{Bearing} = 235^\circ - 57.54^\circ = 177.46^\circ \approx 177.5^\circ\).
(a) [2 marks]: - M1 for \(135^\circ - 55^\circ\) - A1 for showing \(80^\circ\) clearly
(b) [3 marks]: - M1 for substitution into Cosine Rule: \(120^2 + 150^2 - 2 \times 120 \times 150 \times \cos(80^\circ)\) - A1 for \(QR^2 = 30649\) or \(QR = 175.07\) - A1 for \(175\text{ m}\) (cao)
(c) [3 marks]: - M1 for substitution into Sine Rule: \(\frac{\sin(PQR)}{150} = \frac{\sin(80^\circ)}{\text{their } QR}\) - A1 for \(\sin(PQR) \approx 0.8438\) - A1 for \(57.5^\circ\) or FT their \(QR\)
(d) [1 mark]: - B1 for \(177.5^\circ\) (allow FT from their (c))
(e) [2 marks]: - M1 for \(\frac{1}{2} \times 120 \times 150 \times \sin(80^\circ)\) - A1 for \(8860\text{ m}^2\) (accept 8863)
题目 10 · Structured Multi-Part
11 分
A bag contains 6 red cards, 4 blue cards, and 2 green cards. Three cards are chosen at random from the bag without replacement.
(a) Calculate the probability that: (i) all three cards are red, (ii) none of the three cards are green, (iii) exactly two of the three cards are blue.
(b) If this experiment of choosing three cards without replacement is repeated 440 times, calculate the expected number of times that all three cards are red.
查看答案详解收起答案详解
解题
(a) Total number of cards is \(6 + 4 + 2 = 12\).
(i) The probability that all three cards are red is: \(P(\text{RRR}) = \frac{6}{12} \times \frac{5}{11} \times \frac{4}{10} = \frac{120}{1320} = \frac{1}{11}\) (or \(\approx 0.0909\)).
(ii) The probability that none of the three cards are green is equivalent to choosing three cards from the \(6 + 4 = 10\) non-green cards: \(P(\text{no green}) = \frac{10}{12} \times \frac{9}{11} \times \frac{8}{10} = \frac{720}{1320} = \frac{6}{11}\) (or \(\approx 0.545\)).
(iii) To get exactly two blue cards, the selections can be \(B B X\), \(B X B\), or \(X B B\), where \(X\) is not blue (total of 8 cards available). For one of these outcomes, e.g., \(B B X\): \(P(B B X) = \frac{4}{12} \times \frac{3}{11} \times \frac{8}{10} = \frac{96}{1320}\). Since there are 3 possible arrangements: \(P(\text{exactly 2 blue}) = 3 \times \frac{96}{1320} = \frac{288}{1320} = \frac{12}{55}\) (or \(\approx 0.218\)).
(b) The expected number of times all three cards are red in 440 trials is: \(\text{Expected Value} = 440 \times P(\text{RRR}) = 440 \times \frac{1}{11} = 40\).
评分标准
(a)(i) [2 marks]: - M1 for \(\frac{6}{12} \times \frac{5}{11} \times \frac{4}{10}\) - A1 for \(\frac{1}{11}\) or \(0.0909\)
(a)(ii) [3 marks]: - M2 for \(\frac{10}{12} \times \frac{9}{11} \times \frac{8}{10}\) (M1 if one fraction error or using longer addition method) - A1 for \(\frac{6}{11}\) or \(0.545\)
(a)(iii) [3 marks]: - M1 for \(\frac{4}{12} \times \frac{3}{11} \times \frac{8}{10}\) - M1 for multiplying their probability by 3 - A1 for \(\frac{12}{55}\) or \(0.218\)
(b) [3 marks]: - M1 for attempting \(440 \times P(\text{RRR})\) - M1 for \(440 \times \text{their (a)(i)}\) - A1 for 40 (must be an integer)
题目 11 · Structured Multi-Part
11 分
Consider the following two sequences: Sequence A: \(4, 10, 16, 22, 28, \dots\) Sequence B: \(3, 9, 27, 81, 243, \dots\)
(a) For Sequence A: (i) Write down the next two terms. (ii) Find an expression, in terms of \(n\), for the \(n\)th term.
(b) For Sequence B: (i) Write down the next term. (ii) Find an expression, in terms of \(n\), for the \(n\)th term.
(c) A new sequence, Sequence C, is formed by subtracting the \(n\)th term of Sequence A from the \(n\)th term of Sequence B. (i) Write down the first three terms of Sequence C. (ii) Find the \(n\)th term of Sequence C. (iii) Find the 6th term of Sequence C.
查看答案详解收起答案详解
解题
(a)(i) Sequence A is an arithmetic progression with a first term of 4 and a common difference of 6. The next two terms are \(28 + 6 = 34\) and \(34 + 6 = 40\).
(ii) The \(n\)th term is: \(a + (n-1)d = 4 + (n-1)6 = 6n - 2\).
(b)(i) Sequence B is a geometric progression with a first term of 3 and a common ratio of 3. The next term is \(243 \times 3 = 729\).
(ii) The \(n\)th term is: \(a \cdot r^{n-1} = 3 \cdot 3^{n-1} = 3^n\).
(c)(i) The first three terms of Sequence C are: \(C_1 = B_1 - A_1 = 3 - 4 = -1\), \(C_2 = B_2 - A_2 = 9 - 10 = -1\), \(C_3 = B_3 - A_3 = 27 - 16 = 11\).
(ii) The \(n\)th term of Sequence C is the \(n\)th term of Sequence B minus the \(n\)th term of Sequence A: \(C_n = 3^n - (6n - 2) = 3^n - 6n + 2\).
(iii) The 6th term of Sequence C is: \(C_6 = 3^6 - 6(6) + 2 = 729 - 36 + 2 = 695\).
评分标准
(a)(i) [1 mark]: - B1 for 34, 40
(a)(ii) [2 marks]: - M1 for \(6n + c\) (where \(c \neq -2\)) or \(4 + 6(n-1)\) - A1 for \(6n - 2\) (oe)
(b)(i) [1 mark]: - B1 for 729
(b)(ii) [2 marks]: - M1 for \(3^k\) or showing powers of 3 - A1 for \(3^n\) (oe)
(c)(i) [2 marks]: - B1 for any two correct - B1 for all three correct: -1, -1, 11
(c)(ii) [1 mark]: - B1 for \(3^n - 6n + 2\) (accept FT from (a)(ii) and (b)(ii))
(c)(iii) [2 marks]: - M1 for substituting \(n = 6\) into their expression in (c)(ii) - A1 for 695
部分 3: Investigation and Modelling
Answer both Part A (Investigation) and Part B (Modelling). Full working and communication details are required to earn high marks.
2 题目 · 60 分
题目 1 · Extended Investigation
30 分
### INVESTIGATION: DIAGONAL PATHS IN GRID RECTANGLES
This investigation looks at the number of grid squares a diagonal line passes through when drawn from one corner of a grid rectangle to the opposite corner.
In this investigation: - The rectangle has a width of \(W\) units and a height of \(H\) units. - The diagonal is drawn from the bottom-left corner to the top-right corner. - A square is "entered" if the diagonal line passes through any part of its interior (not just touching a corner).
---
#### 1 Rectangles of width 1
(a) A rectangle has width 1 and height 4. Complete the statement: The number of squares entered by the diagonal in a rectangle of width 1 and height 4 is ....................
(b) Complete the statement: The number of squares entered by the diagonal in a rectangle of width 1 and height \(H\) is ....................
---
#### 2 Rectangles of width 2 (where height \(H\) is odd)
(a) A rectangle has width 2 and height 3. Calculate the number of squares entered by the diagonal. ....................
(b) A rectangle has width 2 and height 5. Calculate the number of squares entered by the diagonal. ....................
(c) Complete the table for rectangles of width 2 where the height \(H\) is an odd number.
#### 3 Rectangles of width 3 (where height \(H\) is not a multiple of 3)
(a) Find the number of squares entered by the diagonal when the height is 4. ....................
(b) Find the number of squares entered by the diagonal when the height is 5. ....................
(c) Complete the statement: When the width is 3 and the height \(H\) is not a multiple of 3, the number of squares entered by the diagonal in terms of \(H\) is ....................
---
#### 4 Rectangles with no common factors
In this part, the width \(W\) and height \(H\) of the rectangle have no common factors greater than 1 (they are coprime).
(a) Find the number of squares entered by the diagonal when \(W = 5\) and \(H = 8\). ....................
(b) Find an expression, in terms of \(W\) and \(H\), for the number of squares entered by the diagonal. ....................
---
#### 5 Rectangles with common factors
When the width \(W\) and height \(H\) have a greatest common divisor, \(g\), greater than 1, the diagonal passes through corners of the grid squares inside the rectangle.
(a) Complete the table using drawings on a grid or by finding a pattern.
(b) The general formula for the number of squares entered by the diagonal is: $$\text{Number of squares} = W + H - g$$ Show that this formula is correct for a rectangle with width \(W = 6\) and height \(H = 9\).
(c) Use the formula to calculate the number of squares entered by the diagonal when \(W = 120\) and \(H = 150\). ....................
查看答案详解收起答案详解
解题
### Detailed Solution
#### 1 Rectangles of width 1 (a) For \(W = 1\) and \(H = 4\), the diagonal starts at the bottom-left corner of the first square and passes through each of the 4 stacked squares in turn. Thus, the number of squares entered is **4**. (b) Following this pattern, if the height is \(H\), the diagonal passes through each of the \(H\) squares vertically, so the number of squares entered is **\(H\)**.
#### 2 Rectangles of width 2 (height \(H\) is odd) (a) For \(W = 2, H = 3\), since 2 and 3 have no common factors, the diagonal does not cross any grid intersections except the corners. Drawing this out reveals it passes through **4** squares. (b) For \(W = 2, H = 5\), similarly, the diagonal passes through **6** squares. (c) Looking at the values: - Height 1: 2 squares - Height 3: 4 squares - Height 5: 6 squares - Height 7: **8** squares - Height \(H\): **\(H + 1\)** squares.
#### 3 Rectangles of width 3 (height \(H\) is not a multiple of 3) (a) For \(W = 3, H = 4\), the number of squares entered is \(3 + 4 - 1 = \mathbf{6}\). (b) For \(W = 3, H = 5\), the number of squares entered is \(3 + 5 - 1 = \mathbf{7}\). (c) The pattern shows that the number of entered squares is always \(H + 2\). This matches the formula \(W + H - 1\) where \(W = 3\), giving **\(H + 2\)**.
#### 4 Rectangles with no common factors (a) For \(W = 5, H = 8\), since 5 and 8 are coprime, we use the pattern: \(\text{squares} = W + H - 1 = 5 + 8 - 1 = \mathbf{12}\). (b) The expression is **\(W + H - 1\)**.
(b) For \(W = 6\) and \(H = 9\), the greatest common divisor is \(g = 3\). Using the formula: \(6 + 9 - 3 = 12\). Alternatively, the diagonal crosses 3 identical smaller sub-rectangles of size \(2 \times 3\). In each \(2 \times 3\) sub-rectangle, the diagonal enters \(2 + 3 - 1 = 4\) squares. Total entered squares = \(3 \times 4 = 12\), which verifies the formula.
(c) For \(W = 120\) and \(H = 150\), the greatest common divisor is \(g = 30\). Using the formula: \(120 + 150 - 30 = \mathbf{240}\).
评分标准
#### Marking Scheme
- **Question 1** - (a) **[2 marks]**: M1 for showing/drawing the diagonal for width 1 height 4. A1 for 4. - (b) **[2 marks]**: B2 for \(H\) (or \(1H\)).
- **Question 2** - (a) **[2 marks]**: M1 for attempting to count or draw. A1 for 4. - (b) **[2 marks]**: M1 for attempting to count or draw. A1 for 6. - (c) **[3 marks]**: B1 for 8. B2 for \(H + 1\) (B1 for \(H + k\) where \(k \neq 1\)).
- **Question 3** - (a) **[2 marks]**: A1 for 6. - (b) **[2 marks]**: A1 for 7. - (c) **[3 marks]**: B3 for \(H + 2\) (B2 for \(H + 3 - 1\) unsimplified, B1 for \(H + k\)).
- **Question 4** - (a) **[2 marks]**: M1 for \(5 + 8 - 1\). A1 for 12. - (b) **[3 marks]**: B3 for \(W + H - 1\) (B2 for unsimplified or missing one term, e.g. \(W+H\)).
- **Question 5** - (a) **[3 marks]**: B1 for each correct value: 4, 8, 6. - (b) **[3 marks]**: B1 for identifying \(g = 3\). B1 for applying the formula to get 12. B1 for a clear explanation dividing the rectangle into three \(2 \times 3\) sub-rectangles or equivalent argument. - (c) **[3 marks]**: B1 for finding \(g = 30\). M1 for substituting into \(W + H - g\). A1 for 240.
题目 2 · Extended Modelling
30 分
### **MODELLING: HEAVY DUTY WOODEN CRATES**
**Introduction** A manufacturing company uses open-topped wooden crates to transport heavy machine parts. The base of each crate is a square with external side length $x\text{ cm}$. The external height of each crate is $h\text{ cm}$. The crates are constructed from heavy timber that is $1\text{ cm}$ thick.
---
#### **Question 1**
(a) Show that the internal dimensions of the base of the crate are $(x - 2)\text{ cm}$ by $(x - 2)\text{ cm}$ and the internal height is $(h - 1)\text{ cm}$.
(b) Write down a formula for the internal capacity, $C$, of the crate in terms of $x$ and $h$.
(c) To reinforce the base for very heavy parts, an extra metal plate of thickness $0.5\text{ cm}$ is placed on the inside bottom of the crate. Modify your formula for $C$ in part (b) to include the effect of this metal plate.
(d) The external area of the base of a crate is $1600\text{ cm}^2$. The external height of this crate is $10\text{ cm}$ longer than the external side length of the base.
(i) Using your modified model from part (c), calculate the capacity of this crate. Show your working clearly.
(ii) The total volume of materials used to make the crate is the difference between the external volume of the crate and its internal capacity. Calculate this total volume of materials for this crate.
---
#### **Question 2**
The net of the open-topped crate (excluding the inner metal plate) consists of the square base of side $x\text{ cm}$ and four rectangular sides of dimensions $x\text{ cm}$ by $h\text{ cm}$.
(a) Show that the total external surface area of the crate, $A$, is given by: $$A = x^2 + 4xh$$
(b) A crate is designed to have a fixed external volume of $16\,000\text{ cm}^3$.
(i) Write down an expression for $h$ in terms of $x$.
(ii) Show that the model for the external surface area, $A$, in terms of $x$ is: $$A = x^2 + \frac{64\,000}{x}$$
(iii) Sketch this model for $0 < x \le 80$.
(iv) Find the value of $x$ that gives the minimum external surface area of the crate, and write down this minimum area.
(v) For this minimum-area crate, calculate the external height $h$.
(vi) Find the capacity of this minimum-area crate, using the formula from part (c).
查看答案详解收起答案详解
解题
### **Question 1 Solutions**
(a) - The external base is a square of side $x\text{ cm}$. Since the wood has thickness $1\text{ cm}$ on both opposite sides, the internal base side length is: $$x - 1 - 1 = x - 2\text{ cm}$$ - The external height is $h\text{ cm}$. Since the crate is open-topped, there is only a wooden base of thickness $1\text{ cm}$ at the bottom and no lid at the top. The internal height is: $$h - 1\text{ cm}$$
(c) - The internal metal plate decreases the internal height by $0.5\text{ cm}$, so the new internal height is $(h - 1) - 0.5 = h - 1.5\text{ cm}$. - $$C = (x - 2)^2(h - 1.5)$$
(a) - The open-topped crate has 1 square base of area $x^2$ and 4 rectangular side faces, each of area $xh$. - Total surface area $A = x^2 + 4xh$.
(b) (i) - External Volume $V = x^2 h = 16\,000 \implies h = \frac{16\,000}{x^2}$. (ii) - Substituting $h$ into the surface area equation: $$A = x^2 + 4x\left(\frac{16\,000}{x^2}\right) = x^2 + \frac{64\,000}{x}$$ (iii) - The graph starts high for small values of $x$ (asymptote at $x=0$), decreases to a minimum, and then rises as $x^2$ dominates. At $x=80$, $A = 80^2 + \frac{64\,000}{80} = 6400 + 800 = 7200$. (iv) - Using a GDC to find the minimum of $A = x^2 + \frac{64\,000}{x}$: $$x \approx 31.7\text{ cm} \quad (31.748\text{ cm})$$ $$A \approx 3020\text{ cm}^2 \quad (3023.9\text{ cm}^2)$$ (v) - $$h = \frac{16\,000}{x^2} = \frac{16\,000}{31.748^2} \approx 15.9\text{ cm} \quad (15.874\text{ cm})$$ (vi) - Using $C = (x - 2)^2(h - 1.5)$ with $x \approx 31.748$ and $h \approx 15.874$: $$C = (31.748 - 2)^2(15.874 - 1.5) = (29.748)^2 \times 14.374$$ $$C \approx 884.94 \times 14.374 \approx 12\,720\text{ cm}^3 \approx 12\,700\text{ cm}^3\text{ (to 3 s.f.)}$$
评分标准
**Question 1** - (a) **[2 marks]**: - **M1** for showing base side deduction: $x - 1 - 1 = x - 2$ - **M1** for explaining open top height deduction: $h - 1$ - (b) **[2 marks]**: - **B2** for $C = (x - 2)^2(h - 1)$ (or **B1** for $(x-2) \times (x-2) \times (h-1)$ unsimplified). - (c) **[1 mark]**: - **B1** for $C = (x-2)^2(h-1.5)$ or equivalent correct modification. - (d)(i) **[4 marks]**: - **B1** for finding $x = 40$ and $h = 50$. - **M1** for correct substitution of their $x$ and $h$ into their formula from (c). - **A1** for evaluating $38^2 \times 48.5$ or equivalent intermediate step. - **A1** for $70\,034$ (accept $70\,000$ to 3 s.f.). - (d)(ii) **[3 marks]**: - **M1** for calculating external volume: $1600 \times 50 = 80\,000$. - **M1** for subtracting their capacity from external volume. - **A1** for $9\,966$ (or follow-through correct subtraction).
**Question 2** - (a) **[2 marks]**: - **M1** for identifying 5 faces with correct dimensions: base area $= x^2$, sides area $= 4 \times xh$. - **A1** for convincing algebraic steps showing $A = x^2 + 4xh$. - (b)(i) **[1 mark]**: - **B1** for $h = \frac{16\,000}{x^2}$. - (b)(ii) **[2 marks]**: - **M1** for substituting $h$ expression into $A = x^2 + 4xh$. - **A1** for correct simplification showing $A = x^2 + \frac{64\,000}{x}$. - (b)(iii) **[3 marks]**: - **B1** for correct general shape (U-like curve with single minimum, not touching the vertical axis). - **B1** for showing minimum point in the correct quadrant. - **B1** for indicating correct end point coordinate at approximately $(80, 7200)$. - (b)(iv) **[3 marks]**: - **B1** for $x = 31.7$ (accept $31.7$ to $31.8$). - **B2** for minimum area $A = 3020$ (accept $3020$ to $3024$) (**B1** for $3020$ seen but without correct units). - (b)(v) **[2 marks]**: - **M1** for substituting their minimum $x$ value into $h = \frac{16\,000}{x^2}$. - **A1** for $15.9$ (accept $15.8$ to $15.9$). - (b)(vi) **[5 marks]**: - **M1** for substituting their minimum $x$ and $h$ into capacity formula $(x - 2)^2(h - 1.5)$. - **A1** for internal base dimension: $29.7$ (or $29.748$). - **A1** for internal height dimension: $14.4$ (or $14.374$). - **M1** for evaluating $(29.748)^2 \times 14.374$. - **A1** for final answer of $12\,700$ (accept $12\,700$ to $12\,721$).
想知道自己有几分把握？
thinka 是 DSE 学生在用的 AI 练习应用，提供无限量练习题、即时自动批改和详细解题步骤。超过 100,000 名学生用它确认自己是真的会，而不只是「以为会」。